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Numerical Methods forPartial Differential Equations

Joachim Sch oberl

April 6, 2009

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Contents

1 Introduction 51.1 Classication of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Weak formulation of the Poisson Equation . . . . . . . . . . . . . . . . . . 6

1.3 The Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . 82 The abstract theory 11

2.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Projection onto subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Riesz Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Symmetric variational problems . . . . . . . . . . . . . . . . . . . . . . . . 162.5 Coercive variational problems . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.5.1 Approximation of coercive variational problems . . . . . . . . . . . 202.6 Inf-sup stable variational problems . . . . . . . . . . . . . . . . . . . . . . 21

2.6.1 Approximation of inf-sup stable variational problems . . . . . . . . 23

3 Sobolev Spaces 253.1 Generalized derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.3 Trace theorems and their applications . . . . . . . . . . . . . . . . . . . . . 28

3.3.1 The trace space H 1/ 2 . . . . . . . . . . . . . . . . . . . . . . . . . 343.4 Equivalent norms on H 1 and on sub-spaces . . . . . . . . . . . . . . . . . . 37

4 The weak formulation of the Poisson equation 414.1 Shift theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

5 Finite Element Method 455.1 Finite element system assembling . . . . . . . . . . . . . . . . . . . . . . . 475.2 Finite element error analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 495.3 A posteriori error estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 555.4 Non-conforming Finite Element Methods . . . . . . . . . . . . . . . . . . . 63

6 Linear Equation Solvers 696.1 Direct linear equation solvers . . . . . . . . . . . . . . . . . . . . . . . . . 70

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4 CONTENTS

6.2 Iterative equation solvers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.3 Preconditioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

7 Mixed Methods 957.1 Weak formulation of Dirichlet boundary conditions . . . . . . . . . . . . . 957.2 A Mixed method for the ux . . . . . . . . . . . . . . . . . . . . . . . . . . 967.3 Abstract theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 977.4 Analysis of the model problems . . . . . . . . . . . . . . . . . . . . . . . . 1007.5 Approximation of mixed systems . . . . . . . . . . . . . . . . . . . . . . . 107

8 Applications 1118.1 The Navier Stokes equation . . . . . . . . . . . . . . . . . . . . . . . . . . 1118.2 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1138.3 Maxwell equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

9 Parabolic partial differential equations 1299.1 Semi-discretization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1319.2 Time integration methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

10 Hyperbolic partial differential equations 135

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Chapter 1

Introduction

Differential equations are equations for an unknown function involving differential opera-tors. An ordinary differential equation (ODE) requires differentiation with respect to onevariable. A partial differential equation (PDE) involves partial differentiation with respectto two or more variables.

1.1 Classication of PDEs

The general form of a linear PDE of second order is: nd u : R d R such that

d

i,j =1

x ia i,j (x)

u(x)x j

+d

i=1

bi(x)u (x)

x i+ c(x)u(x) = f (x). (1.1)

The coefficients a i,j (x), bi(x), c(x) and the right hand side f (x) are given functions. Inaddition, certain type of boundary conditions are required. The behavior of the PDEdepends on the type of the differential operator

L :=d

i,j =1

x i

a i,j

x j+

d

i=1

bi

x i+ c.

Replace x i by s i . Thend

i,j =1 s ia i,j s j +

d

i=1 bis i + c = 0

describes a quartic shape in R d. We treat the following cases:

1. In the case of a (positive or negative) denite matrix a = ( a i,j ) this is an ellipse,and the corresponding PDE is called elliptic. A simple example is a = I , b = 0, andc = 0, i..e.

i

2ux 2i

= f.

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6 CHAPTER 1. INTRODUCTION

Elliptic PDEs require boundary conditions.

2. If the matrix a is semi-denite, has the one-dimensional kernel span {v}, and bv = 0,then the shape is a parabola. Thus, the PDE is called parabolic. A simple exampleis

d 1

i=1

2ux 2i

+ux d

= f.

Often, the distinguished direction corresponds to time. This type of equation requiresboundary conditiosn on the d1-dimensional boundary, and initial conditions in thedifferent direction.

3. If the matrix a has d 1 positive, and one negative (or vise versa) eigenvalues, thenthe shape is a hyperbola. The PDE is called hyperbolic. The simplest one is

d 1

i=1

2ux 2i

+ 2ux 2d

= f.

Again, the distinguished direction often corresponds to time. Now, two initial con-ditions are needed.

4. If the matrix a is zero, then the PDE degenerates to the rst order PDE

biux i

+ cu = f.

Boundary conditions are needed at a part of the boundary.These cases behave very differently. We will establish theories for the individual cases.A more general classiccation, for more positive or negative eigenvalues, and systems of PDEs is possible. The type of the PDE may also change for different points x.

1.2 Weak formulation of the Poisson Equation

The most elementary and thus most popular PDE is the Poisson equation

u = f in , (1.2)

with the boundary conditions

u = uD on D ,un = g on N ,

un + u = g onR .

(1.3)

The domain is an open and bounded subset of R d, where the problem dimension d isusually 1, 2 or 3. For d = 1, the equation is not a PDE, but an ODE. The boundary

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1.2. WEAK FORMULATION OF THE POISSON EQUATION 7

:= consists of the three non-overlapping parts D , N , and R . The outer unitnormal vector is called n. The Laplace differential operator is := di=1

2x 2i

, the normal

derivative at the boundary is n

:= di=1

ni

x i. Given are the functions f , u

Dand g in

proper function spaces (e.g., f L2()). We search for the unknown function u, again, ina proper function space dened later.The boundary conditions are called

Dirichlet boundary condition on D . The function value is prescribed,

Neumann boundary condition on N . The normal derivative is prescribed,

Robin boundary condition on R . An affine linear relation between the functionvalue and the normal derivative is prescribed.Exactly one boundary condition must be specied on each part of the boundary.We transform equation (1.2) together with the boundary conditions (1.3) into its weak

form. For this, we multiply (1.2) by smooth functions (called test functions) and integrateover the domain:

uv dx = fvdx (1.4)We do so for sufficiently many test functions v in a proper function space. Next, we applyGauss theorem div pdx = p n ds to the function p := u v to obtain

div( u v) dx = u nvdsFrom the product rule there follows div( uv) = uv + u v. Together we obtain u v dx un v ds = fvdx.

Up to now, we only used the differential equation in the domain. Next, we incorporatethe boundary conditions. The Neumann and Robin b.c. are very natural (and thus arecalled natural boundary conditions). We simply replace un by g and u + g on N andR , respectively. Putting unknown terms to the left, and known terms to the right handside, we obtain

u v dx + R uv ds D un v ds = fvdx + N + R gv ds.Finally, we use the information of the Dirichlet boundary condition. We work brute forceand simple keep the Dirichlet condition in strong sense. At the same time, we only allowtest functions v fullling v = 0 on D . We obtain the

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Weak form of the Poisson equation:Find u such that u = uD on D and

u v dx + R uv ds = fvdx + N + R gv ds (1.5)v such that v = 0 on D .

We still did not dene the function space in which we search for the solution u. A properchoice is

V := {vL2() : u[L2()]d and u|L2( )}.It is a complete space, and, together with the inner product

(u, v)V := ( u, v)L 2 () + ( u, v)L 2 () + ( u, v)L2 ()

it is a Hilbert space. Now, we see that f L2() and gL2() is useful. The Dirichletb.c. uD must be chosen such that there exists an uV with u = uD on D . By denitionof the space, all terms are well dened. We will see later, that the problem indeed has aunique solution in V .

1.3 The Finite Element Method

Now, we are developing a numerical method for approximating the weak form (1.5). Forthis, we decompose the domain into triangles T . We call the set

T =

{T

}triangulation.

The set N = {x j}is the set of nodes. By means of this triangulation, we dene the niteelement space, V h :V h := {vC () : v|T is affine linear T T }

This is a sub-space of V . The derivatives (in weak sense, see below) are piecewise constant,and thus, belong to [ L2()]2. The function vhV h is uniquely dened by its values v(x j )in the nodes x j N . We decompose the set of nodes as

N = N DN f ,where N D are the nodes on the Dirichlet boundary, and N f are all others ( f as free). Thenite element approximation is dened as

Find uh such that uh (x) = uD (x) xN D and

uh vh dx + R u h vh ds = fv h dx + N + R gvh ds (1.6)vhV h such that vh (x) = 0 xN D

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1.3. THE FINITE ELEMENT METHOD 9

Now it is time to choose a basis for V h . The most convenient one is the nodal basis

{i}characterized asi(x j ) = i,j . (1.7)

The Kronecker- is dened to be 1 for i = j , and 0 else. These are the popular hatfunctions. We represent the nite element solution with respect to this basis:

uh (x) = uii(x) (1.8)

By the nodal-basis property (1.7) there holds uh (x j ) = i uii(x j ) = u j . We have todetermine the coefficients ui

R N , with N = |N|. The N D := |N D | values according tonodes on D are given explicitely:u j = uh (x j ) = uD (x j ) x j D

The others have to be determined from the variational equation (1.6). It is equivalent tofulll (1.6) for the whole space {vh V h : vh (x) = 0 x j N D }, or just for its basis{i : xiN f }associated to the free nodes:

i i j dx + R i j ds ui = f j dx + N + R g j ds (1.9) j such that x j N f

We have inserted the basis expansion (1.8). We dene the matrix A = ( A ji )R N N and

the vector f = ( f j )R N as

A ji := i j dx + R i j ds,f j := f j dx + N + R g j ds.

According to Dirichlet- and free nodes they are splitted as

A = ADD ADDAfD Af f and f = f Df f

.

Now, we obtain the system of linear equations for u = ( ui)R N , u = ( uD , u f ):

I 0AfD Af f uDuf

= uDf f . (1.10)

At all, we have N coefficients ui . N D are given explicitely from the Dirichlet values. Theseare N f equations to determine the remaining ones. Using the known uD , we can reformulateit as symmetric system of equations for uf

R N f :

Af f uf = f f AfD uD

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Chapter 2

The abstract theory

In this chapter we develop the abstract framework for variational problems.

2.1 Basic properties

Denition 1. A vector space V is a set with the operations + : V V V and : R V V such that for all u, vV and , R there holds

u + v = v + u (u + v) + w = u + ( v + w)

(u + v) =

u +

v, ( + )

u =

u +

u

Examples are R n , the continuous functions C 0, or the Lebesgue space L2.

Denition 2. A normed vector space (V, ) is a vector space with the operation . :V R being a norm, i.e., for u, vV and R there holds u + v u + v u = || u u = 0u = 0

Examples are ( C 0, sup ), or (C 0, L 2 ).Denition 3. In a complete normed vector space, Cauchy sequences (un )V

N convergeto an uV . A complete normed vector space is called Banach space .

Examples of Banach spaces are ( L2, L 2 ), (C 0, sup ), but not ( C 0, L 2 ).Denition 4. The closure of a normed vector-space (W, V ), denoted as W

V is thesmallest complete space containing W .

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12 CHAPTER 2. THE ABSTRACT THEORY

Example: C L 2 = L2.

Denition 5. A functional or a linear form l() on V is a linear mapping l() : V R .The canonical norm for linear forms is the dual norm

l V := sup0= vV

l(v)v

.

A linear form l is called bounded if the norm is nite. The vector space of all bounded linear forms on V is called the dual space V .

An example for a bounded linear form is l() : L2 R : v v dx.Denition 6. A bilinear form A(, ) on V is a mapping A : V V R which is linear in u and in v. It is called symmetric if A(u, v) = A(v, u) for all u, v

V .

Examples are the bilinear form A(u, v) = uv dx on L2, or A(u, v) := uT Av on R n , whereA is a (symmetric) matrix.Denition 7. A symmetric bilinear form A(, ) is called an inner product if it satises

(v, v) 0vV (v, v) = 0v = 0

Often, is is denoted as (, )A , (, )V , or simply (, ).An examples on R is uT Av, where A is a symmetric and positive denite matrix.

Denition 8. An inner product space is a vector space V together with an inner product (, )V .Lemma 9. Cauchy Schwarz inequality. If A(, ) is a symmetric bilinear form such that A(v, v) 0 for all vV , then there holds

A(u, v) A(u, u )1/ 2A(v, v)1/ 2Proof: For t

R there holds

0

A(u

tv,u

tv) = A(u, u )

2tA(u, v) + t2A(v, v)

If A(v, v) = 0, then A(u, u ) 2tA(u, v) 0 for all tR , which forces A(u, v) = 0, and theinequality holds trivially. Else, if A(v, v) = 0, set t = A(u, v)/A (v, v), and obtain0 A(u, u ) A(u, v)2/A (v, v),

which is equivalent to the statement. 2

Lemma 10. v V := ( v, v)1/ 2V denes a norm on the inner product space (V, (, )V ).

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2.1. BASIC PROPERTIES 13

Denition 11. An inner product space (V, (, )V ) which is complete with respect to V is called a Hilbert space .Denition 12. A closed subspace S of an Hilbert space V is a subset which is a vector space, and which is complete with respect to V .A nite dimensional subspace is always a closed subspace.

Lemma 13. Let T be a continuous linear operator from the Hilbert space V to the Hilbert space W . The kernel of T , ker T := {vV : T v = 0}is a closed subspace of V .

Proof: First we observe that ker T is a vector space. Now, let ( un )ker T N converge

to uV . Since T is continuous, T un T u, and thus T u = 0 and uker T . 2Lemma 14. Let S be a subspace (not necessarily closed) of V . Then

S := {vV : (v, w) = 0 wS }is a closed subspace.

The proof is similar to Lemma 13.

Denition 15. Let V and W be vector spaces. A linear operator T : V W is a linear mapping from V to W . The operator is called bounded if its operator-norm T

V W := sup

0= vV

Tv W v V

is nite.

An example is the differential operator on the according space ddx : (C 1(0, 1), sup +d

dx sup ) (C (0, 1), sup ).Lemma 16. A bounded linear operator is continuous.

Proof. Let vn v, i.e. vn v V 0. Then T vn Tv T V W vn v V convergesto 0, i.e. Tvn Tv. Thus T is continuous.Denition 17. A dense subspace S of V is such that every element of V can be ap-proximated by elements of S , i.e.

> 0uV vS such that u v V .Lemma 18 (extension principle) . Let S be a dense subspace of the normed space V , and let W be a complete space. Let T : S W be a bounded linear operator with respect to thenorm T V W . Then, the operator can be uniquely extended onto V .

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Proof. Let uV , and let vn be a sequence such that vn u. Thus, vn is Cauchy. Tvnis a well dened sequence in W . Since T is continuous, T vn is also Cauchy. Since W is complete, there exists a limit w such that T vn

w. The limit is independent of the

sequence, and thus T u can be dened as the limit w.

Denition 19. A bounded linear operator T : V W is called compact if for every bounded sequence (un )V N , the sequence (Tun ) contains a convergent sub-sequence.Lemma 20. Let V, W be Hilbert spaces. A operator is compact if and only if there existsa complete orthogonal system (un ) and values n 0 such that

(un , um )V = n,m (Tun , Tum )W = n n,m

This is the eigensystem of the operator K : V V : u (Tu,T )W .Proof. (sketch) There exists an maximizing element of (Tv,Tv )W (v,v )V . Scale it to v V = 1and call it u1, and 1 = (T u1 ,T u 1 )W (u1 ,u 1 )V . Repeat the procedure on the V -complement of u1 togenerate u2, and so on.

2.2 Projection onto subspaces

In the Euklidean space R 2 one can project orthogonally onto a line through the origin, i.e.,onto a sub-space. The same geometric operation can be dened for closed subspaces of Hilbert spaces.

Theorem 21. Let S be a closed subspace of the Hilbert space V . Let uV . Then thereexists a unique closest point u0S :

u u0 u v vV There holds

u u0S Proof: Let d := inf vS uv , and let ( vn ) be a minimizing sequence such that uvn d.We rst check that there holds

vn

vm 2 = 2 vn

u 2 + 2 vm

u 2

4 1/ 2(vn + vm )

u 2.

Since 1/ 2(vn + vm )S , there holds 1/ 2(vn + vm )u d. We proof that ( vn ) is a Cauchysequence: Fix > 0, choose N N such that for n > N there holds u vn 2 d2 + 2.Thus for all n,m > N there holdsvn vm 2 2(d2 + 2) + 2( d2 + 2) 4d2 = 42.

Thus, vn converge to some u0V . Since S is closed, u0S . By continuity of the norm,u u0 = d.

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2.3. RIESZ REPRESENTATION THEOREM 15

Fix some arbitrary wS , and dene (t) := uu0 tw

S

2. () is a convex function,it takes its unique minimum d at t = 0. Thus

0 =d(t)

dt |t=0 = {2(u u0, w) 2t(w, w)}|t=0 = 2( u u0, w)We obtained u u0S . If there were two minimizers u0 = u1, then u0 u1 = ( u0 u) (u1 u)S and u0 u1S , which implies u0 u1 = 0, a contradiction. 2

Theorem 21 says that given an uV , we can uniquely decompose it as

u = u0 + u1, u0S u1S

This allows to dene the operators P S : V S and P

S : V S

asP S u := u0 P S u := ( I P S )u = u1

Theorem 22. P S and P S are linear operators.

Denition 23. A linear operator P is called a projection if P 2 = P . A projector iscalled orthogonal , if (Pu,v ) = ( u,Pv ).

Lemma 24. The operators P S and P S are both orthogonal projectors.

Proof: For u

S there holds Pu = u. Since Pu

S , there holds P 2u = P u . It isorthogonal since

(Pu,v ) = ( Pu,v Pv + P v) = ( P u S , v Pv S

) + ( Pu,Pv ) = ( Pu,Pv ).

With the same argument there holds ( u,Pv ) = ( Pu,Pv ). The co-projector P S = I P S is a projector since(I P S )2 = I 2P S + P 2S = I P S .

It is orthogonal since (( I P S )u, v) = ( u, v) (P su, v) = ( u, v) (u, P S v) = ( u, (I P S )v)2

2.3 Riesz Representation Theorem

Let uV . Then, we can dene the related continuous linear functional lu ()V bylu (v) := ( u, v)V vV.

The opposite is also true:

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Theorem 25. Riesz Representation Theorem. Any continuous linear functional l on a Hilbert space V can be represented uniquely as

l(v) = ( ul, v) (2.1)

for some ulV . Furthermore, we have

l V = ul V .

Proof: First, we show uniqueness. Assume that u1 = u2 both fulll (2.1). This leads tothe contradiction

0 = l(u1 u2) l(u1 u2)= ( u1, u1

u2)

(u2, u1

u2) = u1

u2 2.

Next, we construct the ul. For this, dene S := ker l. This is a closed subspace.Case 1: S = {0}. Then, S = V , i.e., l = 0. So take u l = 0.Case 2: S = {0}. Pick some 0 = zS . There holds l(z) = 0 (otherwise, zS S ={0}). Now dene

ul :=l(z)z 2

z S

Then

(ul, v) = ( ul

S

, v

l(v)/l (z)z

S

) + ( ul, l(v)/l (z)z)

= l(z)/ z 2(z, l(v)/l (z)z)= l(v)

Finally, we prove l V = ul V :

l V = sup0= vV

l(v)v

= supv

(ul, v)V v V ul V

and

u =l(z)z 2 z =

l(z)z l V .

2.4 Symmetric variational problems

Take the function space C 1(), and dene the bilinear form

A(u, v) := u v + uv ds

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with bounds 1 and 2 in R + . It is not necessarily symmetric. Let f (.) : V R be acontinuous linear form on V , i.e.,f (v) f V v V .

We are posing the variational problem: nd uV such that

A(u, v) = f (v) vV.

Example 26. Diffusion-reaction equation:

Consider the PDE

div(a(x) u) + c(x)u = f in ,with Neumann boundary conditions. Let V be the Hilbert space generated by the inner

product ( u, v)V := ( u, v)L 2 + ( u, v)L2 . The variational formulation of the PDE involvesthe bilinear form

A(u, v) = (a(x) u) v dx + c(x)uv dx.Assume that the coefficients a(x) and c(x) fulll a(x)

R d d, a(x) symmetric and 1 min (a(x)) max (a(x)) 2, and c(x) such that 1 c(x) 2 almost everywhere. ThenA(, ) is coercive with constant 1 = min {1, 1}and 2 = max {2, 2}.Example 27. Diffusion-convection-reaction equation:

The partial differential equation

u + b u + u = f in with Dirichlet boundary conditions u = 0 on leads to the bilinear form

A(u, v) = u v dx + b uvdx + uv dx.If div b 0, what is an important case arising from incompressible ow elds (div b = 0),then A(, ) is coercive and continuous w.r.t. the same norm as above.

Instead of the linear form f (

), we will often write f

V . The evaluation is writtenas the duality product

f, v V V = f (v).

Lemma 28. A continuous bilinear form A(, ) : V V R induces a continuous linear operator A : V V via Au,v = A(u, v) u, vV.

The operator norm A V V is bounded by the continuity bound 2 of A(, ).

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2.5. COERCIVE VARIATIONAL PROBLEMS 19

Proof: For every uV , A(u, ) is a bounded linear form on V with normA(u,

) V = sup

vV

A(u, v)

v V supvV

2 u V v V

v V = 2 u V

Thus, we can dene the operator A : uV A(u, )V . It is linear, and its operatornorm is bounded byA V V = sup

uV

Au V u V

= supuV

supvV

Au,v V V u V v V

= supuV

supvV

A(u, v)u V v V supuV supvV

2 u V v V u V v V

= 2.

2

Using this notation, we can write the variational problem as operator equation: nduV such that Au = f (in V ).

Theorem 29 (Banachs contraction mapping theorem) . Given a Banach space V and a mapping T : V V , satisfying the Lipschitz condition

T (v1) T (v2) L v1 v2 v1, v2V for a xed L[0, 1). Then there exists a unique uV such that

u = T (u),

i.e. the mapping T has a unique xed point u. The iteration u1V given, compute

uk+1 := T (uk)

converges to u with convergence rate L:

u uk+1 L u ukTheorem 30 (Lax Milgram) . Given a Hilbert space V , a coercive and continuous bilinear form A(, ), and a continuous linear form f (.). Then there exists a unique uV solving

A(u, v) = f (v) vV.

There holdsu V 11 f V (2.4)

Proof: Start from the operator equation Au = f . Let J V : V V be the Rieszisomorphism dened by(J V g, v)V = g(v) vV, gV

.

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Then the operator equation is equivalent to

J V Au = J V f (in V ),

and to the xed point equation (with some 0 = R chosen below)

u = u J V (Au f ). (2.5)We will verify that

T (v) := v J V (Av f )is a contraction mapping, i.e., T (v1)T (v2) V L v1v2 V with some Lipschitz constantL[0, 1). Let v1, v2V , and set v = v1 v2. Then

T (v1) T (v2) 2V = {v1 J V (Av1 f )} {v2 J V (Av2 f )} 2V = v

J V Av 2V

= v 2V 2 (J V Av,v )V + 2 J V Av 2V = v 2V 2 Av,v + 2 Av 2V = v 2V 2A(v, v) + 2 Av 2V v 2V 2 1 v 2V + 222 v 2V = (1 2 1 + 222) v1 v2 2V

Now, we choose = 1/ 22, and obtain a Lipschitz constant

L2 = 1 21/ 22[0, 1).

Banachs contraction mapping theorem state that (2.5) has a unique xed point. Finally,we obtain the bound (2.4) from

u 2V 11 A(u, u ) = 11 f (u) 11 f V u V ,and dividing by one factor u . 2

2.5.1 Approximation of coercive variational problems

Now, let V h be a closed subspace of V . We compute the approximation uh V h by theGalerkin methodA(uh , vh ) = f (vh )

vh

V h . (2.6)

This variational problem is uniquely solvable by Lax-Milgram, since, ( V h , . V ) is a Hilbertspace, and continuity and coercivity on V h are inherited from the original problem on V .

The next theorem says, that the solution dened by the Galerkin method is, up to aconstant factor, as good as the best possible approximation in the nite dimensional space.

Theorem 31 (Cea) . The approximation error of the Galerkin method is quasi optimal

u uh V 2/ 1 inf vV h u vh V

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2.6. INF-SUP STABLE VARIATIONAL PROBLEMS 21

Proof: A fundamental property is the Galerkin orthogonality

A(u

uh , wh ) = A(u, wh )

A(uh , wh ) = f (wh )

f (wh ) = 0

wh

V h .

Now, pick an arbitrary vhV h , and bound

u uh 2V 11 A(u uh , u uh )= 11 A(u uh , u vh ) + 11 A(u uh , vh uh V h

)

2/ 1 u uh V u vh V .Divide one factor u uh . Since vhV h was arbitrary, the estimation holds true also forthe inmum in V h . 2

If A(, ) is additionally symmetric, then it is an inner product. In this case, the coer-civity and continuity properties are equivalent to to1 u 2V A(u, u ) 2 u 2V uV.

The generated norm . A is an equivalent norm to . V . In the symmetric case, we canuse the orthogonal projection with respect to ( ., .)A to improve the bounds to

u uh 2V 11 u uh 2A 11 inf vhV h u vh2A 2/ 1 u vh 2V .

The factor in the quasi-optimality estimate is now the square root of the general, non-symmetric case.

2.6 Inf-sup stable variational problems

The coercivity condition is by no means a necessary condition for a stable solvable system.A simple, stable problem with non-coercive bilinear form is to choose V = R 2, and thebilinear form B(u, v) = u1v1u2v2. The solution of B(u, v) = f T v is u1 = f 1 and u2 = f 2.We will follow the convention to call coercive bilinear forms A(, ), and the more generalones B(, ).Let V and W be Hilbert spaces, and B(, ) : V W R be a continuous bilinear formwith bound

B(u, v) 2 u V v W uV, vW. (2.7)The general condition is the inf-sup condition

inf uV u =0

supvW v =0

B(u, v)u V v W 1. (2.8)

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Dene the linear operator B : V W by Bu,v W W = B(u, v). The inf-supcondition can be reformulated assupvW

Bu,vv W 1 u V , uV and, using the denition of the dual norm,

Bu W 1 u V . (2.9)We immediately obtain that B is one to one, since

Bu = 0u = 0

Lemma 32. Assume that the continuous bilinear form B(, ) fullls the inf-sup condition (2.8). Then the according operator B has closed range.

Proof: Let Bu n be a Cauchy sequence in W . From (2.9) we conclude that also un isCauchy in V . Since V is complete, un converges to some uV . By continuity of B , thesequence Bu n converges to Bu W

. 2The inf-sup condition (2.8) does not imply that B is onto W . To insure that, we can

pose an inf-sup condition the other way around:

inf vW v =0

supuV u =0

B(u, v)u V v W 1. (2.10)

It will be sufficient to state the weaker condition

supuV u =0

B(u, v)u V v W

> 0 vW. (2.11)

Theorem 33. Assume that the continuous bilinear form B(, ) fullls the inf-sup condition (2.8) and condition (2.11). Then, the variational problem: nd uV such that B(u, v) = f (v) vW (2.12)

has a unique solution. The solution depends continuously on the right hand side:

u V 1

1 f W

Proof: We have to show that the range R(B) = W . The Hilbert space W can be splitinto the orthogonal, closed subspaces

W = R(B)R(B).

Assume that there exists some 0 = gR(B). This means that

(Bu,g )W = 0 uV.

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Theorem 36. Assume that B(, ) is continuous with bound 2, and B(, ) fullls thediscrete inf-sup condition with bound 1. Then there holds the quasi-optimal error estimateu uh (1 + 2/ 1) inf vhV h u vh (2.15)

Proof: Again, there holds the Galerkin orthogonality B(u, wh ) = B(uh , wh ) for all whV h .Again, choose an arbitrary vhV h :

u uh V u vh V + vh uh V u vh V + 11h supwhW h

B(vh uh , wh )wh V

= u vh V + 11h supwhW hB(vh u, wh )

wh V

u vh V + 11h supwhW h 2 vh u V wh W wh W

= (1 + 2/ 1h ) u vh V .

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Chapter 3

Sobolev Spaces

In this section, we introduce the concept of generalized derivatives, we dene families of normed function spaces, and prove inequalities between them. Let be an open subset of R d, either bounded or unbounded.

3.1 Generalized derivatives

Let = ( 1, . . . , d)N d0 be a multi-index, || = i , and dene the classical differentialoperator for functions in C ()

D =

x 1

1

x n

d

.

For a function uC (), the support is dened as

supp{u}:= {x : u(x) = 0}.This is a compact set if and only if it is bounded. We say u has compact support in , if supp u. If is a bounded domain, then u has compact support in if and only if uvanishes in a neighbourhood of .

The space of smooth functions with compact support is denoted as

D() := C 0 () := {uC () : u has compact support in }. (3.1)For a smooth function uC

| |(), there holds the formula of integration by parts

D udx = ( 1)| | uD dx D(). (3.2)The L2 inner product with a function u in C () denes the linear functional on D

u() := u, D D := udx.25

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26 CHAPTER 3. SOBOLEV SPACES

We call these functionals in D distributions. When u is a function, we identify it withthe generated distribution. The formula (3.2) is valid for functions uC . The strongregularity is needed only on the left hand side. Thus, we use the less demanding right handside to extend the denition of differentiation for distributions:

Denition 37. For uD, we dene gDto be the generalized derivative D g u of u by g, D D = ( 1)| | u, D D D D

If uC , then D g coincides with D .

The function space of locally integrable functions on is called

Lloc1 () = {u : uK L1(K )compact K }.It contains functions which can behave very badly near . E.g., ee1/x is in L1loc(0, 1). If is unbounded, then the constant function 1 is in Lloc1 , but not in L1.

Denition 38. For uLloc1 , we call g the weak derivative D wu, if gL

loc1 satises

g(x)(x) dx = ( 1)| | u(x)D (x) dx D.The weak derivative is more general than the classical derivative, but more restrictive

than the generalized derivative.

Example 39. Let = (1, 1) and

u(x) = 1 + x x 01 x x > 0Then,

g(x) = 1 x 01 x > 0

is the rst generalized derivative D 1g

of u, which is also a weak derivative. The second generalized derivative h is

h, = 2(0) DIt is not a weak derivative.

In the following, we will focus on weak derivatives. Unless it is essential we will skipthe sub-scripts w and g.

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3.2. SOBOLEV SPACES 27

3.2 Sobolev spaces

For k

N 0 and 1

p 0. Then, we can bound

u(0) = 1 xh

u(x)|x=0 = h0 1 xh u(x) dx= h0 1h u(x) + 1 xh u (x) dx

1h L2

u L 2 + 1 xh L 2

u L 2

h 1/ 2 u L 2 (0 ,h ) + h1/ 2 u L2 (0,h ) .This estimate includes the scaling with the interval length h. If we are not interested in thescaling, we apply Cauchy-Schwarz in R 2, and combine the L2 norm and the H 1 semi-normu L2 to the full H 1 norm and obtain

|u(0)| h 1/ 2 + h1/ 2 u 2L2 + u 2L2 = c u H 1 .Next, we extend the trace operator to the whole Sobolev space H 1:

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3.3. TRACE THEOREMS AND THEIR APPLICATIONS 29

Theorem 44. There is a well dened and continuous trace operator

tr : H 1((0, h))

R

whose restriction to C 1([0, h]) coincides with

u u(0).Proof: Use that C 1([0, h]) is dense in H 1(0, h). Take a sequence u j in C 1([0, h]) con-

verging to u in H 1-norm. The values u j (0) are Cauchy, and thus converge to an u0. Thelimit is independent of the choice of the sequence u j . This allows to dene tr u := u0. 2

Now, we extend this 1D result to domains in more dimensions. Let be bounded, be Lipschitz, and consists of M pieces i of smoothness C 1.

We can construct the following covering of a neighbourhood of in : LetQ = (0 , 1)2

.For 1 i M , let s iC 1(Q, ) be invertible and such that s i L c, (s i) 1 L c,and det s i > 0. The domains S i := s i(Q) are such that s i((0, 1) {0}) = i , and theparameterizations match on s i({0, 1} (0, 1)).Theorem 45. There exists a well dened and continuous operator

tr : H 1() L2( )which coincides with u| for uC 1().

Proof: Again, we prove thattr : C 1() L2( ) : u u|

is a bounded operator w.r.t. the norms . H 1 and L2, and conclude by density. We usethe partitioning of into the pieces i , and transform to the simple square domainQ = (0 , 1)2. Dene the functions ui on Q = (0 , 1)2 as

ui(x) = u(s i(x))

We transfer the L2 norm to the simple domain:

tr u 2L2 ( ) =M

i=1 i u(x)2 dx=

M

i=1 10 u(s i(, 0))2 s i (, 0) d c

M

i=1 10 ui(, 0)2 d

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To transform the H 1-norm, we differentiate with respect to x by applying the chain rule

du i

dx(x) =

du

dx(s i(x))

ds i

dx(x).

Solving for dudx isdudx

(s i(x)) =duidx

(x)dsdx

1

(x)

The bounds onto s and ( s ) 1 imply that

c 1 | x u| | x u| c | x u|We start from the right hand side of the stated estimate:

u 2H 1 () M

i=1 S i | x u|2 dx=

M

i=1 Q | x u(s i(x))|2 det( s ) dx c

M

i=1 Q | x u(x)|2 dxWe got a lower bound for det( s ) = (det( s ) 1) 1 from the upper bound for ( s ) 1.

It remains to prove the trace estimate on Q. Here, we apply the previous one dimen-sional result

|u(, 0)|2 c 10 u(, )2 + u(, ) 2 d (0, 1)The result follows from integrating over

10 |u(, 0)|2 d c 10 10 u(, )2 + u(, ) 2 d d

c u 2H 1 (Q) .

2

Considering the trace operator from H 1() to L2( ) is not sharp with respect to thenorms. We will improve the embedding later.

By means of the trace operator we can dene the sub-space

H 10 () = {uH 1() : tr u = 0}

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It is a true sub-space, since u = 1 does belong to H 1, but not to H 10 . It is a closed sub-space,since it is the kernel of a continuous operator.

By means of the trace inequality, one veries that the linear functional

g(v) := N g tr v dxis bounded on H 1.

Integration by parts

The denition of the trace allows us to perform integration by parts in H 1:

udx = u div dx + tr u n dx [C 1()]2

The denition of the weak derivative (e.g. the weak gradient) looks similar. It allows onlytest functions with compact support in , i.e., having zero boundary values. Only bychoosing a normed space, for which the trace operator is well dened, we can state andprove integration by parts. Again, the short proof is based on the density of C 1() in H 1.

Sobolev spaces over sub-domains

Let consist of M Lipschitz-continuous sub-domains i such that

=M i=1 i i j = if i = j

The interfaces are ij = i j . The outer normal vector of i is n i .Theorem 46. Let uL2() such that

ui := u|i is in H 1(i), and gi = ui is its weak gradient the traces on common interfaces coincide:

tr ij u i = tr ij u j

Then u belongs to H 1(). Its weak gradient g = u fullls g|i = gi .Proof: We have to verify that gL2()

d, dened by g|i = gi , is the weak gradient of u, i.e., g dx = u div dx [C 0 ()]d

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Dene the domain = S 1. . .S M .We dene the extension operator by

(Eu )(x) = u(x) xS i

(Eu )(x) = u(x) x(3.3)

Theorem 47. The extension operator E : H 1() H 1() is well dened and bounded with respect to the normsEu L2 (e) c u L2 ()

and Eu L2 (e) c u L 2 ()

Proof: Let uC 1(). First, we prove the estimates for the individual pieces S i :

S i \ Eu (x)2 dx = S i u(x)

2 detdxdx dx c u 2L 2 (S i )

For the derivatives we usedEu (x)

dx=

du(x(x))dx

=dudx

dxdx

.

Since dxdx and (dxdx )

1 = dxdx are bounded, one obtains

| xEu (x)| | x u(x)|,and

S i \ |xEu |

2dx c S i | u|

2dx

These estimates prove that E is a bounded operator into H 1 on the sub-domains S i \ .The construction was such that for uC 1(), the extension Eu is continuous across ,and also across the individual S i . By Theorem 46, Eu belongs to H 1(), and

Eu 2L2 () = u2 +

M

i=1

u 2S i \ c u 2L 2 () ,

By density, we get the result for H 1(). Let u j C 1() u, than u j is Cauchy, Eu j is

Cauchy in H 1(), and thus converges to u

H 1().

The extension of functions from H 10 () onto larger domains is trivial: Extension by 0is a bounded operator. One can extend functions from H 1() into H 10 (), and further, toan arbitrary domain by extension by 0.

For x = s i(, ), , (0, 1)2, dene the extensionE 0u(x) = (1 ) u(x)

This extension vanishes at

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Theorem 48. The extension E 0 is an extension from H 1() to H 10 (). It is bounded w.r.t.

E 0u H 1 (e)

c u H 1 ()

Proof: ExercisesIn this case, it is not possible to bound the gradient term only by gradients. To see this,

take the constant function on . The gradient vanishes, but the extension is not constant.

3.3.1 The trace space H 1 / 2

The trace operator is continuous from H 1() into L2( ). But, not every gL2( ) isa trace of some uH 1(). We will motivate why the trace space is the fractional order

Sobolev space H 1/ 2( ).We introduce a stronger space, such that the trace operator is still continuous, and

onto. Let V = H 1(), and dene the trace space as the range of the trace operator

W = {tr u : uH 1()}with the norm

tr u W = inf vV

tr u =tr v

v V . (3.4)

This is indeed a norm on W . The trace operator is continuous from V W with norm 1.Lemma 49. The space (W, . W ) is a Banach space. For all gW there exists an uV such that tr u = g and u V = g W

Proof: The kernel space V 0 := {v : tr v = 0}is a closed sub-space of V . If tr u = tr v,then z := u vV 0. We can rewritetr u W = inf

zV 0u z V = u P V 0 u V uV

Now, let gn = tr un W be a Cauchy sequence. This does not imply that un is Cauchy,but P V 0 un is Cauchy in V :

P V 0 (un um ) V = tr ( un um ) W .The P V 0 un converge to some uV

0 , and gn converge to g := tr u. 2The minimizer in (3.4) fullls

tr u = g and (u, v)V = 0 vV 0.

This means that u is the solution of the weak form of the Dirichlet problem

u + u = 0 in u = g on .

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To give an explicit characterization of the norm . W , we introduce Hilbert spaceinterpolation :

Let V 1

V 0 be two Hilbert spaces, such that V 1 is dense in V 0, and the embeddingoperator id : V 1 V 0 is compact. We can pose the eigen-value problem: Find z V 1,R such that

(z, v)V 1 = (z, v)V 0 vV.

There exists a sequence of eigen-pairs ( zk , k) such that k . The zk form an or-thonormal basis in V 0, and an orthogonal basis in V 1.The converse is also true. If zk is a basis for V 0, and the eigenvalues k , then theembedding V 1V 0 is compact.Given u

V 0, it can be expanded in the orthonormal eigen-vector basis:

u =

k=0

ukzk with uk = ( u, zk)V 0

The . V 0 - norm of u is

u 2V 0 = (k

ukzk ,l

ulzl)V 0 =k,l

ukul(zk , zl)V 0 =k

u2k .

If uV 1, then

u 2V 1 = (k

ukzk ,l

ulzl)V 0 =k,l

ukul(zk , zl)V 1 =k,l

ukulk(zk , zl)V 0 =k

u2kk

The sub-space space V 1 consists of all u = ukzk such that k ku2k is nite. This suggests

the denition of the interpolation norm

u 2V s =k

(u, zk)2V 0 sk ,

and the interpolation space V s = [V 0, V 1]s as

V s = {uV 0 : u V s < }.We have been fast with using innite sums. To make everything precise, one rst workswith nite dimensional sub-spaces {u :nN and u = nk=1 ukzk}, and takes the closure.

In our case, we apply Hilbert space interpolation to H 1(0, 1)L2(0, 1). The eigen-valueproblem is to nd zkH 1 and k

R such that

(zk , v)L 2 + ( zk , v )L 2 = k (zk , v)L 2 vH 1

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By denition of the weak derivative, there holds ( zk) = (1 k)zk , i.e., zk H 2. SinceH 2C 0, there holds also zC 2, and a weak solution is also a solution of the strong formzk zk = kzk on (0, 1)zk(0) = zk(1) = 0 (3.5)

All solutions, normalized to zk L2 = 1, are

z0 = 1 0 = 1

and, for kN ,

zk(x) = 2cos(kx) k = 1 + k22.Indeed, expanding uL2 in the cos-basis u = u0 +

k=1 uk2cos(kx), one has

u 2L 2 =

k=0

(u, zk)2L2

and

u 2H 1 =

k=0

(1 + k22)(u, zk)2L2

Differentiation adds a factor k . Hilbert space interpolation allows to dene the fractionalorder Sobolev norm ( s(0, 1))

u 2H s (0,1) =

k=0(1 + k22)s (u, zk)2L 2

We consider the trace tr |E of H 1((0, 1)2) onto one edge E = (0 , 1) {0}. For g W E := tr H 1((0, 1)2), the norm g W is dened byg W = ug H 1 .

Here, ug solves the Dirichlet problem ug|E = g, and (ug, v)H 1 = 0 v H 1 such thattr E v = 0.Since W L

2(E ), we can expand g in the L2-orthonormal cosine basis zk

g(x) = gn zk(x)

The Dirichlet problems for the zk ,

uk + uk = 0 in uk = zk on E u kn = 0 on \ E,

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3.4. EQUIVALENT NORMS ON H 1 AND ON SUB-SPACES 37

have the explicit solutionu0(x, y) = 1

anduk(x, y) = 2cos(kx ) e

k (1 y) + e k (1 y)

ek + e k.

The asymptotic isuk 2L2 (k + 1)

1

anduk 2L2 k

Furthermore, the uk are orthogonal in ( ., .)H 1 . Thus ug = n gn uk has the norm

ug2H 1 = g

2n uk

2H 1 g

2n (1 + k).

This norm is equivalent to H 1/ 2(E ).We have proven that the trace space onto one edge is the interpolation space H 1/ 2(E ).

This is also true for general domains (Lipschitz, with piecewise smooth boundary).

3.4 Equivalent norms on H 1 and on sub-spacesThe intention is to formulate 2 nd order variational problems in the Hilbert space H 1. Wewant to apply the Lax-Milgram theory for continuous and coercive bilinear forms A(., .).We present techniques to prove coercivity.

The idea is the following. In the norm

v 2H 1 = v2L 2 + v

2L2 ,

the L 2 -semi-norm is the dominating part up to the constant functions. The L2 normis necessary to obtain a norm. We want to replace the L2 norm by some different term(e.g., the L2-norm on a part of , or the L2-norm on ), and want to obtain an equivalentnorm.

We formulate an abstract theorem relating a norm . V to a semi-norm . A . An

equivalent theorem was proven by Tartar.Theorem 50 (Tartar) . Let (V, (., .)V ) and (W, (., .)W ) be Hilbert spaces, such that theembedding id : V W is compact. Let A(., .) be a non-negative, symmetric and V -continuous bilinear form with kernel V 0 = {v : A(v, v) = 0}. Assume that

v 2V v2W + v

2A vV (3.6)

Then there holds

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1. The kernel V 0 is nite dimensional. On the factor space V/V 0, A(., .) is an equivalent norm to the quotient norm

u A inf vV 0

u v V uV (3.7)

2. Let B(., .) be a continuous, non-negative, symmetric bilinear form on V such that A(., .) + B(., .) is an inner product. Then there holds

v 2V v2A + v

2B vV

3. Let V 1V be a closed sub-space such that V 0 V 1 = {0}. Then there holdsv V v A vV 1

Proof: 1. Assume that V 0 is not nite dimensional. Then there exists an ( ., .)V -orthonormalsequence uk V 0. Since the embedding id : V W is compact, it has a sub-sequenceconverging in . W . But, since

2 = uk ul 2V uk ul 2W + uk ul A = uk ul 2W for k = l, uk is not Cauchy in W . This is a contradiction to an innite dimensional kernelspace V 0. We prove the equivalence (3.7). To bound the left hand side by the right handside, we use that V 0 = ker A, and norm equivalence (3.6):

u A = inf vV 0

u v A inf vV 0 u v V The quotient norm is equal to P V 0 u . We have to prove that P V 0 u V P V 0 u A forall uV . This follows after proving u V u A for all uV 0 . Assume that this is nottrue. I.e., there exists a V -orthogonal sequence ( uk) such that uk A k 1 uk V . Extracta sub-sequence converging in . W , and call it uk again. From the norm equivalence (3.6)there follows

2 = uk ul2V uk ul W + uk ul A 0

2. On V 0, . B is a norm. Since V 0 is nite dimensional, it is equivalent to . V , say withbounds

c1 v 2V v 2B c2 v 2V vV 0From 1. we know that

c3 v 2V v 2A c4 v 2V vV 0 .

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3.4. EQUIVALENT NORMS ON H 1 AND ON SUB-SPACES 39

Now, we bound

u 2V = P V 0 u2V + P V 0 u

2V

1c1 P V 0 u u P V 0 u2B + P V 0

2V

2c1

u 2B + c2 P V 0 u2V + P V 0 u

2V

=2c1

u 2B +1c2

1 +2c2c1

P V 0 u2A

u 2B + u 2A3. Dene B(u, v) = ( P V 1 u, P V 1 u )V . Then A(., .) + B(., .) is an inner product: A(u, u ) +

B(u, u ) = 0 implies that u V 0 and u V

1, thus u = {0}. From 2. there follows thatA(., .) + B(., .) is equivalent to ( ., .)V . The result follows from reducing the equivalence toV 1.

2

We want to apply Tartars theorem to the case V = H 1, W = L2, and v A = v L2 .The theorem requires that the embedding id : H 1 L2 is compact. This is indeed truefor bounded domains :Theorem 51. The embedding of H k H l for k > l is compact.

We sketch a proof for the embedding H 1 L2. First, prove the compact embeddingH 10 (Q)

L2(Q) for a square Q, w.l.o.g. set Q = (0 , 1)2. The eigen-value problem: Find

zH 10 (Q) and such that

(z, v)L 2 + ( z, v)L 2 = (u, v)L 2 vH 10 (Q)

has eigen-vectors zk,l = sin (kx)sin (ly ), and eigen-values 1 + k22 + l22 . Theeigen-vectors are dense in L2. Thus, the embedding is compact.On a general domain Q, we can extend H

1() into H 10 (Q), embed H 10 (Q) intoL2(Q), and restrict L2(Q) onto L2(). This is the composite of two continuous and acompact mapping, and thus is compact. 2

The kernel V 0 of the semi-norm v is the constant function.

Theorem 52 (Friedrichs inequality) . Let D be of positive measure |D |. Let V D = {vH 1() : trD v = 0}. Then v L2 v L 2 vV D

Proof: The intersection V 0 V D is trivial {0}. Thus, Theorem 50, 3. implies theequivalencev 2V = v

2L2 + v

2L 2 v L2 .

2

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Theorem 53 (Poincare inequality) . There holds

v 2H 1 ()

v 2L 2 + (

v dx)2

Proof: B(u, v) := ( u dx)( v dx) is a continuous bilinear form on H 1, and ( u, v)+B(u, v) is an inner product. Thus, Theorem 50, 2. implies the stated equivalence. 2 Let have positive measure || in R d. Then

u 2H 1 () v2L2 () + v L 2 () ,

Let have positive measure | | in R d 1. Thenu 2H 1 () v

2L 2 () + v L2 ( ) ,

Theorem 54 (Bramble Hilbert lemma) . Let U be some Hilbert space, and L : H k U bea continuous linear operator such that Lq = 0 for polynomials qP k 1. Then there holdsLv U |v|H k .

Proof: The embedding H k H k 1 is compact. The V -continuous, symmetric andnon-negative bilinear form A(u, v) = :| |= k( u, v) has the kernel P p 1. Decomposeu 2H k = u

2H k 1 + A(u, u ). By Theorem 50, 1, there holds

u A inf vV 0

u v H k

The same holds for the bilinear-formA2(u, v) := ( Lu, Lv )U + A(u, v)

Thusu A2 inf vV 0

u v H k uV Equalizing both implies that

(Lu,Lu )U u 2A2 u 2A uV,i.e., the claim.

We will need point evaluation of functions in Sobolev spaces H s . This is possible, weuH

s implies that u is continuous.

Theorem 55 (Sobolevs embedding theorem) . Let R d with Lipschitz boundary. If

uH s with s > d/ 2, then uL with

u L u H s

There is a function in C 0 within the L equivalence class.

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Chapter 4

The weak formulation of the Poissonequation

We are now able to give a precise denition of the weak formulation of the Poisson problemas introduced in Section 1.2, and analyze the existence and uniqueness of a weak solution.

Let be a bounded domain. Its boundary is decomposed as = D N Raccording to Dirichlet, Neumann and Robin boundary conditions.Let

uD H 1/ 2(D ), f L2(),

gL2(N R ), L (D ), 0.Assume that there holds

(a) The Dirichlet part has positive measure |D | > 0,(b) or the Robin term has positive contribution R dx > 0.Dene the Hilbert space

V := H 1(),

the closed sub-spaceV 0 = {v : trD v = 0},

and the linear manifoldV D = {uV : trD u = uD }.

Dene the bilinear form A(., .) : V V R

A(u, v) = u v dx + R uv ds41

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42 CHAPTER 4. THE WEAK FORMULATION OF THE POISSON EQUATION

and the linear formf (v) =

fvdx +

N R

gv dx.

Theorem 56. The weak formulation of the Poisson problem

Find uV D such that

A(u, v) = f (v) vV 0 (4.1)

has a unique solution u.

Proof: The bilinear-form A(., .) and the linear-form f (.) are continuous on V . Tartarstheorem of equivalent norms proves that A(., .) is coercive on V 0.

Since uD is in the closed range of tr D , there exists an uD V D such that

tr uD = uD and uD V uD H 1/ 2 ( D )

Now, pose the problem: Find zV 0 such that

A(z, v) = f (v) A(uD , v) vV 0.The right hand side is the evaluation of the continuous linear form f (.) A(uD , .) onV 0. Due to Lax-Milgram, there exists a unique solution z. Then, u := uD + z solves (4.1).

The choice of uD is not unique, but, the constructed u is unique. 2

4.1 Shift theorems

Let us restrict to Dirichlet boundary conditions uD = 0 on the whole boundary. Thevariational problem: Find uV 0 such that

A(u, v) = f (v) vV 0

is well dened for all f V

0 , and, due to Lax-Milgram there holds

u V 0 c f V 0 .Vice versa, the bilinear-form denes the linear functional A(u, . ) with norm

A(u, .) V 0 c u V 0This dual space is called H 1:

H 1 := [H 10 ()]

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4.1. SHIFT THEOREMS 43

Since H 10 L2, there is L2 H 1(). All negative spaces are dened as H s () :=

[H s0 ](), for sR + . There holds

. . . H 20 H 10L

2H

1H

2 . . .

The solution operator of the weak formulation is smoothing twice. The statements of shift theorem are that for s > 0, the solution operator maps also

f H 1+ s uH 1+ s

with norm boundsu H 1+ s f H 1+ s .

In this case, we call the problem H 1+ s - regular.

Theorem 57 (Shift theorem) .

(a) Assume that is convex. Then, the Dirichlet problem is H 2 regular.

(b) Let s 2. Assume that C s . Then, the Dirichlet problem is H s -regular.We give a proof of (a) for the square (0 , )2 by Fourier series. Let

V N = span {sin(kx)sin( ly) : 1 k, l N }For an u = N k,l =1 ukl sin(kx)sin(ly)V N , there holds

u 2H 2 = u 2L2 + x u 2L 2 + yu 2L2 + 2x u 2L2 + x yu 2L2 + 2y u 2N

k,l =1

(1 + k2 + l2 + k4 + k2l2 + l4)u2kl

N

k,l =1

(k4 + l4)u2kl ,

and, for f = u,

u2L2 =

N

k,l =1(k

2+ l

2)

2u

2kl

N

k,l =1(k

4+ l

4)u

2kl .

Thus we have u H 2 u L2 = f L2 for uV N . The rest requires a closure argument:There is { v : vV N }= V N , and V N is dense in L2. 2Indeed, on non-smooth non-convex domains, the H 2-regularity is not true. Take thesector of the unit-disc

= {(r cos , r sin ) : 0 < r < 1, 0 < < }

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Chapter 5

Finite Element Method

Ciarlets denition of a nite element is:

Denition 59 (Finite element) . A nite element is a triple (T, V T , T ), where

1. T is a bounded set

2. V T is function space on T of nite dimension N T

3. T = {1T , . . . , N T T }is a set of linearly independent functionals on V T .The nodal basis {1T . . .N T T }for V T is the basis dual to T , i.e.,

iT ( jT ) = ij

Barycentric coordinates are useful to express the nodal basis functions.Finite elements with point evaluation functionals are called Lagrange nite elements,

elements using also derivatives are called Hermite nite elements.Usual function spaces on T

R 2 are

P p := span {xiy j : 0 i, 0 j, i + j p}Q p := span {xiy j : 0 i p, 0 j p}

Examples for nite elements are

A linear line segment

A quadratic line segment A Hermite line segment A constant triangle A linear triangle A non-conforming triangle

45

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46 CHAPTER 5. FINITE ELEMENT METHOD

A Morley triangle A Raviart-Thomas triangle

The local nodal interpolation operator dened for functions vC m (T ) is

I T v :=N T

=1

T (v)T

It is a projection.Two nite elements ( T, V T , T ) and ( T , V bT , bT ) are called equivalent if there exists aninvertible function F such that T = F (

T )

V T = {v F 1 : vV bT } T = {T i : V T R : v T i (v F )}

Two elements are called affine equivalent, if F is an affine-linear function.Lagrangian nite elements dened above are equivalent. The Hermite elements are not

equivalent.Two nite elements are called interpolation equivalent if there holds

I T (v) F = I bT (v F )Lemma 60. Equivalent elements are interpolation equivalent

The Hermite elements dene above are also interpolation equivalent.A regular triangulation T = {T 1, . . . , T M }of a domain is the subdivision of a domain into closed triangles T i such that = T i and T i T j is either empty or an common edge of T i and T j or T i = T j in the case i = j .

In a wider sense, a triangulation may consist of different element shapes such as segments,triangles, quadrilaterals, tetrahedra, hexhedra, prisms, pyramids.

A nite element complex {(T, V T , T )}is a set of nite elements dened on the geo-metric elements of the triangulation T .It is convenient to construct nite element complexes such that all its nite elementsare affine equivalent to one reference nite element ( T , V T , T ). The transformation F T issuch that T = F T ( T ).Examples: linear reference line segment on (0 , 1).

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5.1. FINITE ELEMENT SYSTEM ASSEMBLING 47

The nite element complex allows the denition of the global interpolation operator forC m -smooth functions by

I T v|T = I T vT

T

T The nite element space isV T := {v = I T w : wC m ()}

We say that V T has regularity r if V T C r . If V T = C 0, the regularity is dened as 1.Examples:

The P 1 - triangle with vertex nodes leads to regularity 0. The P 1 - triangle with edge midpoint nodes leads to regularity 1.

The P 0 - triangle leads to regularity 1.For smooth functions, functionals T, and eT , sitting in the same location are equiv-

alent. The set of global functionals = {1, . . . , N }is the linearly independent set of functionals containing all (equivalence classes of) local functionals.The connectivity matrix C T

R N N T is dened such that the local functionals arederived from the global ones by

T (u) = C tT (u)

Examples in 1D and 2DThe nodal basis for the global nite element space is the basis in V T dual to the global

functionals j , i.e., j (i) = ij

There holds

i|T = I T i =N T

=1

T (i)T

=N T

=1

(C tT (i))T

=N T

=1(C tT ei)T =

N T

=1C T,i T

5.1 Finite element system assembling

As a rst step, we assume there are no Dirichlet boundary conditions. The nite elementproblem is

Find uhV T such that : A(uh , vh ) = f (vh ) vhV T (5.1)

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The nodal basis and the dual functionals provides the one to one relation between R N andV T :

RN

u uhV T with uh =N

i=1iui and ui = i(uh ).

Using the nodal basis expansion of uh in (5.1), and testing only with the set of basisfunctions, one has

A(N

i=1

uii , j ) = f ( j ) j = 1 . . . N

WithA ji = A(i , j ) and f j = f ( j ),

one obtains the linear system of equations

Au = f

The preferred way to compute the matrix A and vector f is a sum over element contribu-tions. The restrictions of the bilinear and linear form to the elements are

AT (u, v) = T u v dx + T uv dsand

f T (v) = T fvdx + T gv dsThen

A(u, v) =T T

AT (u, v) f (v) =T T

f T (v)

On each element, one denes the N T N T element matrix and element vector in termsof the local basis on T :AT, = AT (T ,

T ) f T, = f T (

T )

Then, the global matrix and the global vector are

A =T T

C T AT C tT

and f =T T

C T f T

Namely,

f i = f (i) =T T

f T (i|T ) =T T

f T (

C T,i T )

=T T

C T,i f T (T ) =T T

C T,i f

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5.2. FINITE ELEMENT ERROR ANALYSIS 49

and

Aij =T T

A(i|T , j |T ) =T T

A(

C T,i T ,

C T,j T )

=T T

C T,i AT, C T,j

On the elements T , the integrands are smooth functions. Thus, numerical integrationrules can be applied.

In the case of Dirichlet boundary conditions, let D {1, . . . , N }correspond to thevertices xi at the Dirichlet boundary, and f = {1, . . . N } \ D .We have the equations

i D

A ji ui +i f

A ji ui = f j j f

Inserting ui = uD (xi) for i i results in the reduced system

i f

A ji ui = f j i D

A ji uD (xi)

An alternative approach is to approximate Dirichlet boundary conditions by Robin b.c.,un + u = u D , with large parameter .

5.2 Finite element error analysis

Let u be the solution of the variational problem, and uh its Galerkin approximation in thenite element sub-space V h . Ceas Lemma bounds the nite element error u uh by thebest approximation error

u uh V C inf vV h u vh V .The constant factor C is the ratio of the continuity bound and the coercivity bound of thebilinear form A(., .).

Provided that the solution u is sufficiently smooth, we can take the nite element

interpolant to bound the best approximation error:

inf vV h

u vh V u I T u V In the following, we will bound the interpolation error.

Lemma 61. Let T and T be d-dimensional domains related by the invertible affine linear transformation F T : T T F T (x) = a + Bx,

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where aR d and B is a regular matrix in R d d. Then there holds:

u F T L2 ( bT ) = (det B) 1/ 2 u L 2 (T ) (5.2)

x im. . .

x i1

(u F T ) =d

jm =1

. . .d

j1 =1

x j m

. . .

x j1u F T B jm ,i m . . . B j1 ,i 1 (5.3)

|u F |H m ( bT ) (det B) 1/ 2 B m |u|H m (T ) (5.4)Proof: Transformation of integrals, chain rule. 2

We dene the diameter of the element T

hT = diam T

A triangulation is called shape regular , if all its elements fulll

|T | h2T with a good constant 1. If one studies convergence, one considers families of triangu-lations with decreasing element sizes hT . In that case, the family of triangulations is calledshape regular, if there is a common constant C such that all elements of all triangulationsfulll |T | Ch2T .Lemma 62. Let F T = a + Bx be the mapping from the reference triangle to the triangleT . Let |T | h2T . Then there holds

BT hT B 1T h 1T The following lemma is the basis for the error estimate. This lemma is the main

application for the Bramble Hilbert lemma. Sometimes, it is called the Bramble Hilbertlemma itself:

Lemma 63. Let (T, V T , T ) be a nite element such that the element space V T containspolynomials up to order P k . Then there holds

v I T v H 1 C |v|H m vH m (T ) for all m > d/ 2, m 1, and m k + 1 .

Proof: First, we prove that id I T is a bounded operator from H m to H 1:v I T v H 1 v H 1 + I T v H 1 = v H 1 +

(v) H 1

v H 1 +

H 1 | (v)| v H m

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The last step used that for H m , with m > d/ 2, point evaluation is continuous. Now, letvP

k(T ). Since P k V T , and I T is a projection on V T , there holds v I T v = 0. TheBramble Hilbert Lemma applied for U = H 1 and L = id

I T proves the result. 2

To bound the nite element interpolation error, we will transform functions from theelements T to the reference element T .

Theorem 64. Let T be a shape regular triangulation of . Let V T be a C 0-regular niteelement space such that all local spaces contain P 1. Then there holdsv I T v L2 ()

T T

h4T |v|H 2 (T )21/ 2

vH 2()

|v I T v|2H 1 () C T T

h2T |v|H 2 (T )2 vH 2()Proof: We prove the H 1 estimate, the L2 one follows the same lines. The interpolation

error on each element is transformed to the interpolation error on one reference element:

|v I T v|2H 1 () =T T

|(id I T )vT |2H 1 (T )

T T

(det BT ) B 1T 2 |(id I T )vT F T |2H 1 ( bT )

=T T

(det BT ) B 1T 2 (id I bT )(vT F T ) 2H 1 ( bT )

On the reference element T we apply the Bramble-Hilbert lemma. Then, we transformback to the individual elements:

|v I T v|2H 1 ()T T

(det BT ) B 1T 2|vT F T |2H m ( bT )

T T

(det BT ) B 1T 2 (det B 1T ) BT

4 |vT |2H 2 (T )

T T

h2T vT 2H 2 (T ) .

2

A triangulation is called quasi

uniform , is all elements are essentially of the same

size, i.e., there exists one global h such that

h hT T T .On a quasi-uniform mesh, there hold the interpolation error estimates

u I T u L2 () h2 |u|H 2|u I T u|H 1 () h |u|H 2

We are interested in the rate of the error in terms of the mesh-size h.

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Theorem 65 (Finite element error estimate) . Assume that

the solution u of the weak bvp is in H 2, the triangulation T is quasi-uniform of mesh-size h, the element spaces contain P 1.

Then, the nite element error is bounded by

u uh H 1 h |u|H 2Error estimates in L2-norm

The above theorem bounds the error in the L2-norm of the function, and the L2-norm of

the derivatives with the same rate in terms of h. This is obtained by the natural norm of the variational formulation.The interpolation error suggests a faster convergence in the weaker norm L2. Under

certain circumstances, the nite element error measured in L2 also decays faster. Theconsidered variational problem is

Find uV : A(u, v) = f (v) vV.

We dene the dual problem as

Find wV : A(v, w) = f (v) vV.

In the case of a symmetric bilinear form, the primal and the dual problem coincide.Theorem 66 (Aubin-Nitsche) . Assume that

the dual weak bvp isH 2 regular the triangulation T is quasi-uniform of mesh-size h, the element spaces contain P 1.

Then, there holds the L2-error estimate

u

uh L 2 h2

|u

|H 2

Proof: Solve the dual problem with the error u uh as right hand side:Find wV : A(v, w) = ( u uh , v)L 2 vV.

Since the dual problem is H 2 regular, there holds w H 2, and w H 2 u uh L2 .Choose the test function v := u uh to obtain the squared norm

A(u uh , w) = ( u uh , u uh )L 2 .

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Using the Galerkin orthogonality A(u uh , vh ) = 0 for all vhV h , we can insert I T w:u

uh 2L 2 = A(u

uh , w

I T w).

Next we use continuity of A(., .) and the interpolation error estimates:

u uh 2L2 u uh H 1 w I T w H 1 u uh H 1 h |w|H 2 .From H 2 regularity:

u uh 2L2 h u uh H 1 u uh L2 ,and, after dividing one factor

u uh L2 h u uh H 1 h2 u H 2 .2

Approximation of Dirichlet boundary conditions

Till now, we have neglected Dirichlet boundary conditions. In this case, the continuousproblem is

Find uV D : A(u, v) = f (v) vV 0,

where

V D =

{v

H 1 : trD

v = uD}

and V 0 =

{v

H 1 : trD

v = 0

}.

The nite element problem is

Find uhV hD : A(uh , vh ) = f (vh ) vhV h0,

whereV hD = {I T v : vV D } and V h0 = {I T v : vV 0}.

The denition of V hD coincides with {vhV h : vh (xi) = uD (xi) vertices xi on D }.There holds V h0V 0, but, in general, there does not hold V hD V D .

Theorem 67 (Error estimate for Dirichlet boundary conditions) . Assume that

A(., .) is coercive on V h0:A(vh , vh ) 1 vh 2V vhV h0

A(., .) is continuous on V :A(u, v) 2 u V v V u, vV

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Then there holds the nite element error estimate

u uh H 1 h|u|H 2Proof: To make use of the coercivity of A(., .), we need an element in V h0. There holds

Galerkin orthogonality A(u uh , vh ) = 0 vhV h0:u uh 2V = u I h u + I h u uh 2V 2 u I h u 2V + 2 I h u uh 2V

2 u I h u 2V +2

1A(I h u uh , I h u uh )

2 u I h u 2V +2

1A(I h u u, I h u uh ) +

21

A(u uh , I h u uh )

2 u I h u 2V +221

I h u u I h u uh + 0

2 u I h u2V +

221 I h u u ( I h u u + u uh )

= (2 +221

) u I h u 2V +221

u I h u V u uh V Next, we apply ab 12 a2 + 12 b2 for a = 2 2 1 u I h u V and b = u uh V :

u uh 2V (2 +221

) u I h u 2V + 2 22 21

u I h u 2V +12

u uh 2V Moving the term 12 u uh to the left, we obtain

u uh 2V u I h u 2V h u H 22

High order elements

One can obtain faster convergence, if the solution is smooth, and elements of higher orderare used:

Theorem 68. Assume that

the solution is smooth: uH m for m 2 all element spaces V T contain polynomials P p for p 1 the mesh is quasi-uniform

Then there holds

h 1 u I h u L 2 + u I h u H 1 hmin {m 1,k } u H mThe proof is analogous to the case m = 2 and k = 1. The constants in the estimates

depend on the Sobolev index m and on the polynomial order p. Nodal interpolation isinstable (i.e., the constant grow with p) for increasing order p. There exist better choicesto bound the best approximation error.

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5.3. A POSTERIORI ERROR ESTIMATES 55

Graded meshes around vertex singularities

On non-convex meshes domains, the solution is in general not in H 2, but in some weighted

Sobolev space. The information of the weight can be used to construct proper locallyrened meshes.On a sector domain with a non-convex corner of angle > , the solution is bounded

in the weighted Sobolev normr D 2u L 2 C,

with = . One may choose a mesh such that

hT hr T , T T where r T is the distance of the center of the element to the singular corner, and h

R + isa global mesh size parameter.

We bound the interpolation error:

u I T u 2H 1T T

h2T |u|H 2 (T )T T

h2 |r D 2u|L2 (T )h2 r D 2u 2L2 () C h

2

The number of elements in the domain can be roughly estimated by the integral overthe density of elements. The density is number of elements per unit volume, i.e., the inverseof the area of the element:

N el

|

T

| 1 dx =

h 2r 2 dx = h 2

r 2 dx Ch 2

In two dimensions, and (0, 1), the integral is nite.Combining the two estimates, one obtains a relation between the error and the numberof elements:

u I T u 2V N 1elThis is the same order of convergence as in the H 2 regular case !

5.3 A posteriori error estimates

We will derive methods to estimate the error of the computed nite element approximation.Such a posteriori error estimates may use the nite element solution uh , and input datasuch as the source term f .

(uh , f )

An error estimator is called reliable, if it is an upper bound for the error, i.e., thereexists a constant C 1 such that

u uh V C 1 (uh , f ) (5.5)

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An error estimator is efficient , if it is a lower bound for the error, i.e., there exists aconstant C 2 such that

u

uh V

C 2 (uh , f ). (5.6)

The constants may depend on the domain, and the shape of the triangles, but may notdepend on the source term f , or the (unknown) solution u.

One use of the a posteriori error estimator is to know the accuracy of the nite elementapproximation. A second one is to guide the construction of a new mesh to improve theaccuracy of a new nite element approximation.

The usual error estimators are dened as sum over element contributions:

2(uh , f ) =T T

2T (uh , f )

The local contributions should correspond to the local error. For the common errorestimators there hold the local efficiency estimates

u uh H 1 (T ) C 2 T (uh , f ).The patch T contains T and all its neighbor elements.

In the following, we consider the Poisson equation u = f with homogenous Dirichletboundary conditions u = 0 on . We choose piecewise linear nite elements on triangles.The Zienkiewicz Zhu error estimator

The simplest a posteriori error estimator is the one by Zienkiewicz and Zhu, the so calledZZ error estimator.

The error is measured in the H 1-semi norm:

u uh L 2Dene the gradient p = u and the discrete gradient ph = uh . The discrete gradient

ph is a constant on each element. Let ph be the p.w. linear and continuous nite elementfunction obtained by averaging the element values of ph in the vertices:

ph (xi) =1

|{T : xiT }|T :x iT p

h |T for all vertices xi

The hope is that the averaged gradient is a much better approximation to the true gradient,i.e.,

p ph L2 p ph L 2 (5.7)holds with a small constant 1. This property is known as super-convergence .It isindeed true on (locally) uniform meshes, and smoothness assumptions onto the sourceterm f .

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The ZZ error estimator replaces the true gradient in the error p ph by the goodapproximation ph :(uh ) = ph

ph L2 ()

If the super-convergence property (5.7) is fullled, than the ZZ error estimator is reli-able:

u uh L2 = p ph L2 ph ph L2 + p ph L2 ph ph L 2 + p ph L2 ,

andu uh L 2

11

ph ph L2 .It is also efficient, a similar short application of the triangle inequality.

There is a rigorous analysis of the ZZ error estimator, e.g., by showing equivalence tothe following residual error estimator.

The residual error estimator

The idea is to compute the residual of the Poisson equation

f + uh ,

in the natural norm H 1. The classical -operator cannot be applied to uh , since the rstderivatives, uh , are non-continuous across element boundaries. One can compute theresiduals on the elements

f |T + uh |T T T ,and one can also compute the violation of the continuity of the gradients on the edgeE = T 1 T 2. We dene the normal-jump term

u hn

:=u hn 1 |T 1 +

u hn 2 |T 2 .

The residual error estimator is

res (uh, f )2 :=

T

resT

(uh, f )2

with the element contributions

resT (uh , f )2 := h2T f + uh

2L 2 (T ) +

E :E T E

hE u hn

2

L2 (E ).

The scaling with hT corresponds to the natural H 1 norm of the residual.

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To show the reliability of the residual error estimator, we need a new quasi -interpolationoperator, the Clement- operator h . In contrast to the interpolation operator, this operatoris well dened for functions in L2.

We dene the vertex patch of all elements connected with the vertex x

x =T :xT

T,

the edge patch consisting of all elemenets connected with the edge E

E =T :E T =

T,

and the element patch consisting of the element T and all its neighbors

T =T :T T =

T .

The nodal interpolation operator I h was dened as

I h v =x iV

v(xi)i ,

where i are the nodal basis functions. Now, we replace the nodal value v(xi) be a localmean value.

Denition 69 (Clement quasi-interpolation operator) . For each vertex x,let vx

be themean value of v on the patch x , i.e.,

vx =1

|x | x v dx.The Clement operator is

h v :=x iV

vx i i .

In the case of homogeneous Dirichlet boundary values, the sum contains only inner vertices.

Theorem 70. The Clement operator satises the following continuity and approximation estimates:

h v L2 (T ) v L2 (T )v h v L2 (T ) hT v L 2 (T )v h v L2 (E ) h

1/ 2E v L2 (E )

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Proof: First, choose a reference patch T of dimension 1. The quasi-interpolationoperator is bounded on H 1(T ):

v h v L 2 ( bT ) + (v h v) L 2 ( bT ) v H 1 ( bT ) (5.8)If v is constant on T , then the mean values in the vertices take the same values, andalso (h v) |T is the same constant. The constant function (on T ) is in the kernel of v h v H 1 (T ) . Due to the Bramble-Hilbert lemma, we can replace the norm on the righthand side of (5.8) by the semi-norm:

v h v L 2 ( bT ) + (v h v) L 2 ( bT ) v L 2 ( bT ) (5.9)The rest follows from scaling. Let F : x hx scale the reference patch T to the actualpatch T . Then

v h v L2 (T ) + h (v h v) L 2 (T ) h v L2 (T )The estimate for the edge term is similar. One needs the scaling of integrals form the

reference edge E to E :v L 2 (E ) = h

1/ 2E v F L 2 (E )

Theorem 71. The residual error estimator is reliable:

u uh res (uh , f )Proof: From the coercivity of A(., .) we get

u uh H 1A(u uh , u uh )

u uh H 1 sup

0= vH 1

A(u uh , v)v H 1

.

The Galerkin orthogonality A(u uh , vh ) = 0 for all vhV h allows to insert the Clementinterpolant in the numerator. It is well dened for vH 1:u uh H 1 sup0= vH 1

A(u uh , v h v)v H 1

.

We use that the true solution u fullls A(u, v) = f (v), and insert the denitions of A(., .)and f (.):

A(u uh , v h v) = f (v h v) A(uh , v h v)= fvdx uh (v h v) dx=

T T T fvdx T T T uh (v h v) dx

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On each T , the nite element function uh is a polynomial. This allows integration by partson each element:

A(u uh , v h v) =T T T fvdx T T T uh (v h v) dx + T u

h

n(v h v) ds

All inner edges E have contributions from normal derivatives from their two adjacenttriangles T E, 1 and T E, 2. On boundary edges, v h v vanishes.

A(u uh , v h v)=

T T (f + uh )(v h v) dx + E E u hn |T E, 1 + u hn |T E, 2 (v h v) ds=

T T (f + uh )(v h v) dx + E E uh

n(v h v) ds

Applying Cauchy-Schwarz rst on L2(T ) and L2(E ), and then in R n :

A(u uh , v h v)

T

f + uh L 2 (T ) v h v L 2 (T ) +E

u hn L 2 (E )

v h v L 2 (E )

= T hT f + uh L2 (T )h

1T v h v L2 (T ) + E h

1/ 2

E

u hn L 2 (E ) h

1/ 2

E v h v L2 (E )

T

h2T f + uh2L2 (T )

1/ 2

T

h 2T v h v 2L2 (T )1/ 2

+

+E

hE u hn

2

L2 (E )

1/ 2

E

h 1E v h v 2L2 (E )1/ 2

We apply the approximation estimates of the Clement operator, and use that only abounded number of patches are overlapping:

T

h 2T v h v 2L 2 (T )T

v 2L 2 (T ) v2L 2 () ,

and similar for the edges

E

h 1E v h v 2L2 (E ) v 2L2 () .

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Combining the steps above we observe

u

uh V sup

vH 1

A(u uh , v h v)v

1H

supV H 1

T h2T f + uh 2L2 (T ) + E hE

u hn

2L2 (E )

1/ 2v L2 ()

v H 1

T

h2T f + uh2L2 (T ) +

E

hE u hn

2

L2 (E )

1/ 2

,

what is the reliability of the error estimator res (uh , f )

Theorem 72. If the source term f is piecewise polynomial on the mesh, then the error estimator res is efficient:

u uh V res (uh , f )Goal driven error estimates

The above error estimators estimate the error in the energy norm V . Some applicationsrequire to compute certain values (such as point values, average values, line integrals, uxesthrough surfaces, ...). These values are descibed by linear functionals b : V R . We wantto design a method such that the error in this goal, i.e.,

b(u)

b(uh )

is small. The technique is to solve additionally the dual problem, where the right handside is the goal functional:

Find wV : A(v, w) = b(v) vV.

Usually, one cannot solve the dual problem either, and one applies a Galerkin method alsofor the dual problem:

Find whV h : A(vh , wh ) = b(vh ) vhV h .

In the case of point values, the solution of the dual problem is the Green function (whichis not in H 1). The error in the goal is

b(u uh ) = A(u uh , w) = A(u uh , w wh ).A rigorous upper bound for the error in the goal is obtained by using continuity of thebilinear-form, and energy error estimates 1 and 2 for the primal and dual problem,respectively:

|b(u uh )| u uh V w wh V 1(uh , f ) 2(wh , b).

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A good heuristic is the following (unfortunately, not correct) estimate

b(u uh ) = A(u uh , w wh )T T

u uh H 1 (T ) wwh H 1 (T )T

1T (uh , f ) 2T (wh , b)

(5.10)The last step would require a local reliability estimate. But, this is not true.

We can interpret (5.10) that way: The local estimators 2T (wh ) provide a way forweighting the primal local estimators according to the desired goal.

Mesh renement algorithms

A posteriori error estimates are used to control recursive mesh renement:

Start with initial mesh T Loopcompute fe solution uh on T compute error estimator T (uh , f )if tolerance then stoprene elements with large T to obtain a new mesh

The mesh renement algorithm has to take care of

generating a sequence of regular meshes generating a sequence of shape regular meshesRed-Green Renement:

A marked element is split into four equivalent elements (called red renement):

But, the obtained mesh is not regular. To avoid such irregular nodes, also neighboringelements must be split (called green closure):

If one continues to rene that way, the shape of the elements may get worse and worse:

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5.4. NON-CONFORMING FINITE ELEMENT METHODS 63

A solution is that elements of the green closure will not be further rened. Instead, removethe green closure, and replace it by red renement.

Marked edge bisection:Each triangle has one marked edge. The triangle is only rened by cutting from the middleof the marked edge to the opposite vertex. The marked edges of the new triangles are theedges of the old triangle.

If there occurs an irregular node, then also the neighbor triangle must be rened.

To ensure nite termination, one has to avoid cycles in the initial mesh. This can beobtained by rst sorting the edges (e.g., by length), end then, always choose the largestedges as marked edge.

Both of these renement algorithms are also possible in 3D.

5.4 Non-conforming Finite Element Methods

In a conforming nite element method, one chooses a sub-space V h V , and denes thenite element approximation as

Find uhV h : A(uh , vh ) = f (vh ) vhV h

For reasons of simpler implementation, or even of higher accuracy, the conforming frame-work is often violated. Examples are:

The nite element space V h is not a sub-space of V = H m . Examples are the non-conforming P 1 triangle, and the Morley element for approximation of H 2.

The Dirichlet boundary conditions are interpolated in the boundary vertices. The curved domain is approximated by straight sided elements The bilinear-form and the linear-form are approximated by inexact numerical inte-gration

The lemmas by Strang are the extension of Ceas lemma to the non-conforming setting.

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The First Lemma of Strang

In the rst step, let V h V , but the bilinear-form and the linear-form are replaced by

mesh-dependent forms Ah (., .) : V h V h Rand

f h (.) : V h R .We do not assume that Ah and f h are dened on V . We assume that the bilinear-formsAh are uniformly coercive, i.e., there exists an 1 independent of the mesh-size such that

Ah (vh , vh ) 1 vh 2V vhV hThe nite element problem is dened as

Find uhV h : Ah (uh , vh ) = f h (vh ) v

hV

h

Lemma 73 (First Lemma of Strang) . Assume that

A(., .) is continuous on V Ah (., .) is uniformly coercive

Then there holds

u uh inf vhV h u vh + supwhV h |A(vh , wh ) Ah (vh , wh )|

wh

+ supwhV h

f (wh ) f h (wh )whProof: Choose an arbitrary vh V h , and set wh := uh vh . We use the uniformcoercivity, and the denitions of u and uh :

1 uh vh 2V Ah (uh vh , uh vh ) = Ah (uh vh , wh )= A(u vh , wh ) + [A(vh , wh ) Ah (vh , wh )] + [Ah (uh , wh ) A(u, wh )]= A(u vh , wh ) + [A(vh , wh ) Ah (vh , wh )] + [f h (wh ) f (wh )]

Divide by uh

vh = wh , and use the continuity of A(., .):

uh vh u vh + |A(vh , wh ) Ah (vh , wh )|

wh+ |f (wh ) f h (wh )|

wh(5.11)

Using the triangle inequality, the error u uh is bounded byu uh inf vhV h u vh + vh uh

The combination with (5.11) proves the result. 2

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5.4. NON-CONFORMING FINITE ELEMENT METHODS 65

Example: Lumping of the L2 bilinear-form:Dene the H 1 - bilinear-form

A(u, v) = u v + uv dx,and perform Galerkin discretization with P 1 triangles. The second term leads to a non-diagonal matrix. The vertex integration rule T v dx |T |3 3 =1 v(xT, )

is exact for vP 1. We apply this integration rule for the term uv dx:

Ah (u, v) = u v + T T |T |33

=1

u(xT, )v(xT, )

The bilinear form is now dened only for u, v V h . The integration is not exact, sinceuvP 2 on each triangle.

Inserting the nodal basis i , we obtain a diagonal matrix for the second term:

i(xT, ) j (xT, ) =1 for xi = x j = xT,0 else

To apply the rst lemma of Strang, we have to verify the uniform coercivity

T

|T |3

3

=1|vh (xT, )|2 1

T T |vh |2 dx vhV h , (5.12)which is done by transformation to the reference element. The consistency error can beestimated by

| T uh vh dx |T |3 3 =1 uh (x )vh (x )| h2T uh L2 (T ) vh L 2 (T ) (5.13)Summation over the elements give

A(uh , vh ) Ah (uh , vh ) h2 uh H 1 () vh H 1 ()The rst lemma of Strang proves that this modication of the bilinear-form preserves theorder of the discretization error:

u uh H 1

inf vhV h u vh H 1

+ supwhV h |A(vh , wh )

Ah (vh , wh )

|wh H 1 u I h u H 1 + supwhV h |

A(I h u, wh ) Ah (I h u, wh )|wh H 1

h u H 2 + supwhV h

h2 I h u H 1 wh H 1wh H 1

h u H 2

A diagonal L2 matrix has some advantages:

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It avoids oscillations in boundary layers (exercises!) In explicit time integration methods for parabolic or hyperbolic problems, one has to

solve linear equations with the L2-matrix. This becomes cheap for diagonal matrices.

The Second Lemma of Strang

In the following, we will also skip the requirement V h V . Thus, the norm . V cannotbe used on V h , and it will be replaced by mesh-dependent norms . h . These norms mustbe dened for V + V h . As well, the mesh-dependent forms Ah (., .) and f h (.) are dened onV + V h . We assume

uniform coercivity:Ah (vh , vh )

1 vh 2h

vh

V h

continuity:Ah (u, vh ) 2 u h vh h uV + V h , vhV h

The error can now be measured only in the discrete norm u uh V h .Lemma 74. Under the above assumptions there holds

u uh h inf vhV h u vh h + supwhV h |Ah (u, wh ) f h (wh )|

wh h(5.14)

Remark : The rst term in (5.14) is the approximation error, the second one is calledconsistency error.Proof: Let vhV h . Again, set wh = uh vh , and use the V h -coercivity:

1 uh vh 2h Ah (uh vh , uh vh ) = Ah (uh vh , wh )= Ah (u vh , wh ) + [ f h (wh ) Ah (u, wh )]

Again, divide by uh vh , and use continuity of Ah (., .):uh vh h u vh h +

Ah (u, wh ) f h (wh )wh h

The r

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