The Þnite elemen t appr o x imation - UPMC · The Þnite elemen t appr o x imation ÓPur e mathematicians sometimes ar e satisÞe d with showing that the non-existenc e of a solution

Chapter 7

The finite element approximation

”Pure mathematicians sometimes are satisfied with showingthat the non-existence of a solution implies a logical contra-diction, while engineers might consider a numerical resultas the only reasonable goal. Such one sided views seem toreflect human limitations rather than objective values. Initself mathematics is an indivisible organism uniting theo-retical contemplation and active application.”Richard Courant (1888-1972)

The aim of this chapter is to introduce the basic theory of finite element methods. Nowadays, finite el-ement methods are widely used in almost every field of engineering analysis. The German mathematicianRichard Courant (1888-1972) shall probably be credited for formulating the essence of what is now calleda finite element [Courant, 1943]. The development of these methods became e!ective with the adventof computers and is now recognized as one of the most powerful and versatile method for constructionapproximations of the solutions of boundary-value problems. We will give here a brief overview of thefundamental mathematical ideas that form the core of the method.

Contents7.1 General principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 997.2 The one dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.3 Triangular finite elements in higher dimensions . . . . . . . . . . . . . . . . . 1207.4 The finite element method for the Stokes problem . . . . . . . . . . . . . . . 132

In Chapter 4 we have seen the principle of the variational approximation of elliptic problems. The mainidea of the finite element method is to replace the Hilbert space V in which the variational formulationis posed by a finite dimensional subspace Vh. We will first briefly return to the internal variationalapproximation principle and present finite elements in one dimension.

7.1 General principle

We consider the variational abstract problem introduced in Chapter 4. More precisely, given a Hilbertspace V , a bilinear continuous and V -elliptic form a(·, ·) defined on V ! V and a continuous linear form

99

100 Chapter 7. The finite element approximation

l(·) defined on V !, we consider the variational formulation of a problem:(P) : Find u " V such that a(u, v) = l(v) , for all v " V ,

and we already have stated that this problem has a unique solution using Lax-Milgram Theorem 4.1.

7.1.1 Internal approximation of a variational problem

The internal approximation of problem (P) consists then in replacing the Hilbert space V by a finitedimensional subspace, denoted Vh, in which to find the solution uh. Here, h > 0 represents a parameterrelated to the discretization of the domain (in space or time) and intended to vanish in the theoreticalresults. We assume that for every v " V , there exists an element rh(v) " Vh such that

limh"0

#rh(v)$ v# = 0 .

The bilinear form a(·, ·) and the linear form l(·) are defined on Vh ! Vh and Vh, respectively and theproblem (P) is replaced by the following discrete problem:

(Ph) : Find uh " Vh such that a(uh, vh) = l(vh) , for all vh " Vh.

We introduce the keyword internal related to the approximation because we suppose here that Vh % V .If N = dimVh, we consider a basis (!i)1#i#N of Vh. The decomposition of uh in the basis of Vh,uh =

!Ni=1 ui!i, leads to rewrite the problem (Ph) in the following form:

N"

j=1

uj a(!j , !i) = l(!i) , &1 ' i ' N . (7.1)

Introducing the sti!ness matrix Ah = (aij) " RN,N of coe"cients aij = a(!j , !i), for all 1 ' i, j ' N ,the vector Uh = (ui)1#i#N and the vector Fh = (fi)1#i#N such that fi = l(!i)1#i#N allow us to concludethat we have the equivalence:

uh solution of (Ph) () AhUh = Fh .

Obviously, the V -ellipticity assumption on the bilinear form a implies the existence and the uniquenessof the solution uh to the problem (Ph). However, this assumption is too strong for our purposes. As weare considering a finite dimensional space Vh, it is su"cient to consider that:

(a(vh, vh) = 0) =) (vh = 0) , &vh " Vh ,

Since the bilinear form a is V -elliptic, then the matrix Ah is positive definite (cf. Proposition 4.2).

7.1.2 A priori error estimates

It is interesting to evaluate the error related to the replacement of V by the finite dimensional subspaceVh. To this end, we assume that the problems (P) and (Ph) are both well-posed, in particular that thebilinear form a(·, ·) is continuous (with a continuity constant M > 0) and that there exists a constant"h > 0 such that:

&vh " Vh , a(vh, vh) * "h#vh#2V ,

and we denote by u and uh the respective solutions of problems (P) and (Ph).

Proposition 7.1 Under the previous assumptions, we have the orthogonality identity:

&vh " Vh , a(u$ uh, vh) = 0 .

7.1. General principle 101

Proof. Since Vh % V , we have trivially that a(u, vh) = l(vh) = a(uh, vh), for all vh " Vh. !If the bilinear form a(·, ·) is symmetric, continuous and V -elliptic on V !V of constant ", then it definesan inner product and an energy norm associated to it as:

#u#e = (a(u, u))1/2 , &u " V ,

equivalent to the norm of V :+

"#u#V ' #u#e '##a# #u#V , &u " V .

The approximate solution uh is thus the orthogonal projection for the inner product a(·, ·), of the solutionu on the subspace Vh.

A strong advantage of the internal variational approximation is that it provides an optimal estimateof the error between the exact solution u of the problem (P) and the approximate solution uh of theproblem (Ph). The error #u$uh#V is comparable to the minimum of #u$ vh#V when vh covers Vh. Thiserror estimate in norm # · #V is given by Cea’s lemma (cf. Chapter 4) that we recall here.

Lemma 7.1 (Cea) Under the previous hypothesis, we have the following error estimates.

(i) If the bilinear form is not V -elliptic, we have:

#u$ uh#V '$

1 +#a#"h

%inf

vh$Vh

#u$ vh#V , where #a# = supvh,wh$Vh

a(vh, wh)#vh#V #wh#W

.

(ii) If the bilinear form a(·, ·) is continuous and V -elliptic with a coercivity constant ", then we have:

#u$ uh#V ' M

"inf

vh$Vh

#u$ vh#V .

(iii) If in addition the bilinear form a(·, ·) is symmetric the previous estimate becomes:

#u$ uh#V '&

M

"inf

vh$Vh

#u$ vh#V .

Proof. Let consider wh " Vh. From the orthogonality identity we deduce that

a(uh $ wh, vh) = a(u$ wh, vh) ,&vh " Vh .

and taking into account the continuity of a, we write:

"h#uh $ wh#V ' supvh!Vh

a(uh $ wh, vh)#vh#V

= supvh!Vh

a(u$ wh, vh)#vh#V

' #a##u$ wh#V .

Similarly, we havea(u$ uh, u$ uh) = a(u$ uh, u$ vh) , &vh " Vh ,

and we conclude with the V -ellipticity and the continuity of a. If the form a is symmetric, we can improve theestimate. Since uh is the orthogonal projection of u onto Vh with respect to the inner product induced by a, wehave the Pythagorean relation:

a(u$ uh, u$ uh) = #u$ uh#2e ' #u$ wh#2e = a(u$ wh, u$ wh) , &wh " Vh ,

and we deduce that for all wh " Vh

"#u$ uh#2V ' a(u$ uh, u$ uh) ' a(u$ wh, u$ wh) ' M#u$ wh#2V .

and the results follows. !


Corollary 7.1 Under the previous hypothesis, let (Vh)h denotes a set of finite dimensional subspaces ofV and let us assume that

&v " V , infvh$Vh

#v $ vh#Vh"0$, 0 .

Then, if infh "h > 0, uh converges toward u in V .

The objective for the “ideal” approximation method is to define suitable approximation spaces Vh toapply the Galerkin approach. To this end, we search for a compromise between the dimension N of Vh

(and thus the dimension of the matrix A) and the accuracy of the numerical solution uh. We shall alsoconsider spaces Vh that allow to compute easily the quantities a(!j , !i) and l(!i). Finally, specific spacesVh may result in sparse matrices A where the number of nonzero elements is small, or well-conditionedmatrices with a small condition number and thus easy to solve. In this spirit, the finite element methodtends to answer all these requirements. Before detailling the main concepts of the finite element methods,we give a few words about Ritz and Petrov-Galerkin methods.

7.1.3 Ritz and Petrov-Galerkin methods

The Ritz method

We consider that the hypothesis of the Lax-Milgram theorem are satisfied and we recall that, if thebilinear form a(·, ·) defined on V ! V is symmetric, solving the problem:

(P) find u " V, such that a(u, v) = l(v) , for all v " Vis indeed equivalent to solving the problem:

( 'P) find u " V, such that J(u) = infv$V

J(v), where J(v) =12a(v, v)$ l(v).

In the Ritz method, the space V is replaced by a finite dimensional subspace Vh % V such thatdimVh = N and the approximate solution uh shall solve:

((Ph) find uh " Vh such that J(uh) = infvh$Vh

J(vh).

Theorem 4.2 ensures the existence of a unique solution to this minimization problem as a consequence ofLax-Milgram theorem.

Since the dimension of the space Vh is N , there exists a basis (!j)1#j#N of Vh and every uh " Vh can

be decomposed as uh =N"

j=1

ui!j and we use the classical notation U = (u1, . . . , uN )t " RN .

We consider the one-to-one mapping # : Vh , RN , uh -, U and we pose J = J . #%1 such that forevery uh " Vh we have:

J (U) = J(uh) ,

or, when replacing J(uh) by its value:

J(uh) =12a

)

*N"

j=1

uj!j ,N"

j=1

uj!j

+

,$ l

)

*N"

j=1

uj!j

+

,

=12

N"

i=1

N"

j=1

uiuja(!i, !j)$N"

i=1

uil(!i) .

This leads to a matrix formulation of the minimization functional J(u):

J(uh) =12U t Ah U $ U t Fh = J (U) ,

7.1. General principle 103

where Ah = (aij) " RN,N with aij = a(!i, !j) and Fh = (fi) " RN is such that fi = l(!i). Hence, solvingthe minimization problem ((Ph) is equivalent to solving the following problem:

( -Ph,R) find U " RN such that J (U) = infV $RN

J (V ) , where J (V ) = 12V t Ah V $ V t Fh.

For obvious reasons, the sti!ness matrix Ah is symmetric positive definite and thus the functionalJ is quadratic on RN . This is su"cient to ensure the existence and uniqueness of U " RN solving theminimization problem ( -Ph,R). Furthermore, the solution U of the minimization problem ( -Ph,R) is alsothe solution of the linear system AhU = Fh.

Remark 7.1 When the bilinear form a(·, ·) is symmetric, the Galerkin and Ritz methods are strictlyequivalent.

The Petrov-Galerkin method

The principles of Petrov-Galerkin and Galerkin methods are very similar in the sense that they bothwill attempt to solve the problem (P). However, in the Petrov-Galerkin approach, we consider twofinite-dimensional approximation subspaces Vh and Wh in V such that

dimVh = dimWh = N .

The approximate solution uh is searched in the space Vh but the test functions in the variational formu-lation are now the shape functions of Wh. For these reasons, Vh is called the approximation space andWh is the space of test functions. The problem to solve is now the following:

(Ph,PG) find uh " Vh , such that a(uh, vh) = l(vh) , for all vh " Wh.Suppose (!j)1#j#N is a basis of Vh and ($j)1#j#N a basis of Wh, then every uh " Vh can be decomposed

as uh =N"

j=1

ui!j and we can rewrite the problem as follows:

find uh " Vh , such thatN"

j=1

uja(!j , $i) = l($i) , i = 1, . . . , N ,

And the linear system to solve is AhU = Fh, where Ah = (aij) " RN,N with aij = a(!j , $i) andFh = (fi) with fi = l($i), for all i = 1, . . . , N .

7.1.4 The finite element method

In the finite element method, the domain # is subdivided into a partition or a mesh Th, i.e., a (poten-tially large) collection of geometrically simple elements, and the approximation space Vh is composed ofpiecewise polynomial functions on each element K of the partition Th. We will see in Chapter 9 how toconstruct a partition Th for a domain # of arbitrary geometric shape. The parameter h represents herethe grain of the discretization, i.e., the elementary size of the elements K in Th as defined by:

h = maxK$Th

diam(K) .

Typically, a basis of Vh will be composed of functions whose support is restricted on one or a few elementsof Th and the polynomials are usually of low degree. Hence, when h , 0 the space Vh will better andbetter approximate the space V and the sti!ness matrix A will be sparse, most of its coe"cients beingzeros.


Finite elements vs. finite di!erences or finite volumes

The principle of finite di!erence or finite volume discretization methods is very similar. Both approachesconsider a partition of the domain into a numerous small pieces, although none of them consider avariational formulation of the problem at hand. For instance, let consider the homogeneous Dirichletboundary-value problem in two dimensions:

$$u = f in #u = 0 on %#

With the finite di!erence method, the domain # is covered by a regular uniform grid. At each internalnode xi,j = (ih, jh), we search a discrete value ui,j to approximate u(xi,j) and we assume for examplethat the Laplacian operator is approximated using a 5 points scheme, thus leading to write:

4ui,j $ ui,j+1 $ ui,j%1 $ ui+1,j $ ui%1,j

h2= f(xi,j) .

In the finite volume method, the partition Th of the domain # is arbitrary and unknowns are associatedwith each element K " Th. Using Green’s identity, we can rewrite the previous equation as follows:

$.

!K

%u

%n=

.

Kf , &K " Th

and we discretize the left-hand side term using a formula mixing the unknowns on K and on the neigh-boring elements. Considering a square domain, we have:

Ki,j = [(i$ 1/2)h, (i + 1/2)h]! [(j $ 1/2)h, (j + 1/2)h] ,

and if ui,j is the approximation of u on Ki,j , the flux integrated on the interface Ki,j / Ki+1,j is thendiscretized by ui+1,l $ uk,l. Repeating the procedure for the other fluxes and approximating the term/Ki,j

f by h2f(xi,j) yields the equation:

4ui,j $ ui,j+1 $ ui,j%1 $ ui+1,j $ ui%1,j = h2f(xi,j) .

The resulting linear system is very similar to that obtained using a finite di!erence scheme.Actually, the vast majority of finite di!erence methods can be deduced from finite element methods if

the problem at hand has a variational formulation. It is less obvious for finite volume methods. Moreover,we have strong theoretical mathematical tools to study finite element methods. In addition, the latterhave several intrinsic advantages:

1. the versatility of the formulation on arbitrarily complex geometries, and the possibility to locallyrefine the partition Th to approximate solutions with singularities,

2. the boundary conditions are naturally taken into account in the space V in the variational formu-lation and in its internal approximation Vh,

3. the general framework of the variational approximations is convenient for the error analysis.

Other variational methods have been developed, like spectral methods, that are especially adapted tothe approximation of smooth solutions but are limited to simple geometries and methods using waveletsbasis.

7.2. The one dimensional case 105

7.2 The one dimensional case

At first, we introduce the general principle of the Lagrange finite element method in one dimension ofspace. Without loss of generality, we can restrict our study to the unit domain # =]0, 1[. To set theideas, we will also consider the following boundary-value problem:

Given f " L2(#) and c " L&(#), find the function u solving:0$u!!(x) + c(x)u(x) = f(x) , x "]0, 1[

u(0) = u(1) = 0 .(7.2)

Here, a mesh is simply a set of points (xj)0#j#N+1 or intervals Kj = [xj , xj+1] such that 0 = x0 <x1 < · · · < xN+1 = 1. The mesh is said to be uniform if the points (xj) are equidistributed along thesegment [0, 1], i.e. such that xj = jh, with h = 1/(N + 1), 0 ' j ' N + 1. More generally, we denote byh = max |xj+1 $ xj | the size parameter.

7.2.1 Lagrange P1 elements

The finite element methods for Lagrange P1 elements involves the space of globally continuous a"nefunctions on each interval:

V 1h = {vh " C0([0, 1]) , vh|Kj " P1 , 0 ' j ' N} ,

and the subspace of V 1h :

V 10,h = {vh " V 1

h , such that vh(0) = vh(1) = 0} ,

More generally, Pk denotes the vector space of polynomials in one variable and of degree less than orequal to k:

Pk =

12

3p(x) =k"

j=0

"jxj "j " R

45

6 .

The finite element method consists in applying the internal variational approximation approach to thespaces V 1

h and V 10,h. In this context, the functions of V 1

h can be represented using very simple shapefunctions.

Lemma 7.2 The space V 1h is a subspace of H1(#) of dimension N + 2. Every function vh of V 1

h isuniquely determined by its values at the mesh vertices (xj)0#j#N+1:

vh(x) =N+1"

j=0

vh(xj)!j(x) , &x " [0, 1] ,

where (!j)0#j#N+1 is the basis of the shape functions !j with compact support in each interval [xj%1, xj+1]defined as:

!j(x) =

172

73

x$ xj%1

hx " [xj%1, xj ]

xj+1 $ x

hx " [xj , xj+1]

such that !j(xi) = &ij . (7.3)


x0 xj+1xj

!j

xj!1 xN+1

!N+1!01

Figure 7.1: Global shape functions for the space V 1h .

Proof. We know that piecewise C1 continuous functions belong to the space H1(#). Hence, Vh is a subspace ofH1(#). Moreover, since we have !j(xi) = &ij , where &ij is the Kronecker symbol, the result follows. !

Remark 7.2 Notice that the functions (!j)0#j#N+1 can be expressed using only two functions:

'0(x) = 1$ x , '1(x) = x .

The basis functions are defined as the composition of a shape function of a reference finite element (i.e.that depends only of the polynomial approximation) and of an a"ne transformation (depending only onthe discretization) as, for all 0 ' j ' N :

!j(x) =

1772

773

'1

$x$ xj%1

xj $ xj%1

%, x " [xj%1, xj ]

'0

$x$ xj

xj+1 $ xj

%, x " [xj , xj+1]

and !N+1(x) = '1 ((x$ xN )/h). This will be useful to compute the coe"cients of the matrix in the linearsystem to solve.

Corollary 7.2 The space V 10,h is a subspace of H1

0 (#) of dimension N and every function vh of V 10,h is

uniquely determined by its values at the mesh vertices (xj)1#j#N :

vh(x) =N"

j=1

vh(xj)!j(x) , &x " [0, 1] ,

Remark 7.3 Notice that functions vh " V 1h are not twice di!erentiable on # and thus it is meaningless

to attempt solving the problem (7.2) as the second derivative of any vh " V 1h is a sum of Dirac masses

at the mesh vertices. However, it is meaningful to solve a variational formulation of this problem withfunctions vh " V 1

h since only the first derivatives are involved.

The variational formulation of problem (7.2) consists in finding u " H10 (#), such that:

.

!u!(x)v!(x) dx +

.

!c(x)u(x)v(x) dx =

.

!f(x)v(x) dx , &v " H1

0 (#) , (7.4)

and the variational formulation of the internal approximation consists in finding uh " V0,h, such that:.

!u!h(x)v!h(x) dx +

.

!c(x)uh(x)vh(x) dx =

.

!f(x)vh(x) dx , &vh " V0,h . (7.5)


Introducing the notation uh(xj)1#j#N for the approximate value of the exact solution at the mesh vertexxj , leads to the approximate problem:

find uh(x1), . . . , uh(xN ) such that for all i = 1, . . . , N

N"

j=1

$.

!!!j(x)!!i(x) dx +

.

!c(x)!j(x)!i(x) dx

%uh(xj) =

.

!f(x)!i(x) dx .

And as expected, this formulation is equivalent to solving in RN the linear system:

AhUh = Fh

where Uh = (uh(xj))1#j#N , Fh =8/

! f(x)!i(x) dx91#i#N

and the matrix Ah is defined as:

Ah =$.

!!!j(x)!!i(x) dx +

.

!c(x)!j(x)!i(x) dx

%

1#i,j#N

.

Coe"cients of the matrix Ah

Actually, the matrix Ah appears as the sum of the sti!ness matrix Kh defined by its coe"cients(kij)1#i,j#N :

kij =.

!!!j(x)!!i(x) dx =

N"

k=0

. xk+1

xk

!!j(x)!!i(x) dx ,

and the mass matrix Mh defined by its coe"cients (mij)1#i,j#N :

mij =.

!c(x)!j(x)!i(x) dx =

N"

k=0

. xk+1

xk

c(x)!j(x)!i(x) dx .

Since the shape functions !j have a small support, most of the coe"cients in Ah are zeros. More precisely,for a given index i, there is only three consecutive values of j such that the coe"cient aij is potentiallynot equal to zero. The structure of the matrice is then easy to deduce: Ah is a tridiagonal matrix. Thecoe"cients of Ah are thus given by:

ajj =. xj+1

xj!1

(!!j(x))2 dx +. xj+1

xj!1

c(x)(!j(x))2 dx

ajj%1 =. xj

xj!1

!!j(x)!!j%1(x) dx +. xj

xj!1

c(x)!j(x)!j%1(x) dx

ajj+1 =. xj+1

xj

!!j(x)!!j+1(x) dx +. xj+1

xj

c(x)!j(x)!j+1(x) dx

For the sake of simplicity, we consider here the function c as being constant, c(x) = c0 for all x " #.Hence, we write:

mjj = c0

. xj

xj!1

!j(x)!j%1(x) dx = c0

. xj

xj!1

'1

$x$ xj%1

h

%'0

$x$ xj%1

h

%

= c0h

. 1

0'1(y)'0(y) dy = c0h

. 1

0(1$ y)y dy = c0

h

6,

where we posed y = (x$ xj%1)/h. Finally, we find the coe"cients of Ah:

ajj%1 = $1h

+ c0h

6ajj =

2h

+ c02h

3and ajj+1 = $1

h+ c0

h

6,


Remark 7.4 Instead of regarding the node contributions, we could have analyzed the elements. Considerthe element Kj = [xj , xj+1]; on this element there is only two non-zero shape functions:

!j |Kj =xj+1 $ x

xj+1 $ xj=

xj+1 $ x

h!j+1|Kj =

x$ xj

xj+1 $ xj=

x$ xj

h

!!j |Kj =$1

xj+1 $ xj=$1h

!!j+1|Kj =1

xj+1 $ xj=

1h

Then, we can arrange the elementary contributions of the element Kj to the sti!ness matrix and to themass matrix as 2! 2 symmetric matrices EKj and EMj:

EKj =

:kj

11 kj12

kj21 kj

22

;and EMj =

:mj

11 mj12

mj21 mj

22

;

with

kj11 =

. xj+1

xj

(!!j(x))2 dx ,

=. xj+1

xj

1h2

dx =1h

kj12 = kj

21 =. xj+1

xj

!!j(x)!!j+1(x) dx ,

=. xj+1

xj

$ 1h2

dx = $1h

kj22 =

. xj+1

xj

(!!j+1(x))2 dx

=. xj+1

xj

1h2

dx =1h

mj11 =

. xj+1

xj

c(x)(!j(x))2 dx , mj12 = mj

21 =. xj+1

xj

c(x)!j(x)!j+1(x) dx , mj22 =

. xj+1

xj

c(x)(!j+1(x))2 dx

and thus to conclude that:

EKj =1h

$1 $1$1 1

%and EMj = c0

h

6

$2 11 2

%

We will see that this point of view is more practical when dealing with the matrix assembly, especially inhigher dimensions.

Matrix assembly

The assembly of the matrix Ah is easy and is obtained algorithmically using a loop over all mesh elementsKj and adding their contributions to the right coe"cients of the global system. Assuming a(i, j) denotesthe coe"cients aij of Ah, a pseudo-code to perform this task would be:

for k=1, N+1 do // loop over all elementsfor i=1,2 do // local loop

for j=1,2 doig = k+i-2 // global indicesjg = k+j-2

A(ig,jg) = A(ig,jg) + a(i,j)end loop j

end loop iend loop k

The numerical resolution of the linear system is by far the most computationally expensive part of themethod. We refer the reader to Chapter 5 for more details about the direct and indirect techniques tosolve this system.


Coe"cients of the right-hand side Fh

Each component fi of the vector Fh " RN is obtained as:

fi =N"

k=0

. xk+1

xk

f(x)!i(x) dx .

Usually, the function f is not known analytically. Hence, we decompose f in the basis of the shapefunctions (!j)1#j#N :

f(x) =N%1"

j=1

fj!j(x) dx

and the problem is reduced to the evaluation of the integrals:. xk+1

xk

!j(x)!i(x), dx .

We use for instance the trapeze formula:. xk+1

xk

((x) dx =xk+1 $ xk

2(((xk+1) + ((xk)) ,

that gives the exact result for polynomial of degree one and leads here fj = hf(xj); or the Simpsonformula: . xk+1

xk

((x) dx =xk+1 $ xk

6(((xk+1) + 4((xk+1/2) + ((xk)) ,

that gives the exact result for polynomials of degree lesser than or equal to 3, and leads here to

fj =h

6(f(xj) + 4f(xj+1/2) + f(xj+1)) .

Neumann boundary-value problem

The finite element method can be applied to solve the Neumann boundary-value problem:Given f " L2(#), c " L&(#) such that c(x) * c0 > 0 almost everywhere in # and ",) " R, find the

function u solving: 0$u!!(x) + c(x)u(x) = f(x) , x " #u!(0) = " u!(1) = ) .

(7.6)

in a very similar manner. Recall that this problem has a unique solution u " H1(#) (cf. Chapter 4).The variational formulation of the internal approximation consists here in finding uh " V 1

h such that.

!u!h(x)v!h(x) dx +

.


.

!f(x)vh(x) dx$"vh(0) + )vh(1) , &vh " V 1

h (#) . (7.7)

The variational formulation consists in solving in RN+2 the linear system:

AhUh = Fh

with Uh = (uh(xj))0#j#N+1 and the sti!ness matrix Ah is defined as:

Ah =$.

!!!j(x)!!i(x) dx +

.

!c(x)!j(x)!i(x)) dx

%

0#i,j#N+1

.


and Fh = (fj)0#j#N+1 such that:

fj =.

!f(x)!j dx 1 ' j ' N

f0 =.

!f(x)!0(x) dx$ "

fN+1 =.

!f(x)!N+1(x) dx + ) .

7.2.2 Convergence of the Lagrange P1 finite element method

Definition 7.1 (Interpolation) The linear mapping %h : H1(#) , V 1h defined for every v " H1(#)

as:

(%hv)(x) =N+1"

j=0

v(xj)!j(x) , &x " [0, 1] ,

is called P1 interpolation operator. Furthermore, for every v " H1(#), the interpolation operator is suchthat:

limh"0

#v $%hv#H1(!) = 0 .

The P1 interpolate of a function v is the unique piecewise a"ne function that coincide with v at the meshvertices xj . The convergence of the finite element method is related to a series of results that we givehere.

Suppose that the function v is su"ciently smooth, i.e. v " H2(#). Since the derivative of the a"nefunctions in Vh is constant on the intervals Kj = [xj , xj+1], we have then:

(%hv)!(x) =v(xj+1)$ v(xj)

h=

1h

. xj+1

xj

v!(t) dt , &x " [xj , xj+1] .

Since we assumed v " H2(#) then v! " H1(#) and thus v is a continuous function. Using Rolle’s theorem,we deduce that there exists a point (j " [xj , xj+1] such that:

v!((j) =1h

. xj+1

xj

v!(t) dt = (%hv)!(x) , &x " [xj , xj+1] .

We will search for an estimate on #v $%hv#H1(!). To this end, we write:

#v $%hv#2H1(!) =. 1

0|v! $%!hv|2 =

N%1"

j=1

. xj+1

xj

|v!(t)$ v!((j)|2 dt , (7.8)

however, for all t " [xj , xj+1] we have:

v!(t)$ v!((j) =. t

"j

v!!(t) dt ,

hence, using Cauchy-Schwarz’s identity, we write:

|v!(t)$ v!((j)|2 '. t

"j

|v!!(t)|2 dt |t$ (j |

'. xj+1

xj

|v!!(t)|2 dt |t$ (j | .


By integrating on the interval [xj , xj+1] yields:

. xj+1

xj

|v!(t)$ v!((j)|2 dt '. xj+1

xj

|t$ (j | dt

:. xj+1

xj

|v!!(t)|2 dt

;

' (xj+1 $ xj)2

2

. xj+1

xj

|v!!(t)|2 dt

' h2

2

. xj+1

xj

|v!!(t)|2 dt .

And going back to equation (7.8), we have now:

#v $%hv#H1(!) 'h2

2

N%1"

j=1

. xj+1

xj

|v!!(t)| dt

' h2

2#v!!#2L2(!) .

This leads to enounce the following result, that we already partly proved.

Lemma 7.3 (Interpolation error) If v " H2(#) then, there exists two constant C1 and C2 indepen-dent of h such that:

#v $%hv#H1(!) ' C1 h2 #v!!#L2(!) and #v! $ (%hv)!#L2(!) ' C2 h #v!!#L2(!) .

And we can establish the convergence of the finite element method for the Dirichlet boundary-valueproblem as follows.

Theorem 7.1 (Convergence) Suppose u " H10 (#) and uh " V0,h are the solutions of (7.2) and (7.5),

respectively. Then, the Lagrange P1 finite element method converges, i.e. we have:

limh"0

#u$ uh#H1(!) = 0 .

Furthermore, if u " H2(#) then, there exists a constant C independent of h such that:

#u$ uh#H1(!) ' Ch#f#L2(!) .

Proof. Here, since the bilinear form a(·, ·) is V0,h-elliptic, we can consider the ellipticity constant " = 1:

a(u, u) =. 1

0u"2(x) + cu2(x) dx *

. 1

0u"2(x) dx = #u#2H1

0 (!) ,

and regarding the continuity of a(·, ·) we write:

|a(u, v)| '<#u"#2L2(!) + c1#u#2L2(!)

=1/2

' (1 + c1)#u#H10 (!) #v#H1

0 (!) ,


where we assumed that 0 ' c1 = supx![0,1] c(x) ' +0. Thus, the continuity constant M is taken here as 1 + c1.Using Cea’s lemma 7.1, we can easily conclude that:

#u$ uh#H10 (!) '

+1 + c1 inf

vh!V0,h

#u$ vh#H10 (!) . (7.9)

If we assume v " H2(#), we can write, according to the previous lemma:

#u$%hu#2H10 (!) '

h2

2#u""#2L2(!)

' h2

2#u#2H2 ' C

h2

2#f#2L2(!) .

Moreover, since %hu " V0,h, we have also:

infvh!V0,h

#u$ vh#H10 (!) ' #u$%hu#H1

0 (!) .

and using the inequality (7.9), we can conclude. !

Lemma 7.4 There exists a constant C independent of h such that for all v " H1(#):

#%hv#H1(!) ' C#v#H1(!) , and #v $%hv#L2(!) ' Ch#v!#L2(!) .

Furthemore, for all v " H1(#), we have:

limh"0

#v! $ (%hv)!#L2(!) = 0 .

Proof. Given v " H1(#), we have:

#%hv#L2(!) ' supx!!

|%hv(x)| ' supx!!

|v(x)| ' C#v#H1(!) .

Moreover, since %hv is an a"ne function, we have by Cauchy-Schwarz’s identity:

. xj+1

xj

|(%hv)"(t)|2 dt =(v(xj+1)$ v(xj))2

h=

1h

:. xj+1

xj

v"(t) dt

;2

'. xj+1

xj

|v"(t)|2 dt .

and we obtain the first identity by summation over j. Similarly, we write:

|v(x)$%hv(x)| ' 2. xj+1

xj

|v"(t)|dt .

We obtain the second identity by using Cauchy-Schwarz, by integrating with respect to x and then by summationover j.Since C#(#) is dense in H1(#), for every v " H1(#) there exists w " C#(#) such that

#v" $ w"#L2(!) ' * , for * > 0 .

Since %h is a linear mapping verifying the first identity, we have then:

#(Pihv)" $ (%hw)"#L2(!) ' C#v" $ w"#L2(!) ' C* .

From Lemma 7.3, we deduce that, for h su"ciently small:

#w" $ (%hw)"#L2(!) ' * .

We can the write, by adding the last identities:

#v" $ (%hv)"#L2(!) ' #v" $ w"#L2(!) + #w" $ (%hw)"#L2(!) + #(%hv)" $ (%hw)"#L2(!) ' C* ,

and the result follows. !


7.2.3 Lagrange P2 elements

Before introducing the Lagrange P2 finite element method, we like to describe the advantage of consideringhigher-order polynomials on an example taken from [Solin, 2005].

Motivation for high-order elements

Consider the simple homogeneous Poisson boundary-value problem in one dimension of space:0

$u!!(x) = f(x) , in # =]$ 1, 1[ ,u(0) = u(1) = 0 .

, with f(x) =+2

4cos

<+x

2

=.

The exact solution to this problem has the form:

u(x) = cos<+x

2

=.

The well-known variational formulation of this problem consists in: finding u " H10 (#) such that:

.

!u!(x)v!(x) dx =

.

!f(x)v(x) dx , &v " H1

0 (#)

Suppose the domain is decomposed into two intervals [$1, 0] and [0, 1] and let consider the finite elementspace V 1

0,h generated by a single piecewise a"ne function vh defined as:

vh(x) =

0x + 1 , x " [$1, 0]1$ x , x " [0, 1]

The exact solution u and the approximate solution uh are given Figure 7.2, left. The approximation errorin H1 seminorm is:

|u$ uh|1,2 =$.

!|u!(x)$ u!h(x)|2 dx

%1/2

1 0.683667 .

On the other hand, assume a single quadratic element covers the domain [$1, 1]. A basis of the finiteelement space V 2

0,h is composed of the function vh(x) = 1$x2. The exact solution u and the approximatesolution uh are given Figure 7.2, right. The approximation error is then:

|u$ uh(x)|1,2 1 0.20275 .

-1 -0.5 0 0.5 10

0.2

0.4

0.6

0.8

1 u

1-x

x+1

-1 -0.5 0 0.5 10

0.2

0.4

0.6

0.8

1 u

1-x^2

Figure 7.2: .Exact solution u with piecewise a"ne approximation uh (left-hand side) and with quadraticapproximation uh (right-hand side).


This is a clear indication that high-order finite elements are better to approximate smooth functions.Conversely, less regular functions can be approximated accurately using lower degree finite elements.This will be emphasized by Theorem 7.2.

Lagrange P2 elements

We return to problem (7.2) and we consider a set of points (xj)0#j#N+1 or intervals Kj = [xj , xj+1]forming a uniform mesh of #. The finite element method for Lagrange P2 elements involves the discretespace:

V 2h = {vh " C0([0, 1]) , vh|Kj " P2 , 0 ' j ' N} ,

and its subspace:V 2

0,h = {vh " V 2h , such that vh(0) = vh(1) = 0} .

These spaces are composed of continuous, piecewise parabolic functions (polynomials of degree less thanor equal to 2). The P2 finite element method consists in applying the internal variational approximationapproach to these spaces.

Lemma 7.5 The space V 2h is a subspace of H1(#) of dimension 2N + 3. Every function vh " V 2

h isuniquely defined by its values at the mesh vertices (xj)0#j#N+1 and at the midpoints (xj+1/2)0#j#N =(xj + h

2 )0#j#N :

vh(x) =N+1"

j=0

vh(xj)!j(x) +N"

j=0

vh(xj+1/2)!j+1/2(x) , &x " [0, 1] .

where (!j)0#j#N+1 is the basis of the shape functions !j defined as:

!j(x) = ,

$x$ xj

h

%, 0 ' j ' N + 1 and !j+1/2(x) = $

$x$ xj+1/2

h

%, 0 ' j ' N ,

with

,(x) =

172

73

(1 + x)(1 + 2x) $ 1 ' x ' 0(1$ x)(1$ 2x) 0 ' x ' 1

0 |x| > 1 ,

and $(x) =

01$ 4x2 |x| ' 1/2

0 |x| > 1/2

Remark 7.5 Notice that we have:

!i(xj) = &ij !i(xj+1/2) = 0

!i+1/2(xj) = 0 !i+1/2(xj+1/2) = &ij

Corollary 7.3 The space V 20,h is a subspace of H1

0 (#) of dimension 2N +1 and every function vh " V 20,h

is uniquely defined by its values at the mesh vertices (xj)1#j#N and at the midpoints (xj+1/2)0#j#N :

vh(x) =N"

j=1

vh(xj)!j(x) +N"

j=0

vh(xj+1/2) !j+1/2(x) , &x " [0, 1] .


xj+1xj!1

!j+1/2

xj!1/2 xj+1/2

1!j

xj

Figure 7.3: Global shape functions for the space V 2h .

The variational formulation of the internal approximation of the Dirichlet boundary-value problem (7.2)consists now in finding uh " V 2

0,h, such that:

.

!u!h(x)v!h(x) dx +

.


.

!f(x)vh(x) dx , &vh " V 2

0,h . (7.10)

Here, it is convenient to introduce the notation (xk/2)1#k#2N+1 for the mesh points and (!k/2)1#k#2N+1

for the basis of V 20,h. Using these notations, we have:

uh(x) =2N+1"

k=1

uh(xk/2)!k/2(x) .

This formulation leads to solve in R2N+1 a linear system:

AhUh = Fh ,

where Uh = (uh(xk/2))1#k#2N+1 and it is easy to see that the matrix Ah of the linear system to solve isnow defined as:

Ah =$.

!!!k/2(x)!!l/2(x) dx +

.

!c(x)!k/2(x)!l/2(x) dx

%

1#k,l#2N+1

,

and the right-hand side term becomes:

Fh =$.

!f(x)!k/2(x) dx

%

1#k#2N+1

.

Since the shape functions !j have a small support, the matrix Ah is mostly composed of zeros. However,the main di!erence with the Lagrange P1 finite element method, the matrix Ah is no longer a tridiagonalmatrix.

Coe"cients of Ah

The coe"cients of the matrix Ah can be coputed more easily by considering the following change ofvariables, for t " [$1, 1]:

x = xj+1 +xj+2 $ xj

2t = xj+1 +

h

2t , &x " [xj , xj+2] , 0 ' j ' 2N $ 1 .


Hence, the shape functions can be reduced to only three basic shape functions (Figure 7.4):

'%1(t) =t(t$ 1)

2'0(t) = $(t$ 1)(t + 1) '1(t) =

t(t + 1)2

,

and their respective derivatives:

d'%1

dt(t) =

2t$ 12

d'0

dt(t) = $2t

d'1

dt(t) =

2t + 12

.

This approach consists in considering all computations on an interval Kj = [xj , xj+2] on the referenceinterval [$1, 1]. Thus, we have:

d!j

dx=

d'k

dt

dt

dx,

where 'k " [$1, 1]. In this case, the elementary contributions of the element Kj to the sti!ness matrixand to the mass matrix are given by the 3! 3 matrices EKj and EMj :

EKj =13h

)

*7 $8 1$8 16 $81 $8 7

+

, EMj = c0h

30

)

*4 2 $12 16 2$1 2 4

+

, .

-1 -0.5 0 0.5 1

0

0.2

0.4

0.6

0.8

1

Figure 7.4: The three quadratic Lagrange P2 shape functions on the reference interval [$1, 1].

Matrix assembly

for k=1, N do // loop over all elementsfor i=1,3 do // local loop

for j=1,3 doig = 2*k+i-3 // global indicesjg = 2*k+j-3

A(ig,jg) = A(ig,jg) + a(i,j)end loop j

end loop iend loop k


Coe"cients of the right-hand side Fh

Usually, the function f is only known by its values at the mesh points (xj)0#j#2N and thus, we use thedecomposition of f in the basis of shape functions (!j)0#j#2N :

f(x) =2N"

j=0

f(xj)!j(x) dx .

Each component fi of the right-hand side vector is obtained as:

fi =N"

k=1

. x2k

x2k!2

f(x)!i(x) dx .

Using the previous decomposition of f , we obtain:

fi =2N"

j=0

fj

:N"

k=1

. x2k

x2k!2

!j(x)!i(x) dx

;,

and the problem is reduced to computing the inmtegrals:. x2k

x2k!2

!j(x)!i(x) dx, .

It is easy to see that we obtain expressions very similar to that of the mass matrix. More precisely, theelement Kj = [xi, xi+2] will contribute to only three components of indices i, i+! and i + 2 as:

)

*fk

ifk

i+1fk

i+2

+

, =h

30

)

*4 2 $12 16 2$1 2 4

+

,

)

*fi

fi+1

fi+2

+

, ,

where fki denotes the contribution of element k to the component i.

7.2.4 Convergence of the Lagrange P2 finite element method

We rely on Cea’s lemma that provides an estimate of the error:

#u$ uh#H1(!) '&

M

"inf

vh$V 10,h

#u$ vh#H1(!) ,

for a continuous and V -elliptic bilinear form a(·, ·) defined on V 1h ! V 1

h . Suppose now that f " H1(#),then u " H3(#) and if the function c is su"ciently smooth then:

#u#H3(!) ' C #f#H1(!) .

In order to find an upper bound on the right-hand side of the previous estimate, we introduce a mappingwh " V 2

0,h such that for all 1 ' i ' N , wh(xi) = u(xi) and such that wh|[xi,xi+1] is a polynomial of degreetwo or less. To this end, on each interval [xi, xi+1], we consider the polynomial functions:

wi,1(x) ="i

2(x$ xi)2 , wi,2(x) = wi,1(x) + )i(x$ xi) ,


where"i =

1h

. xi+1

xi

u!!(t) dt , )i =1h

. xi+1

xi

(u$ wi,1)!(t) dt

By definition, we have:. xi+1

xi

(u$ wi,1)!!(t) dt = 0 ,

. xi+1

xi

(u$ wi,2)!!(t) dt = 0 ,

. xi+1

xi

(u$ wi,2)!(t) dt = 0 .

Hence, from this relation, we deduce that, for every 0 ' i ' N :

u(xi)$ wi,2(xi) = u(xi+1)$ wi,2(xi+1) .

This allow us to define the polynomial function wh on [0, 1] as follows:

wh(x) = wi,2(x) + (u(xi)$ wi,2(xi)) , & 0 ' i ' N ,

and the previous relations show that wh is defined and continuous on [0, 1], that wh(xi) = u(xi) for all0 ' i ' N and that wh is a polynomial of degree 2 on each [xi, xi+1]. We conclude easily that:

#u$ uh#H1(!) '&

M

"inf

vh$V 10,h

#u$ vh#H1(!) '&

M

"#u$ wh#H1(!) .

Introducing the notation rhu = u $ wh, we can see from the previous identities that rhu " H3(#) andrhu|[xi,xi+1] " H1

0 ([xi, xi+1]). Furthermore, we have:. xi+1

xi

(rhu)!!(t) dt = 0 , and. xi+1

xi

(rhu)!(t) dt = 0 .

To achieve the estimate of Rhu, we introduce a result known as Poincare-Wirtinger inequality.

Lemma 7.6 (Poincare-Wirtinger inequality) Given a bounded interval [a, b] of R, we pose

W[a,b] =>

u " H1([a, b]) ,

. b

au(t) dt = 0

?.

Then, we have: . b

au2(t) dt ' b$ a

2

. b

au!(t) dt , & u " W[a,b] .

Using this result and the previous identities, we deduce that r!!hu " W[xj ,xi+1], for all 0 ' i ' N and thus:

. xi+1

xi

|(rhu)!!(t)|2 dt ' h2

2

. xi+1

xi

|(rhu)!!!(t)|2 dt

' h2

2

. xi+1

xi

|u!!!(t)|2 dt

.

Similarly, we can deduce that:. xi+1

xi

|(rhu)!(t)|2 dt ' h2

2

. xi+1

xi

|(rhu)!!(t)|2 dt

' h4

4

. xi+1

xi

|u!!!(t)|2 dt

.


Adding these last two results yields:

#rhu#2H10 (!) =

. xi+1

xi

|(rhu)!(t)| dt ' h4

4#u!!!#L2(!) .

Finally, we can enounce the convergence result as follows.

Theorem 7.2 (Convergence) Suppose u " H10 (#) and uh " V 2

0,h are the solutions of (7.2) and (7.10),respectively. Then, the Lagrange P2 finite element method converges, i.e. we have:

limh"0

#u$ uh#H1(!) = 0 .

Furthermore, if u " H3(#) (i.e. f " H1(#)), then there exists a constant C independent of h such that:

#u$ uh#H1(!) ' C h2 #u!!!#L2(!) .

Remark 7.6 The convergence rate of the P2 finite element method is better than with the P1 finiteelement method. However, the data f must be su"ciently smooth, here f " H1.

7.2.5 Lagrange Pk elements

This section generalizes the concepts introduced in the previous sections to the interpolation of continuousand polynomials functions of degree k * 1. We consider a set of points (xj)0#j#N+1 or intervals Kj =[xj , xj+1] forming a uniform mesh of # =]0, 1[.

Lagrange finite element spaces

For a given integer k * 1, we define the space of globally continuous functions on [0, 1] whose restrictionon each interval Kj = [xj , xj+1] is a polynomial of degree k:

V kh = {vh " C0([0, 1]) , vh|Kj " Pk , 0 ' j ' N} ,

and the subspace of V kh :

V k0,h = {vh " V k

h , such that vh(0) = vh(1) = 0} .

In each interval Kj , a function of V kh is uniquely determined by its values at k + 1 distinct points along

the segment. Hence, on each interval Kj , we introduce a set of nodes:

yj,l = xj +l

k(xj+1 $ xj) = xj +

l

khj , 0 ' l ' k $ 1 ,

and yN+1,0 = xN+1.

Lemma 7.7 The space V kh is a subspace of H1(#) of dimension k(N + 1) + 1. Every function vh of V k

his uniquely determined by its values at the mesh nodes (yj,l)0#j#N,0#l#k%1 and yN+1,0. Furthermore, theshape functions are such that:

!j,l(yj",l") = &jj"&ll" .

The space V k0,h is a subspace of H1

0 (#) of dimension k(N + 1)$ 1.


Convergence of the Lagrange Pk finite element method

We can consider the internal approximation problem of finding uh " V k0,h such that:

.

!u!h(x)v!h(x) dx +

.


.

!f(x)vh(x) dx , & vh " V k

0,h . (7.11)

Following the same analysis than for the Lagrange P2 element, we have the following convergence result.

Theorem 7.3 (Convergence) Suppose u " H10 (#) and uh " V k

0,h are the solutions of (7.2) and (7.11),respectively. Then, the Lagrange Pk finite element method converges, i.e. we have:

limh"0

#u$ uh#H1(!) = 0 .

Furthermore, if u " Hk+1(#) (i.e. f " Hk%1(#)), then there exists a constant C independent of h suchthat:

#u$ uh#H1(!) ' C hk #f#Hk!1(!) .

Remark 7.7 1. The approximate solution uh converges toward the exact solution u in H1(#) whenh , 0. The Lagrange Pk method is of order k in h, if the function f is su"ciently smooth.

2. The matrix assembly becomes more and more di"cult as the value of k increases. Since the size ofthe problem increases as well, this may result in additional di"culties in solving the resulting linearsystem.

3. The computation of the components of the right-hand side vector Fh must be carried out with asu"ciently accurate method.

7.3 Triangular finite elements in higher dimensions

We consider here a boundary-value problem posed in an open bounded domain # % Rd (d = 2, 3, ingeneral). For the sake of simplicity, we restrict our study to domains with piecewise polygonal (resp.polyhedral when d = 3) boundaries, i.e. # can be exactly covered by a finite union of polygons (resp.polyhedra).

To set the ideas, we consider the homogeneous boundary-value problem (7.2) posed here in an openbounded domain # of R2:

Given f " L2(#) and c " L&(#), find u such that:0$$u + cu = f , in #

u = 0 , on %#(7.12)

We already know that this problem has a unique solution u " H10 (#).

7.3.1 Preliminary definitions

We proceed like in one dimension of space. The domain # is decomposed into a set of N finite elements,triangles (Kj)1#j#N in dimension d = 2 and tetrahedra in dimension d = 3. These two types of elementsbelong to the generic class of simplices.

7.3. Triangular finite elements in higher dimensions 121

d-Simplices

Definition 7.2 A d-simplex K is the convex hull (envelope) of d + 1 points (aj)1#j#d+1 in Rd, calledthe vertices of K, that are not all lying in the same hyperplane. It is the smallest convex passing throughall these points.

Remark 7.8 Let consider d + 1 points (aj)1#j#d+1 in Rd and let denote (ai,j)1#i#d the coordinates ofvector (aj). These points are a"nely independent, i.e. not lying in the same hyperplane, if the matrix

M =

)

@@@@@*

a1,1 a1,2 . . . a1,d+1

a2,1 a2,2 . . . a2,d+1...

... . . . ...ad,1 ad,2 . . . ad,d+1

1 1 . . . 1

+

AAAAA,

is invertible. In such case, the simplex is not degenerated. Any d-simplex has the same number of facesand vertices, each face being itself a d$ 1-simplex.

Furthermore, a few geometric parameters characterize a simplex K:

(i) the diameter hK : the length of the largest element edge,

(ii) the roundness -K : the diameter of the largest inscribed ball,

(iii) the aspect ratio .K =hK

-K: a measure of the non-degeneracy of K.

Barycentric coordinates

Any simplex K can be represented by the barycentric coordinates {/j}1#j#d+1 of its vertices. Barycentriccoordinates are a form of homogeneous coordinates.

Definition 7.3 For every 1 ' j ' d+1, the barycentric coordinate /j of a point x " Rd is the first-degreepolynomial:

/j(x) = c1x1 + · · · + cdxd + cd+1 ,

such that for all 1 ' i ' d + 1, /j(ai) = &ij.

For each j, the d + 1 coe"cients of the barycentric coordinate /j are the unknowns of a linear system ofd + 1 equations. All d + 1 systems share the same matrix M t and give a unique solution if the simplexK is not degenerated.

Proposition 7.2 For every point x " Rd, there exists a unique vector (/j(x))1#j#d+1 such that thefollowing identites hold:

x =d+1"

j=1

aj/j(x) , andd+1"

j=1

/j(x) = 1 .


Figure 7.5: Example (left) and counter-example (right) of conforming triangulation, in two dimensions.

Proof. For each point x = (xi)1$i$d, the scalar values (/j(x))1$j$d+1 are the solutions of a d + 1! d + 1 linearsystem that admits M as asociated matrix. Hence, there is a unique solution to this system if the simplex K is nondegenerated. Each function /j is an a"ne function and one can check easily that /j(ai) = &ij will be a solution ofthis system. Since there is only one such a"ne function, it is then the barycentric coordinates function. !

Since the /j are a"ne functions of x, then we can write:

K = {x " Rd , 0 ' /j(x) ' 1 , 1 ' j ' d + 1} ,

the faces of K are the intersections of K with the hyperplans /j(x) = 0, 1 ' j ' d + 1. We observealso that the change from Cartesian coordinates to barycentric coordinates is an a"ne transformation.Hence, a polynomial of total degree k in Cartesian coordinates can be expressed as a polynomial of totaldegree k in barycentric coordinates, and conversely. For a first-degree polynomial p we have:

p(x) =d+1"

j=1

p(aj)/j(x) . (7.13)

Triangulations and meshes

Definition 7.4 A triangulation, also called a triangular mesh of # is a set Th of non degenerated d-simplices (Kj)1#j#N such that:

(i) Kj % # and # =NB

j=1

Kj,

(ii) the intersection Ki / Kj of any two simplices is a m-simplex, 0 ' m ' d $ 1, such that all itsvertices are also vertices of Ki and Kj.

This definition states that the intersection of two triangles, if it is not empty, shall be reduced to eithera common vertex or an edge. Similarly, in three dimensions, the intersection of two tetrahedra can beeither empty, or reduced to a single common entity (vertex, edge or face). Such a mesh is often called aconforming mesh (cf. Figure 7.5). The vertices or nodes of the mesh Th are the vertices of the d-simplicesKj that compose the mesh. Algorithms to construct such triangulations will be described in Chapter 9and we refer the reader to [Frey-George, 2000] for more information on this topic.

We introduce two conditions on the geometry of a triangulation, with respect to the diameter andthe roundness of its elements.


Definition 7.5 Suppose (Th)h>0 is a sequence of meshes of #. This sequence is said to be a sequence ofregular meshes, or a quasi-uniform sequence, if:

1. the sequence h = maxK$Th

hK tends toward 0,

2. there exists a constant C * 1 such that:

&h > 0 , &K " Th ,h

-K' C . (7.14)

Remark 7.9 In dimension two, if K is a triangle, the condition 7.14 is equivalent to the existence ofan angle (0 > 0 such that

&h > 0 , &K " Th , (K * (0 ,

where (K is the smallest vertex angle in triangle K.

A set of points of a simplex K has a specific role as defined hereafter.

Definition 7.6 For every k " N', we call principal lattice of order k the set:

&k =>

x " K , /j(x) ">

0,1k,2k, . . . ,

k $ 1k

, 1?

, for 1 ' j ' d + 1?

. (7.15)

For k = 1, the lattice is simply the set of vertices of K; for k = 2, the principal lattice is composed ofthe vertices and of the midpoints of the edges (Figure 7.6). More generally, a lattice &k is a finite set ofpoints (.j)1#j#Nk .

Polynomial spaces

We introduce the set Pk of the polynomials p with scalar coe"cients of Rd in R of degree less than orequal to k:

Pk =

1772

773p(x) =

"

i1,...,id(0i1+...id#k

"i1,...,idxi11 . . . xid

d , "ij " R , x = (x1, . . . xd)

4775

776.

Hence, in two and three dimensions of space, we will simply denote:

Pk =

12

3p(x, y) ="

0<i+j#k

"ijxiyj , "ij " R

45

6 ,

Pk =

12

3p(x, y, z) ="

0#i+j+l#k

"ijlxiyjzl , "ijl " R

45

6 .

!1 !3!2

Figure 7.6: Principal lattice of order 1, 2 and 3 for a two-dimensional simplex.


It is easy to verify that Pk is a vector space of dimension:

dim(Pk) =k"

l=0

$d + l $ 1

l

%=

$d + k

k

%=

177772

77773

k + 1 d = 112(k + 1)(k + 2) d = 2

16(k + 1)(k + 2)(k + 3) d = 3

The notion of lattice &k of a simplex K allows to define a bijective mapping between a space of polynomialsPk and a set of points (.j)1#j#Nk . The set &k is said to be unisolvent for Pk. We will use this propertyto define a finite element.

Lemma 7.8 Given a simplex K. For k * 1, we consider the lattice &k of order k whose points aredenoted (.j)1#j#Nk . Then, every polynomial p " Pk is uniquely determined by its values at the points(.j)1#j#Nk . There exists a basis (!j)1#j#Nk of Pk such that:

!j(.i) = &ij , 1 ' i, j ' Nk .

Proof. At first, we notice that the cardinal of &k and the dimension of the vector space Pk coincide:

card(&k) = dim(Pk) =(d + k)!

d!k!.

Indeed, we can write the elements of &k as follows:

&k =

12

3

d"

j=1

"j

kai +

81$

d"

j=1

"j

k

9"0 , 0 ' "1 + · · · + "d ' k

45

6 ,

where the "j " N. We know that the mapping associating to every polynomial Pk its values on the lattice &k is alinear mapping. Hence, it is su"cient to show that it is an injection to have the bijective property. We will proveby recurrence on the dimension d that if p " Pk is such that p(x) = 0 for all x " &k the p = 0 on Rd. At first,notice that a polynomial of degree k that vanishes in k + 1 points of R is identically null. Suppose this is also truefor the dimension d $ 1. We use a recurrence on the degree k. For k = 1, an a"ne function that vanishes at thevertices of a non-degenerated simplex K is identically null according to the relation (7.13). Suppose this propertyis true for all polynoials of degree k$ 1 and let consider a degree k polynomial p that vanishes on &k. We observethat &k contains the subset

&"k = {x " &k , /0(x) = 0} ,

that corresponds to the principal lattice of order k of the d$1-simplex of vertices (a1, . . . , ad). Since the restrictionof p to the hyperplane generated by (a1, . . . , ad) is a polynomial of degree k in d$ 1 variables, then p = 0 on thishyperplan, tnaks to the recurrence hypothesis. If we introduce a system of coordinates (x1, . . . , xd) such that thehyperplane is now defined by xd = 0, then

p(x!, . . . , xd) = xdq(x1, . . . , xd%1) ,

where q is a polynomial of degree d$ 1 that vanishes on the set &k $ &"k since xd is non null on this set. The set&k $ &"k is a principal lattice of order k $ 1 and thus the recurrence hypothesis leads to conclude that q = 0 andconsquently that p = 0. The results follows. !

In practice, we consider only polynomials of degree 1 or 2. Equation (7.13) provides the character-ization of a polynomial of degree one. Given a d-simplex K of vertices (aj)1#j#d+1, we define the edgemidpoints (ajj")1#j<j"#d+1 by their barycentric coordinates:

/j(ajj") = /!j(ajj") =12

, /l(ajj") = 0 l 2= j, j! , .


The principal lattice &2 is exactly composed of the vertices and the edge midpoints and every polynomialp " P2 can be written as:

p(x) =d+1"

j=1

p(aj)/j(x)(2/j(x)$ 1) +"

1#j<j"#d+1

4p(ajj")/j(x)/j"(x) , (7.16)

where the (/j(x))1#j#d+1 are the barycentric coordinates of x " Rd.

7.3.2 Triangular Lagrange Pk finite elements

Suppose the domain # is covered by a simplicial mesh Th. The finite element method for triangularLagrange Pk elements involves the discrete finite dimensional functional space:

V kh =

Cv " C0(#) , vh|Kj " Pk , Kj " Th

D,

and its subspace:

V k0,h =

Evh " V k

h , vh = 0 on %#F

.

Definition 7.7 A triangular Lagrange Pk finite element is locally defined by a triad (K, Pk,&k), where:

(i) K is a d-simplex associated with the mesh Th,

(ii) Pk is a vector space of polynomials of degree less than or equal to k on K,

(iii) &k is the principal lattice of order k of the simplex K " Th.

&k is called the set of nodes of the degrees of freedom of the finite element (K, Ph,&k).

We consider the set of points (ai)1#i#Ndof of the principal lattices of order k of each of the simplicesKi " Th, where Ndof is the number of degrees of freedom of the Pk finite element method. We calldegrees of freedom of a function vh " V k

h the set of the values of v at the so-called nodes (ai)1#i#Ndof .

Remark 7.10 We observe that the nodes of the degrees of freedom coincide exactly with the vertices ofthe simplices Ki " Th, when k = 1. The nodes of the degrees of freedom are composed by the mesh verticesand the edge midpoints, when k = 2.

Lemma 7.9 The space V kh is a subspace of the space H1(#) of finite dimension corresponding to the

number of degrees of freedom. Furthermore, there exists a basis (!j)1#j#Ndof of V kh defined by:

!i(aj) = &ij , 1 ' i, j ' Ndof ,

such that every function vh " V kh can be uniquely written as

vh(x) =Ndof"

i=1

vh(ai)!i(x) .


Proof. It is easy to see that the elements of V kh belong to H1(#). The Lemma 7.8 allozs to conclude that each

function vh " V kh is exactly known by assembling on each Ki " Th polynomials of degree k that coincide on the

degrees of dreedom of the d-faces. By assembling the basis of Pk on each Ki, the basis (!j)1$j$Ndof is defined. !

Corollary 7.4 The subspace V k0,h is a subspace of H1

0 (#) of finite dimension corresponding to the numberof internal degrees of freedom, i.e.] not taking the nodes on %# into account.

Definition 7.8 The triad (K, Pk,&k) is said to be unisolvent if and only if the mapping v " Pk -,(!1(v), . . . ,!Ndof (v)) " RNdof is an isomorphism.

The unisolvency property is equivalent to say that every function in the polynomial space Pk is entirelydetermined by its node values.

7.3.3 Finite element approximation of a boundary-value problem

We return to the numerical approximation of the solution of the homogeneous Dirichlet boundary-valueproblem (7.12). The variational formulation of the internal approximation reads:

find uh " V k0,h such that

.

!(3uh ·3vh)(x) +

.


.

!f(x)vh(x) dx , &vh " V k

0,h , (7.17)

By decomposing uh on the canonical basis (!j)1#j#Ndof and considering as test functions vh = !i, weobtain:

Ndof"

j=1

uh(aj)$.

!(3!j ·3!i)(x) dx +

.

!c(x)!j(x)!i(x) dx

%=

.

!f(x)!i(x) dx .

This formulation leads to solving a linbear system in RNdof :

AhUh = Fh ,

where we have introduced the notations Uh = (uh(aj))1#j#Ndof and Fh = (/! f!i dx)1#i#Ndof and

Ah =$.

!(3!j ·3!i)(x) dx +

.

!c(x)!j(x)!i(x) dx

%

1#i,j#Ndof

.

It is easy to see that the matrix Ah can be decomposed as a sum of a sti!ness matrix Kh and a massmatrix Mh. Actually, this result is independent of the dimension.

Coe"cients of Ah

Since the shape functions !j have a small support around a node ai, the intersection of the supports of!j and !i if often the emptyset and thus the resulting matrix Ah will contains lot of zero coe"cients. Itis a sparse matrix. The coe"cients of Ah can be computed via an exact integration formula. Let denote(/i(x))1#i#d+1 the barycentric coordinates of the point x in a simplex K " Th. For every ("i)1#i#d+1 wehave: .

K/1(x)#1 . . . /d+1(x)#d+1 dx = |K| "1 + . . . "d+1! d!< "

1#j#d+1

"j + d=!,


a1

a12

a2

a23

3a

a13

P = P1 P2 ! = { p(ai), 1 ! i ! 3 } ! = { p(ai), 1 !i ! 3; p(aij), 1 ! i < j ! 3 }

P1

K "i(m) 1 ! i ! 3

"i(m) = 1 "aim · ni

aiaj · ni, j #= i,

m = (x, y) ni Kai "i(m) j

i j1 j2 i aj1aj2 · ni = 0"i(m) ai aj

ak (ajak)P2

"i(2"i " 1), 1 ! i ! 3, 4"i"j , 1 ! i < j ! 3.

! $ R2

k = 1Th N V k

h (#1, . . . , #N )A % RN,N

Aij =

!

!&#i ·&#j =

"

K!Th

!

K&#i ·&#j , 1 ! i, j ! N.

NeK(m) 1 ! m ! Ne ($m,1, . . . , $m,nf )

K(m) nf = 3 k = 1 nf = 6 k = 2(#1, . . . , #N )

(K1, . . . , KNe)

($m,1, . . . , $m,nf )

noeud(1 : nf , 1 : Ne),

noeud(ni, m) = #K(m) $m,ni.

Figure 7.7: Nodes of the degrees of freedom on a finite element (K, P2,&2).

where |K| = mes(K) denotes the ”volume” of the simplex K. The integrals of the right-hand side termFh are approximated using quadrature formulas in each simplex K " Th. For instance the generalizationof the trapeze formula for a simplex K is:

.

K!j(x) dx 1 |K|

d + 1

d+1"

i=1

!j(ai) ,

where (ai)#i+. 1 represent the vertices of K. These formulas are exact for a"ne functions. For instance,considering a triangle K " R2 of vertices (ai)1#i#3 and denoting by (aij)1#i,j#3 the midpoints of theedges aiaj (cf. Figure 7.7), the following quadrature formula is exact for !k " P2:

.

K!k(x) 1 |K|

3

"

1#i,j#3

!k(aij) ,

where |K| represents the area of simplex K, while the following formula is exact for !k " P3:

.

K!k(x) dx 1 |K|

60

)

*"

1#i#3

!k(ai) + 8"

1#i<j#3

!k(aij) + 27!k(a0)

+

, ,

where the point a0 = 1d+1

!1#i#d+1 ai denotes here the barycenter of K.

Remark 7.11 As we mentioned in one dimension of space, the analysis can be simplified by consideringan a"ne transformation allowing to consider any d-simplex K " Th as the image of a reference elementGK. Hence, all computations can be performed on this reference simplex.

7.3.4 The reference finite element

By convention, the vertices of the reference simplex GK are given by the origin a0 = (0, . . . , 0) and thepoints ai = (0, . . . , 1, . . . , 0) , for which all coordinates are equal to zero except the dth coordinate thatis equal to one. For a non-degenerated simplex K, we denote by FK : Rd , Rd the unique a"netransformation that maps ai on ai for all i = 0, . . . , d. Hence, we write;

FK(x) = a0 + BK x ,


a2

FKa3

a2a1

K

!Ka1

a3

Figure 7.8: A"ne transformation of the reference linear triangle in two dimensions.

where BK is a d ! d matrix such that the column i is given by the coordinates of ai $ a0. Since thesimplex K is non-degenerated, BK is invertible and FK is a one-to-one mapping that maps GK on K andwe have:

|K| = |det(BK)| | GK| =|det(BK)|

d!.

We use the notation · in the reference element. Moreover, to simplify, we denote q any quantityobtained by transporting a quantity q using the transformation FK . Hence, we denote:

x = F%1K (x) = B%1

K (x$ a0) 4 x = FK(x) .

For every function v defined on K, we define v defined on GK as:

v(x) = v(x) 4 v = v . FK = v(FK(x)) ,

and we remove the · notation when dealing with F%1K , for every function v defined on GK, we write:

v(x) = v . F%1K (x) = v(F%1

K (x)) .

Similarly, if $ is a linear form acting on the functions defined on K, we define the transported linearform $ acting on the functions defined on GK as:

$(v) = $(v) .

Notice that the barycentric coordinates are preserved by the a"ne trasnformation FK :

/i(x) = /i(x) .

This leads to conclude that the principal lattice of order k of the simplex K is the image by FK of theprincipal lattice of order k of GK, denoted by G&k. Finally, we observe that the space of polynoials of degreek is left invariant by the transformation FK .

Proposition 7.3 We denote by # · # the Euclidean norm of Rd and its subordinate norm. Hence, wehave:

#BK# 'hK

- bK, #B%1

K # 'h bK-K

, |det(BK)| =|K|| GK|

.


Proof. by definition,

#BK# = supv!Rd

#BKv##v# = sup

&v&=!cK

#BKv#- bK

,

and the sup is attained. Let v " Rd be a vector of norm - bK that correspond to the sup. Hence, there exists twopoints y and z on the boundary of the inscribed sphere of GK such that v = y $ z. Thus,

BKv = BK(y $ z) = FK(y)$ FK(z) = y $ z

where y, z are tzo points in K. Hence,#BKv# = #y $ z# ' hK .

The second inequality is deduced from the first by interchanging the roles of K and GK. The third equality isobtained by noticing that |det(BK)| is the Jacobian of FK . Hence,

|K| =.

Kdx =

.

bK|det(BK)| dx = |det(BK)| | GK| .

!As a consequence, we deduce that there exists two constants C1 and C2, independent of K such that:

C2-dK ' |det(BK)| ' C1h

dK .

Transformation of the derivatives

We have the following identities:

#v#Lp(K) =.

K|v(x)|p dx =

.

bK|v(FK(x))|p |det(BK)| dx

= d! |K|.

bK|v(x)|p dx

,

thus yielding the identity:#v#Lp(K) = C |K|1/p #v#Lp( bK) , (7.18)

with C = (d!)1/p. Moreover, we observe that if v(x) = v(x), i.e. if v = v . FK , then we have:

3v(x) = BtK3v(x) ,

and we obtain the identity:

|v|2H1(K) =.

K#3v(x)#2 dx =

.

bK#(Bt

K)‘$13v(x)#2 |det(BK)| dx

= C2 d!|K|-2

K

.

bK#3v(x)#2 dx

,

and then:|v|H1(K) ' C

hK

|K|1/2-K |v|H1( bK) , (7.19)

where C depends only on the dimension d. By interchanging the role of K and GK, we obtain similarly:

|v|H1( bK) ' C|K|1/2

-K|v|H1(K) . (7.20)

We give the following generalization result.


Theorem 7.4 If v " Hm(K), then v " Hm( GK) and there exists a constant C1, independent of K, suchthat:

&v " Hm(K) , |v|Hm( bK) ' C1#BK#m|det(BK)|%1/2|v|Hm(K) . (7.21)

If v " Hm( GK), then v " Hm(K) and there exists a constant C2, independent of K, such that:

&v " Hm( GK) , |v|Hm(K) ' C2#B%1K #m|det(BK)|1/2|v|Hm( bK) . (7.22)

Proof. Result admitted here. !

Remark 7.12 Now, we are able to construct finite elements from a unisolvent triad ( GK, GPk, G&k) bydefining directly the finite element (K, Pk,&k) as the transport of these quantities by the transformationFK . It is easy to see that the new finite element is unisolvent.

7.3.5 Convergence of the finite element method

We introduce a preliminary result.

Proposition 7.4 Let consider (Th)h>0 a sequence of regular triangulations of # % Rd (d ' 3). Then,for every v " Hk+1(#), the interpolate rhv is defined and there exists a constant C, independent of hsuch that:

#v $ rhv#H1(!) ' C hk #v#Hk+1(!) .

The following result states the convergence of the Pk finite element method.

Theorem 7.5 Let consider (Th)h>0 a sequence of regular triangulations of # % Rd (d ' 3) where allelements in all triangulations are a"ne equivalent to a same reference element ( GK, GPk, G&k) of class C0

for a given k * 1 such that Pk % GP % H1( GK). Then, the Pk finite element method converges, i.e.the approximate solution uh of the problem (7.17) converges toward the solution u of problem (7.12) inH1(#):

limh"0

#u$ uh#H1(!) = 0 .

Furthermore, if u " Hk+1(#), then there exists a constant C independent of h such that:

#u$ uh#H1(!) ' C hk #u#Hk+1(!) ,

Proof. To show the convergence, we use Theorem (4.11) where we consider V = C#c (#) % Hk+1(#) dense in

H1(#). The estimate in the previous proposition allows to verify the assumption of Theorem (4.11). Then, we useCea’s lemma to write:

#u$ uh#H1(!) ' C infvh!V k

0,h

#u$ vh#H1(!) ' C#u$ rhu#H1(!) ,

if rhu " H1(#). The previous proposition allows to conclude. !

Remark 7.13 This result involves the exact solution uh of the internal approximation problem in V k0,h.

This requires computing all integrals in Ah and Fh exactly. Since numerical integration formulas are usedin practice, this result may not be valid. Nonetheless, if these integration formulas are exact or at leastvery accurate, the order of convergence of the Lagrange finite element method is preserved,


The Theorem 7.5 shows that the regularity of the exact solution has a direct impact on the orderof convergence of the finite element approximation. Hence, it is often important to have an a prioriknowledge of its regularity. The following result gives an indication for a convex domain.

Theorem 7.6 Let # be a convex polygon and let f " R2. Then, the solution u of the homogeneousDirichlet problem:find u such that: .

!3u.3v =

.

!fv , &v " H1

0 (#) ,

is in H2(#) and we have the following estimate:

&f " L2(#) , #u#H2(!) ' #f#L2(!) .

7.3.6 Numerical resolution of the linear system

We remind here that the finite element approximation of the boundary-value problem (7.12) implies thenumerical solving of a linear system in RNdof :

Ah Uh = (Kh + Mh) Uh = Fh ,

where Uh = (uh(aj))1#j#Ndof and Fh = (/! f!i dx)1#j#Ndof . The generic term of the sti!ness matrix Kh

is of the form:kij = a(!j , !i) , 1 ' i, j ' Ndof .

where a(·, ·) is the bilinear form defined on the functional space H10 and (!j)1#j#Ndof is the canonical

basis of the approximation space. We have mentioned already that the sti!ness matrix Kh inherits ofthe properties of the bilinear form a. More precisely, if a is V-elliptic, then Kh is positive definite andif a is symmetric then Kh is symmetric. We have also indicated that the matrix Ah is a sparse matrix,i.e. it contains a large number of zeroes, as well as the matrices Kh and Mh. Hence, iterative methodsare preferred to solve this linear system (cf. Chapter 5).

Let denote 0 < /1(Kh) < · · · < /Ndof (Kh) the eigenvalues of Kh and 0 < /1(Mh) < · · · < /Ndof (Mh)the eigenvalues of Mh. We introduce the notations:

0(Kh) =/Ndof (Kh)

/1(Kh), and 0(Mh) =

/Ndof (Mh)/1(Mh)

,

and we have the following result.

Proposition 7.5 Suppose (Th)h>0 is a quasi-uniform sequence. Then, there exists two constants C1 > 0,C2 > 0 and a constant C3 > 0 such that:

C1 ' h20(Kh) ' C2 , and 0(Mh) ' C3 .

We observe that the mass matrix Mh is always well-conditioned since its condition number 0(Mh) isindependent of the mesh size h, under the condition that the refinement is quasi-uniform. On the otherhand, the sti!ness matrix Kh becomes ill -conditioned, i.e. 0(Kh) 5 1, when the mesh size h is small.

Remark 7.14 We have shown that the approximation error between the exact solution u and the Galerkinfinite element solution uh is bounded by above by a constant times the ”distance” between he spaces Vand Vh. The smaller the distance, the better the approximation error will be. Is seems then natural toincrease the dimension of the approximation space Vh in order to improve the accuracy of the solution.To this end, we can either:


1. increase the number of elements and thus increase the global number of degrees of freedom whilepreserving the local reference element. This strategy is known as h-refinement,

2. increase the degree of the polynomials and thus modifying the local reference element while preservingthe total number of elements. Obviously, this choice can be only be applied if the exact solution uis su!ciently smooth. This strategy is known as p-refinement.

We will discuss these adaptation options and others in the chapter 9.

7.4 The finite element method for the Stokes problem

To illustrate the application of the finite element method to solve a system of partial di!erential equations,we consider the Stokes problem in two dimensions of space.

We consider the flow of a fluid inside a bounded domain # % R2 and subjected to an external forcefield f . The flow is considered as stationary and inertial forces are supposed negligible here, hence theStokes equations introduced in Chapter 4 describe this viscous flow problem:Find (u, p) in appropriate Hilbert spaces such that:

172

73

$1$u +3p = f in #div u = g in #

u = 0 on %#(7.23)

where the unknowns u and p represent the velocity and the pressure of the fluid, respectively and 1 > 0is the viscosity. Usually, in the applications, the flow is incompressible and the function g is equal tozero.

7.4.1 Mixed formulation

We have already seen the mixed formulation of this problem in Chapter 4. We introduce the functionalspace:

L20(#) = {q " L2(#) ,

.

!q dx = 0} ,

and the mixed formulation reads:Given f " L2(#) and g " L2

0(#), find u " H10 (#)2 and p " L2

0(#) such that:0

a(u, v) + b(v, p) = f(v) , & v " H10 (#)2 ,

b(u, q) = g(q) , & q " L20(#)

(7.24)

where we posed a(u, v) = 1

.

!3u : 3v, b(v, p) = $

.

!p div v, f(v) =

.

!fv and g(q) = $

.

!gq. The

functional spaces H10 (#)2 and L2(#)2 are endowed with the canonical norms:

#u#H1(!) =

)

*"

i=1,2

#ui#2H1(!)

+

,1/2

, #f#L2(!) =

)

*"

i=1,2

#fi#2L2(!)

+

,1/2

,

for u = (ui)1#i#2 " H10 (#)2 and f = (fi)1#i#2 " L2(#)2.

Remark 7.15 If f " C0(#)2 and g " C0(#), if # is of class C2 and if the solution (u, p) of theproblem (7.24) is such that u " C2(#)2 and p " C1(#), then (u, p) is the classical solution of the Stokesproblem.

7.4. The finite element method for the Stokes problem 133

The formulation above is the not the only weak formulation possible for the Stokes problem. An alternateview consists in including the divergence constraint (for instance the incompressibility constraint div u =0) directly in the space of test functions. To this end, we introduce the space:

V0 = {v " H10 (#)2 , div v = 0} .

Since the divergence operator is surjective, there exists a function ug " H10 (#)2 such that for every

g " L20(#), we have:

div ug = g , and #ug#H1(!)2 ' c#g#L2(!) ,

where c > 0 is a constant (see [Brezis, 2005] for more details). Introducing the change of variablesu! = u$ ug leads to the condition div u! = 0 and to consider the following weak formulation:find u! " V0 such that:

a(u!, v) = 6f, v7 $ a(ug, v) , &v " V0 . (7.25)

This weak formulation is called a constrained formulation, since the restriction on the test functions tobe in the space V0 leads to discard the term b(v, p) that vanishes here.

Proposition 7.6 The problem (7.25) is a well-posed problem.

Proof. It is a direct application of Lax-Milgam theorem. ! Although this new formulation has sometheoretical advantages, many practical drawbacks prevent its application for solving a discrete problem.It is indeed di"cult to construct divergence-free finite elements. The reader interested is referred to thebook [Brezzi-Fortin, 1991] for more details on this topic.

7.4.2 Numerical approximation

We will now investigate the condition required for the approximation spaces in velocity and pressure todefine a well-posed discrete problem.

Let consider Xh a subspace of the space H10 (#)2 and Mh a subspace of the space L2

0(#). We introducethe discrete Stokes problem:Find uh " Xh and ph " Mh such that:

1772

773

1

.

!3uh : 3vh $

.

!ph div vh =

.

!f · vh , &vh " Xh ,

.

!qh div uh =

.

!gqh , &qh " Mh .

(7.26)

We introduce the space

Ker(Bh) = {vh " Xh ,

.

!qh div vh = 0 , &qh " Mh} ,

and we enounce the fundamental compatibility result.

Theorem 7.7 The discrete problem (7.26) is a well-posed problem if and only if the spaces Xh and Mh

satisfy the following compatibility condition:

8)h > 0 , infqh$Mh

supvh$Xh

.

!qh div vh

#qh#L2(!)#vh#H1(!)* )h . (7.27)


Furthermore, under this condition, we have the following estimates:

#u$ uh#H1(!) ' c1h infvh$Xh

#u$ vh#H1(!) + c2h infqh$Mh

#p$ qh#L2(!) (7.28)

#p$ ph#L2(!) ' c3h infvh$Xh

#u$ vh#H1(!) + c4h infqh$Mh

#p$ qh#L2(!) (7.29)

where the constants (cih)1#i#4 are given by Theorem 4.12.

Proof. We know that the bilinear form a(·, ·) is V -elliptic on H10 (#)2 !H1

0 (#)2 and the result is then a directconsequence of Theorem 4.12. !

The following result provides a practical method for checking whether the compatibility condition issatisfied or not.

Lemma 7.10 (Fortin) Let consider the spaces X = H10 (#)2 and M = L2

0(#) and let b " L(X, M).Suppose there exists ) > 0 such that:

infq$M

supv$X

b(v, q)#v#X#q#M

* ) .

Then, there exists )h > 0 such that:

infqh$Mh

supvh$Xh

b(vh, qh)#vh#Xh#qh#Mh

* )h ,

if and only if there exists 2h > 0 such that for every v " X, there exists a restriction operator rh "L(X,Xh), i.e. rh(v) " Xh, such that:

&qh " Mh , b(v, qh) = b(rh(v), qh) and #rh(v)#X ' 2h#v#X .

Proof. If the criterion is satisfied, for every qh " Mh we write

supvh!Xh

b(vh, qh)#vh#X

* supv!X

b(rh(v), qh)#rh(v)#X

* 12h

supv!X

b(v, qh)#v#X

* )

2h#qh#M .

Conversely, if the discrete condition is satisfied, then the operator Bh : Xh , Mh defined by b(vh, qh) = 6Bhvh, qh7defines an isomorphism (admitted here). Hence, for every v " X, there exists a unique vh = rh(v) such that

&qh " Mh , b(rh(v), qh) = b(v, qh) .

The mapping v -, rh(v) is linear and we have:

#rh(v)#X ' 1)#Bhrh(v)#M !

' 1)

supqh!Mh

b(rh(v), qh)#qh#M

' 1)

supah!Mh

b(v, qh)#qh#M

' 2h

)#v#X .


The operator rh has the desired properties. !This proof provides a technique to verify the discrete compatibility condition; it consists in construct-

ing the operator rh using the solution of the system:

b(rh(v), qh) = b(v, qh) , &qh " Mh ,

and to check if it is bounded in the space X. This is clearly is non-trivial task. On the other hand, itis possible to construct a simpler operator for specific approximation spaces and finite elements and tocheck if the compatibility condition is satisfied.

It is easy to construct an internal variational approximation of the Stokes problem, as we have alreadyseen:find uh " Xh and ph " Mh such that:

0a(uh, vh) + b(vh, ph) = f(vh) , &vh " Xh ,

b(uh, qh) = g(qh) , &qh " Mh ,(7.30)

If we denote by nu (resp. np) the dimension of the space Xh (resp. Mh), we introduce the basis (!j)1#j#nu

of Xh and the basis ($j)1#j#np of Mh defined using the basis shape functions of the finite elements andwe decompose uh and ph on these basis:

uh =nu"

j=1

uh(aj)!j(x) , and ph(x) =np"

j=1

ph(a!j)$j(x) .

We obtain the following linear system to solve:$

Ah Bth

Bh 0

% $Uh

Ph

%=

$Fh

Gh

%. (7.31)

where Uh = (uh(aj))1#j#nu and Ph = (ph(a!j))1#j#np and the matrices Ah = (aij) " Rnu,nu and Bh =(bki) " Rnp,nu are given by:

aij = 1

.

!3!i : 3!j dx , and bki = $

.

!$i div !k dx .

The vectors Fh = (fj)1#j#nu and Gh = (gk)1#k#np are defined as:

fi =.

!f · !i dx , and gk =

.

!g $k dx .

Lemma 7.11 The vector 1 " Rnp is always contained in the kernel Ker(Bth). The discrete pressure ph

is defined up to an additive constant.

Proof. Let rh " Mh and wh " Xh. By definition,

Wh · BthRh = BhWh · Rh =

.

!rh div wh dx .

Since rh = 1 is in Mh and since

&wh " Xh ,

.

!div wh dx =

.

"!wh · n ds = 0 .

we conclude that Rh = 1 " Ker(Bth). !

Finding a pair of approximation spaces (Xh, Mh) for the Stokes problem has been an active researchfield in numerical analysis. We present here a few examples of such pairs.


The P1/P1 finite element

This seems the obvious first choice for implementing the finite element method for the Stokes problem.We consider the unit domain # =]0, 1[2 and we suppose a regular Cartesian mesh is defined on # (Figure7.9). We introduce the approximation spaces:

Xh = {uh " C0(#)2 , &K " Th , uh|K " P21 , uh|!! = 0} ,

Mh = {ph " L20(#) / C0(#) , &K " Th , ph|K " P1} .

For a given element K " Th, we denote (aiK)1#i#3 its vertices. Then, we have:

&vh " Xh ,

.

!ph div vh =

"

K$Th

(div vh)|K.

Kph ,

="

K$Th

(div vh)|K|K|3

3"

j=1

ph(aj,K) ,

Now, let us consider the pressure field ph defined on the triangulation Th by its values 0, $1, 1 at thethree vertices of every triangle K " Th. In this case, we have

3"

j=1

ph(aj,K) = 0 ,

and it is easy to see that:

&vh " Xh ,

.

!ph div vh = 0 .

This is clearly in contradiction with the requirement of the inf-sup condition. Hence, the P1/P1 finiteelement shall be discarded for solving the Stokes problem.

When the dimension of Ker(Bth) is strictly larger than one, the finite element method is unstable. In

such case, solving the linear system using an iterative method will result in oscillations in the pressurevalues. On the other hand, if dim(Ker(Bt

h)) = 1, the discrete pressure can be uniquely determined byfixing its value at a mesh node or by setting its average value on the domain #.

0 1

-1

0

-1

1

-1

0

1

1

0-1

1

1

0

-1 0

-1

1

1

-1

0

1

-10

Figure 7.9: A regular Cartesian mesh of the unit domain # =]0, 1[2 and the unstable pressure mode forthe P1/P1 finite element.


The P1/P0 finite element

This finite element presents the a priori advantage of being very simple to implement. Given a triangu-lation of the domain #, we define a piecewise continuous approximation of the velocity and a piecewiseconstant (and thus discontinuous) approximation for the pressure. Since the velocity is piecewise a"ne,its divergence is piecewise constant and thus by testing the divergence using constant functions, thedivergence of the discrete field is strongly forced to be null. Hence, this finite element allows to obtainexactly a divergence free velocity field. Nonetheless, the P1/P0 finite element does not fulfill the inf -supcondition and Fortin’s criterion.

Again, we provide a counter-example in two dimensions. Suppose the tiangulation Th of the simplyconnected polygonal domain # contains t triangles, vi internal vertices and vb boundary vertices. Weshall have 2vi degrees of freedom for the space Xh, as the velocity must vanish on the boundary and tdegrees of freedom for the pressure leading to t $ 1 independent divergence-free constraints. The Eulerformulas for a simple polygon indicate that:

t = 2vi + vb $ 2 ,

from which we deduce that:(t$ 1) * 2(vi $ 1) .

We observed that dim(Mh) = t$ 1 and that dim(Xh) = 2vi. The rank theorem allows to write:

dim(Ker(Bth)) = dim(Mh)$ dim(Im(Bt

h)) * (t$ 1)$ dim(Xh)= (t$ 1)$ 2vi = vb $ 3

This last relation indicates that there is at least vb $ 3 spurious pressure modes. In other words, thespace Mh is too rich to impose the Bhuh = 0 constraint. A function uh " Xh is thus overconstrained anda locking phenomenon occurs. In general, uh = 0 is the only discrete divergence-free function of Xh suchthat Bhuh = 0; i.e. Bh is not surjective.

The P1-bubble/P1 (mini) finite element

We have seen that with the P1/P1 finite element, the dimension of the space of the degrees of freedomof the velocity is not large enough. To overcome this problem, it seems natural to attempt enriching thevelocity space rather than impoverishing the pressure space. However, it is not necessary to consider aspace of polynomials of degree two for approximating the velocity. Crouzeix and Raviart suggested (in1973) to add one extra degree of freedom to the barycenter of every simplex of the triangulation Th ofthe domain #. Hence, we define in two dimensions:

P1 = (P1(K)9 Span{bK , K " Th})2 ,

with the bubble function bK defined in two dimensions as:

bK = 273H

i=1

/i , and generalized in any dimension as bK = (d + 1)d+1d+1H

i=1

/i ,

where (/i)1#i#3 denote the barycentric coordinates on K. The degrees of freedom associated with P1 aredefined as:

&k = {v -, v(ai) , 1 ' i ' 3} : {v -,.

ei

v · ni d.} .

Then, we consider the approximation spaces:

138 Chapter 7. The finite element approximation244 Chapitre 11

Vitesse Pression

Figure 11.1 – Element Mini (O(h))

Vitesse PressionVitesse

Figure 11.2 – Element Mini en dimension 3 (O(h))

liberte en vitesse et 3 en pression dans le cas bidimensionnel (respectivement 15 et 4 en dimension3). Cet element est tres utilise, particulierement en dimension 3 (voir la figure 11.2) en raison deson faible nombre de degres de liberte. Cet element converge a l’ordre 1 car l’approximation envitesse ne contient que les polynomes de degre 1.

Le deuxieme element illustre est celui de Taylor-Hood souvent appele P2 ! P1 en raison del’approximation quadratique des composantes de vitesse et de l’approximation P1 de la pression(voir les figures 11.3 et 11.4. Les noeuds de calcul sont situes aux sommets (vitesses et pression) etaux milieux des aretes (vitesses seulement). Cet element converge a l’ordre 2 (O(h2)) en vertu dutheoreme 11.14 car l’approximation en vitesse contient les polynomes de degre 2 et l’approximationen pression contient ceux d’ordre 1. Des equivalents quadragulaires et hexaedriques, egalementd’ordre 2, sont illustres aux figures 11.5 et 11.6.

Remarque 11.4Dans le cas d’elements a pression continue, il est parfois necessaire d’imposer une condition sup-plementaire sur le maillage pour avoir l’existence et l’unicite de la solution. On devra par exemplesupposer qu’aucun element n’a deux cotes sur la frontiere du domaine (voir les exercices de fin dechapitre). !

Figure 7.10: The P1-bubble/P1 (mini) element for velocity (left) and pressure (right) approximation.

Xh = {vh " C0(#)2 , &K " Th , vh|K " P1 , uh|!! = 0} ,

Mh = {ph " L20(#) / C0(#) , &K " Th , ph|K " P1} .

(7.32)It is easy to see that Xh % (H1

0 (#))2.Before giving a compatibility result for this finite element, we need to introduce an interpolation

operator rh. Since v " H10 (#)2) is not necessarily continuous, its Lagrange interpolate might not be

defined. Therefore, in each triangle, we define rh as:

rh(v) = Ch(v) + "K bK ,

where Ch(v) = (Ch(v1), Ch(v2)) is the Clement interpolate defined hereafter and "K " R2 is chosen so asto satisfy the equation: .

K(rh(v)$ v) = 0 .

To this end, we define:

"K =$.

KbK

%%1 .

K(v $ Ch(v)) .

For every v " Xh and qh " Mh, we write:.

!qh div(v $ rh(v)) =

.

!3qh(rh(v)$ v) =

"

K$Th

.

K3qh(rh(v)$ v) = 0 ,

since 3qh is piecewise constant on every K " Th.

Definition 7.9 Let Th be a triangulation of #, 'h be the set of the internal vertices of Th. We introduce(h the space of continuous piecewise linear functions on Th which are zero on the boundary and wedenote (!$)$$"h the basis of P1 finite elements with homogeneous boundary conditions. We introduce theoperator Ch " L(H1

0 (#),(h), proposed by Clement in 1975, and defined as follows:

Ch(v) ="

$$"h

:1

|'$ |

.

%!

v

;!$ ,

where the set '$ is defined as the union of all triangles K " Th for which 2 is a vertex.

In this definition, the nodal value v(2) is replaced in the definition of the interpolant rh by the localaverage value around 2. The Clement operator can be applied to any function in L2(#). We have thefollowing result.

Lemma 7.12 If the family of triangulations (Th)h>0 is quasi-uniform, then the operator Ch satisfies thelocal estimates:

&v " H10 (#) , #v $ Ch(v)#L2(K) ' C hK |v|H1(%K) ,


&v " H10 (#) , |Ch(v)|H1(K) ' C |v|H1(%K) ,

where 'K is the union of all triangles sharing at least one of the vertices of K.

Theorem 7.8 If the family of triangulations (Th)h>0 is quasi-uniform, the approximation spaces Xh andMh are uniformely compatible with respect to h, i.e. there exists a constant )h, independent f h, satisfyingthe condition (7.27).

Proof. Let consider v " H10 (#)2. We construct the operator rh(v) " Xh such that

&qh " Mh ,

.

!qh div(rh(v)) =

.

!qh div(v) .

Since Mh % H1(#), the previous identity is equivalent to writing:

&qh " Mh ,"

K!Th

.

Krh(v) ·3qh =

"

K!Th

.

Kv ·3qh .

Since 3qh " Pd0, the previous identity leads to conclude that:

&K " Th ,

.

Krh(v) =

.

Kv .

Since the operator rh is composed of two terms, we analyze them separately. Regarding the Clement interpolateCh, we invoke the previous estimate:

&v " H10 (#) , |Ch(v)|H1(K) ' C |v|H1(#K) .

For the term "KbK , we write:|"KbK |H1(K) ' C|"K | ,

with the constant C independent of hK and K. Then, we write

|"K | =$.

KbK

%%1

|.

K(v $ Ch(v))|

' C1|K| |

.

K(v $ Ch(v))|

' |K|%1/2#v $ Ch(v)#L2(K) .

We use the estimate #v $ Ch(v)#L2(K) ' C hK |v|H1(#K) , to obtain:

|"K | ' C |v|H1(#K) ,

and finally:|rh(v)|H1(K) ' C |v|H1(#K) ,

By taking the square of this expression and summing over all K " Th we conclude that:

#rh(v)#H1(!) ' C#v#H1(!) .

The final result is obtained by invoking Fortin’s lemma. !

Proposition 7.7 Suppose the solution (u, p) of the problem (7.24) is si"ciently smoth, i.e. u " H2(#)2/H1

0 (#)2 and p " H1(#)/L20(#). Then, the solution (uh, ph) of the problem (7.26) with the spaces defined

by (7.32) is such that:

#u$ uh#H1(!) + #p$ ph#L2(!) ' C h (#u#H2(!) + #p#H1(!)) . (7.33)


7.4.3 Numerical resolution

We have seen that the discrete problem (7.30) leads to solve the linear system (7.31). We assume herethe natural hypothesis that the bilinear form a(·, cdot) is V -elliptic on X with a constant " and that theform b(·, ·) satisfies on xh !Mh a inf -sup condition uniformely in h, and we denote the constant ).

The resolution of the linear system using a direct method is likely to be costly because of the numberof degrees of freedom and thus the size of the linear system. Direct methods are appropriate up toNdof = 104. For this reason, iterative methods are favored. However, the matrix of the system (7.31)is neither positive nor definite. It is indeed an indefinite matrix (but symmetric) with that zero blockon the diagonal. It will have positive and also negative eigenvalues. It is possible to make it positive bysolving the system: 0

AhU + BthP = Fh

$BhU = $Gh. (7.34)

However, the new matrix corresponding to this system is still indefinite and is no longer symmetric. Weknow from Chapter 5 that gradient methods are not e"cient on this type of matrix. We present twocommonly used methods for solving such mixed problems.

Resolution of the saddle point by penalization

The following method is gaining popularity for solving the Stokes problem (and saddle point problemsin general) because of its versatility and facility to be implemented. It consists in replacing the problem(7.31) by the following problem: 0

AhU& + BthP& = Fh

$BhU& + *ShP& = $Gh, (7.35)

where the matrix Sh " Rnp,np is such that:

(ShP,Q) = 6ph, qh7M ,

where (·, ·) and 6·, ·7M denote the standard Euclidean scalar product in Rnp and the inner product in M ,respectively. In the system (7.35), * is a penalization coe"cient su"ciently small.

The main idea is simple. Suppose Ah is a positive definite symmetric matrix, the system (7.31) isthen equivalent to:

infBhVh=Gh

>12(AhVh, Vh)$ (Fh, Vh)

?.

If Sh is any positive definite matrix in Rnp,np , the previous expression can be replaced by:

infVh

>12(AhVh, Vh) +

12*

(S%1h (BhVh $Gh, BhVh $Gh)$ (Fh, Vh)

?,

or equivalently, by writing P& = (1/*)S%1h (BhVh $Gh):0

AhU& + BthP& = Fh

BhU& $Gh = *ShP&. (7.36)

By replacing the quantity P& in the first equation, we obtain the linear system:$

Ah +1*Bt

hS%1h Bh

%U& = Fh +

1*Bt

hS%1h Gh . (7.37)


If the matrix S%1h is a sparse matrix, this provides an e"cient technique to solve our problem. This system

can be solved by an iterative technique like the conjugate gradient, since the matrix is now positive definiteand symmetric. Notice however that the term 1/*Bt

hS%1h Bh has a strong negative impact on the condition

number of the linear system (7.4.3).We now investigate the e!ect of changing the constrained problem to a penalized problem. We have

the following error estimate.

Proposition 7.8 Let consider * > 0 and let (U, P ) and (U&, P&) denote the solutions of system (7.34)and (7.36), respectively. Then, we have the error estimate:

")#U $ U&#X + ")2#P $ P&#M ' C*#P#M . (7.38)

Proof. By di!erence, we obtain the system:0

Ah(U $ U$) + Bth(P $ P$) = 0

$Bh(U $ U$)$ *ShP$ = 0.

We introduce on Rnu the norm # · #' defined by:

&U " Rnu , #U#' = supV !Rnu

(U, V )#V #X

.

The continuity and the V -ellipticity of the bilinear form a implies that:

&U " Rnu , #AhU#' ' #a# #U#X , and &U " Rnu , "#U#X ' #AhU#' .

The matrix Bh " Rnp,np satisfies the inf -sup condition:

min&P&M (=0

max&U&X (=0

(BthP,U)

#P#M #U#X* ) ,

or equivalently:&P " Rnp , )#P#M ' #Bt

hP#' .

The continuity of the form b implies that:

&P " Rnp , #BthP#' ' #b# #P#M .

Using these inequalities leads to write the first equation of the system as:

#P $ P$#M ' 1)#Bt

h(P $ P$)#'

=1)#Ah(U $ U$)#' '

C

)#U $ U$#X .

By multiplying the first equation by (U $ U$) and using the V -ellipticity of the form a we obtain for the secondequation:

"#U $ U$#2X ' (Ah(U $ U$), U $ U$) = (Bth(P$ $ P ), U $ U$) = (P$ $ P,Bh(U $ U$)) = $*(P$ $ P, ShP$)

= $*(P$ $ P, Sh(P$ $ P ))$ *(P$ $ P, ShP )' $*(P$ $ P, ShP )' *#P$ $ P#M #P#M .

The error estimate is obtained by combining the two previous inequalities. !

Remark 7.16 Notice that discretizing a penalized problem is not necessarily equivalent to penalize adiscrete problem [Brezzi-Fortin, 1991]. The penalty method is considered in this last case as a procedural(algorithmic) technique, since a choice of spaces Xh % X and Mh % M has been done. On the otherhand, discetizing the penalized problem consists in choosing Qh = BhVh that is in general a poor choice.


Uzawa’s operator

To conclude this section, we introduce an iterative, although very e"cient, algorithm to solve suchproblem when th operator Ah associated with the bilinear form a is invertible. We know since Chapter4 that the Stokes equations are equivalent to solving a problem of energy minimization. More precisely,the resolution of the linear system (7.31) is equivalent to the following minimization:

J(Uh) = minVh$KerBh

J(Vh) with J(Vh) =12(AhVh, Vh)$ (Fh, Vh) . (7.39)

In the optimization terminology, the pressure is the Lagrange multiplier that imposes the constraintdiv u = g when the energy is minimized.

First, we eliminate the variable Uh from the linear system (7.31), if the matrix Ah is invertible, thusleading to:

Uh = A%1h (Fh $Bt

hP )

and then by replacing Uh in the second equation BhUh = Gh, we obtain the system:

BhA%1h Bt

h Ph = BhA%1h Fh $Gh . (7.40)

We consider the Uzawa matrix U = BhA%1h Bh.

Proposition 7.9 If matrix Ah is positive definite, this matrix is also positive definite.

Proof. One has:(BhA%1

h BthP, P ) = (A%1

h BthP,Bt

hP ) * "#BthP# .

It is positive definite if KerBth = {0}. !

Hence, problem (7.40) is more easy to solve than the original problem (7.31) as e"cient methods exist forpositive definite systems. Unfortunately, even if the matrix Ah is a sparse matrix (i.e. contains lot of zerovalues), its inverse A%1

h is likely to be a full matrix. The resolution of (7.40) must be performed using aniterative method since inverting the matrix A%1

h is extremely computationally costly. The convergencerate of iterative methods is strongly related to the condition number of the matrix.

Proposition 7.10 If the matrix U is symmetric, we have the estimate:

0(U) ' C

"2)20(Sh) .

Remark 7.17 If the family of mesh (Th)h>0 is quasi-uniform, the condition number 0(Sh) is a constantindependent of h. Hence, the previous estimate shows that the conditioning of Uzawa’s operator is of orderone (if " and ) are of order one, which is a reasonable hypothesis). The iterative techniques convergefast without any preconditioning.

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