Partial Diﬀerential Equations and the Finite Element Methodcermics.enpc.fr/~legoll/poly_EDP-EF_jan19.pdf · question, we will focus on the Finite Element Method, an extremely popular

Partial Differential Equations

and the Finite Element Method

Frédéric LEGOLL

January 2019

Contents

General introduction 1

1 Introduction to distribution theory 3

1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 First examples of distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Derivation in the sense of distributions . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 The space of test functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.5 Distributions and locally integrable functions . . . . . . . . . . . . . . . . . . . . . 8

1.6 Multiplication by C∞ functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.7 Convergence of distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.8 More on differentiation: the one-dimensional case . . . . . . . . . . . . . . . . . . . 13

1.8.1 The case of C1 functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.8.2 The case of functions that are piecewise C1 . . . . . . . . . . . . . . . . . . 13

1.8.3 Linear differential equations in D′(a, b) . . . . . . . . . . . . . . . . . . . . . 14

1.8.4 Link between the standard derivative and the derivative in the sense of dis-tributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.9 More on differentiation: the multi-dimensional case . . . . . . . . . . . . . . . . . . 15

1.9.1 Schwarz Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.9.2 The case of C1 functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.9.3 Stokes formula and applications . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.10 Principal values and finite parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.10.1 Principal value of 1/x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.10.2 Finite part of H(x)xα . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.11 Distributions with compact support . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.11.1 Definitions and first properties . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.11.2 Distributions with support restricted to a point . . . . . . . . . . . . . . . . 18

1.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Sobolev spaces 21

2.1 The spaces Hk(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 The space H10 (Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2.2 Poincaré inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2.3 On an alternative definition of H10 (Ω) via the use of traces . . . . . . . . . . 23

2.3 The space H−1(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

i

ii CONTENTS

3 Linear elliptic boundary value problems 31

3.1 Lax-Milgram theorem (symmetric version) . . . . . . . . . . . . . . . . . . . . . . . 32

3.2 A first symmetric linear elliptic boundary value problem . . . . . . . . . . . . . . . 33

3.2.1 Variational formulation of (3.5) . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2.2 Checking the assumptions of the Lax-Milgram theorem . . . . . . . . . . . 35

3.3 Poisson equation on a bounded open set . . . . . . . . . . . . . . . . . . . . . . . . 35

3.3.1 Variational formulation of (3.7) . . . . . . . . . . . . . . . . . . . . . . . . . 36


3.4 Lax-Milgram theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.5 A non-symmetric boundary value problem . . . . . . . . . . . . . . . . . . . . . . . 38

3.5.1 Variational formulation of (3.10) . . . . . . . . . . . . . . . . . . . . . . . . 38


3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Inf-sup theory 49

4.1 The finite dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.1.1 Square matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.1.2 Rectangular matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2 The infinite dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2.1 The Hilbert case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2.2 The Banach case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.3 Stokes problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.3.1 Variational formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.3.2 Recasting the variational formulation of the Stokes problem . . . . . . . . . 59

4.3.3 BNB theorem for problems of the type (4.19) . . . . . . . . . . . . . . . . . 60

4.3.4 Application to the Stokes problem . . . . . . . . . . . . . . . . . . . . . . . 62

4.4 Saddle-point problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5 Numerical approximation of boundary value problems (coercive case) 67

5.1 The Galerkin approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.1.1 Principle of the method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.1.2 Error estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.1.3 The linear system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.2 The Lagrange P1 Finite Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.2.1 Meshing the domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.2.2 The space of P1 polynomial functions . . . . . . . . . . . . . . . . . . . . . 73

5.2.3 Approximation space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5.3 Approximation of the Poisson problem by the P1 Finite Element Method . . . . . 77

5.3.1 Discrete problem and error estimation . . . . . . . . . . . . . . . . . . . . . 77

5.3.2 Assembling the stiffness matrix . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.3.3 Solving the linear system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5.3.4 Approximations by quadrature formulas . . . . . . . . . . . . . . . . . . . . 81

5.4 The P2 Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.4.1 Approximation space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.4.2 Approximation of the Poisson problem . . . . . . . . . . . . . . . . . . . . . 84

5.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

CONTENTS iii

6 Numerical approximation of boundary value problems (the non-coercive case) 956.1 Galerkin approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.1.1 Well-posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 956.1.2 Error estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

6.2 The Stokes problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.2.1 Galerkin approximation of the abstract problem (6.7) . . . . . . . . . . . . 986.2.2 Checkerboard instability for the Stokes problem . . . . . . . . . . . . . . . . 1006.2.3 Numerical locking for the Stokes problem . . . . . . . . . . . . . . . . . . . 1016.2.4 A possible choice of Xh ×Mh for the Stokes problem . . . . . . . . . . . . . 101

A Basic facts on the Lebesgue integral and Lp spaces 105A.1 The space L1(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

A.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105A.1.2 Dominated convergence theorem . . . . . . . . . . . . . . . . . . . . . . . . 105

A.2 The space L2(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106A.3 The spaces Lp(Ω) and Lploc(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106A.4 The space L∞(Ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107A.5 Other properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

B Glossary of mathematical terms 109

iv CONTENTS

General introduction

Many phenomena in the engineering sciences (in solid and fluid mechanics, in finance, in trafficflow, . . . ) can be modelled using Partial Differential Equations (PDEs).

The aim of this course is to introduce the students to some modern mathematical tools toaddress these types of equations. In these lecture notes, we first focus on the Poisson equation ina bounded domain, which reads

Find a function u : Ω −→ R such that−∆u = f in Ωu = 0 on ∂Ω

(1)

where Ω is an open bounded subset of Rd, ∂Ω is its boundary and f is a given function, defined onΩ and valued in R. Problem (1) arises for instance in mechanics: the solution u to (1) is the verticaldisplacement of an elastic membrane submitted to the load f , and clamped at its boundary.

Along this course, we will also consider several variants of (1), in different directions:

• non-symmetric problems;

• boundary conditions different from homogeneous Dirichlet boundary conditions.

We will also introduce the students to non-coercive problems (through the example of the Stokesproblem) and to the inf-sup theory.

This course is devoted to the mathematical analysis and the numerical analysis of Problem (1)and its variants.

From the mathematical analysis standpoint, the questions we address are the following:

• Does the problem (1) have a solution? Is it unique?

• If this is the case, is the map f 7→ u continuous?

Problem (1) is a boundary value problem in the sense that it is composed of a Partial DifferentialEquation (PDE) and a boundary condition (u = 0 on ∂Ω). Its mathematical analysis allows toemphasize a generic trend in modern analysis: the use of Geometry tools in infinite dimensionalvector spaces allows to obtain results for Analysis problems such as (1) in a simple and elegantmanner.

In the second part of these lecture notes, we turn to numerical analysis questions. Indeed, inpractice, the exact solution to (1) (just as for any PDE) is not known in closed form. One oftenresorts to numerical approaches (such as the Finite Element Method) to compute an approximationuh of the exact solution u. In this context, we describe general Galerkin approximations of (1),that yield finite dimensional problems, and address the following questions:

• Is the finite dimensional problem well-posed?

1

2 CONTENTS

• Denoting uh its solution, how the error ‖u− uh‖ can be estimated?

• In practice, how is such an approximation implemented on a computer? For that latterquestion, we will focus on the Finite Element Method, an extremely popular technique inthe engineering community.

The overall objective of this course is to give to the students the tools needed for the math-ematical analysis of PDEs, and for the detailed understanding of the Finite Element Method:estimation of the error as a function of the discretization parameters, practical implementation.

These lecture notes are organized as follows. The first part is devoted to the construction ofthe appropriate functional spaces in which the solution to (1) should be looked for. We start by anintroduction to the distribution theory (in Chapter 1), followed by the construction of the Sobolevspaces (in Chapter 2), which are the relevant spaces of functions when dealing with PDEs.

The mathematical analysis of (1) and its variants (in the coercive case) is carried out in Chap-ter 3. To that aim, we introduce the notion of variational formulation of a boundary value problem,the analysis of which is performed using the Lax-Milgram theorem. The analysis of non-coerciveproblems is discussed in Chapter 4.

The numerical analysis of (1) and its variants is carried out in Chapter 5 in the coercive case,and in Chapter 6 in the non-coercive case.

This course is based on notions introduced in two ENPC courses, Outils Mathématiques pourl’Ingénieur and Analyse et Calcul Scientifique. However, the present lecture notes are, as muchas possible, self-contained. Nevertheless, the reader is supposed to be familiar with Banach andHilbert spaces. Some basic facts concerning the Lebesgue integral and Lp spaces have been collectedin Appendix A. For historical reasons, these lectures notes are written in English. A glossary ofthe main mathematical terms can be found in Appendix B.

I am indebted to Eric Cancès and Alexandre Ern, who have been teaching an Analyse courseat ENPC for many years. The structure of this document is very much inspired from their lecturesnotes [2, 5] and from [3].

I am grateful to the colleagues who gave me opportunities to teach, at ENPC and elsewhere.It has always been a wonderful experience, and I am very pleased to thank them. I would also liketo thank Thomas Hudson and Antoine Levitt for their remarks on a draft version of these lecturenotes.

As always, there are certainly still several typos in these notes, despite our best efforts. Anysuggestions on these notes are therefore warmly welcome.

Frédéric Legoll, Champs sur Marne, January 2019

Chapter 1

Introduction to distribution theory

The distribution theory, introduced by Laurent Schwartz in 1946, is a fundamental piece of Anal-ysis. Distributions generalize the notion of functions, and satisfy the following two striking prop-erties, that make them particularly easy to manipulate:

• any distribution is differentiable and its derivative is again a distribution;

• if a sequence of distributions Tn converges to some distribution T , then the derivatives of Tnconverge to the derivatives of T .

In our context, distributions will be useful to build the appropriate functional spaces for the studyof Partial Differential Equations (see Chapter 2).

In what follows, Ω is an open subset (which may be either bounded or unbounded) of Rd,with d ≥ 1. Following the international convention, we denote by (a, b) the one-dimensional openinterval x ∈ R, a < x < b.

1.1 Definitions

Roughly speaking, a distribution is a linear and continuous form on the vector space D(Ω) of thefunctions which are in C∞(Ω) and which have a compact support in Ω. We start by recalling thedefinition of the support of a continuous function.

Definition 1.1. Let Ω be an open subset of Rd and φ : Ω −→ R be a continuous function. Thesupport of φ is

Supp(φ) = x ∈ Ω, φ(x) 6= 0Ω.

We recall that the notation AΩ

means “closure of the set A in Ω”. The set AΩ

is hence theset of points in Ω that are limits of sequences of points of A. Here are some examples of supportcomputations:

1. Ω = R and φ(x) = sinx: then

x ∈ R, φ(x) 6= 0 = R \ πZ, Supp(φ) = R.

2. Ω = (−1, 2) and φ(x) =

x+ 1 if x ∈ (−1, 0],1− x if x ∈ [0, 1],0 otherwise.

Then x ∈ Ω, φ(x) 6= 0 = (−1, 1) and Supp(φ) = (−1, 1].

3

4 CHAPTER 1. INTRODUCTION TO DISTRIBUTION THEORY

3. Ω = (−2, 2) and φ is the same function as above, that is φ(x) =

x+ 1 if x ∈ (−1, 0],1− x if x ∈ [0, 1],0 otherwise.

Then x ∈ Ω, φ(x) 6= 0 = (−1, 1) and Supp(φ) = [−1, 1].

The support of φ is compact only in the last example (we recall that the compact sets of Rd arethe sets which are bounded and closed).

Exercise 1.2. Let f and g be continuous functions from Rd to R.

• Show that Supp(fg) ⊂ Supp(f) ∩ Supp(g) and that the inclusion may be strict.

• Show that Supp(f + g) ⊂ Supp(f) ∪ Supp(g) and that the inclusion may be strict.

Definition 1.3. Let D(Ω) be the vector space of functions defined on Ω, which are infinitelydifferentiable (hence, D(Ω) ⊂ C∞(Ω)), and which have a compact support. The functions in D(Ω)are called test functions. For any compact set K ⊂ Ω, we denote by DK(Ω) the set of test functionswith support contained in K.

We recall the following notation: for any α = (α1, . . . , αd) ∈ Nd, we set |α| =d∑

i=1

αi and

∂αφ =∂|α|φ

∂α1x1 . . . ∂

αdxd

=∂α1+...+αdφ

∂α1x1 . . . ∂

αdxd

. (1.1)

Definition 1.4. A distribution T in the open set Ω is a linear form on D(Ω) which satisfies thefollowing “continuity property”: for any compact set K ⊂ Ω, there exists an integer p and a constantC such that

∀φ ∈ DK(Ω), |〈T, φ〉| ≤ C supx∈K, |α|≤p

|∂αφ(x)|. (1.2)

The vector space of distributions in Ω is denoted D′(Ω).

Remark 1.5. On a vector space E endowed with a norm ‖ · ‖, the continuity property of a linearform T simply reads

∀x ∈ E, |〈T, x〉| ≤ C‖x‖.

The more complex form of the “continuity property” (1.2) stems from the fact that the topology withwhich the vector space D(Ω) should be endowed such that its dual D′(Ω) has good properties is nota topology arising from a norm.

Definition 1.6. When the integer p can be chosen independently of K, one says that the distri-bution T has a finite order. The smallest possible value of p is the order of T .

Remark 1.7. Heuristically, the larger the order of a distribution T is, the more singular T is.

1.2. FIRST EXAMPLES OF DISTRIBUTIONS 5

1.2 First examples of distributions

We now give some classical examples. First, to any function f ∈ L1loc(Ω), one can associate the

distribution Tf (also denoted f) defined by

∀φ ∈ D(Ω), 〈Tf , φ〉 =∫

Ω

f φ. (1.3)

Next, for any a ∈ Ω ⊂ R, we denote by δa the distribution defined by

∀φ ∈ D(Ω), 〈δa, φ〉 = φ(a) (1.4)

and by δ′a the distribution defined by

∀φ ∈ D(Ω), 〈δ′a, φ〉 = −φ′(a). (1.5)

Last, consider a locally bounded measure µ, i.e. a measure µ defined on the set B(Ω) of the boreliansubsets of Ω ⊂ Rd and such that µ(K) < ∞ for any compact subset K ⊂ Ω. We denote Tµ (orsimply µ) the distribution defined by

∀φ ∈ D(Ω), 〈Tµ, φ〉 =∫

Ω

φdµ. (1.6)

Exercise 1.8. The aim of this exercise is to compute the order of some particular distributions.

• Show that the linear forms Tf defined by (1.3), δa defined by (1.4) and Tµ defined by (1.6)are distributions of order 0 on Ω.

• Show that the linear form δ′a defined by (1.5) is a distribution of order 1 on Ω (Hint: considerthe simple case Ω = R, and consider the action of δ′a on the sequence φj(x) = φ0(x) atan(jx),where φ0 ∈ D(R) is an even function that is equal to 1 in the neighborhood of 0; such a testfunction indeed exists in view of Lemma 1.18 below).

Remark 1.9. There does not exist any function f ∈ L1loc(R) such that Tf = δ0. This shows that

the space of distributions is (much) larger than the space of functions. The proof of this statementis not easy. However, it is easy to see that there exists no function f ∈ C0(R) such that Tf = δ0.

1.3 Derivation in the sense of distributions

Let f ∈ C1(R). The integration by parts formula reads

∀φ ∈ D(R),

∫

R

df

dxφ = −

∫

R

fdφ

dx.

Likewise, for any f ∈ C1(Rd), we have, for any 1 ≤ i ≤ d,

∀φ ∈ D(Rd),

∫

Rd

∂f

∂xiφ = −

∫

Rd

f∂φ

∂xi.

The definition of the derivative of a distribution is inspired by this formula. Note that, in contrastto the case of functions, all distributions are differentiable!


Definition 1.10. Let T ∈ D′(Ω). The derivative of T with respect to the variable xi, which is

denoted by∂T

∂xi, is defined by

∀φ ∈ D(Ω), 〈 ∂T∂xi

, φ〉 = −〈T, ∂φ∂xi

〉.

It is an easy exercise to check that the linear form∂T

∂xidefined above is indeed a distribution

(i.e. that it satisfies the continuity property (1.2)).

Exercise 1.11. Compute the first and second derivative (in the sense of distributions) of the “hat”function (see Figure 1.1) defined on R by

f(x) =

x+ 1 if − 1 < x < 0,1− x if 0 < x < 1,0 otherwise.

x

f(x)

1−1

1

Figure 1.1: Hat function

Exercise 1.12. Check that the distribution δ′a defined by (1.5) is the derivative of δa in the senseof distributions.

Exercise 1.13. Consider the Heaviside function H defined on R by H(x) = 1 if x > 0, H(x) = 0otherwise. Why can H be considered to be a distribution on R? Compute its derivative in the senseof distributions.

Remark 1.14. Any function f ∈ L1loc(R) can be considered to be a distribution. It is thus possible

to define its derivative f ′ in the sense of distributions. In general, f ′ is not a function, but adistribution only.

1.4 The space of test functions

The space of test functions plays an important role in the construction and the manipulation ofdistributions. For instance, to show that the distribution δ′a is of order 1 (see the exercise 1.8), weuse a test function φ0 ∈ D(R) that is even and is equal to 1 in the neighborhood of 0. This sectioncollects some results ensuring the existence of such test functions. It is important to know thatthere exist test functions satisfying the properties discussed below. In contrast, the proof of theseresults are often technical. They will all be skipped here.

We start with the following important result:

1.4. THE SPACE OF TEST FUNCTIONS 7

Theorem 1.15. The space D(Ω) is dense in Lp(Ω) for any 1 ≤ p < +∞.

Remark 1.16. Note that the case p = ∞ is not allowed in the above theorem. It turns outthat D(Ω) is not dense in L∞(Ω). Consider indeed the following example: on Ω = (0, 1), letf ∈ L∞(0, 1) be defined by f(x) = 1 almost everywhere. Then, for any φ ∈ D(0, 1), we have‖f − φ‖L∞ ≥ 1.

Corollary 1.17. For any integer k ≥ 0 and any 1 ≤ p <∞, the space Ck(Ω) ∩Lp(Ω) is dense inLp(Ω).

Proof. This is a direct consequence of Theorem 1.15 as D(Ω) ⊂ Ck(Ω) ∩ Lp(Ω) ⊂ Lp(Ω).

We now list several technical results.

Lemma 1.18. Let Ω be an open subset of Rd and let Ba(r) be the ball of center a ∈ Ω and ofradius r > 0. If Ba(r) ⊂ Ω, there exists φ ∈ D(Ω), with support in Ba(r), and satisfying φ ≥ 0

and

∫

Ω

φ = 1.

Lemma 1.19. Let K ⊂ Ω be compact. There exists a test function φ ∈ D(Ω) such that 0 ≤ φ ≤ 1on Ω and φ ≡ 1 on K (see Figure 1.2).

K

Ω

1

x

(x)φ

Figure 1.2: A specific test function

Lemma 1.20 (Partition of unity). Let Ω1, . . . ,Ωn be open subsets of Rd and let K be a compactset of Rd such that

K ⊂n⋃

k=1

Ωk.

There exist test functions α1, . . . , αn such that

• Supp(αk) ⊂ Ωi,

• 0 ≤ αk ≤ 1,

•n∑

k=1

αk = 1 on K.


Lemma 1.21 (Borel lemma). For any sequence (aα)α∈Nd , there exists φ ∈ D(Rd) such that

∀α ∈ Nd, ∂αφ(0) = aα.

Lemma 1.22 (Hadamard lemma). Let φ ∈ D(R) be such that φ(k)(0) = 0 for any integer 0 ≤ k ≤n. Then there exists ψ ∈ D(R) such that, for any x ∈ R, φ(x) = xnψ(x).

Proof. This follows by using the Taylor formula with a remainder in integral form.

Remark 1.23. A similar result holds in dimension higher than d = 1.

1.5 Distributions and locally integrable functions

We have seen above that we can associate a distribution Tf to any function f ∈ L1loc(Ω) (see

formula (1.3)). It turns out that the space of test functions is sufficiently large as to ensure thatthe map f 7→ Tf from L1

loc(Ω) to D′(Ω) is actually injective. This result, which is very important,is formalized in the theorem below.

Theorem 1.24. Let f and g be functions in L1loc(Ω). Then

f = g almost everywhere ⇔ Tf = Tg in D′(Ω).

Proof. The implication ⇒ is straightforward. We show the converse implication in the simplifiedcase when Ω = Rd and when f and g are in L1(Rd).

Consider f and g in L1(Rd) such that Tf = Tg and set h = f − g. We hence have h ∈ L1(Rd)such that

∀φ ∈ D(Rd),

∫

Ω

hφ = 0. (1.7)

Our aim is to show that h = 0 almost everywhere. The proof falls in two steps.

Step 1 (approximation of identity): Consider a non-negative function χ ∈ D(Rd), with supportin the unit ball and with integral equal to 1. Define the sequence (χn)n∈N⋆ by

χn(x) = n−dχ(x/n).

Such a sequence is an approximation of the identity (see Section 1.7). We recall that, for any n,‖χn‖L1(Rd) = ‖χ‖L1(Rd) = 1. Let ψ ∈ D(Rd). We claim that

‖ψ − ψ ⋆ χn‖L1(Rd) −→n→+∞

0, (1.8)

where ψ ⋆ χn is the convolution product of ψ with χn (see Definition A.18). To prove (1.8), wefirst note, using the change of variables z = y/n, that

ψ ⋆ χn(x) = n−d

∫

Rd

ψ(x− y)χ(y/n) dy =

∫

Rd

ψ(x− z/n)χ(z) dz.

Using the dominated convergence theorem (see Theorem A.1), we obtain that, for any x ∈ Rd,ψ ⋆ χn(x) converges to ψ(x) when n→ ∞. We next note that

|ψ ⋆ χn(x)| =∣∣∣∣∫

Rd

ψ(x− y)χn(y) dy

∣∣∣∣ ≤ ‖ψ‖L∞(Rd) ‖χn‖L1(Rd) = ‖ψ‖L∞(Rd),

and thatSupp(ψ ⋆ χn) ⊂ Supp(ψ) + Supp(χn) ⊂ Supp(ψ) + Supp(χ).

1.5. DISTRIBUTIONS AND LOCALLY INTEGRABLE FUNCTIONS 9

This implies that

|ψ − ψ ⋆ χn| ≤ 2 ‖ψ‖L∞(Rd) 1Supp(ψ)+Supp(χ) for any n,

ψ(x) − ψ ⋆ χn(x) −→n→+∞

0 for any x ∈ Rd.

We can hence use the dominated convergence theorem again, and obtain the convergence (1.8).

Step 2: Let ǫ > 0 and ψ ∈ D(Rd) such that

‖h− ψ‖L1(Rd) ≤ ǫ/3. (1.9)

The existence of such a function ψ stems from the density of D(Rd) in L1(Rd) (see Theorem 1.15).Let (hn)n∈N⋆ be defined by

hn(x) = (h ⋆ χn)(x) =

∫

Rd

h(y)χn(x− y) dy,

where χn was defined in Step 1. Since, for any n ∈ N⋆, χn ∈ D(Rd), we observe that hn vanisheson Rd for any n, in view of (1.7). We hence write

‖h‖L1(Rd) = ‖h− h ⋆ χn‖L1(Rd) ≤ ‖h− ψ‖L1(Rd) + ‖ψ − ψ ⋆ χn‖L1(Rd) + ‖(h− ψ) ⋆ χn‖L1(Rd) .

(1.10)By definition of the convolution product (see the Definition A.18), we have

‖(h− ψ) ⋆ χn‖L1(Rd) ≤ ‖h− ψ‖L1(Rd) ‖χn‖L1(Rd) = ‖h− ψ‖L1(Rd) ≤ ǫ/3. (1.11)

Using (1.8), we know that we can choose N such that, for any n ≥ N , we have

‖ψ − ψ ⋆ χn‖L1(Rd) ≤ ǫ/3. (1.12)

Collecting (1.10), (1.9), (1.12) and (1.11), we get that

‖h‖L1(Rd) ≤ ǫ.

This upper bound holds for any ǫ > 0. This implies that h = 0 in L1(Rd).

It is thus possible to identify any function of L1loc(Ω) with its associated distribution, which

can be denoted f instead of Tf . The previous result can be recast as follows.

Theorem 1.25. Let f and g in L1loc(Ω). Then

f = g a.e. ⇔ f = g in D′(Ω).

Remark 1.26. We therefore write L1loc(Ω) ⊂ D′(Ω) just like we write N ⊂ R ⊂ C. The notation

A ⊂ B here means that there exists a canonical injection from A to B.

Remark 1.27. The notion of distributions generalizes the notion of functions, but it does notmean that any function is a distribution. Only functions in L1

loc are distributions. More singularfunctions are not necessarily distributions. For instance, the function f(x) = 1/|x| is a distributionin Rd for any d ≥ 2 but not in R. See Section 1.10 for more details in that direction.


1.6 Multiplication by C∞ functions

Definition 1.28. Let T ∈ D′(Ω) and g ∈ C∞(Ω). The distribution g T is defined by

∀φ ∈ D(Ω), 〈g T, φ〉 = 〈T, g φ〉.

It is an easy exercise to check that the linear form g T is indeed a distribution.

Remark 1.29. The multiplication of two distributions is not defined. Consider for instance the

function f(x) =1√|x|

, which belongs to L1loc(R) and hence defines a distribution. But the function

f2 does not define a distribution (recall that f2(x) = 1/|x| does not belong to L1loc(R)). More

generally, the product T1 T2 with T1 ∈ D′(Ω) and T2 ∈ D′(Ω) is not defined (see for instance theexercise 1.56).

Exercise 1.30. Show that x δ0 = 0.

1.7 Convergence of distributions

Definition 1.31. Let (Tn)n∈N be a sequence of distributions in Ω. This sequence is said to convergeto some T ∈ D′(Ω) when n→ ∞ if

∀φ ∈ D(Ω), 〈Tn, φ〉 −→n→+∞

〈T, φ〉.

The example of the exercise below is often useful.

Exercise 1.32. Let χ ∈ D(Rd) be a non-negative function, with support in the unit ball and ofintegral equal to 1. Set

χn(x) = ndχ(nx).

Show that

χn −→n→∞

δ0 in D′(Rd).

Definition 1.33. A sequence (χn)n∈N of test functions in D(Rd) satisfying

χn −→n→+∞

δ0 in D′(Rd)

is called an approximation of the identity.

Remark 1.34. This name comes from the fact that the distribution δ0 is the identity for theconvolution operation (this operation is defined in Section A.5 for functions in L1(Rd) and it canbe extended to some distributions).

Proposition 1.35. Let 1 ≤ p ≤ +∞. Convergence in Lploc(Ω) implies convergence in D′(Ω).

1.7. CONVERGENCE OF DISTRIBUTIONS 11

Proof. Consider (fn)n∈N that converges to some f in Lploc(Ω) when n→ ∞. Let φ ∈ D(Ω). Usingthe Hölder inequality, we have, for any 1 < p < +∞,

|〈fn, φ〉 − 〈f, φ〉| =

∣∣∣∣∫

Ω

(fn − f)φ

∣∣∣∣

≤∫

Ω

|fn − f | |φ|

≤ ‖φ‖L∞

∫

Supp(φ)

|fn − f |

≤ ‖φ‖L∞

(∫

Supp(φ)

|fn − f |p)1/p (∫

Supp(φ)

1p′

)1/p′

≤ ‖φ‖L∞ ‖fn − f‖Lp(Supp(φ)) |Supp(φ)|1/p′ .

We hence get that limn→∞

|〈fn, φ〉 − 〈f, φ〉| = 0. If p = 1 or p = +∞, then the proof, which is even

simpler, is left to the reader.

Remark 1.36. The fact that a sequence of functions fn converges almost everywhere to somefunction f (i.e. lim

n→∞fn(x) = f(x) a.e. on Ω) does not imply convergence in D′. The converse is

also false. The three situations below are possible:

1. a sequence converges in D′ but not almost everywhere;

2. a sequence converges almost everywhere but not in D′;

3. a sequence converges almost everywhere and in D′ but the limits are different.

We refer to the exercise 1.73 for some examples.

With the above definitions, differentiation is a continuous operation in D′(Ω), in the followingsense.

Proposition 1.37. Consider a sequence (Tn) that converges to T in D′(Ω) when n → ∞. Then,for any α ∈ Nd, ∂αTn converges to ∂αT in D′(Ω) when n→ ∞.

Proof. Let φ ∈ D(Ω). We have

〈∂αTn, φ〉 = (−1)|α|〈Tn, ∂αφ〉 −→ (−1)|α|〈T, ∂αφ〉 = 〈∂αT, φ〉.

We hence get ∂αTn −→n→∞ ∂αT .

The following proposition is a direct corollary of the above result.

Proposition 1.38. Let (Tn)n∈N be a sequence in D′(Ω). Assume that the series∑

n∈N

Tn converges

in D′(Ω) to a distribution T . Then, for any α ∈ Nd, the series∑

n∈N

∂αTn converges in D′(Ω) and

we have

∂αT =∑

n∈N

∂αTn.


In D′, it is thus possible to differentiate a series as soon as

T =∑

n∈N

Tn,

this equality being an equality between distributions. This result is of course in sharp contrastwith the results when manipulating functions.

The following result is very useful, and make the manipulation of distributions easy. We omitits proof.

Theorem 1.39. Consider a sequence (Tn)n∈N of distributions in Ω. Assume that, for any φ ∈D(Ω), the sequence (〈Tn, φ〉)n∈N converges to some limit ℓφ. Then, the map T defined by

〈T, φ〉 = ℓφ

is a distribution on Ω.

Note that T is obviously a linear form on D(Ω). To show that it is a distribution, we are leftwith showing the continuity property (1.2), which is actually not easy. The above theorem statesthat this property indeed holds.

Exercise 1.40. Let a ∈ Rd and e ∈ Rd such that |e| = 1. Identify the limit in D′(Rd), as ǫ tendsto 0, of the family of distributions

Tǫ =1

ǫδa+ǫe −

1

ǫδa.

Exercise 1.41. Let h ∈ L1loc(R

2) and x0 ∈ R. We define the single layer distribution SL(h, x0) ∈D′(R3) by

∀φ ∈ D(R3), 〈SL(h, x0), φ〉 =∫

R2

h(y, z)φ(x0, y, z) dy dz.

Show that SL(h, x0) is indeed a distribution on R3.

Let g ∈ L1loc(R

2). Identify the limit in D′(R3), as ǫ tends to 0, of the family of distributions

Tǫ = SL(g/ǫ, ǫ)− SL(g/ǫ, 0).

Exercise 1.42. Show that∑

n∈Z

δn converges in D′(R).

Exercise 1.43. Does the series+∞∑

n=1

δ1/n

converge in D′(R)? What about in D′(0,+∞)?

1.8. MORE ON DIFFERENTIATION: THE ONE-DIMENSIONAL CASE 13

1.8 More on differentiation: the one-dimensional case

1.8.1 The case of C1 functions

Consider f ∈ C1(a, b). The functions f anddf

dxare in L1

loc(a, b), and hence in D′(a, b). We have

⟨d

dxTf , φ

⟩= −

∫ b

a

fdφ

dx=

∫ b

a

df

dxφ =

⟨T df

dx, φ⟩.

This means thatd

dxTf = T df

dx,

namely that the derivation in the sense of distributions coincides with the standard derivation forfunctions of class C1.

1.8.2 The case of functions that are piecewise C1

Definition 1.44. A function f is piecewise Ck in (a, b) if, for any compact interval [α, β] containedin (a, b), there exists a finite number of points α = a0 < a1 < . . . < aN+1 = β such that

• in each interval (ai, ai+1), f is Ck;

• f and its derivatives up to the order k can be extended by continuity on each side of the pointsa1, . . . , aN (note that the left and right values of f – or its derivatives – at points ai are notnecessarily equal).

Note that a piecewise C1 function has at most a countable number of discontinuity points and thatthe possible accumulation points of the set of discontinuity points are only a and b.

Here are some examples and counter-examples:

• the Heaviside function is piecewise C1;

• the function x 7→ |x| is piecewise C1;

• the function x 7→ tanx is not piecewise C1;

• the function x 7→√|x| is not piecewise C1.

Theorem 1.45 (Jump formula). Let f be a piecewise C1 function on (a, b). With the abovenotation, we have

f ′ = f ′reg +

∑

i∈I

[f(ci + 0)− f(ci − 0)] δci ,

where f ′ is the derivative of f in the sense of distributions, cii∈I are the points where f isdiscontinuous (recall that I is countable and its only possible accumulation points are a and b),f ′reg is the piecewise continuous function defined (except at the points ci) as the standard derivative

of f , f(ci + 0) and f(ci − 0) are the (right and left) limits of f in ci, and δci is the Dirac mass inci.

Proof. The proof is based on the integration by part formula.

Exercise 1.46. Let f be a piecewise C2 function on (a, b). Check that

f ′′ = f ′′reg +

∑

i∈I

[f ′(ci + 0)− f ′(ci − 0)] δci +∑

i∈I

[f(ci + 0)− f(ci − 0)] δ′ci

for some countable set I and some points cii∈I .


1.8.3 Linear differential equations in D′(a, b)

Theorem 1.47. Consider the open interval (a, b).

1. The distributions T on (a, b) satisfying T ′ = 0 in D′(a, b) are the constant functions.

2. For any S ∈ D′(a, b), there exists T ∈ D′(a, b) such that T ′ = S.

The first statement means that, if T ′ = 0 in D′(a, b), then there exists a constant C such that

T = C, which means that, for any φ ∈ D(a, b), we have 〈T, φ〉 = C

∫ b

a

φ.

Proof. We first note that a test function φ ∈ D(a, b) admits a primitive in D(a, b) if and only if∫ b

a

φ = 0 (in this case, the primitive is

∫ x

a

φ). Let ρ ∈ D(a, b) such that

∫ b

a

ρ = 1. Since the

integral of φ−(∫ b

a

φ

)ρ vanishes, there exists ψ ∈ D(a, b) such that

ψ′(x) = φ(x) −(∫ b

a

φ

)ρ(x). (1.13)

Consider now T ∈ D′(a, b) such that T ′ = 0. We write

〈T, φ〉 =(∫ b

a

φ

)〈T, ρ〉+ 〈T, ψ′〉 =

(∫ b

a

φ

)〈T, ρ〉 − 〈T ′, ψ〉 =

(∫ b

a

φ

)〈T, ρ〉.

Denoting C the constant 〈T, ρ〉, we have hence shown that T = C.

Consider now S ∈ D′(a, b). For any φ ∈ D(a, b), we set

〈T, φ〉 = −〈S, ψ〉,

where ψ ∈ D(a, b) is uniquely defined by (1.13). It is then easy to show that T is a distributionand that T ′ = S.

Theorem 1.47 can be used to show existence and uniqueness results for differential equationsin D′. The case of the equation xT ′ + T = 0 is considered in the exercise 1.76.

1.8.4 Link between the standard derivative and the derivative in the

sense of distributions

For a function f ∈ L1loc, the relation between the standard derivative and the derivative in the

sense of distributions is not straightforward:

1. for a C1 function, the two concepts lead to the same object;

2. for a piecewise C1 function, the standard derivative is defined almost everywhere, but doesnot fully represent the variations of f since it ignores jumps; the derivative in the sense ofdistributions is the correct concept;

3. when f is everywhere differentiable without being piecewise C1, the situation is more com-plex. When f ′ is locally integrable, the two concepts coincide (see Exercise 1.77). Otherwise,the derivative in the sense of distributions is somehow the finite part of the standard deriva-tive (see the exercises 1.54 and 1.58).

1.9. MORE ON DIFFERENTIATION: THE MULTI-DIMENSIONAL CASE 15

1.9 More on differentiation: the multi-dimensional case

1.9.1 Schwarz Theorem

Theorem 1.48 (Schwarz Theorem). Let T ∈ D′(Ω). We have

∂α∂βT = ∂β∂αT = ∂α+βT.

Proof. The proof is straightforward.

1.9.2 The case of C1 functions

As in the one-dimensional case, we have the following result. Consider Ω an open subset of Rd andf ∈ C1(Ω). Then, for any 1 ≤ i ≤ d,

∂

∂xiTf = T ∂f

∂xi

.

1.9.3 Stokes formula and applications

Definition 1.49. An bounded open set Ω ⊂ Rd is smooth (of class C1) if, at any x ∈ ∂Ω, it ispossible to define an exterior normal vector, denoted n(x), and if the function x 7→ n(x) from ∂Ωto Rd is continuous.

Examples: the circle (in dimension d = 2) or the sphere (when d = 3) are smooth. The square(when d = 2) or the cube (when d = 3) are not smooth.

Theorem 1.50 (Stokes formula). Let Ω be a bounded smooth open subset of Rd and let X be afield defined in Ω, which is vector valued, and such that all components belong to C1(Ω). Then

∫

Ω

div (X) dx =

∫

∂Ω

X · n dσ.

Corollary 1.51 (Integration by part formula). Let Ω be a bounded smooth open subset of Rd andlet f and g in C1(Ω). Then

∫

Ω

∂f

∂xig dx =

∫

∂Ω

f g (n · ei) dσ −∫

Ω

f∂g

∂xidx.

Proof. Use the Stokes formula for the field X(x) = f(x) g(x) ei.

Corollary 1.52 (Green formula). Let Ω be a bounded smooth open subset of Rd and let f and gin C2(Ω). Then ∫

Ω

f ∆g = −∫

Ω

∇f · ∇g +∫

∂Ω

f∂g

∂n.

Proof. Use the Stokes formula for the field X(x) = f(x)∇g(x).

1.10 Principal values and finite parts

Principal values and finite parts of functions are distributions that naturally appear when onewants to associate a distribution with some singular functions (i.e. functions not in L1

loc). Here,we only consider two specific examples.


1.10.1 Principal value of 1/x

We define the principal value of the function x ∈ R 7→ 1/x by

∀φ ∈ D(R),

⟨PV

(1

x

), φ

⟩= limǫ→0, ǫ>0

∫

|x|>ǫ

φ(x)

xdx. (1.14)

Exercise 1.53. Show that PV

(1

x

)is indeed a distribution, of order 1. Hint: show that the

distribution is of order at most 1, and then that its order is exactly 1. To do this, compute itsaction on the sequence φj(x) = φ0(x) atan(jx), where φ0 ∈ D(R) is even and equal to 1 in theneighborhood of 0.

Exercise 1.54. Show that the function x ∈ R 7→ ln |x| defines a distribution in D′(R) and computeits derivative.

Exercise 1.55. Show that xPV

(1

x

)= 1.

Exercise 1.56. Explain why the quantities(xPV

(1

x

))δ0 and (x δ0)PV

(1

x

).

are well defined in D′(R) and compute them. In view of this computation, does it seem possible todefine the product of any two distributions?

1.10.2 Finite part of H(x)xα

Let φ ∈ D(R) and H be the Heaviside function. The integral∫

R

H(x)xαφ(x) dx =

∫ +∞

0

xαφ(x) dx

is well-defined for α > −1, but not for any α ≤ −1 if φ(0) 6= 0. In the second case, H(x)xα

is therefore not a distribution. It is nevertheless still possible to associate a distribution to thefunction H(x)xα, as we explain now.

Using the Hadamard lemma 1.22, one can show that, for any ǫ > 0,∫ +∞

ǫ

φ(x)xα dx = Pφ(ǫ) +Rφ(ǫ),

where Pφ is a linear combination of negative powers of ǫ (and of ln ǫ when α is a negative integer)and where Rφ(ǫ) converges to a finite limit as ǫ tends to 0. One then defines the finite part of thefunction H(x)xα, which is denoted FP(H(x)xα), by the formula

∀φ ∈ D(R), 〈FP(H(x)xα), φ〉 = limǫ→0, ǫ>0

Rφ(ǫ).

Exercise 1.57. Show that the finite part of the function H(x)xα is a distribution of order theinteger part of α.

Exercise 1.58. Let −1 < α < 0. Show that the derivative (in the sense of distributions) of the

function H(x)xα is αFP(H(x)xα−1

).

1.11. DISTRIBUTIONS WITH COMPACT SUPPORT 17

1.11 Distributions with compact support

1.11.1 Definitions and first properties

Definition 1.59. Let T ∈ D′(Ω).

1. Let ω be an open subset of Ω. The distribution T is said to vanish on ω if, for any φ ∈ D(Ω)such that Supp(φ) ⊂ ω, we have 〈T, φ〉 = 0.

2. The support of T is the complement in Ω of the union of the open subsets of Ω on which Tvanishes.

Note that, by construction, T vanishes on Ω\Supp(T ). The notion of support of a distributiongeneralizes the notion of support of a function, as shown by Exercise 1.60 below.

Exercise 1.60. Consider f ∈ C0(R). Since C0(R) ⊂ L1loc(R), a distribution Tf can naturally be

associated to the function f (see (1.3)). Show that Supp(Tf ) = Supp(f).

Exercise 1.61. Show that the support of δa ∈ D′(R) is a.

Definition 1.62. The vector space of distributions on Ω with compact support is denoted E ′(Ω).

Theorem 1.63. If a distribution T ∈ D′(Ω) has a compact support, then it is of finite order.

Proof. Let K be the support of T and α = d(K,Rd \Ω) be the distance between K and Rd \Ω (ifΩ = Rd, then we take α = +∞). We set β = inf(1, α) and consider the sets

K ′ =

x ∈ Rd, d(x,K) ≤ β

3

, Ω′ =

x ∈ Rd, d(x,K) <

2β

3

.

The set K ′ is compact, the set Ω′ is open and with compact closure and we have

K ⊂ K ′ ⊂ Ω′ ⊂ Ω′ ⊂ Ω.

Let p ∈ N and C ∈ R such that

∀φ ∈ DΩ′ (Ω), |〈T, φ〉| ≤ C supx∈Ω′, |α|≤p

|∂αφ(x)|.

Consider now ρ ∈ D(Ω) which is equal to 1 on K ′ and with support in Ω′. For any φ ∈ D(Ω), wehave

〈T, φ〉 = 〈T, ρφ〉+ 〈T, (1− ρ)φ〉and 〈T, (1 − ρ)φ〉 = 0 since the supports of T and of (1 − ρ)φ are disjoint. Moreover, sinceSupp(ρφ) ⊂ Ω′, we have

∀φ ∈ D(Ω), |〈T, φ〉| ≤ C supx∈Ω, |α|≤p

|∂α(ρφ)(x)|.

Using Leibniz formula, we get

∀φ ∈ D(Ω), |〈T, φ〉| ≤ C′ supx∈Ω, |α|≤p

|∂αφ(x)|

with C′ = C supx∈Ω, |α|≤p, β≤α

α!

β! (α− β)!|∂βρ(x)|. We thus obtain that T has a finite order, lower

than or equal to p.


Exercise 1.64. Let T ∈ D′(Ω) with Ω ⊂ Rd. Then, for any 1 ≤ i ≤ d, we have Supp

(∂T

∂xi

)⊂

Supp(T ).

Remark 1.65. Let T ∈ D′(Ω) and consider φ ∈ D(Ω) such that φ vanishes on Supp(T ). Thisdoes not imply that 〈T, φ〉 = 0.

Consider indeed the case of T = δ′0, the support of which is 0 (in view of Exercises 1.61and 1.64, we have Supp(δ′0) ⊂ 0, and the support cannot be empty as δ′0 is not equal to 0).Consider φ(x) = xρ(x), with ρ ∈ D(R) equal to 1 on a neighborhood of 0. We observe that φvanishes on Supp(δ′0) while 〈δ′0, φ〉 6= 0.

However, if φ vanishes on an open neighborhood of Supp(T ), then 〈T, φ〉 = 0 (see the proof ofProposition 1.66 below where this argument is used).

1.11.2 Distributions with support restricted to a point

Proposition 1.66. Let T ∈ D′(Ω) of order 0 such that Supp(T ) = a. Then there exists aconstant c such that

T = c δa.

Proof. Let ǫ > 0 such that Ba(ǫ) ⊂ Ω. Let K = Ba(ǫ/2) and C be a constant such that

∀φ ∈ DK(Ω), |〈T, φ〉| ≤ C supx∈K

|φ(x)|.

Let 0 < r < ǫ and ρ ∈ D(Ω), with compact support in Ba(r) and such that 0 ≤ ρ ≤ 1 and ρ = 1in K. For any φ ∈ D(Ω), we have

〈T, φ〉 = 〈T, ρφ〉+ 〈T, (1− ρ)φ〉.The function (1 − ρ)φ vanishes in a neighborhood of a = Supp(T ), hence 〈T, (1− ρ)φ〉 = 0. Wealso observe that Supp(ρφ) ⊂ K, hence

|〈T, φ〉| = |〈T, ρφ〉| ≤ C supΩ

|ρφ| ≤ C supBa(r)

|φ|.

The constant C does not depend on r. Taking the limit r → 0, we thus deduce that

|〈T, φ〉| ≤ C|φ(a)|. (1.15)

Now consider an arbitrary function ψ ∈ D(Ω). We write

ψ = ψ(a)ρ+ (ψ − ψ(a)ρ)

and see that the function φ = ψ − ψ(a)ρ is such that φ(a) = 0. Using (1.15), we obtain that〈T, ψ − ψ(a)ρ〉 = 0. Thus

〈T, ψ〉 = 〈T, ψ(a)ρ〉 = 〈T, ρ〉 ψ(a).We hence get that T = cδa, where c = 〈T, ρ〉.

The next proposition (the proof of which is omitted) generalizes Proposition 1.66 by character-izing the distributions (of any order) that have a support equal to the point a.

Proposition 1.67. Let T ∈ D′(Ω) such that Supp(T ) = a. Then T is of the form

T =∑

|α|≤p

cα∂αδa,

where p is the order of T and where cα, |α| ≤ p, are constant.

Exercise 1.68. Show that δa and PV

(1

x

)have compact support and compute their support.

1.12. EXERCISES 19

1.12 Exercises

We collect here some exercises that are, in general, less straightforward that the previous ones.

Exercise 1.69. For the following maps T : D(R) → R, identify which ones are distributions:

• 〈T, φ〉 =∫

R

|φ(t)| dt.

• 〈T, φ〉 =∞∑

n=0

φ(n)(n).

Exercise 1.70. Show that the map T : D(R) → R defined by 〈T, φ〉 =∞∑

n=1

1

n

(φ

(1

n

)− φ(0)

)is

a distribution of order lower or equal to 1.

Exercise 1.71. Consider T : D(R2) → R defined by

〈T, φ〉 =∫ ∞

0

φ(z, 2z) dz.

Show that T is a distribution and that∂T

∂x+ 2

∂T

∂y= δ(0,0) in D′(R2).

Exercise 1.72. Consider vn(x) =n

1 + n2 x2and wn(x) = atan(nx).

1. Show that vn(x) converges to some v(x) almost everywhere on R. Identify v.

2. Show that, for any α > 0, vn converges in L1((−∞,−α) ∪ (α,∞)).

3. Show that wn converges in D′(R) toπ

2sgn(x).

4. Compute the limit of vn in D′(R) by two different methods.

Exercise 1.73. The aim of this exercise is to study the links between convergence almost every-where and convergence in D′.

1. Consider f ∈ D(R) and the sequence fn(x) = f(x−n). Study the almost everywhere conver-gence of fn, the convergence in D′(R) and the convergence in L1(R).

2. Consider fn(x) = einx. Show that fn does not converge almost everywhere but that fnconverges to 0 in D′(R).

3. Consider fn(x) =

n∑

k=1

χ1/n2

(x− 1

k

), where χǫ is an approximation of identity, that is

χǫ(x) = ǫχ(ǫ x) where χ ∈ D(R), with support in the unit ball and with integral equal to 1.Show that fn → 0 almost everywhere but that fn does not converge in D′(R).

4. Consider fn(x) =1

σn√2π

e−x2/2σ2

n with σn → 0. Show that fn → 0 almost everywhere and

that fn → δ0 in D′(R).


Exercise 1.74. Identify all the distributions T ∈ D′(R) such that xT = 0.

Exercise 1.75. For any φ ∈ D(R), we define

⟨1

x+ i0, φ

⟩= lim

ǫ→0, ǫ>0

∫

R

φ(x)

x+ iǫdx and

⟨1

x− i0, φ

⟩= lim

ǫ→0, ǫ>0

∫

R

φ(x)

x− iǫdx.

Show that1

x+ i0and

1

x− i0define distributions of order 1. Show that

1

x+ i0− 1

x− i0= −2iπδ0 and

1

x+ i0+

1

x− i0= 2PV

(1

x

).

Exercise 1.76. Identify all distributions T ∈ D′(R) such that xT ′ + T = 0. (Hint: considerS = xT ).

Exercise 1.77. Let f ∈ L1loc(R) and F (x) =

∫ x

0

f . Show that the derivative (in the sense of

distributions) of F is equal to f .

Chapter 2

Sobolev spaces

Using the distribution theory exposed in Chapter 1, we are now in position to build the appropriatespaces of functions for the study of Partial Differential Equations.

2.1 The spaces Hk(Ω)

Definition 2.1. Let Ω be an open subset of Rd and let k ∈ N. The set of functions of L2(Ω), thederivative (in the sense of distributions) of which, up to order k, are in L2(Ω), is denoted Hk(Ω):

Hk(Ω) =f ∈ L2(Ω), ∀α ∈ Nd, |α| ≤ k, ∂αf ∈ L2(Ω)

.

We recall that we can associate to any f ∈ L2(Ω) a distribution, which is again denoted f .Since f is a distribution, it can be differentiated, and ∂αf is hence a distribution (we recall that,

for α = (α1, . . . , αd) ∈ Nd, we denote ∂αf =∂α1+···+αdf

∂α1x1 . . . ∂αdxd, see (1.1)). In the above definition,

we wrote that ∂αf ∈ L2(Ω): this means that there exists a function Fα ∈ L2(Ω) such that

∀φ ∈ D(Ω), 〈∂αf, φ〉 =∫

Ω

Fαφ.

Theorem 2.2. Hk(Ω) is a vector space. Endowed with the scalar product

(f, g)Hk =∑

|α|≤k

∫

Ω

∂αf(x) ∂αg(x) dx,

it is an Hilbert space. Its norm is denoted ‖ · ‖Hk .

Proof. Obviously, Hk(Ω) is a vector space and (·, ·)Hkis a scalar product on Hk(Ω). We are left

with showing that Hk(Ω) is a complete space for the norm ‖ · ‖Hk . Let (fn)n∈N be a Cauchysequence of Hk(Ω). We first observe that

‖fp − fq‖L2 ≤ ‖fp − fq‖Hk ,

hence (fn) is a Cauchy sequence in L2, hence it converges in L2 to some f ∈ L2. Likewise, for any|α| ≤ k,

‖∂αfp − ∂αfq‖L2 ≤ ‖fp − fq‖Hk ,

hence ∂αfn converges in L2 to some function gα. In addition, convergence in L2 implies convergencein D′. Hence

fn −→ f in D′

21

22 CHAPTER 2. SOBOLEV SPACES

and∂αfn −→ gα in D′.

From the first assertion, we deduce that

∂αfn −→ ∂αf in D′.

By uniqueness of the limit in D′(Ω), we find that ∂αf = gα ∈ L2. Hence f ∈ Hk. Since∂αfn −→ ∂αf in L2, we get that fn −→ f in Hk.

2.2 The space H10(Ω)

2.2.1 Definition

We start by an important density result.

Proposition 2.3.

1. If Ω = Rd, then D(Rd) is dense in H1(Rd).

2. If Ω ⊂ Rd, Ω 6= Rd, then D(Ω) is not dense in H1(Ω).

Proof. The first assertion can be shown following a technical proof using truncation and regular-ization. We omit it here and focus on the second assertion, in dimension 1, when Ω = (0, 1). Letφ ∈ D(0, 1). We have

∀x ∈ (0, 1), φ(x) =

∫ x

0

φ′(t) dt,

hence

∀x ∈ (0, 1), |φ(x)| ≤∫ x

0

|φ′(t)| dt ≤(∫ x

0

|φ′(t)|2 dt)1/2 (∫ x

0

12 dt

)1/2

≤ ‖φ′‖L2(0,1).

We therefore obtain

‖φ‖L2(0,1) =

(∫ 1

0

|φ(x)|2 dx)1/2

≤ ‖φ′‖L2(0,1),

an estimate which is valid for any φ ∈ D(0, 1).If D(0, 1) were dense in H1(0, 1), then this estimate would remain true for any φ ∈ H1(0, 1).

However, consider the function f = 1 on (0, 1). We see that f ∈ H1(0, 1), ‖f‖L2(0,1) = 1 and‖f ′‖L2(0,1) = 0, hence f does not satisfy the above estimate.

Definition 2.4. The closure of D(Ω) in H1(Ω) is denoted H10 (Ω).

Remark 2.5. In view of Proposition 2.3, we have H10 (R

d) = H1(Rd). In contrast, H10 (Ω) is a

strict subspace of H1(Ω) when Ω ⊂ Rd with Ω 6= Rd.

Proposition 2.6. The space H10 (Ω) is a vector space. It is an Hilbert space for the scalar product

(·, ·)H1 .

Proof. It is obvious that H10 (Ω) is a vector space (it is the closure of a vector space for a given

norm). The scalar product of H1(Ω) restricted to H10 (Ω) of course defines a scalar product on

H10 (Ω). Since H1

0 (Ω) is closed in a complete space, it is itself complete.

2.2. THE SPACE H10 (Ω) 23

2.2.2 Poincaré inequality

We now state a very important result, that we will often need in these lecture notes (see e.g.Section 3.3).

Theorem 2.7 (Poincaré inequality). Let Ω be a bounded open subset of Rd. There exists a constantCΩ such that

∀u ∈ H10 (Ω), ‖u‖L2(Ω) ≤ CΩ ‖∇u‖L2(Ω) .

In the language of functional analysis, it is said that the L2 norm of u is controlled by the L2

norm of its gradient.

Proof of Theorem 2.7. Since the open set Ω is bounded, there exists L > 0 such that Ω ⊂ [−L,L]d.Let φ ∈ D(Ω) and ψ the extension of φ by 0 to the whole space Rd. We have that ψ ∈ C∞(Rd)and that

∀x ∈ [−L,L]d, ψ(x) =

∫ x1

−L

∂ψ

∂x1(t, x2, · · · , xd) dt.

We hence have

|ψ(x)|2 =

(∫ x1

−L

∂ψ

∂x1(t, x2, · · · , xd) dt

)2

≤(∫ x1

−L

∣∣∣∣∂ψ

∂x1(t, x2, · · · , xd)

∣∣∣∣2

dt

)(∫ x1

−L

12 dt

)

≤ 2L

∫ L

−L

∣∣∣∣∂ψ

∂x1(t, x2, · · · , xd)

∣∣∣∣2

dt.

Integrating the above inequality on [−L,L]d, we deduce that

∫

[−L,L]d|ψ|2 ≤ 4L2

∫

[−L,L]d

∣∣∣∣∂ψ

∂x1

∣∣∣∣2

≤ 4L2

∫

[−L,L]d|∇ψ|2.

Since Supp (ψ) = Supp (φ) ⊂ Ω, Supp (∇ψ) = Supp (∇φ) ⊂ Ω, and since φ = ψ and ∇φ = ∇ψon Ω, we deduce that

∀φ ∈ D(Ω), ‖φ‖L2(Ω) ≤ 2L ‖∇φ‖L2(Ω) . (2.1)

In addition, the mappings

H1(Ω) −→ R

u 7→ ‖u‖L2(Ω)and

H1(Ω) −→ R

u 7→ ‖∇u‖L2(Ω)

are continuous. Since D(Ω) is dense in H10 (Ω) under the H1 norm, we get that the estimate (2.1)

remains valid for any function in H10 (Ω).

2.2.3 On an alternative definition of H1

0(Ω) via the use of traces

We close this section by making some remarks on the definition of H10 (Ω) via the notion of traces.

It is often said that H10 (Ω) is the space of functions in H1(Ω) that “vanish on the boundary”. To

formalize this, we first introduce the notion of trace. For any function u ∈ C0(Ω), the trace of uon ∂Ω is defined as

γ(u) : ∂Ω −→ R

x 7→ u(x).


Put differently, we have γ(u) = u|∂Ω. Note that the trace mapping

γ : C0(Ω) −→ C0(∂Ω)u 7→ γ(u)

is linear and continuous. The question is now to extend the notion of trace to functions that areless regular than C0(Ω). This is not always possible. Here are some answers:

1. It is not possible to define the trace of a function u ∈ L2(Ω).

Consider indeed the function u : x 7→ sin(1/x) on Ω = (0, 1), which belongs to L∞(Ω) ⊂L2(Ω). The boundary ∂Ω is the union of two points, 0 and 1. The function u is continuousat 1 and we can hence define its trace in 1 (it is the real number sin 1). In contrast, the setof limit points of u at 0 is the whole interval [−1, 1]. Hence, there is no natural way to definethe trace of u (the value of u) at 0.

2. In dimension d = 1, a function in H1(a, b) admits a continuous representation. The proof ofthis statement is carried out in Exercise 2.14. Hence, one can define the trace in a and in bof a function belonging to H1(a, b).

3. In dimension d ≥ 2, a function in H1(Ω) is not necessarily continuous (see Exercise 2.15).However, one can still define its trace on ∂Ω. More precisely, there exists a linear andcontinuous mapping

γ : H1(Ω) −→ L2(∂Ω)u 7→ γ(u)

such that, for any u ∈ H1(Ω) ∩ C0(Ω), we have γ(u) = u|∂Ω. Actually, the function γ(u) ismore regular than simply L2(∂Ω), it belongs to the space H1/2(∂Ω) (we refer to [1] for moredetails).

We have the following characterization, which provides a definition of H10 (Ω) which is equivalent

to that given in Definition 2.4.

Proposition 2.8.

H10 (Ω) =

u ∈ H1(Ω), γ(u) = 0

.

2.3 The space H−1(Ω)

Definition 2.9. Let Ω be an open subset of Rd. We denote by H−1(Ω) the vector space of distri-butions T ∈ D′(Ω) for which there exists a constant C such that

∀φ ∈ D(Ω), |〈T, φ〉| ≤ C‖φ‖H1(Ω).

Remark 2.10. It is clear that L2(Ω) ⊂ H−1(Ω). Indeed, if f ∈ L2(Ω), then

∀φ ∈ D(Ω), |〈f, φ〉| =∣∣∣∣∫

Ω

fφ

∣∣∣∣ ≤ ‖f‖L2(Ω) ‖φ‖L2(Ω) ≤ ‖f‖L2(Ω) ‖φ‖H1(Ω).

Theorem 2.11. It is possible to identify H−1(Ω) with the topological dual of H10 (Ω) (i.e. with the

vector space of linear and continuous forms on H10 (Ω)).

2.3. THE SPACE H−1(Ω) 25

The proof of Theorem 2.11 is given below. We note here that, as a consequence of the aboveresults, we have the following inclusions:

D(Ω) ⊂ H10 (Ω) ⊂ L2(Ω) ⊂ H−1(Ω) ⊂ D′(Ω)

and, for any T ∈ L2(Ω) and φ ∈ D(Ω), we have

〈T, φ〉 = 〈T, φ〉H−1,H10= 〈T, φ〉L2 =

∫

Ω

T φ.

The space L2 is the so-called “pivot” space for all of these dualities: if T ∈ L2(Ω) and φ ∈ D(Ω),then 〈T, φ〉 can be understood in (D′,D), (H−1, H1

0 ) or (L2, L2).

Proof of Theorem 2.11. Let T ∈ H−1(Ω). The linear map

φ 7→ 〈T, φ〉

is continuous on the space D(Ω) endowed with the H1 norm. Since D(Ω) is dense in H10 (Ω) for

that norm, we take as given the fact that the above mapping can be extended (in a unique way)to some linear and continuous mapping on H1

0 (Ω), that we denote

φ 7→ 〈T, φ〉H−1,H10,

and which satisfies∀φ ∈ D(Ω), 〈T, φ〉H−1,H1

0= 〈T, φ〉.

Hence, to any T ∈ H−1(Ω), we can associate an element of the topological dual of H10 (Ω) (i.e. the

vector space of linear and continuous forms on H10 (Ω)). Hence, we define

α : H−1(Ω) −→ (H10 (Ω))

′

T 7→ 〈T, ·〉H−1,H10.

Conversely, let L ∈ (H10 (Ω))

′. There exists a constant C such that

∀φ ∈ H10 (Ω), |L(φ)| ≤ C‖φ‖H1 .

We restrict L to D(Ω) ⊂ H10 (Ω), and we hence get a linear form on D(Ω) that satisfies

∀φ ∈ D(Ω), |L(φ)| ≤ C‖φ‖H1 .

We are left to show that L is a distribution, i.e. that L satisfies the “continuity property” (1.2).Let K be a compact set contained in Ω and let φ ∈ DK(Ω). We see that

|L(φ)| ≤ C‖φ‖H1 ≤ C(‖φ‖2L2 + ‖∇φ‖2L2

)1/2.

Using that ‖φ‖L2 ≤√|K| sup |φ| and ‖∇φ‖L2 ≤

√|K| sup |∇φ|, we obtain that

|L(φ)| ≤ C′ sup|α|≤1, x∈K

|∂αφ|.

Hence L defines a distribution (of order lower or equal to 1). We hence define

β : (H10 (Ω))

′ −→ H−1(Ω)

L 7→ LD(Ω).

It is easy to check that α β = I(H10 (Ω))′ and that β α = IH−1(Ω).


Proposition 2.12 (Characterization of the distributions in H−1). Let Ω be an open subset ofRd. A distribution T belongs to H−1(Ω) if and only if there exist, for any |α| ≤ 1, a functiongα ∈ L2(Ω) such that

T =∑

|α|≤1

∂αgα.

Proof. It is obvious that, if T is of the form

T =∑

|α|≤1

∂αgα

with gα ∈ L2(Ω), we have

∀φ ∈ D(Ω), |〈T, φ〉| =

∣∣∣∣∣∣〈∑

|α|≤1

∂αgα, φ〉

∣∣∣∣∣∣

≤

∣∣∣∣∣∣

∑

|α|≤1

(−1)|α|〈gα, ∂αφ〉

∣∣∣∣∣∣

≤∑

|α|≤1

‖gα‖L2‖∂αφ‖L2

≤

∑

|α|≤1

‖gα‖L2

‖φ‖H1 .

Hence T ∈ H−1(Ω). The converse statement is more challenging to prove. We omit its proof.

2.4 Exercises

Exercise 2.13.

1. For which values of k do the following functions belong to Hk(−1, 1): x 7→ sinx, x 7→ |x| andx 7→ sgn(x)?

2. For which values of k, α and d does the function x 7→ 1

|x|α belong to Hk(B0(1)), where B0(1)

is the unit ball in Rd?

3. For which values of k, α and d does the function x 7→ 1

|x|α belong to Hk(Rd \B0(1)

)?

Exercise 2.14. Consider two real numbers a and b with a < b. Show that any function in H1(a, b)has a continuous representation.

Exercise 2.15. Let Ω be the disk centered at 0 of radius 1/e in R2. Show that the function

u(x, y) = ln[− ln

(√x2 + y2

)]

belongs to H10 (Ω) but that it is not continuous at 0.

2.4. EXERCISES 27

Exercise 2.16. Consider the vector space

V =v ∈ L2(Ω), ∆v ∈ L2(Ω)

endowed with the scalar product (v, w)V = (v, w)L2 + (∆v,∆w)L2 . Show that V is a Hilbert space.

Exercise 2.17. Let p be a real number with 1 ≤ p ≤ +∞. Let k ∈ N. We set

W k,p(Ω) =u ∈ Lp(Ω), ∂αu ∈ Lp(Ω) for any α ∈ Nd, |α| ≤ k

.

This vector space is endowed with the norm

‖u‖Wk,p =∑

α∈Nd, |α|≤k

‖∂αu‖Lp.

Explain why we can consider the derivative (in the sense of distributions) of u, and show that thespace W k,p(Ω) is a Banach space.

Exercise 2.18. Let u ∈ H1(R). Show that

‖u‖2L∞(R) ≤ ‖u‖L2(R) ‖u′‖L2(R).

Hint: use the density of D(R) in H1(R).

Exercise 2.19. Consider two real numbers a and b with a < b. Let f ∈ H1(a, b). Show that|f | ∈ H1(a, b). It will be useful to show that

d|f |dx

(x) =

f(x)

|f(x)|df

dx(x) if f(x) 6= 0,

0 otherwise.

Deduce that, if f and g belong to H1(a, b), then max(f, g) and min(f, g) belong to H1(a, b).

Exercise 2.20. Let u ∈ H1(Rd). We denote (e1, . . . , ed) the canonical basis of Rd. Show that, forany 1 ≤ i ≤ d, ∫

Rd

∣∣∣∣∂u

∂xi

∣∣∣∣2

= limt→0

1

t2

∫

Rd

|u(x+ t ei)− u(x)|2 dx.

Exercise 2.21. Let Ω be a bounded open subset of Rd and f ∈ L2(Ω). On H10 (Ω), consider the

functional J defined by

J(u) =1

2

∫

Ω

|∇u|2 −∫

Ω

f u.

The quantity J(u) is the energy that naturally arises in thermal or elasticity problems.Show that J is infinite at infinity, namely that J(u) → ∞ when ‖u‖H1 → ∞.

Exercise 2.22. The aim of this exercise is to show the Hardy inequality in D(R3).


1. Let f ∈ C∞([0,∞)) and such that there exists A > 0 such that f(r) = 0 for any r ≥ A. Showthat ∫ ∞

0

f2(r) dr = −2

∫ ∞

0

r f ′(r) f(r) dr

and deduce that ∫ ∞

0

f2(r) dr ≤ 4

∫ ∞

0

r2 (f ′(r))2 dr.

2. Deduce that

∀ψ ∈ D(R3),

∫

R3

ψ2(x)

|x|2 dx ≤ 4

∫

R3

|∇ψ(x)|2 dx. (2.2)

Hint: note that, in spherical coordinates,

∫

R3

ψ2(x)

|x|2 dx =

∫ π

0

∫ 2π

0

(∫ ∞

0

ψ2(r, θ, ϕ) dr

)sin(θ) dθdϕ

and that ∇ψ =∂ψ

∂rer +

1

r

∂ψ

∂θeθ +

1

r sin(θ)

∂ψ

∂ϕeϕ, where the spherical basis (er, eθ, eϕ) is

orthonormal.

Exercise 2.23. Consider the space

A =u ∈ H1(R3), ‖u‖L2(R3) = 1

and the function

E(u) = 1

2

∫

R3

|∇u|2 −∫

R3

|u(x)|2|x| dx.

The quantity E represents the electronic energy of the Hydrogen atom in quantum physics.

1. Show that E(ψ) is well-defined whenever ψ ∈ D(R3).

2. Using the inequality (2.2), show that, for any ψ ∈ D(R3), we have

∫

R3

|ψ(x)|2|x| dx ≤ 2‖ψ‖L2(R3) ‖∇ψ‖L2(R3). (2.3)

3. Let u ∈ H1(R3). The aim of this question is to show that

∫

R3

|u(x)|2|x| dx ≤ 2‖u‖L2(R3) ‖∇u‖L2(R3). (2.4)

(a) Recall the reason for which there exists a sequence of functions ψn ∈ D(R3) that con-verges to u in H1(R3).

(b) Let φ ∈ D(R3). Show thatφ(x)

|x| ∈ L2(R3) and that

limn→∞

∫

R3

ψn(x)φ(x)

|x| dx =

∫

R3

u(x)φ(x)

|x| dx.

(c) Using the inequality (2.2), show that the sequence of functionsψn(x)

|x| is a Cauchy se-

quence in L2(R3).

2.4. EXERCISES 29

(d) Deduce that there exists v ∈ L2(R3) such thatψn(x)

|x| converges to v in L2(R3).

(e) Using the questions (3a)–(3d), show thatu(x)

|x| = v(x) in L2(R3).

(f) Show that∣∣∣∣∫

R3

ψn(x)2

|x| dx−∫

R3

u(x)2

|x| dx

∣∣∣∣ ≤∥∥∥∥ψn − u

|x|

∥∥∥∥L2(R3)

‖ψn + u‖L2(R3)

and deduce that

limn→∞

∫

R3

ψn(x)2

|x| dx =

∫

R3

u(x)2

|x| dx.

(g) Using inequality (2.3) on ψn and taking the limit n→ ∞, show that (2.4) holds.

4. Consider the optimization problem

I = inf E(u); u ∈ A .

Using (2.4), show that I > −∞.

Exercise 2.24 (Hardy inequality). Consider the vector space

W 1(R3) =

u ∈ L2

loc(R3),

u(x)

|x| ∈ L2(R3), ∇u ∈(L2(R3)

)3.

We recall that L2loc(R

3) is the vector space of functions u : R3 7→ R that are measurable and such

that

∫

K

|u|2 < ∞ for any compact set K ⊂ R3. Since L2loc(R

3) ⊂ L1loc(R

3), we have that ∇u is

well-defined in the sense of distributions.

For any u and v in W 1(R3), we set

(u, v)W 1 =

∫

R3

u(x) v(x)

|x|2 dx+

∫

R3

∇u · ∇v. (2.5)

1. Show that, for any u and v in W 1(R3), the right-hand side of (2.5) is well-defined, and that(·, ·)W 1 defines a scalar product on W 1(R3).

2. Show that D(R3) ⊂W 1(R3).

3. Using the question (3e) of Exercise 2.23, show that H1(R3) ⊂W 1(R3).

4. We admit that W 1(R3), endowed with the above scalar product, is a Hilbert space and thatD(R3) is dense in W 1(R3) (when the space W 1(R3) is endowed with the norm ‖ · ‖W 1 asso-ciated to the scalar product (·, ·)W 1).

We recall (see Exercise 2.22) that, for any ψ ∈ D(R3),∫

R3

ψ(x)2

|x|2 dx ≤ 4

∫

R3

|∇ψ|2.

Show that, for any u ∈W 1(R3),∥∥∥∥u

|x|

∥∥∥∥L2

≤ 2 ‖∇u‖L2.


Chapter 3

Linear elliptic boundary valueproblems

This chapter is devoted to the mathematical analysis of linear elliptic boundary value problems.A boundary value problem is usually composed of

• one (or several) Partial Differential Equation (PDE),

• boundary conditions (or asymptotic conditions if the domain on which the problem is posedis the whole space Rd).

For time-dependent problems (which will not be considered here), one has to additionally considerinitial conditions.

We restrict ourselves to linear problems, where the solution depends linearly (or in an affinemanner) on the right-hand side of the PDE and on the imposed boundary conditions.

The term elliptic designates a class of partial differential equations, a typical example in thatclass being the Poisson equation −∆u = f , which is studied in Section 3.3. Other classes of PDEs,that will not be considered here, are parabolic problems, such as the heat equation

∂u

∂t−∆u = f (complemented by appropriate boundary and initial conditions),

or hyperbolic problems (such as the wave equation)

∂2u

∂t2−∆u = f (complemented by appropriate boundary and initial conditions).

In this chapter, we show that the Poisson equation

−∆u = f in D′(Ω),u = 0 on ∂Ω,

(3.1)

is well-posed inH1(Ω), in the sense that it has a unique solution, and that this solution depends in acontinuous manner on the right-hand side f . Problem (3.1) appears e.g. in electrostatic and is oftenwritten in the form −∆V = ρ, where V is the electric potential and ρ is the distribution of electriccharges. Problem (3.1) is also encountered in mechanics (it describes the vertical displacement uof a membrane submitted to some forces f and clamped at its boundary). Besides its own interest,Problem (3.1) is also interesting because many boundary value problems in the engineering sciencesare (more or less elaborated) variants of (3.1).

To show the well-posedness of (3.1), we proceed as follows:

31

32 CHAPTER 3. LINEAR ELLIPTIC BOUNDARY VALUE PROBLEMS

1. First, we show that Problem (3.1) is equivalent to another problem, written in the form of avariational formulation of the type

Find u ∈ V such that∀v ∈ V, a(u, v) = b(v)

(3.2)

for some appropriate choices of a Hilbert space V , a bilinear form a and a linear form b (seeSection 3.3 for details in the particular case of Problem (3.1)). Problems (3.1) and (3.2)are equivalent in the sense that any solution in H1(Ω) of Problem (3.1) is a solution toProblem (3.2), and the converse is also true.

2. Second, using the Lax-Milgram theorem, we show that Problem (3.2) is well-posed (i.e. hasa unique solution).

Besides the specific problem (3.1), other problems are also studied in this chapter. Several exercisesguide the reader through more and more complex examples.

As pointed out above, this chapter is devoted to the mathematical analysis of some boundaryvalue problems, establishing that these problems are well-posed. Questions related to the approx-imation of the exact solutions, to the estimation of the error, and more generally to the numericalanalysis of these problems, will be addressed in Chapter 5.

We also underline that we restrict ourselves, in this chapter and in Chapter 5, to coerciveproblems. Non-coercive problems will be studied in Chapters 4 and 6, from the mathematical andnumerical analysis standpoints, respectively.

3.1 Lax-Milgram theorem (symmetric version)

LetH be a vector space endowed with a norm ‖·‖. We recall that a linear form b onH is continuousif and only if there exists c such that

∀v ∈ H, |b(v)| ≤ c‖v‖.

Likewise, a bilinear form a on H ×H is continuous if and only if there exists M such that

∀(u, v) ∈ H ×H, |a(u, v)| ≤M‖u‖ ‖v‖.

Definition 3.1. Let H be a Hilbert space and a be a bilinear form on H. We say that a is coerciveon H if there exists a real number α > 0 such that

∀u ∈ H, a(u, u) ≥ α ‖u‖2.

Theorem 3.2 (Lax-Milgram theorem, symmetric version). Let H be a Hilbert space, a be a bilinearform on H, symmetric, continuous and coercive. Let b be a continuous linear form on H. Thenthe problem

Find u ∈ H such that∀v ∈ H, a(u, v) = b(v)

(3.3)

has a unique solution. This unique solution is also the unique solution to the minimization problem

Find u ∈ H such thatJ(u) = inf

v∈HJ(v), (3.4)

where the functional J(v) (the so-called energy functional) is defined by

J(v) =1

2a(v, v)− b(v).

3.2. A FIRST SYMMETRIC LINEAR ELLIPTIC BOUNDARY VALUE PROBLEM 33

Proof. The proof falls in two steps.

Step 1 (well-posedness of (3.3)): Introduce

(u, v)a = a(u, v).

It is clear that (·, ·)a is a scalar product on H . In addition, the norms ‖·‖H and ‖·‖a are equivalent.Indeed, using the coercivity and the continuity of the bilinear form a, we obtain that there existsα > 0 and M such that

∀v ∈ H, α‖v‖2 ≤ ‖v‖2a = a(v, v) ≤M‖v‖2.

The space H , which is complete for the norm ‖ · ‖, is thus also complete for the norm ‖ · ‖a. Thespace H , endowed with the scalar product (·, ·)a, is hence a Hilbert space. The bilinear form b iscontinuous for the norm ‖ · ‖, thus also for the equivalent norm ‖ · ‖a. Using the Riesz theorem,we thus obtain that there exists a unique u ∈ H such that

∀v ∈ H, a(u, v) = (u, v)a = b(v).

Step 2 (well-posedness of (3.6)): Consider now the solution u to (3.3) and let v ∈ H . We seth = v − u and compute, using the symmetry of a, that

J(v) = J(u + h)

=1

2a(u+ h, u+ h)− b(u+ h)

= J(u) + a(u, h)− b(h) +1

2a(h, h)

= J(u) +1

2a(h, h)

≥ J(u).

Hence u is a solution to (3.4).

Conversely, consider a solution u to (3.4). Then, for any v ∈ H and any λ ∈ R, we have

J(u) ≤ J(u+ λv) = J(u) + λ (a(u, v)− b(v)) +1

2λ2 a(v, v).

Hence, for any λ ∈ R,

λ(a(u, v) − b(v)) +1

2λ2a(v, v) ≥ 0.

This implies that a(u, v) = b(v), an equality that holds for any v ∈ H . Hence u is a solutionto (3.3).

3.2 A first symmetric linear elliptic boundary value problem

Let Ω be a subset of Rd, λ be a positive real number (λ > 0) and f ∈ L2(Ω). We look for u ∈ H1(Ω)solution to

−∆u+ λu = f in D′(Ω),u = 0 on ∂Ω.

(3.5)

Remark 3.3. All what follows can be extended to the case f ∈ H−1(Ω). We have chosen f ∈ L2(Ω)only for the sake of simplicity.


3.2.1 Variational formulation of (3.5)

We set

• H = H10 (Ω),

• a(u, v) =

∫

Ω

∇u · ∇v + λ

∫

Ω

u v,

• b(v) =

∫

Ω

f v,

and consider the problem Find u ∈ H such that∀v ∈ H, a(u, v) = b(v).

(3.6)

Proposition 3.4. Problems (3.5) and (3.6) are equivalent.

Proof. We first show that a is continuous on H ×H and that b is continuous on H :

∀v ∈ H = H10 (Ω), |b(v)| =

∣∣∣∣∫

Ω

f v

∣∣∣∣ ≤ ‖f‖L2 ‖v‖H1 ,

which means that b is continuous on H . Furthermore, we have

∀(u, v) ∈ H ×H, |a(u, v)| =

∣∣∣∣∫

Ω

∇u · ∇v + λ

∫

Ω

u v

∣∣∣∣

≤∣∣∣∣∫

Ω

∇u · ∇v∣∣∣∣+ λ

∣∣∣∣∫

Ω

u v

∣∣∣∣

≤(∫

Ω

|∇u|2)1/2(∫

Ω

|∇v|2)1/2

+ λ

(∫

Ω

u2)1/2(∫

Ω

v2)1/2

≤ (1 + λ)‖u‖H1 ‖v‖H1 ,

which means that a is continuous on H ×H .

We now show that (3.6) implies (3.5). Let u be a solution to (3.6). We already have thatu ∈ H1

0 (Ω), hence u ∈ H1(Ω) and u = 0 on ∂Ω. Let now φ ∈ D(Ω). We have φ ∈ H10 (Ω), thus

〈f, φ〉 =

∫

Ω

f φ

= b(φ)

= a(u, φ)

=

∫

Ω

∇u · ∇φ + λ

∫

Ω

uφ

= 〈∇u,∇φ〉 + λ〈u, φ〉= 〈−∆u+ λu, φ〉.

Therefore −∆u+ λu = f in D′(Ω). Thus u is a solution to (3.5).

3.3. POISSON EQUATION ON A BOUNDED OPEN SET 35

Conversely, let u be a solution to (3.5). We already have that u ∈ H1(Ω) and u = 0 on ∂Ω,thus u ∈ H1

0 (Ω). Let φ ∈ D(Ω). We have

b(φ) =

∫

Ω

f φ

= 〈f, φ〉= 〈−∆u+ λu, φ〉= 〈∇u,∇φ〉+ λ〈u, φ〉= a(u, φ).

Hence,

∀φ ∈ D(Ω), a(u, φ) = b(φ).

Since a(u, ·) and b are continuous on H = H10 (Ω) and since D(Ω) is dense in H1

0 (Ω), we concludethat

∀v ∈ H, a(u, v) = b(v).

Therefore, u is a solution to (3.6).

3.2.2 Checking the assumptions of the Lax-Milgram theorem

We now check that Problem (3.6) falls within the assumptions of the Lax-Milgram theorem. Weverify that

• H = H10 (Ω) is a Hilbert space;

• b is linear and continuous on H ;

• a is bilinear, symmetric and continuous on H ×H . In addition, a is coercive: for any v ∈ H ,

a(v, v) =

∫

Ω

|∇v|2 + λ

∫

Ω

v2

≥ min(1, λ)

(∫

Ω

|∇v|2 +∫

Ω

v2)

= min(1, λ)‖v‖2H1 .

We are thus in position to state that Problem (3.6) has a unique solution, in view of the Lax-Milgram theorem 3.2. The problems (3.5) and (3.6) are equivalent. The problem (3.5) thus has aunique solution.

We note that the energy functional associated to Problem (3.5) is

J(v) =1

2

∫

Ω

|∇v|2 + λ

2

∫

Ω

|v|2 −∫

Ω

f v.

3.3 Poisson equation on a bounded open set

Let Ω be a bounded subset of Rd and f ∈ L2(Ω). We look for u ∈ H1(Ω) solution to

−∆u = f in D′(Ω),u = 0 on ∂Ω.

(3.7)

Remark 3.5. Again, as in Section 3.2, it is possible to take f ∈ H−1(Ω) rather than f ∈ L2(Ω).



We set

• H = H10 (Ω),

• a(u, v) =

∫

Ω

∇u · ∇v,

• b(v) =

∫

Ω

f v,


(3.8)


Proof. The proof follows the same steps as the proof of Proposition 3.4.





• a is bilinear, symmetric and continuous on H ×H .

We are left with showing that a is coercive. Using the Poincaré inequality (see Theorem 2.7), wesee that, for any v ∈ H1

0 (Ω),

‖v‖2H1 = ‖v‖2L2 + ‖∇v‖2L2 ≤ (1 + C2Ω) ‖∇v‖2L2 = (1 + C2

Ω) a(v, v).

Hence

a(v, v) ≥ 1

1 + C2Ω

‖v‖2H1 .

In view of Lax-Milgram theorem 3.2, we deduce that Problem (3.8) has a unique solution. Theproblems (3.7) and (3.8) are equivalent. The problem (3.7) thus has a unique solution.

We note that the energy functional associated to Problem (3.7) is

J(v) =1

2

∫

Ω

|∇v|2 −∫

Ω

f v.

3.4 Lax-Milgram theorem

The symmetry assumption on the bilinear form a is restrictive as several boundary value problemsare not symmetric. This is in particular the case when some advection (i.e. transport) terms arepresent in the problem. It turns out that the symmetry assumption can be removed.

3.4. LAX-MILGRAM THEOREM 37

Theorem 3.7 (Lax-Milgram theorem). Let H be a Hilbert space, a be a bilinear form on H,continuous and coercive. Let b be a continuous linear form on H. Then the problem

Find u ∈ H such that∀v ∈ H, a(u, v) = b(v)

(3.9)

has a unique solution.

Of course, the symmetric version (Theorem 3.2) is a particular case of this more general version.Note also that there is no minimization problem associated to the variational formulation (3.9), incontrast to the symmetric case.

Remark 3.8. The proof of Theorem 3.7 in the finite dimensional case is easy. Consider the Hilbertspace H = Rn and the bilinear form a(u, v) = vTAu for any u and v in H, where A is a n × nmatrix.

We first show that Ker A = 0. Let u ∈ H such that Au = 0. Then a(u, u) = uTAu = 0.Thanks to the coercivity assumption on a, this implies that u = 0.

Using the fact that dim Ker A + dim Im A = dim H, we get that dim Im A = dim H, andhence the matrix A is invertible.

Proof of Theorem 3.7. Consider the mapping

Φ : H −→ Hu 7→ b such that a(u, ·) = 〈b, ·〉H .

Note that we have used the Riesz theorem to represent the continuous and linear form v ∈ H 7→a(u, v) ∈ R by an element of H , denoted b = Φ(u). The mapping Φ is of course linear, and it isalso continuous:

‖b‖2 = 〈b, b〉H = a(u, b) ≤M‖u‖ ‖b‖hence ‖Φ(u)‖ = ‖b‖ ≤M‖u‖. We wish to show that Φ is bijective. The proof falls in three steps.

Step 1: Φ is injective: Let u ∈ H such that Φ(u) = 0. For any v ∈ H , we have a(u, v) = 0, thusa(u, u) = 0. Using the coercivity of a, we deduce that

0 = a(u, u) ≥ α‖u‖2.

Hence u = 0.

Step 2: Φ is surjective: Let V = Im(Φ) ⊂ H . We show that V = H .

(a) Let us first show that X = Im(Φ) is closed. Let u ∈ H , u 6= 0. We have

supv∈H, v 6=0

〈Φ(u), v〉H‖v‖ = sup

v∈H, v 6=0

a(u, v)

‖v‖ ≥ a(u, u)

‖u‖ ≥ α‖u‖.

In addition, for any v 6= 0, we have

〈Φ(u), v〉H‖v‖ ≤ ‖Φ(u)‖.

We thus obtain that‖Φ(u)‖ ≥ α‖u‖

and this estimate also holds if u = 0. Consider now a sequence (bn)n∈N in X that convergesin H to some b ∈ H . We want to show that b ∈ X . Let un ∈ H such that Φ(un) = bn. Since(bn)n∈N is a Cauchy sequence, we see that (un)n∈N is also a Cauchy sequence. Indeed,

‖bp − bq‖ = ‖Φ(up)− Φ(uq)‖ = ‖Φ(up − uq)‖ ≥ α‖up − uq‖.


Therefore, (un)n∈N converges in H to some u ∈ H . In addition, Φ is continuous, so

bn = Φ(un) −→ Φ(u) in H.

Hence b = Φ(u) ∈ X , by uniqueness of the limit in H . The vector space X is hence closed.

(b) Let b0 ∈ H . Since X is closed in H , we can consider the orthogonal projection of b0 on X ,that we denote b1 ∈ X . For any v ∈ X , we have 〈v, b0 − b1〉H = 0. Hence, for any w ∈ H ,we have

0 = 〈Φ(w), b0 − b1〉H = a(w, b0 − b1).

Taking w = b0 − b1 and using the coercivity of a, we deduce that b0 = b1 ∈ X , which impliesthat H = X .

Step 3: Conclusion: Since b is a continuous linear form on H , it can be represented by somef ∈ H . The problem (3.9) can thus be recast as finding u ∈ H such that a(u, ·) = 〈f, ·〉H , that isfinding u ∈ H such that Φ(u) = f . In view of the above two steps, the mapping Φ is bijective, sothis problem has a unique solution.

3.5 A non-symmetric boundary value problem

Let Ω be a bounded subset of Rd and f ∈ L2(Ω). Let c : Ω → Rd be a vector field of class C1(Ω)which is divergence-free:

div c =d∑

i=1

∂ci∂xi

= 0 in Ω.

We look for u ∈ H1(Ω) solution to

−∆u+ c · ∇u = f in D′(Ω),u = 0 on ∂Ω.

(3.10)

Note that the product c · ∇u is well-defined in D′(Ω). This comes from the fact that (i) ∇u isa function in L2(Ω), hence in L1(Ω) since Ω is bounded, and (ii) c is a bounded function (it isa continuous function on a compact set). The product c · ∇u is thus in L1(Ω), and therefore inD′(Ω).


We set

• H = H10 (Ω),

• a(u, v) =

∫

Ω

∇u · ∇v +∫

Ω

(c · ∇u) v,

• b(v) =

∫

Ω

f v,


(3.11)


3.6. EXERCISES 39

Proof. In the proof of Proposition 3.4, we have already shown the continuity of b. We now showthe continuity of a. For any (u, v) ∈ H ×H , we compute

|a(u, v)| =

∣∣∣∣∫

Ω

∇u · ∇v +∫

Ω

(c · ∇u) v∣∣∣∣

≤∣∣∣∣∫

Ω

∇u · ∇v∣∣∣∣+∣∣∣∣∫

Ω

(c · ∇u) v∣∣∣∣

≤ ‖∇u‖L2‖∇v‖L2 + supΩ

|c| ‖∇u‖L2‖v‖L2

≤ (1 + supΩ

|c|) ‖u‖H1‖v‖H1 .

Thus a is continuous in H ×H .

The sequel of the proof follows the arguments of the proof of Proposition 3.4.





• a is bilinear and continuous.

We are left with showing that a is coercive. We note that, for any φ ∈ D(Ω),

∫

Ω

(c · ∇φ)φ =

∫

Ω

c · ∇(φ2

2

)=

∫

∂Ω

(c · n)(φ2

2

)−∫

Ω

(φ2

2

)(div c)

and the first term in the right-hand side vanishes as φ vanishes in the neighborhood of ∂Ω. Wethus have, using that c is divergence-free and the Poincaré inequality,

a(φ, φ) =

∫

Ω

|∇φ|2 +∫

Ω

(c · ∇φ)φ

=

∫

Ω

|∇φ|2 − 1

2

∫

Ω

(φ2

2

)(div c)

=

∫

Ω

|∇φ|2

≥ 1

1 + C2Ω

‖φ‖H1 .

By continuity of a and density of D(Ω) in H = H10 (Ω), we deduce that a is coercive on H .

In view of Lax-Milgram theorem 3.7, we deduce that Problem (3.11) has a unique solution.The problems (3.10) and (3.11) are equivalent. The problem (3.10) thus has a unique solution.

3.6 Exercises

Exercise 3.10. Let Ω be a bounded open subset of Rd. We consider

• a matrix field A : Ω −→ Rd×d that belongs to (L∞(Ω))d×d,

• two vector fields b and c in (L∞(Ω))d,


• a scalar-valued field d ∈ L∞(Ω).

Write the variational formulation of the following boundary value problem: Find u ∈ H10 (Ω) such

that−div (A∇u) + div (bu) + c · ∇u+ du = f in D′(Ω).

We now assume that the matrix field is uniformly coercive on Ω, that is that there exists α > 0such that

∀ξ ∈ Rd, ξTA(x)ξ ≥ αξT ξ almost everywhere on Ω.

Identify conditions on A, b, c and d under which the above boundary value problem has a uniquesolution in H1

0 (Ω).

Exercise 3.11 (Regularization). Let Ω be an open subset of Rd and f ∈ L2(Ω).

1. Let ε > 0. We look for uε ∈ H1(Ω) solution to

−ε∆uε + uε = f in D′(Ω),

uε = 0 on ∂Ω.

Show that this problem is well-posed and that

∀ε, ‖uε‖L2(Ω) ≤ ‖f‖L2(Ω).

2. We denote un the solution in H1(Ω) to

− 1

n∆un + un = f in D′(Ω),

un = 0 on ∂Ω.(3.12)

(a) Let φ ∈ D(Ω), q ∈ N⋆ and bn = 〈un, φ 〉. Show that

|bn+q − bn| ≤ 1

n+ q‖un+q‖L2(Ω)‖∆φ‖L2(Ω) +

1

n‖un‖L2(Ω)‖∆φ‖L2(Ω)

and that

|bn+q − bn| ≤ 2

n‖f‖L2(Ω)‖∆φ‖L2(Ω).

(b) Deduce that un converges in D′(Ω) to some distribution that we denote T .

(c) Using (3.12), show that T = f .

3. We now assume that f ∈ D(Ω). Let vn = un − f .

(a) Write the problem that vn satisfies.

(b) Using its variational formulation, show that

‖∇vn‖L2(Ω) ≤ ‖∇f‖L2(Ω).

(c) Deduce thatlimn→∞

‖vn‖L2(Ω) = 0

(d) What does this imply on the convergence of un?

3.6. EXERCISES 41

Exercise 3.12. Let u0 ∈ L2(Rd) and τ > 0. Consider the sequence un ∈ H1(Rd) recursivelydefined by: for all n ≥ 0,

un+1 − unτ

= ∆un+1. (3.13)

1. Show that the sequence unn≥1 is well-defined.

2. In the case d = 1, consider u defined by

u(x) =

cos(ωx) if x ≤ −2π/ω,

1 if −2π/ω ≤ x ≤ 2π/ω,

cos(ωx) if x ≥ 2π/ω,

where ω = 1/√τ . Compute the second derivative (in the sense of distributions) of u.

3. Using the previous question, build a function u0 ∈ L2(R) such that the problem: Find u1 ∈H1(R) such that

u1 − u0τ

= u′′0

does not have any solution.

4. In which context does (3.13) appear? What conclusion can you draw from the above question?

Exercise 3.13. Let f ∈ L2(R2) and λ ∈ R. Consider the problem:

Find u ∈ V such that∀v ∈ V, aλ(u, v) = b(v)

(3.14)

where V = H2(R2), b(v) =

∫

R2

f v and

aλ(u, v) =

∫

R2

∆u∆v + λ

∫

R2

∇u · ∇v +∫

R2

u v.

1. Show that, for any u and v in H2(R2), we have

∫

R2

∆u∆v =2∑

i,j=1

∫

R2

∂2u

∂xi∂xj

∂2v

∂xi∂xj.

2. Show that, if λ > 0, then Problem (3.14) is well-posed.

3. Show that, for any φ ∈ D(R2),

∫

R2

|∇φ|2 = −∫

R2

φ∆φ,

and deduce from this that, for any u in H2(R2),

∫

R2

|∇u|2 ≤ 1

2

(∫

R2

u2 +

∫

R2

|∆u|2).

Deduce that, if λ > −2, then Problem (3.14) is well-posed.


Exercise 3.14 (Non homogeneous Dirichlet boundary conditions). Let Ω be a bounded open subsetof Rd, f ∈ L2(Ω), u0 ∈ H1(Ω) (this assumption will be made stronger below). We consider thefollowing problem:

Find u ∈ H1(Ω) such that−∆u = f in D′(Ω),u = u0 on ∂Ω.

(3.15)

The functions u and u0 are in H1(Ω), thus they both have a trace on ∂Ω. The last equation abovesimply means that the two traces are equal.

Note also that the function u0 is defined on purpose on Ω, although only its trace on the boundary∂Ω appears in (3.15).

1. Introduce w = u−u0, and show that u is a solution to (3.15) if and only if w is a solution to

Find w ∈ H1(Ω) such that−∆w = g in D′(Ω),w = 0 in ∂Ω,

(3.16)

for a distribution g that will be made precise. The interest of (3.16) is that it is a problemwith homogeneous boundary conditions.

2. We assume for now that u0 ∈ H2(Ω). Deduce that g ∈ L2(Ω). Write a variational formulation(with unknown w) equivalent to (3.16), and show the existence and uniqueness of a solutionto (3.15).

3. Using the Green formula, show that the solution u to (3.15) satisfies

u ∈ H1(Ω), u = u0 on ∂Ω, and

∀ϕ ∈ H10 (Ω),

∫

Ω

∇u · ∇ϕ =

∫

Ω

fϕ.

Note that the solution u and the test function ϕ do not belong to the same space.

The following exercise aims at proving the Theorem 3.15 below, which is a functional inequalitythat belongs to the same family as the Poincaré inequality of Theorem 2.7:

Theorem 3.15. Let Ω ⊂ Rd be a bounded, open and connected subset of Rd, and let β be a realnumber β > 0. There exists C > 0 such that

∀v ∈ H1(Ω), ‖∇v‖2L2(Ω) + β‖v‖2L2(∂Ω) ≥ C‖v‖2H1(Ω). (3.17)

Exercise 3.16. Our aim is to prove Theorem 3.15. We proceed by contradiction and assume thatthere exists no constant C such that (3.17) holds.

1. Show that there exists a sequence unn∈N⋆ such that ‖un‖H1(Ω) = 1 and

‖∇un‖2L2(Ω) + β‖un‖2L2(∂Ω) ≤1

n. (3.18)

What can be said of the limits (when n→ ∞) of ‖∇un‖L2(Ω) and of ‖un‖L2(∂Ω)?

2. We admit that the bound ‖un‖H1(Ω) = 1 implies that there exists u⋆ ∈ H1(Ω) and a subse-quence un′ such that un′ converges to u⋆ in L2(Ω).

Using (3.18), show that ∇u⋆ = 0. We then deduce from the fact that Ω is connected that u⋆is a constant.

3.6. EXERCISES 43

3. Show that un′ converges to u⋆ in H1(Ω), and next that the constant u⋆ is different from 0.

4. Using that the trace mapping is continuous from H1(Ω) to L2(∂Ω), deduce that un′ convergesto u⋆ in L2(∂Ω).

5. Reach a contradiction.

Exercise 3.17 (Robin boundary conditions). Let Ω be a bounded open subset of Rd, f ∈ L2(Ω),g ∈ C∞(∂Ω) and a real number β > 0. We look for u ∈ H1(Ω) solution to

−∆u = f in D′(Ω)∂u

∂n+ βu = g on ∂Ω,

(3.19)

where∂u

∂n= ∇u · n (n is the outward normal vector to ∂Ω). The precise interpretation of the

boundary condition will be done below.

1. As a first step, we wish to write a variational formulation associated to (3.19). We thusassume that all functions are smooth, that all integrations by parts are allowed, . . .

Show that, if u is a sufficiently regular solution to (3.19), then, for any smooth functionv : Ω → R, we have ∫

Ω

∇u · ∇v + β

∫

∂Ω

u v =

∫

Ω

f v +

∫

∂Ω

g v.

2. We now establish and study a variational formulation of the problem. We recall that, for anyfunction u ∈ H1(Ω), the trace of u on the boundary ∂Ω is well-defined, that this is a functionof class L2(∂Ω), and that there exists C (that only depends on Ω) such that

∀u ∈ H1(Ω), ‖u‖L2(∂Ω) ≤ C‖u‖H1(Ω).

The above question leads to consider the bilinear form

a(u, v) =

∫

Ω

∇u · ∇v + β

∫

∂Ω

u v

and the linear form

b(v) =

∫

Ω

f v +

∫

∂Ω

g v.

(a) Show that a and b are well-defined on H1(Ω) and that they are continuous.

(b) Show that a is coercive on H1(Ω) (use Theorem 3.15). Check also that, if β ≤ 0, thena is not coercive on H1(Ω).

(c) Deduce from the above questions that the problem

Find u ∈ H1(Ω) such that∀v ∈ H1(Ω), a(u, v) = b(v)

(3.20)

is well-posed.

3. We now go back to the boundary value problem, and establish in which sense the solution uto (3.20) satisfies the boundary value problem (3.19).

(a) Show that the solution u to (3.20) satisfies

−∆u = f in D′(Ω). (3.21)


(b) We admit that, if g is sufficiently smooth, then the solution u to (3.20) is in H2(Ω). We

keep this assumption for the sequel of the exercise. Show that∂u

∂n∈ L2(∂Ω).

(c) Multiplying (3.21) by v ∈ H1(Ω) and using a Green formula, show that u is such that,for any v ∈ H1(Ω), we have

∫

∂Ω

v

(∂u

∂n+ βu

)=

∫

∂Ω

g v.

Deduce from this equality that u satisfies the boundary condition∂u

∂n+ βu = g (to that

aim, admit that the image of the trace mapping γ : H1(Ω) → L2(∂Ω) is dense inL2(∂Ω)).

Exercise 3.18 (Poincaré-Wirtinger inequality). We show here a functional inequality that belongsto the same family as the Poincaré inequality of Theorem 2.7.

1. Show that there exists C > 0 such that, for any ϕ ∈ C∞([0, 1]), we have

‖ϕ− 〈ϕ〉‖L2(0,1) ≤ C‖ϕ′‖L2(0,1),

where 〈ϕ〉 =∫ 1

0

ϕ is the mean of ϕ. Hint: start by showing that there exists x0 ∈ [0, 1] such

that 〈ϕ〉 = ϕ(x0).

2. We admit that C∞([0, 1]) is dense in H1(0, 1). Show that there exists C > 0 such that

∀v ∈ H1(0, 1), ‖v − 〈v〉‖L2(0,1) ≤ C‖v′‖L2(0,1). (3.22)

The result of Exercise 3.18, shown in dimension d = 1, can actually be generalized to thefollowing statement.

Theorem 3.19. Let Ω ⊂ Rd be a bounded, open and connected subset of Rd. There exists C > 0such that

∀v ∈ H1(Ω), ‖v − 〈v〉‖L2(Ω) ≤ C‖∇v‖L2(Ω), (3.23)

with 〈v〉 = 1

|Ω|

∫

Ω

v.

Exercise 3.20 (Neumann boundary conditions). Let Ω be a regular, bounded, open and connectedsubset of Rd, f ∈ L2(Ω) and g ∈ C∞(∂Ω). We look for u ∈ H1(Ω) solution to

−∆u = f in D′(Ω)∂u

∂n= g on ∂Ω,

(3.24)

where∂u

∂n= ∇u · n (n is the outward normal vector to ∂Ω). The precise interpretation of the

boundary condition will be done below.

Verify that Problem (3.24) is ill-posed, in the sense that, if u is solution to (3.24), then u + cis also solution, for any constant c.

We thus introduce the set

H1m =

u ∈ H1(Ω),

1

|Ω|

∫

Ω

u = 0

of the H1 functions with vanishing mean, and look for u ∈ H1m(Ω) that satisfies (3.24).

3.6. EXERCISES 45

1. As a first step, we wish to write a variational formulation associated to (3.24). We thusassume that all functions are smooth, that all integrations by parts are allowed, . . .

Show that, if u is a sufficiently regular solution to (3.24), then, for any smooth functionv : Ω → R, we have ∫

Ω

∇u · ∇v =

∫

Ω

fv +

∫

∂Ω

gv.

Deduce from this computation that, if (3.24) admits a smooth solution, then

∫

Ω

f +

∫

∂Ω

g = 0. (3.25)

We will keep this assumption for the sequel of the exercise.

2. We now establish and study a variational formulation of the problem. We recall that, for anyfunction u ∈ H1(Ω), the trace of u on the boundary ∂Ω is well-defined, that this is a functionof class L2(∂Ω), and that there exists C (that only depends on Ω) such that

∀u ∈ H1(Ω), ‖u‖L2(∂Ω) ≤ C‖u‖H1(Ω).

The above question leads to consider the bilinear form

a(u, v) =

∫

Ω

∇u · ∇v

and the linear form

b(v) =

∫

Ω

fv +

∫

∂Ω

gv.

(a) Show that H1m(Ω) is a Hilbert space for the H1 scalar product.

(b) Show that a and b are well-defined on H1m(Ω) and that they are continuous.

(c) Using Theorem 3.19, show that a is coercive on H1m(Ω).

(d) Deduce from the above questions that the problem

Find u ∈ H1

m(Ω) such that∀v ∈ H1

m(Ω), a(u, v) = b(v)(3.26)

is well-posed.

3. We now go back to the boundary value problem, and establish in which sense the solution uto (3.26) satisfies the boundary value problem (3.24). We recall that we assume that f and gsatisfy (3.25).

(a) Show that the solution u to (3.26) satisfies

∀v ∈ H1(Ω), a(u, v) = b(v),

where the functions v can now be chosen in H1(Ω) and not only in H1m(Ω).

(b) Show that the solution u to (3.26) satisfies

−∆u = f in D′(Ω). (3.27)

(c) We admit that, if g is sufficiently smooth, then the solution u to (3.26) is in H2(Ω). We

keep this assumption for the sequel of the exercise. Show that∂u

∂n∈ L2(∂Ω).


(d) Multiplying (3.27) by v ∈ H1(Ω) and using a Green formula, show that u is such that,for any v ∈ H1(Ω), we have ∫

∂Ω

v∂u

∂n=

∫

∂Ω

g v.

Deduce from this equality that u satisfies the boundary condition∂u

∂n= g (to that aim,

admit that the image of the trace mapping γ : H1(Ω) → L2(∂Ω) is dense in L2(∂Ω)).

Exercise 3.21 (Periodic boundary conditions). Consider the open set Q =

(−1

2,1

2

)d, a function

f ∈ L2(Q) and the problem

Find u ∈ H1(Q) such that−∆u = f in D′(Q),u is periodic at the boundary of Q.

(3.28)

This problem is ill-posed. Of course, if u is a solution, then u + c is also a solution for anyconstant c. But the situation is actually worse. Consider the particular case d = 2, f = 0 andu(x, y) = x2 − y2: this function is a solution to (3.28), just as u(x, y) = 0. The sequel of thisexercise aims at properly setting the problem.

1. First variational formulation.

We recall that

L2per(Q) =

u ∈ L2

loc(Rd), ∀k ∈ Zd, u(x+ k) = u(x) almost everywhere

and that

H1per(Q) =

u ∈ L2

per(Q),∂u

∂xi∈ L2

per(Q) for any 1 ≤ i ≤ d

.

To remove the indetermination of a solution by the addition of a constant, we consider thespace

H1per,m(Q) =

u ∈ H1

per(Q),

∫

Q

u = 0

.

(a) We recall that H1per(Q) is a Hilbert space for the scalar product

(u,w) =

∫

Q

uw +

∫

Q

∇u · ∇w.

Show that H1per,m(Q) is also a Hilbert space for the same scalar product.

(b) On the space H1per,m(Q), we define the bilinear form

a(u, v) =

∫

Q

∇u · ∇v

and the linear form

b(v) =

∫

Q

f v.

Show that a and b are continuous, and that a is coercive on H1per,m(Q) (to that aim, use

Theorem 3.19).

3.6. EXERCISES 47

(c) Deduce from the above questions that the problem

Find u ∈ H1

per,m(Q) such that∀v ∈ H1

per,m(Q), a(u, v) = b(v)(3.29)

is well-posed.

(d) Let ϕ ∈ D(Q). Is it possible to choose ϕ as test function in (3.29)?

(e) Let ϕ ∈ D(Q) and set ψ = ϕ−〈ϕ〉. Is it possible to choose ψ as test function in (3.29)?Deduce that the solution u to (3.29) satisfies

∀ϕ ∈ D(Q),

∫

Q

∇u · ∇ϕ =

∫

Q

fϕ− 〈ϕ〉∫

Q

f.

Check that, if f satisfies ∫

Q

f = 0, (3.30)

then the unique solution u to (3.29) is solution to (3.28).

Remark 3.22. This function u actually satisfies more boundary conditions than those writ-ten in (3.28) (we have seen that this problem is ill-posed). Actually, the normal derivative ofu is also periodic, in the following sense. Consider the restriction v of u to Q (by definition,

u ∈ H1per,m(Q), hence u is defined on Rd). Then the trace of

∂v

∂nis skew-periodic. In the

case of the function u(x, y) = x2 − y2 that we mentioned above when Q = (−1/2, 1/2)2 andf = 0 (i.e. the Q-periodic function u that is equal to u(x, y) = x2 − y2 on Q), we have,

on the boundary edge corresponding to x = 1/2, that∂v

∂n= 2x = 1 and, on the boundary

edge corresponding to x = −1/2, we have∂v

∂n= −2x = 1. Hence the trace of

∂v

∂nis not

skew-periodic.

2. Second variational formulation.

We have worked with H1per,m(Q) to remove the indetermination by the addition of a constant.

There are several ways to remove this indetermination, and we now show another one.

Define on H1per(Q) the bilinear form

a(u, v) =

∫

Q

∇u · ∇v + 〈u〉〈v〉

and the linear form

b(v) =

∫

Q

f v,

where we recall that 〈v〉 = 1

|Ω|

∫

Ω

v.

(a) Show that a and b are continuous, and that a is coercive on H1per(Q).

(b) Deduce that the problem

Find u ∈ H1

per(Q) such that∀v ∈ H1

per(Q), a(u, v) = b(v)(3.31)

is well-posed.


(c) Again assuming (3.30), and using an appropriate choice of v in (3.31), show that thesolution u to (3.31) satisfies 〈u〉 = 0. Deduce that the unique solution u to (3.31) is alsothe unique solution to (3.29).

3. Show that, if (3.28) has a smooth solution (so that all integrations by parts are correct), thenf satisfies (3.30).

Chapter 4

Inf-sup theory

The previous chapter relies on the Lax-Milgram theorem 3.7. The problem

Find u ∈ V such that∀w ∈ V, a(u,w) = b(w)

(4.1)

is studied there under a coercivity assumption on the bilinear form a (and in a Hilbert space V ).

The coercivity assumption is restrictive. In the finite dimensional setting (that is when V = Rn),consider the bilinear form a(u,w) = wTAu for some matrix A ∈ Rn×n. The well-posedness of (4.1)is equivalent to the invertibility of A. We then note that the matrices

A =

(1 00 −1

)and A =

(0 1−1 0

)

are invertible, but that the associated bilinear form is not cercive.In the infinite dimensional setting, consider the advection-diffusion problem (3.10), that is

−∆u+ c · ∇u = f in D′(Ω),u = 0 on ∂Ω,

for smooth functions f and c, without any smallness assumptions on c or on its divergence. Thebilinear form associated to this problem is

a(u,w) =

∫

Ω

∇u · ∇w +

∫

Ω

(c · ∇u)w,

and it is possible to find vector fields c such that infu∈H1

0 (Ω)

a(u, u)

‖u‖2H1(Ω)

< 0. In this case, a is not

coercive and one cannot use the Lax-Milgram theorem to study the above PDE. It turns out that,in some cases, the above problem is well-posed.

This chapter is devoted to the introduction of a more general theory, the inf-sup theory, tohandle problems such as (4.1) without any coercivity assumptions on a.

4.1 The finite dimensional case

We first establish a necessary and sufficient condition for the “invertibility” of a square matrix A,namely the well-posedness of a problem of the type Au = b (see Lemma 4.7 below). The interestof that condition is that, in the infinite dimensional case, the same condition can be used to ensurethe well-posedness of a problem of the type Au = b, where u belongs to some Banach space V andb belongs to the dual of some Banach space W (see Theorem 4.13 below).

49

50 CHAPTER 4. INF-SUP THEORY

4.1.1 Square matrices

Let us start by considering Problem (4.1) in a finite dimensional setting. We thus set V = Rn,and a(u,w) = wTAu for some matrix A ∈ Rn×n. We now consider a vector b ∈ Rn and look foru ∈ Rn such that (4.1) holds, that is

∀w ∈ Rn, wTAu = wT b,

which is of course equivalent to Au = b. This problem is well-posed if and only if A is invertible.

The coercivity assumption amounts to requesting that there exists α > 0 such that, for any

u ∈ Rn, we have uTAu ≥ αuTu. Since uTAu =1

2uT(A+AT

)u, this amounts to requesting that

the eigenvalues of the symmetric matrix A+AT are all positive. This condition is sufficient for Ato be invertible (as the Lax-Milgram theorem shows), but it is not necessary. Consider indeed theparticular case of a symmetric matrix A. This matrix is invertible if and only if its eigenvalues donot vanish. But they do not have to be positive.

Consider now the following assumption on A: we assume that there exists α > 0 such that

infu∈Rn, u6=0

supw∈Rn, w 6=0

wTAu

‖u‖ ‖w‖ ≥ α, (4.2)

which can be rephrased as

∀u ∈ Rn, supw∈Rn, w 6=0

wTAu

‖w‖ ≥ α‖u‖. (4.3)

We note that the above supremum is attained for w = Au (equality case in the Cauchy-Schwarzinequality), that is

supw∈Rn, w 6=0

wTAu

‖w‖ =√uTATAu. (4.4)

The eigenvectors of ATA form an orthonormal basis of Rn. Denoting λ2i the non-negative eigen-values of ATA, we see that the condition (4.3) is equivalent to

∀u ∈ Rn,

√√√√n∑

i=1

λ2iu2i ≥ α

√√√√n∑

i=1

u2i ,

which is equivalent to∀1 ≤ i ≤ n, |λi| ≥ α. (4.5)

Remark 4.1. The λi introduced above are called the singular values of A.

Remark 4.2. Observing that√uTATAu = ‖Au‖, we deduce from (4.3) and (4.4) that the inf-sup

condition (4.2) can be written as ‖Au‖ ≥ α‖u‖. This is to be compared with a coercivity assumptionon A, which would read uTAu ≥ α‖u‖2.

Suppose for a moment that A is symmetric. Then the condition (4.5) means that the eigenvaluesof A do not vanish, which is exactly a necessary and sufficient condition for A to be invertible.

Suppose now that A is not necessarily symmetric. Let us show that (4.3) implies that A isinvertible. Consider first u ∈ Ker A. Then (4.3) implies that

supw∈Rn, w 6=0

wTAu

‖w‖ = 0 ≥ α‖u‖,

4.1. THE FINITE DIMENSIONAL CASE 51

thus u = 0, hence A is injective. Owing to the fact that dim Ker A + dim Im A = dim V , weget that dim Im A = dim V , hence A is surjective. The condition (4.3) hence implies that A isbijective.

Conversely, if A is invertible, then (4.3) holds. Consider indeed the symetric matrix ATA, whichhas non-negative eigenvalues, and let λ = inf Sp ATA. If λ = 0, then there exists e ∈ Rn of unitnorm such that ATAe = 0. We thus have eTATAe = 0, hence Ae = 0. Since A is invertible, we gete = 0, which is in contradiction with ‖e‖ = 1. We have hence shown that λ = inf Sp ATA > 0,and hence that condition (4.5) holds.

We have thus shown the following result:

Lemma 4.3. Let A be a square matrix. Then the condition (4.3) is a necessary and sufficientcondition for A to be invertible.

Remark 4.4. Assume that A is coercive, namely that there exists α > 0 such that uTAu ≥ αuTufor any u ∈ Rn. Then the condition (4.3) is obviously satisfied.

4.1.2 Rectangular matrices

We now go one step further and consider the following generalization of Problem (4.1):

Find u ∈ V such that∀w ∈ W, a(u,w) = b(w)

where V and W are two Banach spaces. Note that the space in which we look for the solution u isa priori different from the space to which the test functions w belong. In addition, V and W arenow Banach spaces.

Again, we consider the finite dimensional case, where V = Rn and W = Rm. We thus considera matrix A ∈ Rm×n, a vector b ∈ Rm, and look for u ∈ Rn such that

∀w ∈ Rm, wTAu = wT b,

which is of course equivalent to Au = b.

Remark 4.5. Of course, a necessary condition for the well-posedness of the problem Au = b isthat u and b belong to spaces of identical dimension, and hence that n = dim V = dim W = m.

We however do not wish to write this assumption, since there is no equivalent of it in a infinitedimensional context. We rather wish to write necessary and sufficient conditions for the invertibilityof A which can easily be translated in a infinite dimensional context. This will be the case of theconditions (4.6) and (4.7) below. We will then check (see Remark 4.9 below) that these conditionsimply, in a finite dimensional context, that dim V = dim W .

Consider now the following assumptions on A. We assume that there exists α > 0 such that

infu∈Rn, u6=0

supw∈Rm, w 6=0

wTAu

‖u‖ ‖w‖ ≥ α (4.6)

and we assume that

If w ∈ Rm is such that wTAu = 0 for any u ∈ Rn, then w = 0. (4.7)

Note that (4.6) can be rephrased as

∀u ∈ Rn, supw∈Rm, w 6=0

wTAu

‖w‖ ≥ α‖u‖.


The supremum being again attained for w = Au, we get

supw∈Rm, w 6=0

wTAu

‖w‖ =√uTATAu.

The eigenvectors of ATA form an orthonormal basis of Rn. Denoting λ2i the non-negative eigen-values of ATA, we see that the condition (4.6) is equivalent to

∀u ∈ Rn,

√√√√n∑

i=1

λ2iu2i ≥ α

√√√√n∑

i=1

u2i ,

which is equivalent to∀1 ≤ i ≤ n, |λi| ≥ α.

We have the following result:

Lemma 4.6. (i) The condition (4.6) is equivalent to the fact that Ker A = 0.(ii) The condition (4.7) is equivalent to the fact that dim Im A = dim W .

We recall that, in finite dimension, A is surjective if and only if AT is injective.

Proof of Lemma 4.6. Let us prove the assertion (i). Assume that condition (4.6) holds and consideru ∈ Ker A. We have

supw∈Rm, w 6=0

wTAu

‖w‖ = 0 ≥ α‖u‖,

thus u = 0. Conversely, assume that Ker A = 0 and that condition (4.6) does not hold. Since

we have that supw∈Rm, w 6=0

wTAu

‖w‖ ≥ 0 for any u ∈ Rn, this means that, for any integer p, there exists

up ∈ Rn with ‖up‖ = 1 such that

supw∈Rm, w 6=0

wTAup‖w‖ ≤ 1

p.

The sequence up is bounded in Rn. It thus converges to some u⋆, up to the extraction of asubsequence, which satisfies ‖u⋆‖ = 1. For any w ∈ Rm, we have

wTAup‖w‖ ≤ sup

w∈Rm, w 6=0

wTAup‖w‖ ≤ 1

p,

which yields, passing to the limit p→ ∞, that

wTAu⋆

‖w‖ ≤ 0.

Taking w = Au⋆, we get Au⋆ = 0. Since Ker A = 0, this yields u⋆ = 0, which is in contradictionwith ‖u⋆‖ = 1. We have thus shown that condition (4.6) holds.

We now prove the assertion (ii). Assume that condition (4.7) holds and consider w ∈ Ker AT .Then, for any u ∈ Rn, we have wTAu = 0, hence w = 0. We thus get that Ker AT = 0. SinceAT is injective, we get, thanks to the finite dimension setting, that A is surjective, and hence thatdim Im A = dim W .

Assume conversely that dim Im A = dim W . Then A is surjective, hence AT is injective. Thisdirectly shows the condition (4.7).

4.2. THE INFINITE DIMENSIONAL CASE 53

We then have the following result:

Lemma 4.7. Assume that V and W are finite dimensional spaces. Let b ∈ W . The condi-tions (4.6)–(4.7) are necessary and sufficient conditions for the problem Au = b to be well-posed inV .

Proof. Assume that the conditions (4.6)–(4.7) hold. Then, in view of Lemma 4.6, we see that A isinjective and that Im A = W . The problem Au = b is thus well-posed in V .

Assume now that the problem Au = b is thus well-posed in V . The existence of a solution tothat problem implies that Im A = W , hence that condition (4.7) holds, in view of Lemma 4.6(ii).The uniqueness of a solution to that problem implies that Ker A = 0, hence that condition (4.6)holds, in view of Lemma 4.6(i).

Remark 4.8. Assume that A is a square matrix, that is dim V = dim W , as in Section 4.1.1.The condition (4.6) implies that Ker A = 0. Since dim Ker A+ dim Im A = dim V , we deducethat dim Im A = dim V = dim W , hence the condition (4.7). The converse is also true. Wethus see that, if dim V = dim W , then conditions (4.6) and (4.7) are equivalent. We thus recoverLemma 4.3 as a consequence of Lemma 4.7.

Remark 4.9. Assuming that conditions (4.6)–(4.7) hold, let us show that dim V = dim W . Werecall that V = Rn and W = Rm.

In view of condition (4.6) and Lemma 4.6(i), we get that Ker A = 0, and hence that m ≥ n(there are more equations in the system Au = 0 than unknowns).

In view of condition (4.7) and Lemma 4.6(ii), we get that Im A = W . In addition, we alwayshave dim Im A ≤ dim V=n. We thus get that m ≤ n.

4.2 The infinite dimensional case

This section is devoted to the study of the problem

Find u ∈ V such that∀w ∈ W, a(u,w) = b(w)

(4.8)

where V and W are two Banach spaces, a is a bilinear continuous form on V ×W and b is linearcontinuous form on W .

4.2.1 The Hilbert case

We start by the simple case when W is a Hilbert space. We then have the following result:

Theorem 4.10. Assume that W is a Hilbert space. We assume that the two following conditionsare satisfied:

• there exists α > 0 such that

infu∈V, u6=0

supw∈W,w 6=0

a(u,w)

‖u‖V ‖w‖W≥ α; (4.9)

• the bilinear form a is such that

If w ∈W is such that a(u,w) = 0 for any u ∈ V , then w = 0. (4.10)

Then the problem (4.8) is well-posed (i.e. admits one and only one solution).


Assume that the conditions (4.9) and (4.10) hold. Then the problem (4.8) has a unique solutionu ∈ V . Furthermore, we have the a priori estimate

α‖u‖V ≤ supw∈W,w 6=0

a(u,w)

‖w‖W= sup

w∈W,w 6=0

b(w)

‖w‖W= ‖b‖W ′ . (4.11)

Remark 4.11. As we will see below (see Theorem 4.13), the conditions (4.9) and (4.10) are notonly sufficient but also necessary conditions for (4.8) to be well-posed.

Remark 4.12. Consider the particular case when V = W is a Hilbert space, and a is a bilinearcontinuous, coercive form on V . Then a satisfies the two properties (4.9) and (4.10). Indeed, thecoercivity of a implies that

supw∈V,w 6=0

a(u,w)

‖u‖V ‖w‖V≥ a(u, u)

‖u‖2V≥ α,

from which we deduce (4.9). Consider now w ∈ V such that a(u,w) = 0 for any u ∈ V . Choosingu = w, we get a(w,w) = 0, which implies, again using the coercivity of a, that w = 0. Thisproves (4.10).

Proof of Theorem 4.10. The proof follows the same arguments as the proof of the Lax-Milgramtheorem (in the general case), see Theorem 3.7. We assume that W is a Hilbert space and denote〈·, ·〉W its scalar product. The norm of V is denoted ‖ · ‖V .

Consider the mapping

Φ : V −→ Wu 7→ b such that a(u, ·) = 〈b, ·〉W .

Note that we have used the Riesz theorem in W to represent the continuous and linear formw ∈ W 7→ a(u,w) ∈ R by an element of W . The mapping Φ is of course linear, and it is alsocontinuous:

‖b‖2W = 〈b, b〉W = a(u, b) ≤M‖u‖V ‖b‖Whence ‖Φ(u)‖W = ‖b‖W ≤ M‖u‖V . We wish to show that, under assumptions (4.9)–(4.10), themap Φ is bijective. The proof falls in three steps.

Step 1: Φ is injective: Let u ∈ V such that Φ(u) = 0, that is such that a(u,w) = 0 for anyw ∈W . In view of (4.9), we deduce that u = 0.

Step 2: Φ is surjective: Let X = Im(Φ) ⊂W . We show that X =W .

(a) Let us first show that X = Im(Φ) is closed. Let u ∈ V , u 6= 0. Using (4.9), we have

supw∈W, w 6=0

〈Φ(u), w〉W‖w‖W

= supw∈W, v 6=0

a(u,w)

‖w‖W≥ α‖u‖V .

In addition, for any w ∈W , w 6= 0, we have

〈Φ(u), w〉W‖w‖W

≤ ‖Φ(u)‖W .

We thus obtain that

‖Φ(u)‖W ≥ α‖u‖Vand this estimate also holds if u = 0. Consider now a sequence (bn)n∈N in X that convergesin W to some b ∈ W . We want to show that b ∈ X . Let un ∈ V such that Φ(un) = bn.

4.2. THE INFINITE DIMENSIONAL CASE 55

Since (bn)n∈N is a Cauchy sequence in W , we see that (un)n∈N is also a Cauchy sequence inV . Indeed,

‖bp − bq‖W = ‖Φ(up)− Φ(uq)‖W = ‖Φ(up − uq)‖W ≥ α‖up − uq‖V .Therefore, since V is a complete space, we get that (un)n∈N converges in V to some u ∈ V .In addition, Φ is continuous, so

bn = Φ(un) −→ Φ(u) in W.

Hence b = Φ(u) ∈ X , by uniqueness of the limit in W . The vector space X is hence closed.

(b) Let b0 ∈ W . Since X is closed in W , we can consider the orthogonal projection of b0 on X ,that we denote b1 ∈ X . For any v ∈ X , we have 〈b0 − b1, v〉W = 0. Hence, for any u ∈ V , wehave

0 = 〈b0 − b1,Φ(u)〉W = a(u, b0 − b1).

In view of the condition (4.10), we deduce that b0 = b1 ∈ X , which implies that W = X .

Step 3: Conclusion: Since b is a continuous linear form on W , it can be represented by somef ∈ W . The problem (4.8) can thus be recast as finding u ∈ V such that a(u, ·) = 〈f, ·〉W , thatis finding u ∈ V such that Φ(u) = f . In view of the above two steps, the mapping Φ is bijectivefrom V to W , so this problem has a unique solution.

4.2.2 The Banach case

We now turn to the general case when W is a Banach space. We admit the following result, theproof of which is difficult (see e.g. [3, 4]). This result provides a statement similar to Lemma 4.7,but for infinite dimensional spaces.

Theorem 4.13 (Banach–Necas–Babuska). Assume that W is reflexive. The problem (4.8) iswell-posed (i.e. admits one and only one solution) if and only if the two following conditions aresatisfied:

• there exists α > 0 such that

infu∈V, u6=0

supw∈W,w 6=0

a(u,w)

‖u‖V ‖w‖W≥ α; (4.12)

• the bilinear form a is such that

If w ∈W is such that a(u,w) = 0 for any u ∈ V , then w = 0. (4.13)

This theorem will be henceforth called the BNB theorem.Assume that the conditions (4.12) and (4.13) hold. Then the problem (4.8) has a unique

solution u ∈ V , which again satisfies the a priori estimate (4.11).

Remark 4.14. The notion of reflexivity goes beyond the scope of these lecture notes. It is definedas follows. Let V be a Banach space. Let J : V → V ′′ be defined by: for any v ∈ V , we set

∀v′ ∈ V ′, 〈Jv, v′〉V ′′,V ′ = 〈v′, v〉V ′,V ,

where we recall that V ′ = L(V,R) is the dual of V . It can be shown that J is always injective. IfJ is surjective, then the space V is said to be reflexive.

We point out the following useful facts:

• If the dimension of V is finite, then V is reflexive.

• If V is a Hilbert space, then it is reflexive.

• For any p ∈ (1,+∞) (note that we exclude the cases p = 1 and p = ∞), the Banach spaceLp(Ω) is reflexive.


4.3 Stokes problem

Consider a viscous and incompressible fluid. The flow of that fluid is described by the Navier-Stokesequations

∂u

∂t+ (u · ∇)u+∇p−∆u = f, (4.14)

div u = 0, (4.15)

where u : [0,+∞[×Rd → Rd is the velocity field and p : [0,+∞[×Rd → R is the pressure field.All physical parameters have been set to 1 in (4.14). Of course, these equations should be comple-mented by appropriate boundary conditions and initial conditions. The function f : [0,+∞[×Rd →Rd represents the volumic forces applied on the fluid (for instance, gravity). The equation (4.14)encodes the conservation of the momentum, while the equation (4.15) encodes the conservation ofthe mass.

When the velocity is small, the nonlinear term (u ·∇)u may be neglected. We consider in what

follows stationary problems, hence the term∂u

∂tvanishes.

In what follows, we assume that Ω is an open connected bounded subset of Rd. Our aim isto study the Stokes problem, complemented (for simplicity) by homogeneous Dirichlet boundaryconditions for the velocity. We hence look for a velocity field u : Ω → Rd and a pressure fieldp : Ω → R such that

−∆u+∇p = f in [D′(Ω)]d, div u = 0 in D′(Ω), u = 0 on ∂Ω.

Of course, the pressure is only determined up to the addition of a constant. We thus restrict

ourselves to pressures of vanishing mean, i.e. satisfying

∫

Ω

p = 0.

More precisely, we introduce the space

L20(Ω) =

q ∈ L2(Ω),

∫

Ω

q = 0

,

and we consider the following problem:

Find (u, p) ∈ (H10 (Ω))

d × L20(Ω) such that

−∆u+∇p = f in [D′(Ω)]d,

div u = 0 in D′(Ω),

(4.16)

for some f ∈ (L2(Ω))d. When u ∈ (H10 (Ω))

d, we recall that ∆u is a vector in Rd, the i-thcomponent of which is equal to ∆ui, where ui is the i-th component of u.

4.3.1 Variational formulation

The variational formulation of (4.16) reads as follows. We introduce

• the Hilbert space V = (H10 (Ω))

d × L20(Ω), endowed with the scalar product

〈x, y〉V = 〈p, q〉L2(Ω) +d∑

i=1

〈ui, vi〉H1(Ω)

for any x = (u, p) ∈ V and y = (v, q) ∈ V .

4.3. STOKES PROBLEM 57

• the bilinear form a defined on V × V by: for any x = (u, p) ∈ V and y = (v, q) ∈ V , we set

a(x, y) =

∫

Ω

∇u · ∇v − p div v + q div u,

where ∇u · ∇v is defined by ∇u · ∇v =d∑

i=1

∇ui · ∇vi =d∑

i=1

d∑

j=1

∂ui∂xj

∂vi∂xj

.

• the linear form b defined on V by: for any y = (v, q) ∈ V , we set

b(y) =

∫

Ω

f · v =

d∑

i=1

∫

Ω

fi vi.

Consider the problem: Find x ∈ V such that∀y ∈ V, a(x, y) = b(y).

(4.17)


We note that Problem (4.17) falls within the scope of Problem (4.8). In addition, the space V isa Hilbert space. As will be shown below, the bilinear form a is continuous on V ×V and the linearform b is continuous on V . However, Problem (4.17) cannot be studied using the Lax-Milgramtheorem. Indeed, for any x = (u, p) ∈ V , we have, thanks to the Poincaré inequality, that

a(x, x) =

d∑

i=1

‖∇ui‖2L2(Ω) ≥1

1 + C2Ω

d∑

i=1

‖ui‖2H1(Ω).

However, the pressure p does not appear in the quantity a(x, x), and therefore a(x, x) cannot bea upper-bound (up to a multiplicative constant) of ‖x‖2V . We will have to resort to the BNBtheorem 4.13 to study the well-posedness of (4.17).

Proof of Proposition 4.15. The proof falls in three steps.

Step 1. We first show that a is continuous on V × V and that b is continuous on V . For anyy = (v, q) ∈ V , we have

|b(y)| ≤d∑

i=1

∣∣∣∣∫

Ω

fi vi

∣∣∣∣ ≤d∑

i=1

‖fi‖L2(Ω ‖vi‖H1(Ω) ≤ C‖y‖V ,

which means that b is continuous on V . Furthermore, for any x = (u, p) ∈ V and y = (v, q) ∈ V ,we have

|a(x, y)| ≤d∑

i=1

∣∣∣∣∫

Ω

∇ui · ∇vi∣∣∣∣+∣∣∣∣∫

Ω

p div v

∣∣∣∣+∣∣∣∣∫

Ω

q div u

∣∣∣∣

≤d∑

i=1

‖ui‖H1(Ω)‖vi‖H1(Ω) + ‖p‖L2(Ω)‖div v‖L2(Ω) + ‖q‖L2(Ω)‖div u‖L2(Ω)

≤ C‖x‖V ‖y‖V ,

which means that a is continuous on V × V .


Step 2. We now show that (4.17) implies (4.16). Let x = (u, p) ∈ V be a solution to (4.17). Letφ ∈ (D(Ω))d. We see that y = (φ, 0) ∈ V , thus

d∑

i=1

〈fi, φi〉D′,D =

d∑

i=1

∫

Ω

fi φi

= b(y)

= a(x, y) [Equation (4.17)]

=

∫

Ω

∇u · ∇φ − p div φ

=d∑

i=1

〈∇ui,∇φi〉D′,D −d∑

i=1

〈p, ∂φi∂xi

〉D′,D

=d∑

i=1

〈−∆ui, φi〉D′,D + 〈 ∂p∂xi

, φi〉D′,D.

The functions φi being independent, we obtain, for any 1 ≤ i ≤ d, that −∆ui+∂p

∂xi= fi in D′(Ω).

This yields the first equation in (4.16).

Let now ψ ∈ D(Ω) and ψ = ψ −m with m =1

|Ω|

∫

Ω

ψ. We see that y = (0, ψ) ∈ V , thus

0 = b(y)

= a(x, y) [Equation (4.17)]

=

∫

Ω

ψ div u

=

∫

Ω

ψ div u−m

∫

Ω

div u.

By integration by part and using that u vanishes on ∂Ω, we see that

∫

Ω

div u =

∫

∂Ω

u ·n = 0. The

last term above thus vanishes, and we get that

∫

Ω

ψ div u = 0 for any ψ ∈ D(Ω), which is exactly

the second equation in (4.16).

Step 3. Conversely, let x = (u, p) be a solution to (4.16). Let φ ∈ (D(Ω))d, ψ ∈ D(Ω) and

ψ = ψ −m with m =1

|Ω|

∫

Ω

ψ. We see that y = (φ, ψ) ∈ V , and that

b(y) =

∫

Ω

f · φ

=

d∑

i=1

〈fi, φi〉D′,D

=

d∑

i=1

〈−∆ui +∂p

∂xi, φi〉D′,D [First equation of (4.16)]

=

d∑

i=1

〈∇ui,∇φi〉D′,D −d∑

i=1

〈p, ∂φi∂xi

〉D′,D

= a(x, y)−∫

Ω

ψ div u

= a(x, y) +m

∫

Ω

div u−∫

Ω

ψ div u.


The term

∫

Ω

div u vanishes by integration by part and using that u vanishes on ∂Ω. The term∫

Ω

ψ div u vanishes in view of the second equation of (4.16). We hence get that

∀φ ∈ (D(Ω))d, ∀ψ ∈ D(Ω), a(x, y) = b(y). (4.18)

We eventually proceed by density and continuity. Let y = (v, q) ∈ V . By density of (D(Ω))d in(H1

0 (Ω))d, there exists φn ∈ (D(Ω))d such that lim

n→∞‖v − φn‖(H1(Ω))d = 0. By density of D(Ω) in

L2(Ω), there exists ψn ∈ D(Ω) such that limn→∞

‖q − ψn‖L2(Ω) = 0. We then have

∣∣∣∣1

|Ω|

∫

Ω

ψn

∣∣∣∣ =∣∣∣∣1

|Ω|

∫

Ω

(ψn − q)

∣∣∣∣ ≤ C‖q − ψn‖L2(Ω).

Setting ψn = ψn −mn with mn =1

|Ω|

∫

Ω

ψn, we hence write

‖q − ψn‖L2(Ω) ≤ ‖q − ψn‖L2(Ω) + |mn| ≤ C‖q − ψn‖L2(Ω),

and thus limn→∞

‖q−ψn‖L2(Ω) = 0. Sstting yn = (φn, ψn), we thus have yn ∈ V and limn→∞

‖y−yn‖V =

0. Furthermore, writing (4.18) for (φn, ψn), we have a(x, yn) = b(yn). We then pass to the limitn→ ∞ and get that a(x, y) = b(y). Therefore, x is a solution to (4.17).

4.3.2 Recasting the variational formulation of the Stokes problem

The Stokes problem (4.17) has a specific structure, which we exhibit here. In Section 4.3.3, wewill state a specific version of the BNB Theorem 4.13 well adapted to that specific structure. Thisspecific structure (namely that of a saddle point problem) will be further discussed in Section 4.4.

We set

1. X = (H10 (Ω))

d and M = L20(Ω);

2. let a0 be the bilinear form defined on X ×X by: for any u and v in X ,

a0(u, v) =

∫

Ω

∇u · ∇v =

d∑

i=1

∫

Ω

∇ui · ∇vi.

3. let b be the bilinear form defined on X ×M by: for any v ∈ X and any p ∈M ,

b(v, p) = −∫

Ω

p div v.

4. for any v ∈ X , we set

g1(v) =

∫

Ω

f · v =

d∑

i=1

∫

Ω

fi vi.

5. for any q ∈M , we set g2(q) = 0.

Then the variational formulation (4.17) of the Stokes problem can be recast in the following form:

Find (u, p) ∈ X ×M such that∀v ∈ X, a0(u, v) + b(v, p) = g1(v),∀q ∈M, b(u, q) = g2(q).

(4.19)

The problem (4.19) has a specific structure in the sense that


• the unknown function p does not appear in the second equation;

• the unknown functions u and p are coupled through the same bilinear form in the twoequations in (4.19);

• the space in which the solution is searched is the same as the space in which test functionsare considered.

4.3.3 BNB theorem for problems of the type (4.19)

We now state a specific version of the BNB Theorem 4.13 well adapted to the structure of Prob-lem (4.19). Note that we do not assume a0 to be symmetric.

Theorem 4.16. Let X and M be two Hilbert spaces. We assume that the bilinear (resp. linear)forms a0 and b (resp. g1 and g2) are continuous. We also assume that the bilinear form a0 iscoercive on X. Then Problem (4.19) is well-posed if and only if

∃β > 0, ∀q ∈M, supv∈X, v 6=0

b(v, q)

‖v‖X≥ β‖q‖M . (4.20)

Proof. Let α0 be the coercivity constant of a0 on X :

∀v ∈ X, a0(v, v) ≥ α0‖v‖2X .

We set V = X ×M that we endow with the norm ‖(v, q)‖V = ‖v‖X + ‖q‖M . We introduce thebilinear form a, defined on V × V by

∀x = (u, p) ∈ V, ∀y = (v, q) ∈ V, a(x, y) = a0(u, v) + b(v, p) + b(u, q).

Let g ∈ V ′ be defined by

∀y = (v, q) ∈ V, g(y) = g1(v) + g2(q).

It is obvious that Problem (4.19) is equivalent to the following problem:

Find x ∈ V such that

∀y ∈ V, a(x, y) = g(y).

Problem (4.19) is hence well-posed if and only if the bilinear form a satisfies the conditions (4.12)and (4.13).

Assume that the conditions (4.12) and (4.13) are satisfied. Let q ∈M . From (4.12), we deducethat

α‖q‖M ≤ sup(v,q)∈V, (v,q) 6=0

a((0, q), (v, q)

)

‖v‖X + ‖q‖M

= sup(v,q)∈V, (v,q) 6=0

b(v, q)

‖v‖X + ‖q‖M

= supv∈X, v 6=0

b(v, q)

‖v‖X,

thus the condition (4.20). Note that we have not used the condition (4.13).

Conversely, assume that the condition (4.20) holds and let us show that the conditions (4.12)and (4.13) are satisfied.


1. The proof of (4.12) relies on tools that go beyond the scope of these lecture notes (it actuallyrelies on the tools that one uses to show Theorem 4.13, that we have admitted). We hereadmit that the condition (4.20) implies (4.12).

2. We now show (4.13). Let (v, q) ∈ V such that, for any (u, p) ∈ V , we have a((u, p), (v, q)

)=

0. Taking (u, p) = (0, q), we get that b(v, q) = 0. Taking (u, p) = (v, 0), we get thata0(v, v) + b(v, q) = 0, hence a0(v, v) = 0, and thus v = 0 thanks to the coercivity of a0 on X .

The condition a((u, p), (v, q)

)= 0 for any (u, p) ∈ V hence yields that b(u, q) = 0 for any

u ∈ X . Using (4.20), we deduce that q = 0. This proves (4.13).

This completes the proof.

Remark 4.17. A complete proof of Theorem 4.16 can be given in the finite dimensional setting.We then have that a0(u, v) = vTAu and b(v, q) = vTBq for some square matrix A and some matrixB. We introduce G1 ∈ X and G2 ∈M such that g1(v) = vTG1 and g2(q) = qTG2. Problem (4.19)can then be written as: find u ∈ X and p ∈M such that, for any v ∈ X and any q ∈M ,

vTAu+ vTBp = vTG1, uTBq = qTG2,

that is

Au +Bp = G1, BTu = G2.

The bilinear form a0 is coercive, hence, thanks to the Lax-Milgram theorem, the matrix A is invert-ible. We then write that u = A−1(G1−Bp), and thus look for p ∈M such that BTA−1(G1−Bp) =G2, that is

BTA−1Bp = BTA−1G1 −G2.

Problem (4.19) is well-posed if and only if the square matrix BTA−1B is invertible.

Assume that (4.20) holds, that is

∀q ∈M, supv∈X, v 6=0

vTBq

‖v‖ ≥ β‖q‖.

Let q ∈ Ker BTA−1B, that is BTA−1Bq = 0. Introduce u = A−1Bq. We then see that

a0(u, u) = uTAu = qTBTA−TAA−1Bq = qTBTA−TBq =(qTBTA−TBq

)T= qTBTA−1Bq = 0.

Using the coercivity of a0, we deduce that u = 0, hence Bq = Au = 0. We infer from (4.20) thatq = 0. We hence have shown that Ker BTA−1B = 0. The square matrix BTA−1B is henceinvertible.

Assume conversely that the matrix BTA−1B is invertible and that (4.20) does not hold. Thismeans that, for any integer n, there exists qn ∈M with ‖qn‖ = 1 such that

supv∈X, v 6=0

vTBqn‖v‖ ≤ 1

n.

The sequence qn is bounded in the finite dimensional space M . It thus converges to some q⋆, up tothe extraction of a subsequence, which satisfies ‖q⋆‖ = 1. For any v ∈ X, we have

vTBqn‖v‖ ≤ sup

v∈X, v 6=0

vTBqn‖v‖ ≤ 1

n,


which yields, passing to the limit n→ ∞, that

vTBq⋆

‖v‖ ≤ 0.

Taking v = Bq⋆, we get Bq⋆ = 0, and thus BTA−1Bq⋆ = 0, which implies that q⋆ = 0 sinceBTA−1B is invertible. This is in contradiction with ‖q⋆‖ = 1. We have thus shown that condi-tion (4.20) holds.

4.3.4 Application to the Stokes problem

The mathematical analysis of the Stokes problem relies on the following result, the proof of whichgoes beyond the scope of these lecture notes (see e.g. [6, Lemma 3.1]).

Theorem 4.18. Let Ω be an open, bounded and connected subset of Rd. Then the mapping

div : (H10 (Ω))

d −→ L20(Ω)

is surjective.

The above theorem means that, for any q ∈ L20(Ω), there exists vq ∈ (H1

0 (Ω))d such that

div vq = q. Using the Open Mapping theorem (see e.g. [4]), we obtain the existence of a constantρ > 0 such that ρ‖vq‖H1(Ω) ≤ ‖q‖L2(Ω).

Corollary 4.19. Under the assumptions of Theorem 4.18, the condition (4.20) is satisfied forthe Stokes problem, which corresponds to the choices X = (H1

0 (Ω))d, M = L2

0(Ω) and b(v, p) =

−∫

Ω

p div v.

Proof. For any q ∈ L20(Ω), we have

supv∈X, v 6=0

b(v, q)

‖v‖X= sup

v∈X, v 6=0

∫

Ω

q div v

‖v‖H1(Ω)

≥

∫

Ω

q div vq

‖vq‖H1(Ω)

≥ ρ

∫

Ω

q2

‖q‖L2(Ω)

= ρ‖q‖M .

We thus obtain (4.20).

We are now in position to conclude:

Theorem 4.20. Under the assumptions of Theorem 4.18, the Stokes problem (4.16) is well-posed.In addition, there exists c > 0 such that

∀f ∈ (L2(Ω))d, ‖u‖H1(Ω) + ‖p‖L2(Ω) ≤ c ‖f‖L2(Ω). (4.21)

Proof. We recall that the PDE (4.16) is equivalent to the variational formulation (4.17) (see Propo-sition 4.15), which we have recast in the form (4.19).

We now verify the assumptions of Theorem 4.16. We have that X = (H10 (Ω))

d and M = L20(Ω)

are two Hilbert spaces. In view of the Poincaré inequality, the bilinear form a0 is coercive onX . The condition (4.20) is satisfied thanks to Corollary 4.19. We thus deduce that (4.19), andhence (4.16), is well-posed. The estimate (4.21) is a rewriting of (4.11).

4.4. SADDLE-POINT PROBLEMS 63

4.4 Saddle-point problems

We discuss here in more generality problems of the type (4.19), where we assume that X and Mare two Hilbert spaces, that a0 is a continuous bilinear form on X ×X , b is a continuous bilinearform on X ×M , and g1 (resp. g2) is a continuous form on X (resp. M).

Definition 4.21. If the bilinear form a0 is symmetric and positive on X ×X, in the sense thata0(v, v) ≥ 0 for any v ∈ X, then the problem (4.19) is said to be a saddle-point problem.

To understand this definition, introduce the so-called lagrangian L : X ×M → R defined by

L(v, q) =1

2a0(v, v) + b(v, q)− g1(v) − g2(q).

Definition 4.22. The point (u, p) ∈ X ×M is called a saddle point of L if

∀(v, q) ∈ X ×M, L(u, q) ≤ L(u, p) ≤ L(v, p). (4.22)

Proposition 4.23. Let X and M be two Hilbert spaces. Assume that the bilinear form a0 issymmetric and positive on X × X. Then, (u, p) ∈ X ×M is a saddle point of L in the senseof (4.22) if and only if (u, p) is a solution to (4.19).

Proof. We first observe that

• for any u ∈ X , the function q ∈M 7→ L(u, q) is affine;

• for any p ∈M , the function v ∈ X 7→ L(v, p) is convex: for any v1 and v2 in X and α ∈ (0, 1),we indeed have, using the symmetry of a0, that

2(L(αv1 + (1− α)v2, p)− αL(v1, p)− (1− α)L(v2, p)

)

= a0(αv1 + (1 − α)v2, αv1 + (1− α)v2)− αa0(v1, v1)− (1− α)a0(v2, v2)

= α2a0(v1, v1) + (1 − α)2a0(v2, v2) + 2α(1− α)a0(v1, v2)− αa0(v1, v1)− (1− α)a0(v2, v2)

= 2α(1− α)a0(v1, v2)− α(1− α)a0(v1, v1)− α(1 − α)a0(v2, v2)

= −α(1− α)a0(v1 − v2, v1 − v2).

Since a0 is positive, we get that

L(αv1 + (1− α)v2, p) ≤ αL(v1, p) + (1− α)L(v2, p),

which means that v ∈ X 7→ L(v, p) is convex.

• Since a0 is symmetric, we compute that

d(v,q)L(u, p)(u, p) = a0(u, u) + b(u, p) + b(u, p)− g1(u)− g2(p).

We thus see that (u, p) is such that d(v,q)L(u, p) = 0 if and only if

∀u ∈ X, a0(u, u) + b(u, p) = g1(u)

and

∀p ∈M, b(u, p) = g2(p).

Hence, (u, p) is such that d(v,q)L(u, p) = 0 if and only if (u, p) is a solution to (4.19).


We now prove the proposition. Assume first that (u, p) is a solution to (4.19). Then, thanks tothe third observaiton above, we have d(v,q)L(u, p) = 0. Define J1 on X by J1(v) = L(v, p). Wethen get that dvJ1(u) = 0. Using the convexity of J1 (second observation above), we deduce thatJ1(u) ≤ J1(v) for any v ∈ X , which we write as L(u, p) ≤ L(v, p) for any v ∈ X . Define now J2 onM by J2(q) = L(u, q). We also have that dqJ2(p) = 0. Since J2 is affine (first observation above),we deduce that J2 is constant on M , which we write as L(u, p) = L(u, q) for any q ∈M .

Assume conversely that (u, p) ∈ X ×M is a saddle point of L in the sense of (4.22), and defineJ1 and J2 as above. We hence have that J1(u) ≤ J1(v) for any v ∈ X and that J2(q) ≤ J2(p) forany q ∈M , and hence dvJ1(u) = 0 and dqJ2(p) = 0. We then compute that

d(v,q)L(u, p)(u, p) = a0(u, u) + b(u, p) + b(u, p)− g1(u)− g2(p)

= dvJ1(u)(u) + dqJ2(p)(p)

= 0.

Using the third observation above, we deduce that (u, p) is a solution to (4.19).

4.5 Exercises

Exercise 4.24 (Advection-diffusion problem in a non-coercive setting). We consider the advection-diffusion problem studied in Section 3.5, that is: Find u ∈ H1

0 (Ω) such that

−∆u+ c · ∇u = f in D′(Ω), (4.23)

where Ω is a bounded open subset of Rd, c : Ω 7→ Rd is a smooth vector field and f ∈ L2(Ω).In contrast to Section 3.5, we make no assumptions on c (in particular, its divergence is not

small, it is not non-negative, and c is not small). We wish to establish that (4.23) is well-posed.

1. Write the variational formulation of (4.23).

2. We assume that there exists a function σ ∈ C1(Ω) such that

−div (∇σ + c σ) = 0

and such that σ(x) ≥ cσ > 0 almost everywhere on Ω.

(a) In the case when c = ∇V for a smooth function V , show that such a function σ existsby giving its expression in terms of V .

(b) We admit that, as soon as Ω and c are sufficiently smooth, there exists σ satisfying theabove assumptions. Using Theorem 4.13, show that (4.23) is well-posed.

Exercise 4.25 (Another viewpoint on the Stokes equation). Rather than considering the vari-ational formulation (4.17) of the Stokes problem (4.16), which takes into account the equationdiv u = 0, it can be tempting to include this equation in the space in which we work. We thusintroduce the space

V =u ∈ (H1

0 (Ω))d, div u = 0

and the variational formulation:

Find u ∈ V such that∀w ∈ V, a(u,w) = g(w)

(4.24)

where

a(u,w) =

∫

Ω

∇u · ∇w, g(w) =

∫

Ω

f · w.

4.5. EXERCISES 65

1. We first establish the well-posedness of (4.24).

(a) Recall why the space V is a Hilbert space when endowed with the H1 scalar product.

(b) Recall why the bilinear form a is continuous on V × V and why the linear form g iscontinuous on V .

(c) Show that (4.24) is well-posed.

(d) Show that (4.24) is equivalent to a minimization problem of the form inf J(v), v ∈ V for some functional J : V → R.

2. Show that, if x = (u, p) is a solution to (4.17), then u is a solution to (4.24).

Since we know that (4.17) and (4.24) are well-posed, this implies that, if u is a solutionto (4.24), then there exists p ∈ L2

0(Ω) such that x = (u, p) is a solution to (4.17). We explainbelow how to obtain this p in practice.

3. Let u be a solution to (4.24). Our aim here is to compute p ∈ L20(Ω) such that x = (u, p)

is a solution to (4.17). Let L(w) = g(w) − a(u,w) =

∫

Ω

f · w −∫

Ω

∇u · ∇w. We know that

L(w) = 0 for any w ∈ V ⊂ (H10 (Ω))

d, and we wish to

Find p ∈ L20(Ω) such that, for all w ∈ (H1

0 (Ω))d, we have L(w) = b(w, p), (4.25)

where we recall that b(w, p) = −∫

Ω

p div w.

(a) Using Corollary 4.19, show that there exists at most one solution to (4.25).

(b) One could think of using the general BNB theorem 4.13 to solve (4.25). The Corol-lary 4.19 show that Condition (4.12) is satisfied. Show that Condition (4.13) is notsatisfied. This is why we have to argue in a different manner.

(c) On the space L20(Ω), we introduce the bilinear form A(p, q) =

∫

Ω

p q and the linear form

G(q) = −L(vq), where vq ∈ (H10 (Ω))

d is the velocity field defined by Theorem 4.18:div vq = q and ‖vq‖H1

0 (Ω) ≤ C‖q‖L2(Ω).

Show that G is independent of the choice of vq, as soon as div vq = q.

(d) Show that A is continuous and coercive on L20(Ω), and that G is continuous on L2

0(Ω).We note that G is only defined on L2

0(Ω), and not on L2(Ω).

(e) Show that the problem

Find p ∈ L20(Ω) such that, for all q ∈ L2

0(Ω), we have A(p, q) = G(q)

is well posed.

(f) Let w ∈ (H10 (Ω))

d. We set q = div w ∈ L20(Ω). Check that

b(w, p) = −A(p, q) = −G(q) = L(vq)

and that L(vq) = L(w). Conclude.

We remark that the formulation (4.24) directly provides the velocity u, but that the pressure pshould be computed in a second stage, in a non-trivial manner. In practice, it is often the case thatthe quantity of interest in the Stokes problem is the pressure. This is why alternatives to (4.24)(such as (4.17)) have been proposed, despite the fact that their analysis needs more advanced toolsthan the Lax-Milgram lemma.

The formulation (4.24) is also challenging when it comes to its discretization. A natural choiceis to introduce a conformal discretization Vh ⊂ V . However, it is not simple to manipulate (globally)


divergence-free functions that are piecewise polynomials (especially in 3D). Furthermore, to recoverthe pressure, a possibility is to solve (4.25) in a discrete setting. To that aim, one needs theequivalent of Theorem 4.18 in a discrete setting. These are other reasons to prefer (4.17), thediscretization of which is somewhat simpler.

Chapter 5

Numerical approximation ofboundary value problems (coercivecase)

This chapter is devoted to the numerical approximation of linear elliptic boundary value problems,such as those considered in Chapter 3. We focus here on a simple case, namely the Poisson problemwith homogeneous Dirichlet boundary conditions: we look for u ∈ H1(Ω) solution to

−∆u = f in D′(Ω),u = 0 on ∂Ω,

(5.1)

where Ω is a bounded subset of Rd and f ∈ L2(Ω). We have proved in Section 3.3 the existenceand uniqueness of a solution to (5.1).

Several types of numerical strategies can be used to approximate this solution. In what follows,we focus on the Finite Element Method. It is a very popular method, often used by engineerswhen addressing continuum mechanics problems, fluid mechanics problems, . . . Although we focushere on a simple linear, elliptic, stationary (i.e. time independent) boundary value problem, weunderline that the Finite Element Method can be used for many boundary value problems, notnecessarily elliptic, and that may be time-dependent and/or nonlinear. It is indeed a very robustapproach, that can be used for domains Ω of arbitrary geometry.

In all what follows, we will argue on the variational formulation associated to (5.1), which reads

Find u ∈ V such that∀v ∈ V, a(u, v) = b(v),

(5.2)

where

• V = H10 (Ω) is a Hilbert space;

• the bilinear form a, which is defined by

a(u, v) =

∫

Ω

∇u · ∇v,

is continuous on V × V and coercive;

67

68CHAPTER 5. NUMERICAL APPROXIMATION OF BOUNDARY VALUE PROBLEMS (COERCIVE CASE)

• the linear form b, which is defined by

b(v) =

∫

Ω

f v,

is continuous on V .

We recall that the problems (5.1) and (5.2) are equivalent, and that Problem (5.2) is well-posed inview of the Lax-Milgram theorem.

Remark 5.1. In the present case, the bilinear form a is also symmetric. We do not make thisassumption in general and will clearly underline the arguments that are based on this additionalproperty.

5.1 The Galerkin approach

Many approximation methods for (5.1) belong to the class of Galerkin methods. This is the caseof Spectral Methods, and of the Finite Element Method on which we focus here. In contrast, finitedifference approaches are not Galerkin methods.

In this section, we detail the general principle of Galerkin approaches. We consider the abstractproblem (5.2), assuming that

V is a Hilbert space, (5.3)

that the linear form b is continuous on V , namely that there exists C such that

∀v ∈ V, |b(v)| ≤ C‖v‖V , (5.4)

and that the bilinear form a is continuous on V × V and coercive: there exist M and α > 0 suchthat

∀u, v ∈ V, |a(u, v)| ≤M ‖u‖V ‖v‖V and a(v, v) ≥ α‖v‖2V . (5.5)

We recall that we do not assume the bilinear form a to be symmetric.

5.1.1 Principle of the method

The Galerkin1 approach consists in replacing, in (5.2), the space V (which is usually of infinitedimension) by a space Vh of finite dimension. The numerical approximation uh of u ∈ V will besearched in Vh. The space Vh is thus called the approximation space. The presence of the subscripth encodes the fact that Vh is of finite dimension. For instance, in the Finite Element Method,it directly refers to the size of the mesh which is used to build Vh (see Section 5.2.1 below). InSpectral Methods, it relates in a more implicit manner to some characteristic of the approximationspace Vh.

In what follows, we assume thatVh ⊂ V.

In that case, the approximation method is said to be conformal. In the sequel, we approximateProblem (5.2) by the following problem:

Find uh ∈ Vh such that∀vh ∈ Vh, a(uh, vh) = b(vh).

(5.6)

1After the name of the russian mathematician Boris Grigorievich Galerkin, 1871–1945.

5.1. THE GALERKIN APPROACH 69

Remark 5.2. It is also possible to introduce non-conformal methods. When V = H1(Ω), a typicalexample is to introduce a partition ∪K of Ω (as is usually done in a Finite Element Method), and

to consider Vh = ⊕H1(K), i.e. vh ∈ Vh if and only if vh =∑

K

vKh for some vKh ∈ H1(K).

Consider an edge Γ shared by two elements K1 and K2. Since no contraints are imposed on thevalues of vK1

h and vK2

h on that edge, the function vh does not belong to H1(K1 ∪K2) in general.As a consequence, in general, Vh 6⊂ V .

Remark 5.3. In some cases, it is difficult to work in practice with the exact forms a and b,because they e.g. involve integrals that are difficult to exactly compute. In such cases, it is possibleto further approximate (5.6) by considering the problem

Find uh ∈ Vh such that∀vh ∈ Vh, ah(uh, vh) = bh(vh),

where ah (resp. bh) is an approximation of a (resp. b). For instance, integrals arising in thedefinition of the bilinear form a may be approximated by numerical quadrature rules. We refer toExercise 5.31 for more details.

In some other cases, it is actually advantageous to modify the forms a and b, in order to improvethe numerical approximation. We refer to Exercise 5.34 for an example.

Remark 5.4. In the problem (5.2), the space V is both the space to which the solution u belongs,and the space in which the test functions v are. When introducing a finite dimensional problem, itis possible to use two different approximations of this space, and to define the discrete problem as

Find uh ∈ Vh such that∀vh ∈Wh, a(uh, vh) = b(vh),

where, for instance, Vh and Wh are two finite dimensional subspaces of V . Such approaches arecalled Petrov-Galerkin approaches. Their analysis goes beyond the scope of these lecture notes.

Proposition 5.5. Under assumptions (5.3), (5.4) and (5.5), the problem (5.6) has one and onlyone solution.

Proof. We consider the space Vh ⊂ V endowed with the scalar product 〈·, ·〉V of the Hilbert spaceV . Since Vh is of finite dimension, it is closed in V , and thus Vh is a Hilbert space. The forms aand b are continuous on Vh. The form a is coercive on V , and thus on Vh. We are thus in positionto use the Lax-Milgram theorem on (5.6) and conclude that this problem is well-posed.

5.1.2 Error estimation

We wish to bound from above the error e = u − uh in the ambiant norm ‖ · ‖V . We first observean important property:

Lemma 5.6 (Galerkin orthogonality). The error e = u− uh satisfies

∀vh ∈ Vh, a(e, vh) = 0. (5.7)


Proof. The exact solution u satisfies a(u, v) = b(v) for any v ∈ V . Since Vh ⊂ V , we can take a testfunction in Vh and thus deduce that a(u, vh) = b(vh) for any vh ∈ Vh. Using that a(uh, vh) = b(vh)for any vh ∈ Vh, we immediately get the result.

We are now in position to establish the celebrated Céa lemma, which states that the error u−uhis bounded from above (up to an explicit multiplicative constant) by the best approximation error:

Lemma 5.7 (Céa lemma). We suppose that assumptions (5.3), (5.4) and (5.5) hold. Let u be thesolution to (5.2) and uh be the solution to (5.6). Then

‖u− uh‖V ≤ M

αinf

vh∈Vh

‖u− vh‖V . (5.8)

Proof. For any vh ∈ Vh, we deduce from the Galerkin orthogonality (5.7) that a(u−uh, uh−vh) = 0.Therefore, we have

α‖u− uh‖2V ≤ a(u − uh, u− uh) = a(u − uh, u− vh) ≤M ‖u− uh‖V ‖u− vh‖V .

We can now divide by ‖u− uh‖V to obtain

‖u− uh‖V ≤ M

α‖u− vh‖V .

As the function vh is arbitrary in Vh, we deduce (5.8).

The quantity infvh∈Vh

‖u − vh‖V is usually called the best approximation error. It is an upper

bound (up to a multiplicative constant) of the error u− uh. Obviously, it is also a lower bound ofthe error, since inf

vh∈Vh

‖u− vh‖V ≤ ‖u−uh‖V . If the best approximation error converges to 0, then

so does the error u− uh, at the same rate.We also note that the best approximation error quantifies how well the approximation space

Vh is suited to approximate u. It is natural that this best approximation error bounds from abovethe error. An important consequence of the Céa lemma is the fact that, if the approximation spaceis well-chosen, namely if the best approximation error is small, then the numerical approach isaccurate.

Remark 5.8 (Geometric interpretation of the Galerkin orthogonality). In the specific case whena is symmetric, it induces a scalar product on V . The norm associated to this scalar product isequivalent to the norm ‖ ·‖V , as a consequence of the continuity and coercivity of a: for any v ∈ V ,

α‖v‖2V ≤ a(v, v) ≤M‖v‖2V .

In that case, one can understand (5.7) as the fact that uh is the orthogonal projection on Vh of theexact solution u ∈ V (orthogonal with respect to the scalar product a(·, ·)).

Remark 5.9 (Error estimate, the symmetric case). When a is symmetric, it is possible to get abetter error estimate than (5.8). Indeed, we have, for any vh ∈ Vh,

a(u− vh, u− vh) = a(u− uh + wh, u− uh + wh)

with wh = uh − vh, hence

a(u− vh, u− vh) = a(u− uh, u− uh) + a(wh, wh) + a(u− uh, wh) + a(wh, u− uh).

5.1. THE GALERKIN APPROACH 71

Using the symmetry of a, the last two terms are equal. Using the Galerkin orthogonality (5.7),they both vanish, thus

a(u− vh, u− vh) = a(u− uh, u− uh) + a(wh, wh) ≥ a(u− uh, u− uh),

where we used the coercivity of a in the last bound. We thus get that

α‖u− uh‖2V ≤ a(u− uh, u− uh) ≤ a(u− vh, u− vh) ≤M ‖u− vh‖2V ,

hence

‖u− uh‖V ≤√M

αinf

vh∈Vh

‖u− vh‖V .

This bound is better than (5.8) as M/α ≥ 1.

5.1.3 The linear system

The space Vh is of finite dimension, therefore the problem (5.6) can actually be recasted as a linearsystem, as we now show. Denote N the dimension of Vh and let (ϕ1, . . . , ϕN ) be a basis of Vh. Weexpand uh ∈ Vh on the basis of Vh as

uh =

N∑

j=1

uj ϕj .

By considering ϕi as test function, we see that the problem (5.6) is equivalent to finding U =(u1, . . . , uN) ∈ RN such that

∀1 ≤ i ≤ N,

N∑

j=1

a(ϕj , ϕi)uj = b(ϕi).

We thus introduce the matrix A ∈ RN×N defined by

Aij = a(ϕj , ϕi)

and the vector B ∈ RN defined by

Bi = b(ϕi),

and recast the above problem as the linear system

AU = B. (5.9)

The matrix A is called the stiffness matrix, by reference to problems in mechanics where it hasfirst been introduced. This matrix has properties that directly come from the properties satisfiedby the bilinear form a.

Lemma 5.10. If the bilinear form a is symmetric, then the matrix A is symmetric. If the bilinearform a is coercive, then the matrix A is positive definite.


Proof. The first assertion is obvious. We only show the second. Let ξ = (ξ1, . . . , ξN ) ∈ RN and set

u =

N∑

i=1

ξi ϕi (note that u ∈ Vh). Denoting by (·, ·)RN the scalar product in RN , we see that

(ξ, Aξ)RN =

N∑

i,j=1

ξiAij ξj

=

N∑

i,j=1

ξi ξja(ϕj , ϕi)

= a

N∑

j=1

ξj ϕj ,

N∑

i=1

ξi ϕi

= a(u, u).

Assume that a is coercive. Then, for any ξ ∈ RN , we have (ξ, Aξ)RN ≥ 0. Furthermore, if

(ξ, Aξ)RN = 0, we get a(u, u) = 0, hence u = 0 by coercivity, hence ξ = 0. The matrix A is indeed

positive definite.

5.2 The Lagrange P1 Finite Element

We have described above the Galerkin approach in general terms. We now detail it in a specificcase of approximation space Vh. More precisely, we now describe the simplest Finite ElementMethod, based on the so-called Lagrange P1 Finite Element. We assume that the domain Ω is apolygon (see Remark 5.21 below).

For simplicity, we assume that the dimension d is equal to 2. The approach is however notrestricted to this case.

In the sequel, we will use the notation | · |H2(Ω) that we define here. We have previouslyintroduced the H2 norm ‖ · ‖H2(Ω): for any v ∈ H2(Ω),

‖v‖H2(Ω) =√‖v‖2L2(Ω) + ‖∇v‖2L2(Ω) + ‖∇2v‖2L2(Ω).

For any v ∈ H2(Ω), we define

|v|H2(Ω) = ‖∇2v‖L2(Ω)

so that

|v|2H2(Ω) =

d∑

i,j=1

∥∥∥∥∂2v

∂xi∂xj

∥∥∥∥2

L2(Ω)

.

5.2.1 Meshing the domain

Definition 5.11. A mesh of Ω is a covering of the polygon Ω by triangles (by convention, thesetriangles are considered as closed sets, whereas Ω is an open set). Denoting Th = K1, . . . ,KNe

these triangles (where Ne is the number of triangles in the mesh), we thus have

Ω = ∪Ne

i=1Ki.

The mesh is said to be admissible if, for any i 6= j, the set Ki ∩Kj is either empty, or restrictedto a point which is a vertex of Ki and Kj, or equal to a segment which is an edge of Ki and Kj.

5.2. THE LAGRANGE P1 FINITE ELEMENT 73

Figure 5.1: Example (left) and counter-example (right) of an admissible mesh

An example and a counter-example of an admissible mesh are shown on Figure 5.1.

For any 1 ≤ i ≤ Ne, we denote hi the diameter of the triangle Ki (defined as the longuest edgeof Ki) and we set

h = max1≤i≤Ne

hi.

This parameter quantifies how fine the mesh is. We denote si,1, si,2, si,3 the three vertices of thetriangle Ki. Collecting these vertices over the different triangles, we obtain the set s1, . . . , sNs

of the mesh vertices. It is useful to distinguish the vertices inside the open set Ω (there are N int

s ofthem) and the vertices on ∂Ω (there are Nbound

s of them). By construction, Ns = N ints +Nbound

s .

5.2.2 The space of P1 polynomial functions

In dimension d = 2, we set

P1 =p : R2 → R, p(x, y) = α+ βx + γy for some real numbers α, β and γ

.

This is a vector space of dimension 3. The following result is of paramount importance whenbuilding the P1 Lagrange Finite Element:

Proposition 5.12. A function p ∈ P1 is completely determined by its value at three non-alignedpoints. Furthermore, its value on a segment (not restricted to a point) is completely determined byits value at the two end points of this segment.

Consider a triangle K in R2, that we assumed to be non-degenerate (i.e. its three vertices arenot aligned). Let a1, a2 and a3 be its three vertices. Thanks to the above proposition, there existsa unique function λ1 ∈ P1 such that

λ1(a1) = 1, λ1(a

2) = 0, λ1(a3) = 0.

Likewise, there exists a unique function λ2 ∈ P1 such that λ2(a1) = 0, λ2(a

2) = 1 and λ2(a3) = 0,

and a unique function λ3 ∈ P1 such that λ3(a1) = 0, λ3(a

2) = 0 and λ3(a3) = 1. The functions

λ1, λ2 and λ3 are called the barycentric coordinates of the triangle K. They satisfy the following(straightforward) properties:

• λ1 + λ2 + λ3 = 1;

• for any i ∈ 1, 2, 3, λi is vanishing on the edge of K which is opposite to the vertex ai;

• for any x ∈ K and any i ∈ 1, 2, 3, we have 0 ≤ λi(x) ≤ 1;

• denoting G the barycenter of K, we have λi(G) = 1/3 for any i ∈ 1, 2, 3.


5.2.3 Approximation space

Set

V(1)h =

vh ∈ C0(Ω), vh|K ∈ P1 for any K ∈ Th, vh = 0 on ∂Ω

.

The functions in V(1)h are globally continuous over Ω, and piecewise C1. Using the jump formula,

we get that V(1)h ⊂ H1(Ω), the derivative (in the sense of distributions) being equal (on each

triangle) as the standard derivative. Note that the gradient of vh ∈ V(1)h is constant on each

triangle. Since the functions in V(1)h vanish on ∂Ω, we see that this approximation is conformal:

Lemma 5.13. We have V(1)h ⊂ H1

0 (Ω).

We now identify a basis of the space V(1)h . Let s ∈ Ω be a vertex of the mesh (note that s is

not on the boundary ∂Ω). Let K(s) ⊂ Th be the set of elements which have s as a vertex. Forany element K ∈ K(s), we note λK,s the barycentric coordinate associated to the vertex s in theelement K. We then set

ϕs(x, y) =

λK,s(x, y) if (x, y) ∈ K for K ∈ K(s),

0 otherwise.

An exemple of such function ϕs is shown on Figure 5.2. On that example, the support of ϕs consistsof 6 triangles. By construction, ϕs is equal to 1 at the vertex s and vanishes at all the other verticesof the mesh. Furthermore, the value of a function in P1 on an edge is fully determined by its valueat the end-points of the edge. We thus see that ϕs ∈ C0(Ω). In addition, we have that ϕs|∂Ω = 0and that, for any element K ∈ Th, ϕs|K ∈ P1. This implies that

ϕs ∈ V(1)h .

Enumerating the internal vertices of the mesh ass1, . . . , sN int

s

, we order the functions built above

asϕ1, . . . , ϕN int

s

, where ϕj is the function that is equal to 1 at vertex sj and vanishes at all other

vertices.

Figure 5.2: Example of function ϕs in dimension 2

The functions ϕj built above form a basis of V(1)h :

Proposition 5.14. The setϕ1, . . . , ϕN int

s

is a basis of the vector space V

(1)h , which is of dimen-

sion N ints .

5.2. THE LAGRANGE P1 FINITE ELEMENT 75

Proof. The familyϕ1, . . . , ϕN int

s

is free: if we have

∑

j

aj ϕj = 0

for some real numbers aj, then by evaluating this expression on the node si of the mesh, we getai = 0 since ϕj(si) = δij .

Consider now some vh ∈ V(1)h . We consider the function wh defined by

wh =∑

j

vh(sj)ϕj .

On any element K, the functions vh and wh are affine and they are equal at the three vertices ofK. They thus are equal on K. We thus have vh = wh on Ω, which implies that any function in

V(1)h can be written as a linear combination of the functions ϕj . This concludes the proof.

Our aim is now to estimate the best approximation error in V(1)h of any function u ∈ H1

0 (Ω),namely to understand how the quantity inf

vh∈V(1)h

‖u− vh‖H1(Ω) scales with respect to the mesh size

h. To that aim, we introduce the interpolation operator

I(1)h : C0(Ω) → V

(1)h

v 7→ vh =

N ints∑

i=1

v(si)ϕi.

We see that I(1)h v is the unique function in V

(1)h that takes the same values as v on all the inter-

nal nodes of the mesh. In the sequel, we estimate the interpolation errors∥∥∥v − I

(1)h v

∥∥∥H1(Ω)

and∥∥∥v − I

(1)h v

∥∥∥L2(Ω)

for any v ∈ H2(Ω) (we recall that, in dimension d ≤ 3, any function v ∈ H2(Ω)

has a continuous representation, hence I(1)h is well-defined on H2(Ω); see [1] for more details).

In dimension d ≥ 2, the interpolation error depends on two parameters:

• the diameter of the elements K ∈ Th (as in the one-dimensional case); we denoted hi thediameter of Ki.

• the radius of the largest cercle included in Ki, that we denote ρi.

We set

σTh= max

1≤i≤Ne

hiρi,

which is larger than 1, by construction. The ratio hi/ρi increases when the smallest angle in thetriangle Ki decreases (and the ratio diverges when this angle tends to 0). We admit the followingresult:

Theorem 5.15. Let d ≤ 3. There exists a constant cinter, independent of the mesh, such that, forany v ∈ H2(Ω) ∩H1

0 (Ω), we have

∥∥∥v − I(1)h v

∥∥∥H1(Ω)

≤ cinter σThh |v|H2(Ω) and

∥∥∥v − I(1)h v

∥∥∥L2(Ω)

≤ cinter h2 |v|H2(Ω). (5.10)


Remark 5.16. The fact that v = I(1)h v = 0 on ∂Ω plays no role to obtain (5.10). The same

estimate holds for any v ∈ H2(Ω), where I(1)h v is now defined by I

(1)h v =

Ns∑

i=1

v(si)ϕi (note that the

sum in i goes over all nodes of the mesh, and not only over the internal nodes).

Remark 5.17. In dimension d ≥ 4, a function in H2(Ω) is not necessarily continuous, and hence

I(1)h v is not defined. However, the following result, which is practice sufficient to perform the

numerical analysis, holds: there exists a constant C such that, for any v ∈ H2(Ω) ∩ H10 (Ω), we

haveinf

vh∈V(1)h

‖v − vh‖H1(Ω) ≤ C h |v|H2(Ω), (5.11)

where C is independent of v and h, but may depend on the mesh geometry.

Remark 5.18. In (5.10), the H1 bound may be attained in some simple situations (and thus∥∥∥v − I(1)h v

∥∥∥H1(Ω)

indeed depends on σTh). Consider the case when the mesh consists of a unique

triangle, with vertices (0, 0), (1, 0) and (−1, ε). The diameter h is of the order of 1 whereas the

parameter ρ is of the order of ε. Let v(x, y) = x2. We then check that[I(1)h v

](x, y) = x+2y/ε, so

that

∥∥∥v − I(1)h v

∥∥∥H1(Ω)

|v|H2(Ω)∼ ε−1 when ε → 0. The same situation may occur in a general mesh (i.e.

not restricted to a single triangle), for a thin triangle with two angles close to 0 and the third angleclose to π.

Remark 5.19. The Céa lemma 5.7 is written in the H1 norm, which is the one for which thebilinear form a is coercive. It cannot be written in the L2 norm, in the sense that the inequality ‖u−uh‖L2(Ω) ≤ C inf

vh∈Vh

‖u−vh‖L2(Ω) does not hold in general. An estimate on the best approximation

error in L2 norm (such as the estimate∥∥∥v − I

(1)h v

∥∥∥L2(Ω)

≤ cinter h2 |v|H2(Ω) of (5.10)) hence does

not directly imply that ‖u−uh‖L2(Ω) ≤ Ch2. We refer to Exercise 5.37 for L2 estimates on u−uh.

In practice, we do not work with a single mesh Th, but with a sequence of meshes obtained onefrom the other by sequential refinements (e.g., each triangle of a coarse mesh is subdivided intoseveral triangles to obtain a finer mesh). We denote

Thj

j≥1

this sequence of meshes.

Definition 5.20. Let σ0 ∈ R be fixed. The sequence of meshesThj

j≥1

is said to be regular of

parameter σ0 if, for any j, we haveσThj

≤ σ0.

Considering a regular sequence of meshes, we directly see that, for any v ∈ H2(Ω)∩H10 (Ω) and

any h,∥∥∥v − I

(1)h v

∥∥∥H1(Ω)

≤ cinter σ0 h |v|H2(Ω) and∥∥∥v − I

(1)h v

∥∥∥L2(Ω)

≤ cinter h2 |v|H2(Ω).

The interpolation error in norm H1 (resp. in norm L2) is therefore of the order of h (resp. of theorder of h2). This estimate is sharp with respect to h, in the sense that there exists a function

v ∈ H2(Ω)∩H10 (Ω) such that the interpolation error v− I

(1)h v is of the order of h (resp. h2) in the

H1 norm (resp. in the L2 norm).

5.3. APPROXIMATION OF THE POISSON PROBLEM BY THE P1 FINITE ELEMENT METHOD77

Remark 5.21. If Ω is not a polygon, then it cannot be meshed by triangles Ki such that Ω =∪Ne

i=1Ki. One possibility is to consider a polygon Ωpoly included in Ω, to mesh Ωpoly by the trian-gulation Th, and to consider the discrete space

V(1)h =

vh ∈ C0(Ω), vh|K ∈ P1 for any K ∈ Th, vh = 0 on ∂Ωpoly, vh = 0 on Ω \ Ωpoly

,

which satisfies V(1)h ⊂ H1

0 (Ω). The interpolation error between v ∈ H10 (Ω) and some vh ∈ V

(1)h has

two contributions: the first one comes from the error in Ωpoly and can be estimated as above, whilethe second one comes from the fact that vh = 0 in Ω \Ωpoly whereas v does not necessarily vanishthere.

Another possibility is to use an approximation space Vh which is not included in H10 (Ω), in the

spirit of Remark 5.2.

We do not pursue in these directions in these lecture notes.

5.3 Approximation of the Poisson problem by the P1 FiniteElement Method

Let f ∈ L2(Ω). We wish to approximate the unique solution u ∈ H1(Ω) to the Poisson prob-lem (5.1) (which is also the unique solution to the variational formulation (5.2)) by a Galerkin

approach on the finite dimensional space V(1)h built in Section 5.2.

5.3.1 Discrete problem and error estimation

In view of (5.6), the discrete problem is:

Find uh ∈ V

(1)h such that

∀vh ∈ V(1)h , a(uh, vh) = b(vh).

(5.12)

We have seen in Section 5.1.3 that solving this discrete problem amounts to solving a linearsystem of the form AU = B (see (5.9)), where the stiffness matrix A is of size N int

s × N ints ,

where N ints = dimV

(1)h is the number of internal vertices in the mesh. Its generic term, for any

1 ≤ i, j ≤ N ints , is given by

Aij = a(ϕj , ϕi) =

∫

Ω

∇ϕi · ∇ϕj , (5.13)

where the basis functions ϕj of V(1)h have been built in Section 5.2.3. The right-hand side is given,

for any 1 ≤ i ≤ N ints , by

Bi = b(ϕi) =

∫

Ω

f ϕi. (5.14)

In view of Lemma 5.10, the matrix A is symmetric definite positive, hence the linear system (5.9)

has a unique solution U . It is related to the unique solution uh of (5.12) by uh =∑N int

s

i=1 Ui ϕi.

We now state an error estimate:

Theorem 5.22. Let u ∈ H1(Ω) be the solution to (5.2) and uh ∈ V(1)h be the solution to (5.12).

We assume that the sequence of meshes is regular of parameter σ0 and that the domain Ω is convex.Then there exists a constant c, that may depend on Ω and σ0, but that is independent of h and f ,such that

‖u− uh‖H1(Ω) ≤ c h ‖f‖L2(Ω).


The estimate stated in the above theorem is sharp: there exist Poisson problems such that‖u− uh‖H1(Ω) is indeed of the order of h. Another way to prove this sharpness is to recall that (i)the interpolation error estimate provided in Theorem 5.15 is sharp, and (ii) the error ‖u−uh‖H1(Ω)

is bounded from above and from below (up to multiplicative constants that are independent of h)by the interpolation error.

Proof. Using the Céa lemma 5.7, we get

‖u− uh‖H1(Ω) ≤M

αinf

vh∈V(1)h

‖u− vh‖H1(Ω), (5.15)

where, for the problem of interest here, the continuity constant is equal to M = 1 and the coercivityconstant is such that α−1 = 1 + C2

Ω, where CΩ is the Poincaré constant of Ω (see Theorem 2.7).

We next wish to bound from above the best approximation error by the interpolation error

(that is, to take vh = I(1)h u in (5.15)). For this vh to be well-defined, it is sufficient (in dimension

d ≤ 3) that u ∈ H2(Ω). This is indeed the case, due to the following regularity result: if Ωis convex and f ∈ L2(Ω), then the unique solution u to (5.2) belongs to H2(Ω) and satisfies|u|H2(Ω) ≤ CΩ‖f‖L2(Ω), where the constant CΩ only depends on Ω.

The remainder of the proof is straightforward: taking vh = I(1)h u, we deduce from (5.15) that

‖u− uh‖H1(Ω) ≤ (1 + C2Ω) inf

vh∈V(1)h

‖u− vh‖H1(Ω) ≤ (1 + C2Ω) ‖u− I

(1)h u‖H1(Ω).

We are now in position to use the estimate (5.10) of the interpolation error:

‖u− uh‖H1(Ω) ≤ (1 + C2Ω) cinter σ0 h |u|H2(Ω) ≤ (1 + C2

Ω) cinter σ0 CΩ h ‖f‖L2(Ω).

This concludes the proof.

Remark 5.23. In dimension d ≥ 4, the fact that u ∈ H2(Ω) does not imply that I(1)h u is well-

defined. However, the proof can be performed by inserting (5.11) rather than (5.10) in (5.15).

Remark 5.24 (Non-convex domains). In the case when the domain Ω is not convex, the solution uto (5.2) generally does not belong to H2(Ω). It instead belongs to H3/2+ε(Ω) for some ε satisfying0 < ε ≤ 1/2. The parameter ε depends on how much Ω is not convex. More precisely, since Ωis not convex, there are angles between two consecutive edges of ∂Ω that are larger than π (in theextreme case of a crack, there is an angle of 2π between two consecutive edges). The parameter εdepends on the value of the largest of these (larger than π) angles.

The interpolation I(1)h u is not defined, but it is still possible to introduce a variant of this

interpolation and estimate its distance to u in terms of h. The Finite Element approximation (5.12)turns out to be again converging when h → 0, but with a smaller rate. More precisely, we have‖u− uh‖H1(Ω) ≤ c h1/2+ε where c is independent of h.

Remark 5.25. It is of course possible to deduce from Theorem 5.22 a bound on the error inthe L2 norm: ‖u − uh‖L2(Ω) ≤ c h ‖f‖L2(Ω). However, this estimate is not optimal. We show inExercise 5.37 how to obtain a better (and actually optimal) bound.


5.3.2 Assembling the stiffness matrix

The structure of the stiffness matrix, namely the position of non-zero coefficients, is very differentbetween the one-dimensional case and the multi-dimensional case. We recall that, in the formercase, the stiffness matrix is tridiagonal. In the multi-dimensional case, there are two differentsituations:

• a first possibility is to consider a structured mesh, obtained as the tensorial product of aone-dimensional mesh in the x direction by a one-dimensional mesh in the y direction. Thevertices are hence located on a tensorial grid. These meshes are well-adapted for rectangulardomains Ω. If the same meshsize is used in both directions, one says that the mesh is uniform.

• another possibility (the so-called non-structured meshes) consists in considering (for the meshvertices) a cloud of points without any geometrical structure. In practice, such meshes aremore often used, as they can be used for domains Ω of arbitrary shape. It is also possible tolocally refine the mesh, in order to locally improve the approximation. When all the elementsare roughly of the same shape (i.e. when the parameters hi and ρi introduced in Section 5.2.3are essentially the same over all elements), one says that the mesh is quasi-uniform.

On Figure 5.3, we show an example of non-structured mesh (which is quasi-uniform) and twoexamples of structured, uniform meshes for the domain Ω = (0, 1)× (0, 1). The internal nodes areshown by black circles. Both structured meshes have been built from a one-dimensional mesh ofthe segment (0, 1) with the nodes 0, 1/4, 1/2, 3/4, 1. The domain Ω is thus paved with 16squares. Each of them is next cut in two (resp. four) triangles to obtain the mesh shown on thecenter (resp. the right-hand side) of Figure 5.3.

Figure 5.3: An example of a non-structured, quasi-uniform mesh (left) and two examples of struc-tured, uniform meshes (center and right)

On Figure 5.4, we show a non-structured mesh of Ω = (0, 1) × (0, 1), with a local refinementaround the center of the square. This mesh has been built to approximate with a good accuracythe solution to a problem where the exact solution is singular at the center of the square.

In dimension d ≥ 2, the stiffness matrix A defined by (5.13) is not tridiagonal. When usingstructured meshes, this matrix has nevertheless some structure. For instance, for the mesh shown atthe center of Figure 5.3, the stiffness matrix is block-tridiagonal, and its coefficients only take a fewdifferent values (as in the one-dimensional case on a uniform mesh, where all diagonal coefficients –except possibly the first and the last – are equal, and where all off-diagonal coefficients are equal).

For general meshes, the stiffness matrix still has some particular property, related to the Lapla-cian operator: it is sparse. This property is interesting in terms of storage and resolution of thelinear system (5.9). We first introduce the notion of sparse matrices.


Figure 5.4: An example of non-structured mesh with local refinement

Definition 5.26. Let A be a matrix of size N ×N . Let N⋆ be the number of coefficients in A thatdo not vanish. The matrix A is sparse if N⋆ ≪ N2.

We first observe that, for the matrix A defined by (5.13), if Aij 6= 0, then there exists anelement K ∈ Th such that the two vertices si and sj belong to K. Thus, if i 6= j, a necessarycondition for Aij 6= 0 is that the vertices si and sj form an edge in the mesh. Thus, on a given linei of the matrix A, there are only a few coefficients that do not vanish, namely as many coefficientsas edges starting from i. Furthermore, the diagonal coefficients of A do not vanish. We thereforehave that

N⋆ ≤ N ints + 2Nf ,

where Nf is the number of edges in the mesh. The number of edges is roughly proportional to thenumber of vertices, hence

N⋆(N int

s )2≤ c

N ints

where c is a universal constant. The matrix A is hence sparse as soon as the mesh is sufficientlyfine (i.e. N int

s sufficiently large).

5.3.3 Solving the linear system

We now briefly discuss how to solve in practice the linear system (5.9), that reads AU = B, thatwe obtained in Section 5.1.3. We denote by N the size of this linear system.

When the stiffness matrix A is symmetric definite positive, a very efficient method to solve thelinear system (5.9) is to use the Conjugate Gradient algorithm, which is an iterative algorithm basedon the computation of quantities of the form vTAw for given vectors v and w. Its convergence isoften fast. We recall that this algorithm is based on the fact that the solution to the linear systemAU = B is also the unique minimizer of the quadratic functional

J(v) =1

2vTAv − vTB, v ∈ RN .

In the cases when the matrix A is small (e.g. in one-dimensional cases, or when the two-dimensional mesh to solve the problem is coarse), it is possible to use a direct method to solve thelinear system. The most general method is based on the so-called LU factorization of A, whichconsists in finding a lower triangular matrix Llow (Llow

ij = 0 as soon as j > i) and a upper triangular

matrix Uup such that A = Llow Uup. The resolution of the linear system then amounts to the twofollowing steps:


• find v ∈ RN such that Llow v = B.

• find U ∈ RN such that Uup U = v.

These two steps are easy to implement, due to the fact that the matrices Llow and Uup aretriangular. For instance, to find v, we first identify v1 by Llow

11 v1 = B1, next v2 by Llow21 v1+L

low22 v2 =

B2, . . .

When the matrix A is symmetric, it is also possible to use the so-called Choleski decompositionof A in the form A = Llow (Llow)T for a lower triangular matrix Llow. Once this decompositionhas been computed, solving the linear system AU = B is again easy.

These various approaches can be compared in terms of cost, which is here measured in termsof the number of elementary operations (additions, multiplications, . . . ) that are needed. In theregime when N ≫ 1 (which is the relevant regime for our applications), the cost to compute thematrices Llow and Uup such that A = Llow Uup scales as N3/3. If A is symmetric, the cost tocompute its Choleski decomposition A = Llow (Llow)T scales as N3/6. The cost for solving thelower or upper triangular linear systems scales as N2, and is therefore negligible. Thus, the costto compute U solution to AU = B by a direct method is of the order of N3. Note also that, ifseveral right-hand sides B have to be considered, the cost for the second, third, . . . , resolutions isnegligible, as the matrix A has already been factorized as A = Llow Uup or A = Llow (Llow)T .

We now turn to the cost of the Conjugate Gradient algorithm. The cost of each iteration isof the order of N2 (it may even be smaller if A is sparse). The number of iterations is relatedto the condition number ρ of the matrix A, which is equal to ρ = λN/λ1, where λj1≤j≤N arethe eigenvalues of A (recall that we use the Conjugate Gradient algorithm in the case when Ais symmetric definite positive) that we sorted in the increasing order: 0 < λ1 ≤ · · · ≤ λN . Thenumber of iterations increases when ρ increases. If the number of iterations is smaller than N ,then the Conjugate Gradient algorithm is less expensive than a direct approach.

In practice, stiffness matrices stemming from a Finite Element discretization of the Laplacianoperator have a poor condition number. If used directly, the Conjugate Gradient algorithm mayneed many iterations to converge, and be less efficient than a direct method.

To circumvent this difficulty, it is possible to use a preconditionner. Consider an invertiblematrix D. Instead of considering the linear system AU = B, we consider the equivalent systemDADT (D−T U) = DB. We thus wish to find V such that AV = B, with A = DADT andB = DB, and next compute U by U = DT V . Note that the matrix A is again symmetric definitepositive. The Conjugate Gradient algorithm may thus be used to solve the system AV = B. If D issufficiently close to A−1/2 is the sense that the condition number of A is small, then the procedure isefficient. We are thus left with finding a good matrix D, which is called here a precondionner. Thisquestion is problem-dependent. One can consider it from a linear algebra viewpoint (forgettingthat the matrix A comes from a variationnal formulation) or from a PDE viewpoint (arguing onthe variational formulation rather than on the linear system). A basic choice is to pick a diagonalmatrix D with Dii = 1/

√Aii. Anyhow, this question has been thoroughly studied and there exists

nowadays very efficient preconditionners, for many types of matrices A.

5.3.4 Approximations by quadrature formulas

In order to compute the right-hand side B of the linear system to solve, which is given by (5.14),it is often needed to use a quadrature formula, as the integral cannot be exactly computed (thesame difficulty also arises for the computation of the stiffness matrix, when the operator is morecomplex than the Laplacian operator, or when the basis functions are not as simple as P1 basisfunctions).

Such quadrature formulas are defined as follows. We first give ourselves an integer parameterℓg, which is the number of points that are going to be used on an element K. We next give ourselves


ℓg real numbers ω1, . . . , ωg which are the weights and ℓg points ξ1, . . . , ξg in K which are calledthe nodes of the quadrature formula. The integral of any function χ on K is then approximated as

∫

K

χ ≈ℓg∑

ℓ=1

ωℓ χ(ξℓ).

To quantify the accuracy of such quadrature formulas, we introduce the vector space of poly-nomial functions in x and y of total degree lower or equal than k:

Pk =

p : R

2 → R, p(x, y) =∑

0≤m,n≤k, m+n≤k

αmn xm yn for some real numbers αmn

.

The vector space Pk is of dimension (k + 1)(k + 2)/2. The degree of the quadrature formula, thatwe denote kg, is defined as the largest integer k such that the quadrature formula is exact on Pk,that is

∀p ∈ Pk,

∫

K

p =

ℓg∑

ℓ=1

ωℓ p(ξℓ).

The degree kg is directly related to the accuracy of the quadrature formula as explained by thefollowing result, the proof of which is based on a Taylor expansion.

Proposition 5.27. On a triangle K ⊂ R2 of diameter h, consider a quadrature formula (of nodesξℓ and of weights ωℓ, with 1 ≤ ℓ ≤ ℓg) of degree kg. Then, there exists a constant cq, independentof the element K, such that

∀χ ∈ C1+kg (K),

∣∣∣∣∣∣

∫

K

χ−ℓg∑

ℓ=1

ωℓ χ(ξℓ)

∣∣∣∣∣∣≤ cqh

3+kg‖χ‖C1+kg (K).

As a consequence, for χ ∈ C1+kg (Ω), we have

∣∣∣∣∫

Ω

χ− I(χ)

∣∣∣∣ ≤ cqh1+kg‖χ‖C1+kg (Ω),

where I(χ) is the quadrature formula.

We eventually note that the cost of a quadrature formula is directly related to ℓg, as the costis dominated by the cost of the evaluations of the function χ.

In dimension d = 2, it is convenient to locate the nodes of the quadrature by their barycentriccoordinates. We collect in Table 5.1 some examples of quadrature formulas on a triangle K. Themultiplicity is the number of circular permutations that one should do on the node that is given inthe table to obtain all the nodes of the quadrature formula. For instance, the quadrature formula ofdegree 2 (shown in the third line) is based on 3 nodes, with the same weight |K|/3, the barycentric

coordinates of which are

(1

2,1

2, 0

),

(0,

1

2,1

2

)and

(1

2, 0,

1

2

).

Using the first quadrature formula with 3 nodes (the one shown in the second line of the table),we obtain the following approximation:

Bi =

∫

Ω

f ϕi =∑

K∈K(si)

∫

K

f ϕi ≈∑

K∈K(si)

3∑

ℓ=1

ωℓ f(ξKℓ )ϕi(ξ

Kℓ ),

5.4. THE P2 FINITE ELEMENT METHOD 83

ℓg weights ωℓ nodes ξℓ multiplicity degree kg of thequadrature formula

1 |K|(1

3,1

3,1

3

)1 1

3 |K|/3 (1, 0, 0) 3 1

3 |K|/3(1

2,1

2, 0

)3 2

−9|K|/16(1

3,1

3,1

3

)1

4 4

25|K|/48(1

5,1

5,3

5

)3

Table 5.1: Quadrature formulas on a triangle K of surface |K|

where ξKℓ 3ℓ=1 are the 3 quadrature nodes inside K (we recall that K(si) is the set of elementshaving si as a vertex). For this quadrature formula, the quadrature nodes are the element vertices,thus ϕi(ξ

Kℓ ) vanishes on two nodes and is equal to 1 on the third node:

Bi ≈∑

K∈K(si)

1

3|K| f(si) =

1

3|K(si)| f(si)

where |K(si)| is the surface of the polygon K(si).

It is possible to show that the rate of convergence of the P1 Finite Element method is preservedas soon as, to approximate the right-hand side B, a quadrature formula of degree kg ≥ 0 isused. More precisely, consider the approximation Bq of B obtained by a quadrature formula ofdegree kg ≥ 0. Let U q be the solution to the linear system AU q = Bq and let uqh =

∑j U

qj ϕj

be the associated discrete solution (note that the stiffness matrix A is assumed to be exactlycomputed). Then, there exists a constant c, which is independent of h, such that ‖u− uqh‖H1(Ω) ≤c h. Comparing this estimate with the one obtained in Theorem 5.22, we observe that the errorconverges to 0 at the same rate as if the right-hand side B was exactly computed.

5.4 The P2 Finite Element Method

We now briefly present a variant of the P1 Finite Element Method presented above, namely the P2

Finite Element Method. This variant can easily be generalized to the Pk Finite Element Methodfor any integer k.

Remark 5.28. There exists other Finite Element Methods, besides the ones based on the vectorspace Pk. We refer to the Exercises 5.36 and 5.39 for some examples.

We again assume (for simplicity) that the dimension d is equal to 2, and that the domain Ω isa polygon. We again mesh Ω by triangles as in Section 5.2.1:

Ω = ∪Ne

i=1Ki.

5.4.1 Approximation space

We introduce the vector space of functions that are piecewise equal to polynomial functions oftotal degree lower or equal to 2:

P2 =p : R2 → R, p(x, y) = α1 + α2x+ α3y + α4x

2 + α5y2 + α6xy for some real numbers αj6j=1

.


We next set

V(2)h =

vh ∈ C0(Ω), vh|K ∈ P2 for any K ∈ Th, vh = 0 on ∂Ω

,

and we see that this leads to a conformal approximation:

V(2)h ⊂ H1

0 (Ω).

A function vh ∈ V(2)h is fully determined on an element K by its value at the three vertices and its

value at the middle of the three edges. The following set of functions thus forms a basis of V(2)h :

• functions in V(2)h that are equal to 1 on one vertex of the mesh, and that vanish on all other

vertices and at all middle points of all edges;

• functions in V(2)h that vanish on all vertices of the mesh, that are equal to 1 at the middle

point of one edge, and that vanish at all other middle points.

The best approximation error in V(2)h is again quantified by introducing a suitable interpolation

operator I(2)h : C0(Ω) → V

(2)h . We admit the following result:

Theorem 5.29. Consider a regular sequence of meshesThj

j≥1

, in the sense of Definition 5.20.

There exists a constant cinter, independent of the mesh, such that, for any v ∈ H3(Ω)∩H10 (Ω), we

have∥∥∥v − I

(2)h v

∥∥∥H1(Ω)

≤ cinter h2 |v|H3(Ω) and

∥∥∥v − I(2)h v

∥∥∥L2(Ω)


Note that the P2 interpolation error is bounded for functions that are more regular than inTheorem 5.15 where we consider the P1 interpolation error: we now ask that v ∈ H3(Ω) insteadof v ∈ H2(Ω). For these more regular functions, a smaller interpolation error is obtained, as it isof order h2 in the H1 norm (in contrast to the P1 case, where it is of order h in the H1 norm).

We thus note that using a P2 approach instead of a P1 approach may be useful only if thesolution is sufficiently regular. Note also that, for the same mesh size h, the number of degreesof freedom in a P2 Finite Element space is larger than in a P1 Finite Element space. It is thusimportant to compare accuracies not in terms of h, but in terms of the number of degrees offreedom, which is directly related to the cost of the approach.

5.4.2 Approximation of the Poisson problem

Let f ∈ L2(Ω). The Galerkin approximation of the Poisson problem (5.2) on the finite dimensional

space V(2)h is:

Find uh ∈ V(2)h such that

∀vh ∈ V(2)h , a(uh, vh) = b(vh).

(5.16)

As for the P1 approach, this lead to a linear system of the form AU = B (see (5.9)), where the

stiffness matrix A is of size dimV(2)h × dimV

(2)h .

We now state an error estimate:

Theorem 5.30. Let u ∈ H1(Ω) be the solution to (5.2) and uh ∈ V(2)h be the solution to (5.16).

We assume that the sequence of meshes is regular of parameter σ0 and that the solution u belongsto H3(Ω). Then there exists a constant c, independent of h, such that

‖u− uh‖H1(Ω) ≤ c h2.

5.5. EXERCISES 85

Note that, if u is only in H2(Ω), then the above error bound does not hold. Since V(1)h ⊂ V

(2)h ,

we get in this case that ‖u − uh‖H1(Ω) ≤ c h. For a given mesh size h, we thus obtain the sameaccuracy as with the P1 method, for a larger cost. The P2 method is thus not interesting in thiscase.

If the solution u is in H3(Ω), then the P2 method is actually a better approach than the P1

method. For a P1 method, we indeed have ‖u − u(1)h ‖H1(Ω) ≤ c1 h and the number of degrees of

freedom (say on a structured mesh of Ω = (0, 1)2 made of squares divided in two triangles) isN = 1/h2. We therefore get

‖u− u(1)h ‖H1(Ω) ≤

c1√N.

For a P2 method, we have ‖u− u(2)h ‖H1(Ω) ≤ c2 h

2 (note that the constant c2 is in general differentfrom c1) and the number of degrees of freedom (on the same structured mesh of Ω = (0, 1)2) isN = 1/(h/2)2. We therefore get

‖u− u(2)h ‖H1(Ω) ≤

4c2N.

Therefore, if N is large enough (i.e. the mesh sufficiently fine), the P2 method provides a solutionwith a better accuracy than the P1 method, for an equal number of degrees of freedom.

5.5 Exercises

Exercise 5.31 (Variational crimes). Let V be a Hilbert space, b be a continuous linear form onV , and a be a continuous bilinear form on V × V that is coercive. The problem

Find u ∈ V such that∀v ∈ V, a(u, v) = b(v)

has a unique solution. In the light of Remark 5.3, we approximate this problem by the followingvariational formulation:

Find uh ∈ Vh such that∀vh ∈ Vh, ah(uh, vh) = bh(vh),

where ah (resp. bh) is an approximation of a (resp. b), and where Vh ⊂ V . We assume that bh isa continuous linear form on Vh, that ah is a continuous bilinear form on Vh × Vh, that is coerciveuniformly in h: there exists α > 0 independent of h such that, for any h,

∀v ∈ Vh, ah(v, v) ≥ α‖v‖2.

The above discrete problem is thus well-posed and we wish to estimate ‖u− uh‖.

1. Let vh ∈ Vh. Show that

ah(uh − vh, uh − vh) = bh(uh − vh)− ah(vh, uh − vh),

then

ah(uh − vh, uh − vh) = bh(uh − vh)− b(uh − vh) + a(u, uh − vh)− ah(vh, uh − vh),

and then

ah(uh − vh, uh − vh) ≤ a(u− vh, uh − vh) + Eb ‖uh − vh‖+ Ea ‖vh‖ ‖uh − vh‖

where

Eb = supwh∈Vh, wh 6=0

|bh(wh)− b(wh)|‖wh‖


and

Ea = supwh∈Vh, wh 6=0

supsh∈Vh, sh 6=0

|ah(wh, sh)− a(wh, sh)|‖wh‖ ‖sh‖

.

2. Denoting M the continuity constant of the bilinear form a on V × V , deduce that, for anyvh ∈ Vh, we have

α‖uh − vh‖ ≤M‖u− vh‖+ Eb + Ea ‖vh‖.

3. Show that

‖u− uh‖ ≤ Ebα

+ infvh∈Vh

[(1 +

M

α

)‖u− vh‖+

Eaα

‖vh‖].

Exercise 5.32 (Discrete maximum principle). Let A be the stiffness matrix of the discrete prob-lem (5.12), in the one-dimensional case, for Ω = (a, b), when using P1 finite elements on a uniformmesh of size h = (b− a)/(N +1). We recall that A is of size N ×N and is given (using (5.13)) by

A =1

htridiag (−1, 2,−1).

Let B ∈ RN be the right-hand side of the linear system equivalent to (5.12). The components of Bare given by (5.14).

1. For a vector V ∈ RN , we say that V ≤ 0 if Vi ≤ 0 for any 1 ≤ i ≤ N . Let V ∈ RN such thatAV ≤ 0. Show that V ≤ 0.

2. Deduce that, if the function f in (5.12)–(5.13)–(5.14) is such that f ≤ 0 on Ω, then thediscrete solution uh to (5.12) satisfies uh ≤ 0 on Ω. This property is called the discretemaximum principle.

3. We denote αij , with 1 ≤ i, j ≤ N the coefficients of the matrix A−1. Show that αij > 0 forany 1 ≤ i, j ≤ N .

4. Show that, for any 1 ≤ i ≤ N ,N∑

j=1

αij ≤1

8h.

Hint: consider the function w(x) =1

2x (1 − x).

5. Deduce that ‖uh‖L∞(Ω) ≤1

8‖f‖L∞(Ω).

Exercise 5.33 (Interpolation error). The aim of this exercise is to prove the interpolation errorestimate (5.10), in the one-dimensional case. Let Ω = (a, b), that we mesh by the nodes a = x0 <x1 < · · · < xi < · · · < xn+1 = b. For any 0 ≤ i ≤ n, we set Ki = [xi, xi+1] and hi = xi+1 − xi. Letv ∈ H2(Ω).

5.5. EXERCISES 87

1. We consider the element Ki and we define wi : Ki → R by

∀s ∈ Ki, wi(s) = v′(s)− v(xi+1)− v(xi)

xi+1 − xi.

Check that this function is such that v(x) − (I(1)h v)(x) =

∫ x

xi

wi(s) ds for any x ∈ Ki. Show

that‖v − I

(1)h v‖L2(Ki) ≤ hi‖wi‖L2(Ki).

2. Show that there exists some yi ∈ Ki such that wi(yi) = 0. Deduce that

‖wi‖L2(Ki) ≤ hi‖w′i‖L2(Ki).

3. Show that there exists a constant cinter, independent of the mesh, such that, for any v ∈H2(Ω) ∩H1

0 (Ω), we have

∥∥∥v − I(1)h v

∥∥∥H1(Ω)

≤ cinter h |v|H2(Ω) and∥∥∥v − I

(1)h v

∥∥∥L2(Ω)


Exercise 5.34 (Convection-diffusion problems). We consider the convection-diffusion problem (3.10),that reads

−∆u+ c · ∇u = f in D′(Ω),u = 0 on ∂Ω,

where Ω is a bounded subset of Rd and f ∈ L2(Ω). We assume that c : Ω → Rd is a vector field ofclass C1(Ω) which is divergence-free:

div c =

d∑

i=1

∂ci∂xi

= 0 in Ω.

Under these assumptions, we have shown that there exists a unique u ∈ H1(Ω) solution to thatproblem. To do so, we have considered the variational formulation

Find u ∈ H1

0 (Ω) such that∀v ∈ H1

0 (Ω), a(u, v) = b(v).

where

a(u, v) =

∫

Ω

∇u · ∇v +∫

Ω

(c · ∇u) v, b(v) =

∫

Ω

f v.

The aim of this exercise is to study the discretization of that problem, in particular in the casewhen the convection field c is large.

For the sake of simplicity, we consider here the one-dimensional case: we set Ω = (0, 1) andwe look for u ∈ H1(Ω) such that

−νu′′ + βu′ = f in D′(Ω),u(0) = 0, u(1) = 0,

(5.17)

where ν and β are two positive real numbers (note that the convection field is here constant, andthus indeed divergence-free). We take f ∈ L2(Ω).


1. Recall the variational formulation of (5.17) and recall why the problem is well-posed. Showalso that u ∈ H2(Ω).

2. We introduce a discretization of (5.17) based on P1 finite elements and denote uh the discretesolution. Give the expression of the stiffness matrix A in terms of the ratio ν/h and of theso-called Péclet number

γ =hβ

ν.

3. Compute the continuity constant M and the coercivity constant α of the bilinear form a, interms of β, ν and the Poincaré constant CΩ of Ω. Using the Céa estimate (5.8), write anerror bound on ‖u− uh‖H1(Ω).

4. This estimate can actually be improved by going over the proof of (5.8) and carefully boundingthe bilinear form. Show that

ν|u− uh|2H1(Ω) = a(u− uh, u− I

(1)h u

)

and deduce that there exists a constant c, independent of u, ν, β and h, such that

|u− uh|H1(Ω) ≤ c

(1 +

hβ

ν

)h |u|H2(Ω). (5.18)

Which estimate on ‖u − uh‖H1(Ω) do we obtain? Show that this estimate is better than theone obtained in the previous question.

5. What happens in (5.18) when β increases or when ν decreases?

Check that, when γ > 2, the coefficient Ai,i+1 becomes positive. It turns out that the discretesolution is polluted by spurious oscillations (see Figure 5.5 below).

A first possibility to prevent this phenomenon is to use a sufficiently small mesh size h, sothat γ < 2. However, if β increases (or if ν decreases), this may lead to very small valuesfor h and thus a too large computational cost.

Another solution consists in working with a modified bilinear form, that is given by

a⋆(u, v) = a(u, v) + ah(u, v), ah(u, v) =hβ

2

∫

Ω

u′ v′.

Along with that modification, we also modify the linear form b and work with

b⋆(v) = b(v) + bh(v), bh(v) =h

2

∫

Ω

f v′.

We consider the discrete problem

Find u⋆h ∈ V

(1)h such that

∀v ∈ V(1)h , a⋆(u

⋆h, v) = b⋆(v).

(5.19)

Give the expression of the new stiffness matrix and check that its off-diagonal coefficientsalways remain negative. In contrast to the previous approximation uh, it turns out that u⋆hhas no spurious oscillations (see Figure 5.5 below).

5.5. EXERCISES 89

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

ReferenceP1P1 SUPG

0.96 0.98 1.000.0

0.2

0.4

0.6

0.8

1.0

1.2

ReferenceP1P1 SUPG

Figure 5.5: Exact and numerical solutions to (5.17) for ν = 1/256, β = 1, f = 1 and h = 1/16.Left: Plot on the whole domain. Right: Close-up on the boundary layer at the right-end of thedomain. The ’Reference’ solution (in blue) is the exact solution to (5.17). The ’P1’ solution (ingreen) is the P1 approximation using the original bilinear form a (see Question 2). The ’P1 SUPG’solution (in red) is the P1 approximation using the modified variational formulation (5.19). In thiscase, γ = 16, and we indeed observe spurious oscillations on the ’P1’ approximation. The ’P1SUPG’ approximation has no such oscillations.


Note: the method that we have introduced here is not restricted to the one-dimensional case. Itis called the SUPG method, which stands for Streamline-Upwind/Petrov-Galerkin. We refer e.g.to [7] for more details.

Exercise 5.35 (Approximation errors at the mesh nodes). We consider the one-dimensional prob-lem

Find u ∈ H10 (Ω) such that

∀v ∈ H10 (Ω),

∫

Ω

u′ v′ =

∫

Ω

f v,

where Ω = (a, b) and f ∈ L2(Ω). We approximate this problem by a P1 finite element method on amesh based on the nodes xi, 1 ≤ i ≤ n. We denote uh the discrete solution. Our aim is to showthat uh(xi) = u(xi) for any i: the discrete solution is equal to the exact solution at the mesh nodes.We already underline that this miracle only happens in the one-dimensional case.

For any 1 ≤ i ≤ n, we introduce the function Gi defined by

Gi(x) =

b− xib− a

(x− a) if a ≤ x ≤ xi,

xi − a

b− a(b− x) if xi ≤ x ≤ b.

1. For any v ∈ H10 (Ω) and any 1 ≤ i ≤ n, show that

∫

Ω

G′i v

′ = v(xi).

2. Deduce that uh(xi) = u(xi) for any 1 ≤ i ≤ n.

Exercise 5.36 (Mixed finite elements). Let Ω be a bounded domain in Rd and f ∈ L2(Ω). Weconsider the problem: find (u, p) ∈ H1

0 (Ω)× (L2(Ω))d solution to

−div p = f in D′(Ω),∇u = p in D′(Ω).

(5.20)

Of course, if (u, p) is a solution to (5.20), then u ∈ H10 (Ω) is solution to

−∆u = f in D′(Ω), (5.21)

and we have seen various ways to solve that problem. However, if the quantity of interest is prather than u, it may be interesting to keep p as part of the unknowns and to address (5.20) ratherthan (5.21). Our aim here is to directly discretize (5.20).

1. Show that (5.20) is equivalent to the following variational formulation:

Find (u, p) ∈ H10 (Ω)× (L2(Ω))d such that

∀q ∈ (L2(Ω))d, a(p, q) + b(q, u) = 0,∀v ∈ H1

0 (Ω), b(p, v) = c(v),(5.22)

where a and b are bilinear forms and c is a linear form that will be made precise.

5.5. EXERCISES 91

2. We consider a Galerkin approximation of (5.22), using an admissible mesh Th of Ω. To

approximate p, we use P0 finite elements, and thus work with the approximation space V(0)h

(piecewise constant functions). To approximate u, we use P1 finite elements, and thus work

with the approximation space V(1)h (piecewise affine functions). Write the discrete variational

formulation and show that it is equivalent to solving a linear system of the form

(A BBT 0

) (PU

)=

(0F

).

Give the expressions of the matrices A and B and of the vector F using the basis functions

of V(0)h and V

(1)h .

3. In the one-dimensional case, compute explicitly the matrices A and B.

4. Check that the matrix B is injective. Deduce that the matrix BT A−1B is positive definiteand next that the matrix (

A BBT 0

)

is invertible.

Exercise 5.37 (Error estimation in the L2 norm). Our aim here is to establish an optimal errorestimate in the L2 norm for the Poisson problem discretized using the P1 Finite Element Method(see Remark 5.25).

We set f ∈ L2(Ω) and assume that the polygon Ω is convex. We recall that, under this assump-tion, the unique solution u to (5.2) belongs to H2(Ω) and satisfies |u|H2(Ω) ≤ CΩ‖f‖L2(Ω), wherethe constant CΩ only depends on Ω.

Let uh ∈ V(1)h be the solution to (5.12). We assume that the sequence of meshes is regular of

parameter σ0. We recall (see Theorem 5.22) that there exists a constant c, that may depend on Ωand σ0, but that is independent of h and f , such that

‖u− uh‖H1(Ω) ≤ c h ‖f‖L2(Ω).

1. Let ξ be the unique solution to the problem

Find ξ ∈ H10 (Ω) such that

∀v ∈ H10 (Ω),

∫

Ω

∇ξ · ∇v =

∫

Ω

(u− uh) v.

Show that ξ ∈ H2(Ω) and that

‖u− uh‖2L2(Ω) =

∫

Ω

∇(u− uh) · ∇(ξ − I

(1)h ξ

),

where I(1)h ξ is the interpolation of ξ on V

(1)h .

2. Using the fact that |ξ|H2(Ω) ≤ CΩ‖u− uh‖L2(Ω), show that

‖u− uh‖L2(Ω) ≤ c h2,

where c may depend on Ω, σ0, f but is independent of h. This result is the so-called Aubin-Nitsche lemma. This estimate is sharp in the sense that there exist Poisson problems suchthat ‖u− uh‖L2(Ω) is indeed of the order of h2.


Exercise 5.38 (L2 projection). We consider a polygon Ω ⊂ R2 and an admissible mesh Th of Ω

with N internal vertices. Let V(1)h be the P1 finite element space associated with this mesh. We

denote (ϕ1, . . . , ϕN ) the basis of V(1)h .

We consider here the L2 projection on V(1)h , namely the operator Πh : L2(Ω) → V

(1)h such that,

for any v ∈ L2(Ω), we have

‖v −Πhv‖L2(Ω) = infvh∈V

(1)h

‖v − vh‖L2(Ω).

The existence and uniqueness of Πhv comes from the orthogonal projection theorem in the Hilbert

space L2(Ω). As usual, Πhv is the function in V(1)h which is the closest (in the L2 norm) to v. It

also satisfies

∀1 ≤ i ≤ N,

∫

Ω

(v − Πhv) ϕi = 0.

1. Let M ∈ RN×N be the matrix defined by

∀1 ≤ i, j ≤ N, Mij =

∫

Ω

ϕi ϕj .

Show that M is symmetric definite positive.

2. Show that Πhv may be computed by solving a linear system of the form M X = V for aright-hand side V ∈ RN that will be made precise. The relation between Πhv and X ∈ RN

will also be explicited.

Exercise 5.39 (Q1 finite element in dimension 2). We present here the Q1 finite element method.We assume that the dimension d is equal to 2, although our construction carries over to anydimension. The main idea is to mesh the domain with quadrangles, in contrast to the Pk FiniteElement Methods where the domain is meshed with triangles.

Let Ω = (α1, β1) × (α2, β2). We consider two one-dimensional meshes xi0≤i≤N+1 andyj0≤j≤M+1 such that

α1 = x0 < x1 < · · · < xi < · · · < xN+1 = β1 and α2 = y0 < y1 < · · · < yj < · · · < yM+1 = β2

and we mesh Ω by the rectangles Kij defined by Kij = [xi, xi+1] × [yj , yj+1]. More generally, Ωcould be meshed by quadrangles that are not necessarily rectangles. We denote by Th this mesh.Note that the mesh is again admissible in the sense that the intersection of any two elements iseither empty, or restricted to a point which is a vertex of both elements, or equal to a segmentwhich is an edge of the two elements. The mesh size is defined as

h = max

(max

0≤i≤Nxi+1 − xi, max

0≤j≤Myj+1 − yj

).

We introduce the vector space of polynomial functions of partial degree lower or equal to 1:

Q1 =p : R2 → R, p(x, y) = α+ βx+ γy + δxy for some real numbers α, β, γ and δ

.

Note that, in contrast to P1, the function (x, y) ∈ R2 7→ xy belongs to Q1. A function in Q1 isfully determined by its value at the four vertices of the non-degenerate quadrangle.

5.5. EXERCISES 93

1. We define the space

W(1)h =

vh ∈ C0(Ω), vh|K ∈ Q1 for any K ∈ Th, vh = 0 on ∂Ω

.

For any (i, j) with 1 ≤ i ≤ N and 1 ≤ j ≤ M , we consider the function ϕij ∈ W(1)h such

that, for any 1 ≤ ℓ ≤ N , 1 ≤ m ≤M ,

ϕij(xℓ, ym) =

1 if (ℓ,m) = (i, j),0 otherwise.

Show that the set ϕij , 1 ≤ i ≤ N, 1 ≤ j ≤M is a basis of the vector space W(1)h . Is the

space W(1)h is vector subspace of H1

0 (Ω)?

2. Let f ∈ L2(Ω). The Galerkin approximation of the Poisson problem (5.2) on the finite

dimensional space W(1)h is:

Find uh ∈W

(1)h such that

∀vh ∈ W(1)h , a(uh, vh) = b(vh).

(5.23)

Let θi1≤i≤N (resp. ξj1≤j≤M ) be the basis functions of the one-dimensional P1 discrete

space based on the mesh xi0≤i≤N+1 (resp. yj0≤j≤M+1) of the segment (α1, β1) (resp.(α2, β2)). Let Mθ and Aθ be the mass and the stiffness matrices associated to the functionsθi1≤i≤N :

∀1 ≤ i1, i2 ≤ N,[Mθ

]i1 i2

=

∫ α2

α1

θi1 θi2 and[Aθ]i1 i2

=

∫ α2

α1

θ′i1 θ′i2 .

We proceed likewise with the functions ξj1≤j≤M and introduce the mass matrix Mξ and

the stiffness matrix Aξ.

Show that the stiffness matrix of (5.23) can be written in terms of Mθ, Aθ, Mξ and Aξ.

3. The approximation properties in W(1)h are similar as in the P1 space V

(1)h . We introduce the

interpolation operator

I(1)h : C0(Ω) → W

(1)h

v 7→ vh =

N∑

i=1

M∑

j=1

v(xi, yj)ϕij .

We see that I(1)h v is the unique function in W

(1)h that takes the same values as v on all the

internal nodes of the mesh. We admit the following result: there exists cinter such that, forany h,

∀v ∈ H2(Ω) ∩H10 (Ω),

∥∥∥v − I(1)h v

∥∥∥H1(Ω)

≤ cinter h |v|H2(Ω).

Prove that the solution uh to (5.23) converges to the solution u to (5.2) and state an errorestimate.

Note: the interest of using a Q1 approach (rather than a P1 approach) is that it is sometimeseasier to mesh Ω using quadrangles rather than triangles.


Chapter 6

Numerical approximation ofboundary value problems (thenon-coercive case)

This chapter is devoted to the numerical approximation of linear elliptic boundary value problems,in a non-coercive setting. We consider here problems such as those studied in Chapter 4, thewell-posedness of which has been established using the inf-sup theory.

6.1 Galerkin approximation

In this section, we consider the variational formulation (4.8), which reads

Find u ∈ V such that∀w ∈W, a(u,w) = b(w),

(6.1)

where V and W are two Banach spaces, a is a bilinear continuous form on V ×W and b is linearcontinuous form on W . We assume that a satisfies the inf-sup conditions (4.9) and (4.10). Inaddition, for the sake of simplicity, we assume that W is a Hilbert space. In view of Theorem 4.10,the problem (6.1) is well-posed.

6.1.1 Well-posedness

Our aim here is to approximate (6.1) in a finite dimensional setting. For the sake of simplicity, werestrict ourselves to a conformal approximation. We thus introduce the finite dimensional spacesVh ⊂ V and Wh ⊂W , and consider the problem

Find uh ∈ Vh such that∀wh ∈ Wh, a(uh, wh) = b(wh).

(6.2)

In view of Lemma 4.7, this problem is well-posed if and only if the two following conditions aresatisfied:

There exists αh > 0 such that infuh∈Vh, uh 6=0

supwh∈Wh, wh 6=0

a(uh, wh)

‖uh‖V ‖wh‖W≥ αh (6.3)

andIf wh ∈Wh is such that a(uh, wh) = 0 for any uh ∈ Vh, then wh = 0. (6.4)

95

96CHAPTER 6. NUMERICAL APPROXIMATION OF BOUNDARY VALUE PROBLEMS (THE NON-COERCIVE CASE)

Note that αh may depend on h (but should be positive).

We first observe that the inf-sup conditions (4.9)–(4.10) on V ×W do not imply the inf-supconditions (6.3)–(6.4) on Vh ×Wh. This is in sharp contrast with the coercivity assumption: ifthere exists α > 0 such that a(u, u) ≥ α‖u‖2V for any u ∈ V , then, for any Vh ⊂ V and any uh ∈ Vh,we have a(uh, uh) ≥ α‖uh‖2V , thus the coercivity of a on Vh (with the same coercivity constant).

It is actually easy to build an example of a bilinear form a such that (4.9)–(4.10) hold and suchthat (6.3)–(6.4) do not hold. Consider for instance the case V =W = R2 and a(u, v) = vTAu with

A =

(0 11 0

).

The matrix A is symmetric, its two eigenvalues are 1 and −1, and it is invertible. The condi-tions (4.9)–(4.10) are thus satisfied. Consider now the subspace

Vh =Wh =

(s0

), s ∈ R

⊂ V.

For any uh ∈ Vh and wh ∈Wh, we have a(uh, wh) = 0. The condition (6.3) is hence not satisfied.

In general, there is no alternative to checking that (6.3)–(6.4) indeed hold on Vh ×Wh. Werecall that (6.3)–(6.4) imply that dim Vh = dim Wh (see Remark 4.9). We also recall that, ifdim Vh = dim Wh, then the conditions (6.3) and (6.4) are equivalent (see Remark 4.8). In thatcase, it is thus sufficient to check that either (6.3), or (6.4), holds for (6.2) to be well-posed.

6.1.2 Error estimation

We wish to bound from above the error e = u−uh. The equivalent of the Céa lemma in the inf-supsetting is the following result:

Lemma 6.1. We suppose that assumptions (4.9)–(4.10) hold. Consider the subspaces Vh ⊂ V andWh ⊂W , and assume that (6.3)–(6.4) hold as well.

Let u be the solution to (6.1) and uh be the solution to (6.2). Then

‖u− uh‖V ≤(1 +

Caαh

)inf

vh∈Vh

‖u− vh‖V , (6.5)

where αh is the constant appearing in (6.3) and Ca is the continuity constant of a on V ×W : forany v ∈ V and w ∈W , we have |a(v, w)| ≤ Ca‖v‖V ‖w‖W .

Note that the assumptions (4.9), (4.10) and (6.4) are only mentioned to ensure the existenceand uniqueness of u and uh. Given some u (resp. uh) satisfying (6.1) (resp. (6.2)), the onlyassumption used to prove (6.5) is (6.3).

Proof. Let vh ∈ Vh. We deduce from (6.3) that

αh‖uh − vh‖V ≤ supwh∈Wh, wh 6=0

a(uh − vh, wh)

‖wh‖W.

Since a(uh, wh) = a(u,wh) for any wh ∈Wh, we get that

αh‖uh − vh‖V ≤ supwh∈Wh, wh 6=0

a(u− vh, wh)

‖wh‖W≤ Ca‖u− vh‖V .

We now use the triangle inequality:

‖u− uh‖V ≤ ‖u− vh‖V + ‖uh − vh‖V ≤(1 +

Caαh

)‖u− vh‖V .

6.2. THE STOKES PROBLEM 97

Since vh ∈ Vh is arbitrary, we deduce that

‖u− uh‖V ≤(1 +

Caαh

)inf

vh∈Vh

‖u− vh‖V ,

which is (6.5).

As in the coercive setting (see the Céa lemma 5.7), the best approximation error infvh∈Vh

‖u−vh‖Vis an upper bound (up to a multiplicative constant) of the error u−uh. Obviously, it is also a lowerbound of the error, since inf

vh∈Vh

‖u− vh‖V ≤ ‖u− uh‖V . If the best approximation error converges

to 0, and provided that there exists α > 0 such that αh ≥ α, then so does the error u− uh, at thesame rate.

As in the coercive case, we see that the error estimation is based on two ingredients:

• the quality of the approximation space, which is quantified by the best approximation errorinf

vh∈Vh

‖u − vh‖V , and which is independent of the equation which is considered (and of the

fact that the equation is, or is not, coercive). This best approximation error can be estimatedin terms of h in view of Theorem 5.15 (for P1 finite elements), Theorem 5.29 (for P2 finiteelements), . . .

• the stability of the problem, which is quantified by the constant 1 +Ca/αh, where Ca is thecontinuity constant of a and αh is the inf-sup constant appearing in (6.3) (in the coercivesetting, this stability is quantified by the constant Ca/α, where Ca is again the continuityconstant of a, and α is its coercivity constant). We note that αh depends on the problemwhich is studied (namely the bilinear form a) and also on the discretization spaces Vh andWh.

6.2 The Stokes problem

We now consider the Stokes problem (4.16), which reads

Find (u, p) ∈ (H10 (Ω))

d × L20(Ω) such that

−∆u+∇p = f in [D′(Ω)]d,

div u = 0 in D′(Ω).

(6.6)

We recall that a possible variational formulation is (4.17), which falls within the general frame-work (6.1). Another variational formulation, more specific to the Stokes problem, is (4.19), whichreads, we recall

Find (u, p) ∈ X ×M such that∀v ∈ X, a0(u, v) + b(v, p) = g1(v),∀q ∈M, b(u, q) = g2(q),

(6.7)

where X = (H10 (Ω))

d, M = L20(Ω) =

q ∈ L2(Ω),

∫

Ω

q = 0

, a0 is the bilinear form defined on

X ×X by

a0(u, v) =

∫

Ω

∇u · ∇v =

d∑

i=1

∫

Ω

∇ui · ∇vi,

b is the bilinear form defined on X ×M by

∀v ∈ X, ∀p ∈M, b(v, p) = −∫

Ω

p div v,


g1 is the linear form defined on X by

g1(v) =

∫

Ω

f · v =

d∑

i=1

∫

Ω

fi vi,

and g2 is the linear form on M defined by g2(q) = 0.

We assume in this section that X and M are two Hilbert spaces, that the bilinear (resp. linear)forms a0 and b (resp. g1 and g2) are continuous, and that the bilinear form a0 is coercive on X .

The well-posedness of (6.7) is established by Theorem 4.16. We have seen there that (6.7) iswell-posed if and only if b satisfies the inf-sup condition (4.20).

6.2.1 Galerkin approximation of the abstract problem (6.7)

Our aim here is to approximate (6.7) in a finite dimensional setting. For the sake of simplicity, weagain restrict ourselves to a conformal approximation. We thus introduce Xh ⊂ X and Mh ⊂ Mand consider the problem

Find (uh, ph) ∈ Xh ×Mh such that∀vh ∈ Xh, a0(uh, vh) + b(vh, ph) = g1(vh),∀qh ∈Mh, b(uh, qh) = g2(qh).

(6.8)

In view of Theorem 4.16 (this time applied in a finite dimensional setting), this problem is well-posed if and only b satisfies an inf-sup condition in Xh ×Mh, that is

∃βh > 0, ∀qh ∈Mh, supvh∈Xh, vh 6=0

b(vh, qh)

‖vh‖X≥ βh‖qh‖M . (6.9)

Note that βh may depend on h (but should be positive).

We again observe that the inf-sup condition (4.20) on X × M does not imply the inf-supcondition (6.9) on Xh ×Mh. In general, there is thus no alternative to checking that (6.9) indeedholds on Xh ×Mh. We will come back to this in the subsequent sections.

We now turn to estimating the error between (u, p) and (uh, ph). We have the following result:

Lemma 6.2. We suppose that the assumption (4.20) holds. Consider the subspaces Xh ⊂ X andMh ⊂ M , and assume that (6.9) holds as well for some βh satisfying βh ≥ β > 0 where β isindependent of the mesh size h.

Let (u, p) be the solution to (6.7) and (uh, ph) be the solution to (6.8). Then there exists aconstant C independent of h such that

‖u− uh‖X + ‖p− ph‖M ≤ C

(inf

vh∈Xh

‖u− vh‖X + infqh∈Mh

‖p− qh‖M). (6.10)

Note that the assumption (4.20) is only mentioned to ensure the existence and uniqueness of(u, p). Given some (u, p) (resp. (uh, ph)) satisfying (6.7) (resp. (6.8)), the only assumption used toprove (6.10) is (6.9).

Proof. Let vh ∈ Xh and qh ∈Mh. We write

‖p− ph‖M ≤ ‖p− qh‖M + ‖qh − ph‖M

≤ ‖p− qh‖M + β−1h sup

vh∈Xh, vh 6=0

b(vh, qh − ph)

‖vh‖X[using (6.9)]

= ‖p− qh‖M + β−1h sup

vh∈Xh, vh 6=0

b(vh, qh − p) + b(vh, p− ph)

‖vh‖X

≤ ‖p− qh‖M + β−1h sup

vh∈Xh, vh 6=0

b(vh, qh − p)

‖vh‖X+ β−1

h supvh∈Xh, vh 6=0

b(vh, p− ph)

‖vh‖X. (6.11)


We infer from the first line of (6.7) and the first line of (6.8) that

∀vh ∈ Xh, a0(u− uh, vh) + b(vh, p− ph) = 0. (6.12)

Likewise, the second line of (6.7) and the second line of (6.8) yield that

∀qh ∈Mh, b(u− uh, qh) = 0. (6.13)

Collecting (6.11) and (6.12), we obtain

‖p− ph‖M ≤ ‖p− qh‖M + β−1h sup

vh∈Xh, vh 6=0

b(vh, qh − p)

‖vh‖X+ β−1

h supvh∈Xh, vh 6=0

a0(u− uh, vh)

‖vh‖X

≤(1 +

Cbβh

)‖p− qh‖M +

Caβh

‖u− uh‖X , (6.14)

where Cb (resp. Ca) is the continuity constant of the bilinear form b (resp. the bilinear form a0).Now using the coercivity of a0 (with the constant α), we write

α‖u− uh‖2X ≤ a0(u− uh, u− uh)

= a0(u− uh, u− vh) + a0(u− uh, vh − uh)

= a0(u− uh, u− vh)− b(vh − uh, p− ph) [using (6.12)]

= a0(u− uh, u− vh)− b(vh − u, p− ph)− b(u− uh, p− ph)

= a0(u− uh, u− vh)− b(vh − u, p− ph)− b(u− uh, p− qh) [using (6.13)]

≤ Ca‖u− uh‖X‖u− vh‖X + Cb‖u− vh‖X‖p− ph‖M + Cb‖u− uh‖X‖p− qh‖M .

Collecting the first and the third term, we get

α‖u− uh‖2X ≤ (Ca + Cb)‖u− uh‖X (‖u− vh‖X + ‖p− qh‖M ) + Cb‖u− vh‖X‖p− ph‖M≤ (Ca + Cb)‖u− uh‖X (‖u− vh‖X + ‖p− qh‖M )

+ Cb

(1 +

Cbβh

)‖u− vh‖X‖p− qh‖M + Cb

Caβh

‖u− vh‖X‖u− uh‖X

≤(Ca + Cb +

Ca Cbβh

)‖u− uh‖X (‖u− vh‖X + ‖p− qh‖M )

+ Cb

(1 +

Cbβh

)‖u− vh‖X‖p− qh‖M , (6.15)

where we have used (6.14) at the second line. Let

τh = infvh∈Xh

‖u− vh‖X + infqh∈Mh

‖p− qh‖M

be the best approximation error. Since vh ∈ Xh and qh ∈Mh are arbirarty, we deduce from (6.15)that

‖u− uh‖2X ≤ C‖u− uh‖X τh + Cτ2h

where C is a constant depending on Ca, Cb, α and βh. This implies that

‖u− uh‖X ≤ Cτh.

Inserting this estimate in (6.14), we obtain

‖p− ph‖M ≤ Cτh.

This concludes the proof of (6.10).


6.2.2 Checkerboard instability for the Stokes problem

In this section, we consider a specific choice of Xh ⊂ X and Mh ⊂ M such that the inf-supcondition (6.9) does not hold. The choice we describe here is the most well-known bad choice forXh and Mh.

We consider the 2D situation, with Ω = (0, 1)2, and we mesh Ω by quadrangles. We thenset Xh = Q1 (i.e. each component of the velocity uh is assumed to be a Q1 polynomial, seeExercice 5.39) and

Mh =qh ∈ L2

0(Ω), qh is constant on each element.

For any v ∈ Xh and q ∈ Mh, we denote by vi,j the value of the velocity at node (i, j), and qKi,j

the value of the pressure in the element Ki,j (for the sake of simplicity, we drop the subscript hfor v and q). We then compute

b(v, q) = −∫

Ω

q div v = −∑

K

∫

K

q div v = −∑

i,j

qKi,j

∫

Ki,j

div v = −∑

i,j

qKi,j

∫

∂Ki,j

v · n.

On each edge of Ki,j , the field v = (vx, vy) is affine. We hence have

b(v, q)

= −1

2h∑

i,j

qKi,j

((vxi+1,j + vxi+1,j+1) + (vyi,j+1 + vyi+1,j+1)− (vxi,j + vxi,j+1)− (vyi,j + vyi+1,j)

)

=1

2h∑

i,j

(qKi,j− qKi−1,j )(v

xi,j + vxi,j+1) +

1

2h∑

i,j

(qKi,j− qKi,j−1)(v

yi,j + vyi+1,j)

= h2∑

i,j

∇xqij vxi,j + h2

∑

i,j

∇yqij vyi,j , (6.16)

with the notations

2h∇xqij = (qKi,j− qKi−1,j ) + (qKi,j−1 − qKi−1,j−1),

2h∇yqij = (qKi,j− qKi,j−1) + (qKi−1,j − qKi−1,j−1).

Note that we have used above, when reordering the terms, the fact that v vanishes on the boundaryof Ω.

We wish to build a non-zero field q ∈ Mh such that b(v, q) = 0 for all v ∈ Xh. Since vxi,j andvyi,j in (6.16) are arbitrary, the field q should satisfy ∇xqij = 0 and ∇yqij = 0 for any i and j.

Subtracting the two equations, we get qKi,j−1 = qKi−1,j . Adding the two equations, we getqKij

= qKi−1,j−1 . This means that, when moving along each “diagonal” of the cartesian mesh, thevalue of q should remain the same. The set of solutions is thus completely determined by twocontants, say qK0,0 = α and qK1,0 = β.

In addition, q should satisfy the constraint of zero-average. This imposes that α = −β. Thepressure field q has thus the structure of a checkerboard, alternating two opposite values. For sucha field (called spurious mode), we have b(v, q) = 0 for any v ∈ Xh.

Remark 6.3. It could be tempting to remove this spurious mode of the discretization space Mh,by considering

Mh =

qh ∈ L2

0(Ω), qh is constant on each element,

∫

Ω

qh qspurious = 0

where qspurious is the spurious pressure mode we have identified above. This does not properly fixthe issue, as this leads to an inf-sup property (6.9) with some positive βh, which unfortunatelyscales as O(h) when h→ 0 (see [4, Section 4.2.3] for details).

In order to fix the problem and obtain approximation spaces that satisfy the inf-sup condi-tion (6.9), we have either to enrich Xh or to consider a smaller space Mh.


6.2.3 Numerical locking for the Stokes problem

The fact that the inf-sup condition (6.9) does not hold for Xh = Q1 and piecewise constant pressurefields is not related to the fact that we use quadrangles. A similar issue arises if we use a meshmade of triangles, as we now show.

We again consider the 2D situation, and we mesh Ω by triangles. We then set Xh = P1 (i.e.each component of the velocity uh is assumed to be a P1 polynomial) and

Mh =qh ∈ L2

0(Ω), qh is constant on each element.

For simplicity, we consider a mesh made of isocele triangles (but the argument carries over togeneral meshes). We denote N2 the number of internal nodes (so that each “side” of Ω has beencut into N + 1 elements). We see that dim Xh = 2N2 (two degrees of freedom per internal node)and dim Mh = 2(N + 1)2 − 1 (there are 2(N + 1)2 elements, one degree of freedom per elementand we take into account the mean-free constraint), that is dim Mh = 2N2 + 4N + 1.

Consider the operatorD : uh ∈ Xh −→ div uh ∈Mh.

If it is surjective, then the inf-sup condition (6.9) is satisfied for the Stokes problem (we just have tofollow the arguments of the proof of Theorem 4.18). We have dim Ker D + dim Im D = dim Xh.If D is surjective, then dim Ker D + dim Mh = dim Xh. But this is not possible since dim Mh >dim Xh.

Another (equivalent) way to conclude is to introduce the matrix B : Rx → Rm such thatb(vh, qh) = QThBVh, where Qh (resp. Vh) is the vector collecting the degrees of freedom of qh ∈Mh

(resp. of vh ∈ Xh), and m = dim Mh, x = dim Xh. The inf-sup condition (6.9) reads

∃βh > 0, ∀Qh ∈ Rm, supVh∈Rx, Vh 6=0

QThBVh‖Vh‖

≥ βh‖Qh‖,

which is equivalent to the fact that BT is injective (see Lemma 4.6(i)). We have dim Ker BT +dim Im BT = dim Mh, hence

dim Ker BT = dim Mh − dim Im BT ≥ dim Mh − dim Xh = 4N + 1 > 0.

The matrix BT thus cannot be injective.We note that the second line of (6.8) can be recast as QThBUh = 0 for any Qh ∈ Rm. In some

cases, the space Mh is so rich (and its dimension so large compared to that of Xh) that the onlysolution to that equation is Uh = 0 (i.e. the matrix B is injective). This motivates the name oflocking for such a choice of Xh and Mh.

6.2.4 A possible choice of Xh ×Mh for the Stokes problem

We now build an example of spaces Xh ⊂ X and Mh ⊂ M such that the inf-sup condition (6.9)holds on Xh ×Mh. With the aim to obtain approximation spaces that are easy to manipulate,we wish to only slightly modify the choice of Section 6.2.3 (piecewise affine velocities, piecewiseconstant pressures).

Consider first the choice Xh = P1 and

Mh =qh ∈ L2

0(Ω), qh ∈ P1

. (6.17)

For simplicity, we again consider a mesh made of isocele triangles. We denote N2 the number ofinternal nodes (so that each “side” of Ω has been cut into N + 1 elements). The space Xh is thesame as in Section 6.2.3. The space Mh is now of dimension (N +2)2 − 1 = N2 +4N +3, which is


smaller (as soon as N ≥ 2) than the dimension of the space Mh in Section 6.2.3. It however turnsout that this couple of spaces still does not satisfy the inf-sup condition (6.9) (see [4, Section 4.2.3]for details).

Keeping the pressure space Mh as defined by (6.17), we now enrich the velocity space Xh. Ineach element K, and for each component j of vh ∈ Rd, we assume that

v(j)h ∈ Span

λK0 , λ

K1 , . . . , λ

Kd+1

, 1 ≤ j ≤ d,

where, for any 1 ≤ i ≤ d+1, we have λKi ∈ P1 and λKi (xj) = δij (where xj are the nodes of K) andwhere λK0 ∈ H1

0 (K), 0 ≤ λK0 ≤ 1 on K and λK0 (xK) = 1, where xK is the barycenter of K. Thefunctions λKi (1 ≤ i ≤ d+1) are the usual piecewise affine functions on K, which are equal to 1 atone node of K and vanish at the other nodes of K. The function λK0 is a so-called bubble function:it vanishes on ∂K, is non-negative on K and is equal to 1 somewhere (e.g. at the barycenter) inK. For instance, up to a multiplicative constant, we can take λK0 = Πd+1

i=1 λKi .

Theorem 6.4. Let Ω be an open, bounded and connected subset of Rd. Assume that the pressurespace Mh is defined by (6.17) and that Xh is defined as above. Then the inf-sup condition (6.9)holds for some constant β > 0 independent of h.

This choice of spaces Xh and Mh is sometimes called the mini-element. Other choices canbe made (with polynomial functions of higher degree) such that (6.9) is satisfied. We refer to [4,Section 4.2] for details.

Proof. The idea of the proof is as follows. Consider qh ∈ Mh ⊂ M . We know from Theorem 4.18that there exists v ∈ X such that div v = −qh with ρ‖v‖X ≤ ‖qh‖M for some ρ > 0.

Assume that we can find some Πh(v) ∈ Xh such that

∀qh ∈Mh, b(Πh(v), qh) = b(v, qh), (6.18)

‖Πh(v)‖X ≤ c‖v‖X . (6.19)

We then write that

supvh∈Xh, vh 6=0

b(vh, qh)

‖vh‖X≥ b(Πh(v), qh)

‖Πh(v)‖X

=b(v, qh)

‖Πh(v)‖X[using (6.18)]

≥ b(v, qh)

c‖v‖X[using (6.19)]

= − 1

c‖v‖X

∫

Ω

qh div v

=‖qh‖2Mc‖v‖X

[by def. of v in terms of qh]

≥ ρ

c‖qh‖M ,

which is exactly the inf-sup condition (6.9) for some constant β > 0 independent of h.

We are thus left with showing (6.18), which reads

∀qh ∈Mh,

∫

Ω

qh div Πh(v) =

∫

Ω

qh div v.

Since qh ∈Mh belongs to H1(Ω), this condition reads

∀qh ∈Mh,

∫

Ω

Πh(v) · ∇qh =

∫

Ω

v · ∇qh.


Assume that Πh(v) ∈ Xh is such that

∀K,∫

K

Πh(v) =

∫

K

v. (6.20)

For any qh ∈Mh, we compute

∫

Ω

Πh(v) · ∇qh =∑

K

∇qh|K ·∫

K

Πh(v) =∑

K

∇qh|K ·∫

K

v =

∫

Ω

v · ∇qh

and thus obatin (6.18). We are thus left with finding Πh(v) ∈ Xh satisfying (6.20) and (6.19).

We note that ensuring (6.20) for some Πh(v) which would be in P1 is challenging, since adjustingthe average of Πh(v) on K requires to adjust the values of Πh(v) at the nodes of K, which modifiesthe average of Πh(v) on the neighboring elements.

Here, the space Xh is richer. In particular, we can adjust the average of Πh(v) on K by simplyadjusting the coefficient in front of the bubble function λK0 , the support of which is included in K(this adjustment is hence local and does not modify Πh(v) on the other elements). We look forΠh(v) under the form

ei · Πh(v) = ei · Ihv +∑

K

γKi λK0 ,

where (ei)1≤i≤d is the canonical basis of Rd and Ihv is an accurate approximation of v in the P1

space (we can take the nodal interpolant of v if v is continuous and thus has well-defined values atthe nodes), chosen such that ‖v − Ihv‖L2(Ω) ≤ Ch‖v‖H1(Ω).

The parameters γKi are chosen such that (6.20) holds. We hence request that, for any K and

i, we have

∫

K

ei · Πh(v) =∫

K

ei · v, which implies

∫

K

ei · v =

∫

K

ei · Ihv + γKi

∫

K

λK0

and hence

γKi =

∫K ei · (v − Ihv)∫

K λK0

.

We admit that, with such a definition of Πh(v), the estimate (6.19) is satisfied. This concludes theproof.


Appendix A

Basic facts on the Lebesgue integraland Lp spaces

We briefly collect here some basic facts concerning the Lebesgue integral and Lp spaces. For moredetails, we refer to the first year ENPC course Outils Mathématiques pour l’Ingénieur.

In what follows, Ω is an open subset (which may or may not be bounded) of Rd (with d ≥ 1)and p is a real number, p ≥ 1.

We recall that two functions f and g defined on Ω are said to be equal almost everywhere onΩ if the measure of the set x ∈ Ω, f(x) 6= g(x) vanishes.

A.1 The space L1(Ω)

A.1.1 Definition

We first recall that the space L1(Ω) is defined by

L1(Ω) =

f : Ω → R, f is measurable,

∫

Ω

|f(x)| dx < +∞.

The vector space L1(Ω) is obtained as the quotient of the space L1(Ω) with respect to the equiva-lence relation “f ∼ g if f = g almost everywhere”. When endowed with the norm

‖f‖L1(Ω) =

∫

Ω

|f(x)| dx,

the vector space L1(Ω) is a Banach space.

A.1.2 Dominated convergence theorem

The following result is of paramount importance.

Theorem A.1. Let fn be a sequence of integrable functions in Ω (meaning fn ∈ L1(Ω)) and let gbe a non-negative function which is integrable on Ω. We assume that

• for almost every x ∈ Ω, the sequence (fn(x))n converges to some f(x) when n→ ∞;

• for any n ∈ N, and for almost every x ∈ Ω, we have |fn(x)| ≤ g(x).

105

106 APPENDIX A. BASIC FACTS ON THE LEBESGUE INTEGRAL AND LP SPACES

Then the function f is integrable on Ω and limn→∞

∫

Ω

|f(x)− fn(x)| dx = 0, which implies that

limn→∞

∫

Ω

fn(x) dx =

∫

Ω

f(x) dx.

We admit the following result:

Theorem A.2. Consider a sequence of functions fn ∈ L1(Ω) that converge to some f in L1(Ω)when n → ∞. Then there exists a subsequence of fnn∈N

that converges to f almost everywherewhen n→ ∞.

A.2 The space L2(Ω)

Definition A.3. The space L2(Ω) is defined by

L2(Ω) =

f : Ω → R, f is measurable,

∫

Ω

|f(x)|2 dx < +∞.

The vector space L2(Ω) is obtained as the quotient of the space L2(Ω) with respect to the equivalencerelation “f ∼ g if f = g almost everywhere”.

Theorem A.4. When endowed with the scalar product

(f, g)L2 =

∫

Ω

f(x) g(x) dx,

the vector space L2(Ω) is a Hilbert space.

A.3 The spaces Lp(Ω) and Lploc(Ω)

Definition A.5. The space Lp(Ω) is defined as

Lp(Ω) =f : Ω → R, f is measurable,

∫

Ω

|f(x)|p dx < +∞.

The vector space Lp(Ω) is obtained as the quotient of the space Lp(Ω) with respect to the equivalencerelation “f ∼ g if f = g almost everywhere”.

Theorem A.6. When endowed with the norm

‖f‖Lp(Ω) =

(∫

Ω

|f(x)|p dx)1/p

,

the vector space Lp(Ω) is a Banach space.

Theorem A.7 (Hölder inequality). Let p ∈ R with 1 < p < +∞, and let q ∈ R with 1/p+1/q = 1(note that 1 < q < +∞). Let f ∈ Lp(Ω) and g ∈ Lq(Ω). Then f g ∈ L1(Ω) and we have thefollowing Hölder inequality:

‖f g‖L1(Ω) ≤ ‖f‖Lp(Ω) ‖g‖Lq(Ω).

A.4. THE SPACE L∞(Ω) 107

Lemma A.8. Assume that the open set Ω is bounded. Let p and q two real numbers with q > p ≥ 1.Then Lq(Ω) ⊂ Lp(Ω).

Definition A.9. We say that f is locally integrable on Ω if, for any compact set K contained inΩ, the function f is integrable on K. We denote

L1loc(Ω) =

f ; f ∈ L1(K) for any compact subset K ⊂ Ω

the vector space of locally integrable functions. Likewise,

Lploc(Ω) = f ; f ∈ Lp(K) for any compact subset K ⊂ Ω .

Lemma A.10. We have that Lp(Ω) ⊂ Lploc(Ω). For any real numbers p and q such that q > p ≥ 1,we have Lqloc(Ω) ⊂ Lploc(Ω).

Definition A.11. Let fn be a sequence of functions in Lploc(Ω) and f ∈ Lploc(Ω). We say thatfn converges to f in Lploc(Ω) when n → ∞ if, for any compact subset K ⊂ Ω, the sequence fn|Kconverges to f |K in Lp(K) when n→ ∞.

A.4 The space L∞(Ω)

Definition A.12. A function f is said to be essentially bounded on Ω if there exists a non-negativereal number M such that

µ(x ∈ Ω; |f(x)| ≥M

)= 0,

where µ is the Lebesgue measure. Hence, except on a null set (that is, a set the measure of whichvanishes), we have |f(x)| < M .

The set of functions that are essentially bounded on Ω is denoted L∞(Ω).

Definition A.13. The vector space L∞(Ω) is obtained as the quotient of the space L∞(Ω) withrespect to the equivalence relation “f ∼ g if f = g almost everywhere”.

Theorem A.14. When endowed with the norm

‖f‖L∞(Ω) = infM ≥ 0; µ

(x ∈ Ω; |f(x)| ≥M

)= 0,

the vector space L∞(Ω) is a Banach space.

Lemma A.15. If f is essentially bounded on Ω, then |f(x)| ≤ ‖f‖L∞(Ω) almost everywhere on Ω.If in addition f is continuous on Ω, then |f(x)| ≤ ‖f‖L∞(Ω) for any x ∈ Ω.

Theorem A.16. If f ∈ L1(Ω) and g ∈ L∞(Ω), then the product f g belongs to L1(Ω) and

‖f g‖L1(Ω) ≤ ‖f‖L1(Ω) ‖g‖L∞(Ω).

108 APPENDIX A. BASIC FACTS ON THE LEBESGUE INTEGRAL AND LP SPACES

A.5 Other properties

Theorem A.17 (Interpolation). Let f ∈ Lp(Ω) ∩ Lq(Ω) with 1 ≤ p ≤ q ≤ ∞. Then, for any rsuch that p ≤ r ≤ q, we have that f ∈ Lr(Ω) and

‖f‖Lr ≤ ‖f‖αLp ‖f‖1−αLq

with α ∈ (0, 1) such that1

r=α

p+

1− α

q.

Definition - Theorem A.18. Let f and g in L1(Rd). The function f ⋆ g, defined by

(f ⋆ g)(x) =

∫

Rd

f(x− y) g(y) dy

is called the convolution of f and g. It belongs to L1(Rd) and

‖f ⋆ g‖L1(Rd) ≤ ‖f‖L1(Rd) ‖g‖L1(Rd).

Appendix B

Glossary of mathematical terms

English Frencharithmetic (operation) (opération) arithmétique

axiom of choice axiome du choixalmost everywhere (a.e.) presque partout (p.p.)

ball boulebasis base

Banach space espace de Banachbilinear bilinéaire

binomial coefficient coefficient binomialBorel set ensemble borélien

Borel sigma-algebra tribu borélienneboundary value problem problème aux limites

bounded set ensemble bornébounded function fonction bornée

bounded above / bounded from above majorébounded below / bounded from below minoré

boundary bordto cancel s’annulercanonical canonique

Cauchy sequence suite de CauchyCauchy-Schwarz inequality inégalité de Cauchy-Schwarz

celestial motion dynamique célestechain rule dérivation des fonctions composées

characterisation caractérisationcharacteristic function fonction charactéristique

closed set ensemble ferméclosure fermeture

compact set ensemble compactcompactness compacitécomplement complémentaireconformal conformeconsistency consistanceconcavity concavité

completeness complétudecontinuity continuité

109

110 APPENDIX B. GLOSSARY OF MATHEMATICAL TERMS

English Frenchcontraction contraction, fonction contractante

convolution (of functions) (fonctions) convoléesconvolution (as an operation) convolution (comme opération)

to convolve convoluercoordinates coordonnéescorollary corollairecountable dénombrable

counter-example contre-exemplecritical point point critique

to deduce déduiredeterminant déterminantderivative différentielle, dérivée

differentiable différentiabledifferentiation dérivationdiffeomorphism difféomorphisme

Dirac mass, Dirac delta masse de DiracDirichlet problem problème de Dirichlet

discretisation discrétisationdistribution distribution

dominated convergence convergence dominéedynamics dynamiquedynamical dynamique

error analysis analyse d’erreuressentially bounded essentiellement bornée

to establish établirEuclidean space espace euclidienexplicit (scheme) (schéma) explicite

exponent exposantexponentially de manière exponentielle

extended prolongéfactor facteur

field (algebraic object) corpsfield (function) champ

finite element method méthode des éléments finisfiniteness finitude

finite-dimensional de dimension finiefixed point point fixe

form formeFubini’s theorem théorème de Fubini

function application, fonctionGalerkin method méthode de GalerkinHardy inequality inégalité de Hardy

Hamiltonian HamiltonienneHermitian space espace hermitien

Heaviside function fonction de HeavisideHessian (matrix) (matrice) hessienne

Hilbert space espace de HilbertHilbertian hilbertien

Hölder inequality inégalité de Hölderhomeomorphism homéomorphisme

hyperplane hyperplan

111

English Frenchhypothesis hypothèse

implicit (scheme) (schéma) impliciteto imply impliquer

in the sense of au sens deinduced engendré, induit

inner product produit scalaireintegral intégraleinteger entier

integrable intégrableintegration by parts intégration par parties

isomorphism isomorphismeisometry isométrie

index, indices indice, indicesJacobian matrix matrice jacobienne

Jacobian jacobienLax-Milgram (theorem) théorème de Lax-Milgram

lemma lemmeLet . . . be Soit . . .

linear linéaireLipschitz Lipschitzien

lower bound minorationLyapunov function fonction de Lyapunov

mapping application, fonctionmean value theorem théorème des accroissements finis

measurable set ensemble mesurablemeasure mesure

measure space espace mesurémesh maillage

Minkowski inequality inégalité de Minkowskimonotonicity monotonie

neighbourhood voisinagenorm norme

normed norménull (set) (ensemble) négligeableopen set ensemble ouvert

ordinary differential equation équation différentielle ordinaireorder (of approximation) ordre (d’approximation)

orthogonal orthogonalorthonormal orthonormé

outward normal vector vecteur normal sortantparallelogram law/identity formule de la médiane

partial derivative dérivée partiellepartial differential equation équation aux dérivées partielles

partial ordering order partielpiecewise par morceaux

Poincaré inequality inégalité de PoincaréPoisson problem problème de Poisson

positivity positivitépotential energy énergie potentielle

power(s) puissance(s)preceding précédant

probability measure probabilité

112 APPENDIX B. GLOSSARY OF MATHEMATICAL TERMS

English Frenchprobability space espace probabiliséprincipal value valeur principale

quadratic quadratiqueto quotient quotienter

real number, real réelregular régulier

regularity régularitéremainder reste

Riesz representation theorem Théorème de Rieszrobustness robustesse

roots racinesrounding error erreur d’arrondiscalar product produit scalaire

scale échelle(numerical) scheme schéma (numérique)

sequence suite(a) series (une) série

sesquilinear sesquilinéaireset ensemble

sigma-algebra tribusingularity singularité

smooth régulier, lisseSobolev space espace de Sobolev

square (matrix) (matrice) carréesquare-integrable de carré intégrable

stability stabilité(time)step pas (de temps)

step function fonction etagéestiffness matrix matrice de rigidité

subset sous-ensemblesuch that (s.t.) tel que (t.q.)

to suffice suffiresufficiently suffisamentsummable sommablesymmetric symétriquesymmetry symétrie

to tend (to/towards) tendre, converger (vers)test function fonction test

topology topologietrajectory trajectoire

trapezium rule méthode des trapèzestriangle inequality inégalité triangulaire

truncation troncatureuniform uniforme

uniqueness unicitéunits unités

upper bound majorationvariational problem problème variationnel

vector space espace vectorielwith values in à valeurs danswith respect to par rapport à

Young’s inequality inégalité de Young

Bibliography

[1] H. Brézis, Analyse fonctionnelle: théorie et applications, Dunod, 1999.

[2] E. Cancès and A. Ern, Analyse, ENPC lecture notes, Sept. 2011.

[3] E. Cancès and A. Ern, Analyse: Distributions, transformée de Fourier et équationsaux dérivées partielles, in preparation.

[4] A. Ern and J.-L. Guermond, Theory and practice of finite elements, Texts in AppliedMathematics, Vol. 159, Springer, New York, 2004.

[5] A. Ern and G. Stoltz, Calcul Scientifique, ENPC lecture notes, Dec. 2014.

[6] G. Galdi, An introduction to the mathematical theory of the Navier–Stokes equa-tions, volume I, Springer Tracts in Natural Philosophy, Vol. 38, Springer, New York, 1994.

[7] A. Quarteroni and A. Valli, Numerical approximation of partial differential equations,Springer Series in Computational Mathematics, vol. 23, Springer-Verlag, Berlin, 1994.

113

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