5 The ﬁnite element approximation · 2017. 5. 9. · 210 5 The ﬁnite element approximation (iii) If in addition the bilinear form a(·,·) is symmetric the previous estimate becomes:

5

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 result asthe only reasonable goal. Such one sided views seem to reflecthuman limitations rather than objective values. In itself math-ematics is an indivisible organism uniting theoretical contem-plation and active application.”

Richard Courant (1888-1972)

The aim of this chapter is to introduce the basic theory of finite elementmethods. Nowadays, finite element methods are widely used in almost everyfield of engineering analysis. The German mathematician Richard Courant(1888-1972) shall probably be credited for formulating the essence of what isnow called a finite element [?]. The development of these methods becameeffective with the advent of computers and is now recognized as one of themost powerful and versatile method for construction approximations of thesolutions of boundary-value problems. We will give here a brief overview ofthe fundamental mathematical ideas that form the core of the method.

In Chapter ?? we have seen the principle of the variational approximationof elliptic problems. The main idea of the finite element method is to replacethe Hilbert space V in which the variational formulation is posed by a finitedimensional subspace Vh. We will first briefly return to the internal variationalapproximation principle and present finite elements in one dimension.

5.1 General principle

We consider the variational abstract problem introduced in Chapter ??. Moreprecisely, given a Hilbert space V , a bilinear continuous and V -elliptic form

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208 5 The finite element approximation

a(·, ·) defined on V × V and a continuous linear form l(·) defined on V ′, weconsider 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 usingLax-Milgram Theorem ??.

5.1.1 Internal approximation of a variational problem

The internal approximation of problem (P) consists then in replacing theHilbert space V by a finite dimensional subspace, denoted Vh, in which tofind the solution uh. Here, h > 0 represents a parameter related to the dis-cretization of the domain (in space or time) and intended to vanish in thetheoretical results. We assume that for every v ∈ V , there exists an elementrh(v) ∈ Vh such that

limh→0

‖rh(v)− v‖ = 0 .

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

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

We introduce the keyword internal related to the approximation becausewe suppose here that Vh ⊂ V . If N = dimVh, we consider a basis (ϕi)1≤i≤N

of Vh. The decomposition of uh in the basis of Vh, uh =∑N

i=1 uiϕi, leads torewrite the problem (Ph) in the following form:

N∑

j=1

uj a(ϕj , ϕi) = l(ϕi) , ∀1 ≤ i ≤ N . (5.1)

Introducing the stiffness matrix Ah = (aij) ∈ RN,N of coefficients aij =a(ϕj , ϕi), for all 1 ≤ i, j ≤ N , the vector Uh = (ui)1≤i≤N and the vectorFh = (fi)1≤i≤N such that fi = l(ϕi)1≤i≤N allow us to conclude that we havethe equivalence:

uh solution of (Ph) ⇐⇒ AhUh = Fh .

Obviously, the V -ellipticity assumption on the bilinear form a implies theexistence and the uniqueness of the solution uh to the problem (Ph). However,this assumption is too strong for our purposes. As we are considering a finitedimensional space Vh, it is sufficient 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 ??).


5.1 General principle 209

5.1.2 A priori error estimates

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

∀vh ∈ Vh , a(vh, vh) ≥ αh‖vh‖2V ,

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

Proposition 5.1.1. Under the previous assumptions, we have the orthogo-nality identity:

∀vh ∈ Vh , a(u− uh, vh) = 0 .

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 ofconstant α, then it defines an inner product and an energy norm associatedto it as:

‖u‖e = (a(u, u))1/2 , ∀u ∈ V ,

equivalent to the norm of V :√

α‖u‖V ≤ ‖u‖e ≤√‖a‖ ‖u‖V , ∀u ∈ V .

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

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

Lemma 5.1.1 (Cea). Under the previous hypothesis, we have the followingerror estimates.

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

‖u−uh‖V ≤(

1 +‖a‖αh

)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 coercivityconstant α, then we have:

‖u− uh‖V ≤ M

αinf

vh∈Vh

‖u− vh‖V .



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

‖u− uh‖V ≤√

M

αinf

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:

αh‖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 have

a(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 issymmetric, we can improve the estimate. Since uh is the orthogonal projection of uonto Vh with respect to the inner product induced by a, we have the Pythagoreanrelation:

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

α‖u− uh‖2V ≤ a(u− uh, u− uh) ≤ a(u− wh, u− wh) ≤ M‖u− wh‖2V .

and the results follows. !

Corollary 5.1.1. Under the previous hypothesis, let (Vh)h denotes a set offinite dimensional subspaces of V and let us assume that

∀v ∈ V , infvh∈Vh

‖v − vh‖Vh→0−→ 0 .

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

The objective for the “ideal” approximation method is to define suitableapproximation spaces Vh to apply the Galerkin approach. To this end, wesearch for a compromise between the dimension N of Vh (and thus the dimen-sion of the matrix A) and the accuracy of the numerical solution uh. We shallalso consider spaces Vh that allow to compute easily the quantities a(ϕj , ϕi)and l(ϕi). Finally, specific spaces Vh may result in sparse matrices A wherethe number of nonzero elements is small, or well-conditioned matrices witha small condition number and thus easy to solve. In this spirit, the finite el-ement method tends to answer all these requirements. Before detailling themain concepts of the finite element methods, we give a few words about Ritzand Petrov-Galerkin methods.



5.1.3 Ritz and Petrov-Galerkin methods

The Ritz method

We consider that the hypothesis of the Lax-Milgram theorem are satisfiedand we recall that, if the bilinear 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 sub-space Vh ⊂ V such that dimVh = N and the approximate solution uh shallsolve:

(Ph) find uh ∈ Vh such that J(uh) = infvh∈Vh

J(vh).

Theorem ?? ensures the existence of a unique solution to this minimizationproblem as a consequence of Lax-Milgram theorem.

Since the dimension of the space Vh is N , there exists a basis (ϕj)1≤j≤N

of Vh and every uh ∈ Vh can be decomposed as uh =N∑

j=1

uiϕj and we use the

classical notation U = (u1, . . . , uN )t ∈ RN .We consider the one-to-one mapping ξ : Vh → RN , uh -→ U and we pose

J = J ξ−1 such that for every uh ∈ Vh we have:

J (U) = J(uh) ,

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

J(uh) =12a

N∑

j=1

ujϕj ,N∑

j=1

ujϕj

− l

N∑

j=1

ujϕj

=12

N∑

i=1

N∑

j=1

uiuja(ϕi, ϕj)−N∑

i=1

uil(ϕi) .

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

J(uh) =12U t Ah U − U t Fh = J (U) ,

where Ah = (aij) ∈ RN,N with aij = a(ϕi, ϕj) and Fh = (fi) ∈ RN is suchthat fi = l(ϕi). Hence, solving the minimization problem (Ph) is equivalentto 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 stiffness matrix Ah is symmetric positive definiteand thus the functional J is quadratic on RN . This is sufficient to ensurethe existence and uniqueness of U ∈ RN solving the minimization problem(Ph,R). Furthermore, the solution U of the minimization problem (Ph,R) isalso the solution of the linear system AhU = Fh.

Remark 5.1.1. When the bilinear form a(·, ·) is symmetric, the Galerkin andRitz methods are strictly equivalent.

The Petrov-Galerkin method

The principles of Petrov-Galerkin and Galerkin methods are very similar inthe sense that they both will attempt to solve the problem (P). However, in thePetrov-Galerkin approach, we consider two finite-dimensional approximationsubspaces Vh and Wh in V such that

dimVh = dimWh = N .

The approximate solution uh is searched in the space Vh but the test functionsin the variational formulation are now the shape functions of Wh. For thesereasons, Vh is called the approximation space and Wh is the space of testfunctions. 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 (ϕj)1≤j≤N is a basis of Vh and (ψj)1≤j≤N a basis of Wh, then every

uh ∈ Vh can be decomposed as uh =N∑

j=1

uiϕj and we can rewrite the problem

as follows:

find uh ∈ Vh , such thatN∑

j=1

uja(ϕj , ψi) = l(ψi) , i = 1, . . . , N ,

And the linear system to solve is AhU = Fh, where Ah = (aij) ∈ RN,N

with aij = a(ϕj , ψi) and Fh = (fi) with fi = l(ψi), for all i = 1, . . . , N .

5.1.4 The finite element method

In the finite element method, the domain Ω is subdivided into a partition or amesh Th, i.e., a (potentially large) collection of geometrically simple elements,and the approximation space Vh is composed of piecewise polynomial functionson each element K of the partition Th. We will see in Chapter 6 how toconstruct a partition Th for a domain Ω of arbitrary geometric shape. Theparameter h represents here the grain of the discretization, i.e., the elementarysize 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 re-stricted on one or a few elements of Th and the polynomials are usually of lowdegree. Hence, when h → 0 the space Vh will better and better approximatethe space V and the stiffness matrix A will be sparse, most of its coefficientsbeing zeros.

Finite elements vs. finite differences or finite volumes

The principle of finite difference or finite volume discretization methods is verysimilar. Both approaches consider a partition of the domain into a numeroussmall pieces, although none of them consider a variational formulation ofthe problem at hand. For instance, let consider the homogeneous Dirichletboundary-value problem in two dimensions:

−∆u = f in Ω

u = 0 on ∂Ω

With the finite difference method, the domain Ω is covered by a regular uni-form grid. At each internal node xi,j = (ih, jh), we search a discrete valueui,j to approximate u(xi,j) and we assume for example that the Laplacianoperator 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 Ω is arbitrary andunknowns are associated with 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 unknownson K and on the neighboring 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 interfaceKi,j ∩Ki+1,j is then discretized by ui+1,l − uk,l. Repeating the procedure forthe other fluxes and approximating the term

∫Ki,j

f by h2f(xi,j) yields theequation:

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 finitedifference scheme.



Actually, the vast majority of finite difference methods can be deducedfrom finite element methods if the problem at hand has a variational formu-lation. It is less obvious for finite volume methods. Moreover, we have strongtheoretical mathematical tools to study finite element methods. In addition,the latter have several intrinsic advantages:

1. the versatility of the formulation on arbitrarily complex geometries, andthe possibility to locally refine the partition Th to approximate solutionswith singularities,

2. the boundary conditions are naturally taken into account in the space Vin the variational formulation and in its internal approximation Vh,

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

Other variational methods have been developed, like spectral methods, that areespecially adapted to the approximation of smooth solutions but are limitedto simple geometries and methods using wavelets basis.

5.2 The one dimensional case

At first, we introduce the general principle of the Lagrange finite elementmethod in one dimension of space. Without loss of generality, we can restrictour study to the unit domain Ω =]0, 1[. To set the ideas, we will also considerthe following boundary-value problem:

Given f ∈ L2(Ω) and c ∈ L∞(Ω), find the function u solving:−u′′(x) + c(x)u(x) = f(x) , x ∈]0, 1[

u(0) = u(1) = 0 .(5.2)

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 uniformif the points (xj) are equidistributed along the segment [0, 1], i.e. such thatxj = jh, with h = 1/(N + 1), 0 ≤ j ≤ N + 1. More generally, we denote byh = max |xj+1 − xj | the size parameter.

5.2.1 Lagrange P1 elements

The finite element methods for Lagrange P1 elements involves the space ofglobally continuous affine functions 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 ,


5.2 The one dimensional case 215

More generally, Pk denotes the vector space of polynomials in one variableand of degree less than or equal to k:

Pk =

p(x) =k∑

j=0

αjxj αj ∈ R

.

The finite element method consists in applying the internal variational approx-imation approach to the spaces V 1

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

V 1h can be represented using very simple shape functions.

Lemma 5.2.1. The space V 1h is a subspace of H1(Ω) of dimension N + 2.

Every function vh of V 1h is uniquely determined by its values at the mesh

vertices (xj)0≤j≤N+1:

vh(x) =N+1∑

j=0

vh(xj)ϕj(x) , ∀x ∈ [0, 1] ,

where (ϕj)0≤j≤N+1 is the basis of the shape functions ϕj with compact supportin each interval [xj−1, xj+1] defined as:

ϕj(x) =

x− xj−1

hx ∈ [xj−1, xj ]

xj+1 − x

hx ∈ [xj , xj+1]

such that ϕj(xi) = δij . (5.3)

Proof.

We know that piecewise C1 continuous functions belong to the space H1(Ω). Hence,

Vh is a subspace of H1(Ω). Moreover, since we have ϕj(xi) = δij , where δij is the

Kronecker symbol, the result follows. !

Remark 5.2.1. Notice that the functions (ϕj)0≤j≤N+1 can be expressed usingonly two functions:

ω0(x) = 1− x , ω1(x) = x .

The basis functions are defined as the composition of a shape function of areference finite element (i.e. that depends only of the polynomial approxima-tion) and of an affine transformation (depending only on the discretization)as, for all 0 ≤ j ≤ N :

ϕj(x) =

ω1

(x− xj−1

xj − xj−1

), x ∈ [xj−1, xj ]

ω0

(x− xj

xj+1 − xj

), x ∈ [xj , xj+1]

and ϕN+1(x) = ω1 ((x− xN )/h). This will be useful to compute the coeffi-cients of the matrix in the linear system to solve.



x0 xj+1xj

ϕj

xj−1 xN+1

ϕN+1ϕ01

Fig. 5.1. Global shape functions for the space V 1h .

Corollary 5.2.1. The space V 10,h is a subspace of H1

0 (Ω) of dimension N andevery function vh of V 1

0,h is uniquely determined by its values at the meshvertices (xj)1≤j≤N :

vh(x) =N∑

j=1

vh(xj)ϕj(x) , ∀x ∈ [0, 1] ,

Remark 5.2.2. Notice that functions vh ∈ V 1h are not twice differentiable on Ω

and thus it is meaningless to attempt solving the problem (5.2) as the secondderivative of any vh ∈ V 1

h is a sum of Dirac masses at the mesh vertices.However, it is meaningful to solve a variational formulation of this problemwith functions vh ∈ V 1

h since only the first derivatives are involved.

The variational formulation of problem (5.2) consists in finding u ∈ H10 (Ω),

such that:∫

Ωu′(x)v′(x) dx +

∫

Ωc(x)u(x)v(x) dx =

∫

Ωf(x)v(x) dx , ∀v ∈ H1

0 (Ω) ,

(5.4)and the variational formulation of the internal approximation consists in find-ing uh ∈ V0,h, such that:∫

Ωu′h(x)v′h(x) dx+

∫

Ωc(x)uh(x)vh(x) dx =

∫

Ωf(x)vh(x) dx , ∀vh ∈ V0,h .

(5.5)Introducing the notation uh(xj)1≤j≤N for the approximate value of the exactsolution at the mesh vertex xj , leads to the approximate problem:

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

N∑

j=1

(∫

Ωϕ′j(x)ϕ′i(x) dx +

∫

Ωc(x)ϕj(x)ϕi(x) dx

)uh(xj) =

∫

Ωf(x)ϕi(x) dx .

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



AhUh = Fh

where Uh = (uh(xj))1≤j≤N , Fh =(∫

Ω f(x)ϕi(x) dx)1≤i≤N

and the matrix Ah

is defined as:

Ah =(∫


∫


)

1≤i,j≤N

.

Coefficients of the matrix Ah

Actually, the matrix Ah appears as the sum of the stiffness matrix Kh definedby its coefficients (kij)1≤i,j≤N :

kij =∫

Ωϕ′j(x)ϕ′i(x) dx =

N∑

k=0

∫ xk+1

xk

ϕ′j(x)ϕ′i(x) dx ,

and the mass matrix Mh defined by its coefficients (mij)1≤i,j≤N :

mij =∫

Ωc(x)ϕj(x)ϕi(x) dx =

N∑

k=0

∫ xk+1

xk

c(x)ϕj(x)ϕi(x) dx .

Since the shape functions ϕj have a small support, most of the coefficients inAh are zeros. More precisely, for a given index i, there is only three consecutivevalues of j such that the coefficient aij is potentially not equal to zero. Thestructure of the matrice is then easy to deduce: Ah is a tridiagonal matrix.The coefficients of Ah are thus given by:

ajj =∫ xj+1

xj−1

(ϕ′j(x))2 dx +∫ xj+1

xj−1

c(x)(ϕj(x))2 dx

ajj−1 =∫ xj

xj−1

ϕ′j(x)ϕ′j−1(x) dx +∫ xj

xj−1

c(x)ϕj(x)ϕj−1(x) dx

ajj+1 =∫ xj+1

xj

ϕ′j(x)ϕ′j+1(x) dx +∫ xj+1

xj

c(x)ϕj(x)ϕj+1(x) dx

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

mjj = c0

∫ xj

xj−1

ϕj(x)ϕj−1(x) dx = c0

∫ xj

xj−1

ω1

(x− xj−1

h

)ω0

(x− xj−1

h

)

= c0h

∫ 1

0ω1(y)ω0(y) dy = c0h

∫ 1

0(1− y)y dy = c0

h

6,

where we posed y = (x− xj−1)/h. Finally, we find the coefficients of Ah:

ajj−1 = − 1h

+ c0h

6ajj =

2h

+ c02h

3and ajj+1 = − 1

h+ c0

h

6,



Remark 5.2.3. Instead of regarding the node contributions, we could have an-alyzed the elements. Consider the element Kj = [xj , xj+1]; on this elementthere is only two non-zero shape functions:

ϕj |Kj =xj+1 − x

xj+1 − xj=

xj+1 − x

hϕj+1|Kj =

x− xj

xj+1 − xj=

x− xj

h

ϕ′j |Kj =−1

xj+1 − xj=−1h

ϕ′j+1|Kj =1

xj+1 − xj=

1h

Then, we can arrange the elementary contributions of the element Kj to thestiffness matrix and to the mass matrix as 2×2 symmetric matrices EKj andEMj :

EKj =(

kj11 kj

12

kj21 kj

22

)and EMj =

(mj

11 mj12

mj21 mj

22

)

with

kj11 =

∫ xj+1

xj

(ϕ′j(x))2 dx ,

=∫ xj+1

xj

1h2

dx =1h

kj12 = kj

21 =∫ xj+1

xj

ϕ′j(x)ϕ′j+1(x) dx ,

=∫ xj+1

xj

− 1h2

dx = − 1h

kj22 =

∫ xj+1

xj

(ϕ′j+1(x))2 dx

=∫ xj+1

xj

1h2

dx =1h

mj11 =

∫ xj+1

xj

c(x)(ϕj(x))2 dx , mj12 = mj

21 =∫ xj+1

xj

c(x)ϕj(x)ϕj+1(x) dx , mj22 =

∫ xj+1

xj

c(x)(ϕj+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 thematrix assembly, especially in higher dimensions.

Matrix assembly

The assembly of the matrix Ah is easy and is obtained algorithmically usinga loop over all mesh elements Kj and adding their contributions to the rightcoefficients of the global system. Assuming a(i, j) denotes the coefficients 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 computation-ally expensive part of the method. We refer the reader to Chapter 3 for moredetails about the direct and indirect techniques to solve this system.

Coefficients 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)ϕi(x) dx .

Usually, the function f is not known analytically. Hence, we decompose f inthe basis of the shape functions (ϕj)1≤j≤N :

f(x) =N−1∑

j=1

fjϕj(x) dx

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

xk

ϕj(x)ϕi(x), dx .

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

xk

θ(x) dx =xk+1 − xk

2(θ(xk+1) + θ(xk)) ,

that gives the exact result for polynomial of degree one and leads here fj =hf(xj); or the Simpson formula:

∫ xk+1

xk

θ(x) dx =xk+1 − xk

6(θ(xk+1) + 4θ(xk+1/2) + θ(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(Ω), c ∈ L∞(Ω) such that c(x) ≥ c0 > 0 almost everywherein Ω and α,β ∈ R, find the function u solving:



−u′′(x) + c(x)u(x) = f(x) , x ∈ Ω

u′(0) = α u′(1) = β .(5.6)

in a very similar manner. Recall that this problem has a unique solutionu ∈ H1(Ω) (cf. Chapter ??). The variational formulation of the internal ap-proximation consists here in finding uh ∈ V 1

h such that∫


∫


∫

Ωf(x)vh(x) dx−αvh(0)+βvh(1) , ∀vh ∈ V 1

h (Ω) .

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

AhUh = Fh

with Uh = (uh(xj))0≤j≤N+1 and the stiffness matrix Ah is defined as:

Ah =(∫


∫

Ωc(x)ϕj(x)ϕi(x)) dx

)

0≤i,j≤N+1

.

and Fh = (fj)0≤j≤N+1 such that:

fj =∫

Ωf(x)ϕj dx 1 ≤ j ≤ N

f0 =∫

Ωf(x)ϕ0(x) dx− α

fN+1 =∫

Ωf(x)ϕN+1(x) dx + β .

5.2.2 Convergence of the Lagrange P1 finite element method

Definition 5.2.1 (Interpolation). The linear mapping Πh : H1(Ω) → V 1h

defined for every v ∈ H1(Ω) as:

(Πhv)(x) =N+1∑

j=0

v(xj)ϕj(x) , ∀x ∈ [0, 1] ,

is called P1 interpolation operator. Furthermore, for every v ∈ H1(Ω), theinterpolation operator is such that:

limh→0

‖v −Πhv‖H1(Ω) = 0 .

The P1 interpolate of a function v is the unique piecewise affine function thatcoincide with v at the mesh vertices xj . The convergence of the finite elementmethod is related to a series of results that we give here.



Suppose that the function v is sufficiently smooth, i.e. v ∈ H2(Ω). Sincethe derivative of the affine functions in Vh is constant on the intervals Kj =[xj , xj+1], we have then:

(Πhv)′(x) =v(xj+1)− v(xj)

h=

1h

∫ xj+1

xj

v′(t) dt , ∀x ∈ [xj , xj+1] .

Since we assumed v ∈ H2(Ω) then v′ ∈ H1(Ω) and thus v is a continuousfunction. Using Rolle’s theorem, we deduce that there exists a point θj ∈[xj , xj+1] such that:

v′(θj) =1h

∫ xj+1

xj

v′(t) dt = (Πhv)′(x) , ∀x ∈ [xj , xj+1] .

We will search for an estimate on ‖v −Πhv‖H1(Ω). To this end, we write:

‖v −Πhv‖2H1(Ω) =∫ 1

0|v′ −Π ′

hv|2 =N−1∑

j=1

∫ xj+1

xj

|v′(t)− v′(θj)|2 dt , (5.8)

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

v′(t)− v′(θj) =∫ t

θj

v′′(t) dt ,

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

|v′(t)− v′(θj)|2 ≤∫ t

θj

|v′′(t)|2 dt |t− θj |

≤∫ xj+1

xj

|v′′(t)|2 dt |t− θj | .

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

xj

|v′(t)− v′(θj)|2 dt ≤∫ xj+1

xj

|t− θj | 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 (5.8), we have now:

‖v −Πhv‖H1(Ω) ≤h2

2

N−1∑

j=1

∫ xj+1

xj

|v′′(t)| dt

≤ h2

2‖v′′‖2L2(Ω) .

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



Lemma 5.2.2 (Interpolation error). If v ∈ H2(Ω) then, there exists twoconstant C1 and C2 independent of h such that:

‖v−Πhv‖H1(Ω) ≤ C1 h2 ‖v′′‖L2(Ω) and ‖v′−(Πhv)′‖L2(Ω) ≤ C2 h ‖v′′‖L2(Ω) .

And we can establish the convergence of the finite element method for theDirichlet boundary-value problem as follows.

Theorem 5.2.1 (Convergence). Suppose u ∈ H10 (Ω) and uh ∈ V0,h are the

solutions of (5.2) and (5.5), respectively. Then, the Lagrange P1 finite elementmethod converges, i.e. we have:

limh→0

‖u− uh‖H1(Ω) = 0 .

Furthermore, if u ∈ H2(Ω) then, there exists a constant C independent of hsuch that:

‖u− uh‖H1(Ω) ≤ Ch‖f‖L2(Ω) .

Proof.

Here, since the bilinear form a(·, ·) is V0,h-elliptic, we can consider the ellipticityconstant α = 1:

a(u, u) =

Z 1

0

u′2(x) + cu2(x) dx ≥Z 1

0

u′2(x) dx = ‖u‖2H10 (Ω) ,

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

|a(u, v)| ≤`‖u′‖2L2(Ω) + c1‖u‖2L2(Ω)

´1/2

≤ (1 + c1)‖u‖H10 (Ω) ‖v‖H1

0 (Ω) ,

where we assumed that 0 ≤ c1 = supx∈[0,1] c(x) ≤ +∞. Thus, the continuity con-stant M is taken here as 1 + c1. Using Cea’s lemma 5.1.1, we can easily concludethat:

‖u− uh‖H10 (Ω) ≤

√1 + c1 inf

vh∈V0,h

‖u− vh‖H10 (Ω) . (5.9)

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

‖u−Πhu‖2H10 (Ω) ≤

h2

2‖u′′‖2L2(Ω)

≤ h2

2‖u‖2H2 ≤ C

h2

2‖f‖2L2(Ω) .

Moreover, since Πhu ∈ V0,h, we have also:

infvh∈V0,h

‖u− vh‖H10 (Ω) ≤ ‖u−Πhu‖H1

0 (Ω) .

and using the inequality (5.9), we can conclude. !



Lemma 5.2.3. There exists a constant C independent of h such that for allv ∈ H1(Ω):

‖Πhv‖H1(Ω) ≤ C‖v‖H1(Ω) , and ‖v −Πhv‖L2(Ω) ≤ Ch‖v′‖L2(Ω) .

Furthemore, for all v ∈ H1(Ω), we have:

limh→0

‖v′ − (Πhv)′‖L2(Ω) = 0 .

Proof.

Given v ∈ H1(Ω), we have:

‖Πhv‖L2(Ω) ≤ supx∈Ω

|Πhv(x)| ≤ supx∈Ω

|v(x)| ≤ C‖v‖H1(Ω) .

Moreover, since Πhv is an affine function, we have by Cauchy-Schwarz’s identity:

Z xj+1

xj

|(Πhv)′(t)|2 dt =(v(xj+1)− v(xj))

2

h=

1h

Z xj+1

xj

v′(t) dt

!2

≤Z xj+1

xj

|v′(t)|2 dt .

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

|v(x)−Πhv(x)| ≤ 2

Z xj+1

xj

|v′(t)|dt .

We obtain the second identity by using Cauchy-Schwarz, by integrating with respectto x and then by summation over j.Since C∞(Ω) is dense in H1(Ω), for every v ∈ H1(Ω) there exists w ∈ C∞(Ω) suchthat

‖v′ − w′‖L2(Ω) ≤ ε , for ε > 0 .

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

‖(Pihv)′ − (Πhw)′‖L2(Ω) ≤ C‖v′ − w′‖L2(Ω) ≤ Cε .

From Lemma 5.2.2, we deduce that, for h sufficiently small:

‖w′ − (Πhw)′‖L2(Ω) ≤ ε .

We can the write, by adding the last identities:

‖v′−(Πhv)′‖L2(Ω) ≤ ‖v′−w′‖L2(Ω)+‖w′−(Πhw)′‖L2(Ω)+‖(Πhv)′−(Πhw)′‖L2(Ω) ≤ Cε ,

and the result follows. !

5.2.3 Lagrange P2 elements

Before introducing the Lagrange P2 finite element method, we like to describethe advantage of considering higher-order polynomials on an example takenfrom [?].



Motivation for high-order elements

Consider the simple homogeneous Poisson boundary-value problem in onedimension of space:

−u′′(x) = f(x) , in Ω =]− 1, 1[ ,u(0) = u(1) = 0 .

, with f(x) =π2

4cos

(πx

2

).

The exact solution to this problem has the form:

u(x) = cos(πx

2

).

The well-known variational formulation of this problem consists in: findingu ∈ H1

0 (Ω) such that:∫

Ωu′(x)v′(x) dx =

∫

Ωf(x)v(x) dx , ∀v ∈ H1

0 (Ω)

Suppose the domain is decomposed into two intervals [−1, 0] and [0, 1] and letconsider the finite element space V 1

0,h generated by a single piecewise affinefunction vh defined as:

vh(x) =

x + 1 , x ∈ [−1, 0]1− x , x ∈ [0, 1]

The exact solution u and the approximate solution uh are given Figure 5.2,left. The approximation error in H1 seminorm is:

|u− uh|1,2 =(∫

Ω|u′(x)− u′h(x)|2 dx

)1/2

≈ 0.683667 .

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

0,h is composed of the functionvh(x) = 1 − x2. The exact solution u and the approximate solution uh aregiven Figure 5.2, right. The approximation error is then:

-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

Fig. 5.2. .Exact solution u with piecewise affine approximation uh (left-hand side)and with quadratic approximation uh (right-hand side).



|u− uh(x)|1,2 ≈ 0.20275 .

This is a clear indication that high-order finite elements are better to approx-imate smooth functions. Conversely, less regular functions can be approxi-mated accurately using lower degree finite elements. This will be emphasizedby Theorem 5.2.2.

Lagrange P2 elements

We return to problem (5.2) and we consider a set of points (xj)0≤j≤N+1 orintervals Kj = [xj , xj+1] forming a uniform mesh of Ω. The finite elementmethod for Lagrange P2 elements involves the discrete space:

V 2h = vh ∈ C0([0, 1]) , vh|Kj ∈ P2 , 0 ≤ j ≤ N ,

and its subspace:

V 20,h = vh ∈ V 2

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

These spaces are composed of continuous, piecewise parabolic functions (poly-nomials of degree less than or equal to 2). The P2 finite element methodconsists in applying the internal variational approximation approach to thesespaces.

Lemma 5.2.4. The space V 2h is a subspace of H1(Ω) of dimension 2N + 3.

Every function vh ∈ V 2h is uniquely defined by its values at the mesh vertices

(xj)0≤j≤N+1 and at the midpoints (xj+1/2)0≤j≤N = (xj + h2 )0≤j≤N :

vh(x) =N+1∑

j=0

vh(xj)ϕj(x) +N∑

j=0

vh(xj+1/2) ϕj+1/2(x) , ∀x ∈ [0, 1] .

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

ϕj(x) = φ

(x− xj

h

), 0 ≤ j ≤ N+1 and ϕj+1/2(x) = ψ

(x− xj+1/2

h

), 0 ≤ j ≤ N ,

with

φ(x) =

(1 + x)(1 + 2x) − 1 ≤ x ≤ 0(1− x)(1− 2x) 0 ≤ x ≤ 1

0 |x| > 1 ,

and ψ(x) =

1− 4x2 |x| ≤ 1/2

0 |x| > 1/2

Remark 5.2.4. Notice that we have:

ϕi(xj) = δij ϕi(xj+1/2) = 0ϕi+1/2(xj) = 0 ϕi+1/2(xj+1/2) = δij



xj+1xj−1

ϕj+1/2

xj−1/2 xj+1/2

1ϕj

xj

Fig. 5.3. Global shape functions for the space V 2h .

Corollary 5.2.2. The space V 20,h is a subspace of H1

0 (Ω) of dimension 2N +1and every function vh ∈ V 2

0,h is uniquely defined by its values at the meshvertices (xj)1≤j≤N and at the midpoints (xj+1/2)0≤j≤N :

vh(x) =N∑

j=1

vh(xj)ϕj(x) +N∑

j=0

vh(xj+1/2)ϕj+1/2(x) , ∀x ∈ [0, 1] .

The variational formulation of the internal approximation of the Dirichletboundary-value problem (5.2) consists now in finding uh ∈ V 2

0,h, such that:∫


∫


∫

Ωf(x)vh(x) dx , ∀vh ∈ V 2

0,h .

(5.10)Here, it is convenient to introduce the notation (xk/2)1≤k≤2N+1 for the meshpoints and (ϕk/2)1≤k≤2N+1 for the basis of V 2

0,h. Using these notations, wehave:

uh(x) =2N+1∑

k=1

uh(xk/2)ϕk/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 ofthe linear system to solve is now defined as:

Ah =(∫

Ωϕ′k/2(x)ϕ′l/2(x) dx +

∫

Ωc(x)ϕk/2(x)ϕl/2(x) dx

)

1≤k,l≤2N+1

,

and the right-hand side term becomes:

Fh =(∫

Ωf(x)ϕk/2(x) dx

)

1≤k≤2N+1

.

Since the shape functions ϕj have a small support, the matrix Ah is mostlycomposed of zeros. However, the main difference with the Lagrange P1 finiteelement method, the matrix Ah is no longer a tridiagonal matrix.



Coefficients of Ah

The coefficients of the matrix Ah can be coputed more easily by consideringthe following change of variables, 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 5.4):

ω−1(t) =t(t− 1)

2ω0(t) = −(t− 1)(t + 1) ω1(t) =

t(t + 1)2

,

and their respective derivatives:

dω−1

dt(t) =

2t− 12

dω0

dt(t) = −2t

dω1

dt(t) =

2t + 12

.

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

dϕj

dx=

dωk

dt

dt

dx,

where ωk ∈ [−1, 1]. In this case, the elementary contributions of the elementKj to the stiffness matrix and to the mass matrix are given by the 3 × 3matrices 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

Fig. 5.4. 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

Coefficients 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 the decomposition of f in the basis of shapefunctions (ϕj)0≤j≤2N :

f(x) =2N∑

j=0

f(xj)ϕj(x) dx .

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

fi =N∑

k=1

∫ x2k

x2k−2

f(x)ϕi(x) dx .

Using the previous decomposition of f , we obtain:

fi =2N∑

j=0

fj

(N∑

k=1

∫ x2k

x2k−2

ϕj(x)ϕi(x) dx

),

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

x2k−2

ϕj(x)ϕi(x) dx, .

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

fk

i

fki+1

fki+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.



5.2.4 Convergence of the Lagrange P2 finite element method

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

‖u− uh‖H1(Ω) ≤√

M

αinf

vh∈V 10,h

‖u− vh‖H1(Ω) ,

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

h . Supposenow that f ∈ H1(Ω), then u ∈ H3(Ω) and if the function c is sufficientlysmooth then:

‖u‖H3(Ω) ≤ C ‖f‖H1(Ω) .

In order to find an upper bound on the right-hand side of the previous es-timate, we introduce a mapping wh ∈ 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 degree two orless. To this end, on each interval [xi, xi+1], we consider the polynomial func-tions:

wi,1(x) =αi

2(x− xi)2 , wi,2(x) = wi,1(x) + βi(x− xi) ,

where

αi =1h

∫ xi+1

xi

u′′(t) dt , βi =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 all 0 ≤ i ≤ N and that wh is a polynomial of degree2 on each [xi, xi+1]. We conclude easily that:

‖u− uh‖H1(Ω) ≤√

M

αinf

vh∈V 10,h

‖u− vh‖H1(Ω) ≤√

M

α‖u− wh‖H1(Ω) .

Introducing the notation rhu = u−wh, we can see from the previous identitiesthat rhu ∈ H3(Ω) and rhu|[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 5.2.5 (Poincare-Wirtinger inequality). Given a bounded inter-val [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 (Ω) =

∫ xi+1

xi

|(rhu)′(t)| dt ≤ h4

4‖u′′′‖L2(Ω) .

Finally, we can enounce the convergence result as follows.

Theorem 5.2.2 (Convergence). Suppose u ∈ H10 (Ω) and uh ∈ V 2

0,h arethe solutions of (5.2) and (5.10), respectively. Then, the Lagrange P2 finiteelement method converges, i.e. we have:

limh→0

‖u− uh‖H1(Ω) = 0 .

Furthermore, if u ∈ H3(Ω) (i.e. f ∈ H1(Ω)), then there exists a constant Cindependent of h such that:

‖u− uh‖H1(Ω) ≤ C h2 ‖u′′′‖L2(Ω) .

Remark 5.2.5. The convergence rate of the P2 finite element method is bet-ter than with the P1 finite element method. However, the data f must besufficiently smooth, here f ∈ H1.



5.2.5 Lagrange Pk elements

This section generalizes the concepts introduced in the previous sections tothe interpolation of continuous and polynomials functions of degree k ≥ 1.We consider a set of points (xj)0≤j≤N+1 or intervals Kj = [xj , xj+1] forminga uniform mesh of Ω =]0, 1[.

Lagrange finite element spaces

For a given integer k ≥ 1, we define the space of globally continuous functionson [0, 1] whose restriction on each interval Kj = [xj , xj+1] is a polynomial ofdegree 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 , weintroduce 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 5.2.6. The space V kh is a subspace of H1(Ω) of dimension k(N +1)+

1. Every function vh of V kh is uniquely determined by its values at the mesh

nodes (yj,l)0≤j≤N,0≤l≤k−1 and yN+1,0. Furthermore, the shape functions aresuch that:

ϕj,l(yj′,l′) = δjj′δll′ .

The space V k0,h is a subspace of H1

0 (Ω) 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:∫


∫


∫

Ωf(x)vh(x) dx , ∀ vh ∈ V k

0,h .

(5.11)Following the same analysis than for the Lagrange P2 element, we have thefollowing convergence result.



Theorem 5.2.3 (Convergence). Suppose u ∈ H10 (Ω) and uh ∈ V k

0,h arethe solutions of (5.2) and (5.11), respectively. Then, the Lagrange Pk finiteelement method converges, i.e. we have:

limh→0

‖u− uh‖H1(Ω) = 0 .

Furthermore, if u ∈ Hk+1(Ω) (i.e. f ∈ Hk−1(Ω)), then there exists a constantC independent of h such that:

‖u− uh‖H1(Ω) ≤ C hk ‖f‖Hk−1(Ω) .

Remark 5.2.6. 1. The approximate solution uh converges toward the exactsolution u in H1(Ω) when h → 0. The Lagrange Pk method is of order kin h, if the function f is sufficiently smooth.

2. The matrix assembly becomes more and more difficult as the value of kincreases. Since the size of the problem increases as well, this may resultin additional difficulties in solving the resulting linear system.

3. The computation of the components of the right-hand side vector Fh mustbe carried out with a sufficiently accurate method.

5.3 Triangular finite elements in higher dimensions

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

To set the ideas, we consider the homogeneous boundary-value prob-lem (5.2) posed here in an open bounded domain Ω of R2:

Given f ∈ L2(Ω) and c ∈ L∞(Ω), find u such that:−∆u + cu = f , in Ω

u = 0 , on ∂Ω(5.12)

We already know that this problem has a unique solution u ∈ H10 (Ω).

5.3.1 Preliminary definitions

We proceed like in one dimension of space. The domain Ω is decomposedinto a set of N finite elements, triangles (Kj)1≤j≤N in dimension d = 2 andtetrahedra in dimension d = 3. These two types of elements belong to thegeneric class of simplices.


5.3 Triangular finite elements in higher dimensions 233

d-Simplices

Definition 5.3.1. A d-simplex K is the convex hull (envelope) of d+1 points(aj)1≤j≤d+1 in Rd, called the vertices of K, that are not all lying in the samehyperplane. It is the smallest convex passing through all these points.

Remark 5.3.1. Let consider d + 1 points (aj)1≤j≤d+1 in Rd and let denote(ai,j)1≤i≤d the coordinates of vector (aj). These points are affinely indepen-dent, 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

is invertible. In such case, the simplex is not degenerated. Any d-simplex hasthe same number of faces and 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 ρK : the diameter of the largest inscribed ball,

(iii) the aspect ratio σK =hK

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

Barycentric coordinates

Any simplex K can be represented by the barycentric coordinates λj1≤j≤d+1

of its vertices. Barycentric coordinates are a form of homogeneous coordinates.

Definition 5.3.2. For every 1 ≤ j ≤ d + 1, the barycentric coordinate λj ofa point x ∈ Rd is the first-degree polynomial:

λj(x) = c1x1 + · · · + cdxd + cd+1 ,

such that for all 1 ≤ i ≤ d + 1, λj(ai) = δij.

For each j, the d + 1 coefficients of the barycentric coordinate λj are theunknowns of a linear system of d+1 equations. All d+1 systems share the samematrix M t and give a unique solution if the simplex K is not degenerated.

Proposition 5.3.1. For every point x ∈ Rd, there exists a unique vector(λj(x))1≤j≤d+1 such that the following identites hold:

x =d+1∑

j=1

ajλj(x) , andd+1∑

j=1

λj(x) = 1 .



Fig. 5.5. Example (left) and counter-example (right) of conforming triangulation,in two dimensions.

Proof.

For each point x = (xi)1≤i≤d, the scalar values (λj(x))1≤j≤d+1 are the solutions of

a d + 1× d + 1 linear system that admits M as asociated matrix. Hence, there is a

unique solution to this system if the simplex K is non degenerated. Each function

λj is an affine function and one can check easily that λj(ai) = δij will be a solution

of this system. Since there is only one such affine function, it is then the barycentric

coordinates function. !Since the λj are affine functions of x, then we can write:

K = x ∈ Rd , 0 ≤ λj(x) ≤ 1 , 1 ≤ j ≤ d + 1 ,

the faces of K are the intersections of K with the hyperplans λj(x) = 0,1 ≤ j ≤ d + 1. We observe also that the change from Cartesian coordinatesto barycentric coordinates is an affine transformation. Hence, a polynomialof total degree k in Cartesian coordinates can be expressed as a polynomialof total degree k in barycentric coordinates, and conversely. For a first-degreepolynomial p we have:

p(x) =d+1∑

j=1

p(aj)λj(x) . (5.13)

Triangulations and meshes

Definition 5.3.3. A triangulation, also called a triangular mesh of Ω is aset Th of non degenerated d-simplices (Kj)1≤j≤N such that:

(i) Kj ⊂ Ω and Ω =N⋃

j=1

Kj,

(ii) the intersection Ki ∩ Kj of any two simplices is a m-simplex, 0 ≤ m ≤d− 1, such that all its vertices 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 either a common vertex or an edge. Similarly, in three



dimensions, the intersection of two tetrahedra can be either empty, or reducedto a single common entity (vertex, edge or face). Such a mesh is often calleda conforming mesh (cf. Figure 5.5). The vertices or nodes of the mesh Th

are the vertices of the d-simplices Kj that compose the mesh. Algorithms toconstruct such triangulations will be described in Chapter 6 and we refer thereader to [?] for more information on this topic.

We introduce two conditions on the geometry of a triangulation, withrespect to the diameter and the roundness of its elements.

Definition 5.3.4. Suppose (Th)h>0 is a sequence of meshes of Ω. This se-quence is said to be a sequence of regular meshes, or a quasi-uniform se-quence, 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

ρK≤ C . (5.14)

Remark 5.3.2. In dimension two, if K is a triangle, the condition 5.14 is equiv-alent to the existence of an angle θ0 > 0 such that

∀h > 0 , ∀K ∈ Th , θK ≥ θ0 ,

where θK is the smallest vertex angle in triangle K.

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

Definition 5.3.5. For every k ∈ N∗, we call principal lattice of order k theset:

Σk =

x ∈ K , λj(x) ∈

0,1k

,2k

, . . . ,k − 1

k, 1

, for 1 ≤ j ≤ d + 1

.

(5.15)

For k = 1, the lattice is simply the set of vertices of K; for k = 2, theprincipal lattice is composed of the vertices and of the midpoints of the edges(Figure 5.6). More generally, a lattice Σk is a finite set of points (σj)1≤j≤Nk .

Σ1 Σ3Σ2

Fig. 5.6. Principal lattice of order 1, 2 and 3 for a two-dimensional simplex.



Polynomial spaces

We introduce the set Pk of the polynomials p with scalar coefficients of Rd inR of degree less than or equal to k:

Pk =

p(x) =

∑

i1,...,id≥0i1+...id≤k

αi1,...,idxi11 . . . xid

d , αij ∈ R , x = (x1, . . . xd)

.

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

Pk =

p(x, y) =∑

0<i+j≤k

αijxiyj , αij ∈ R

,

Pk =

p(x, y, z) =∑

0≤i+j+l≤k

αijlxiyjzl , αijl ∈ R

.

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

)=

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

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

The notion of lattice Σk of a simplex K allows to define a bijective mappingbetween a space of polynomials Pk and a set of points (σj)1≤j≤Nk . The setΣk is said to be unisolvent for Pk. We will use this property to define a finiteelement.

Lemma 5.3.1. Given a simplex K. For k ≥ 1, we consider the lattice Σk oforder k whose points are denoted (σj)1≤j≤Nk . Then, every polynomial p ∈ Pk

is uniquely determined by its values at the points (σj)1≤j≤Nk . There exists abasis (ϕj)1≤j≤Nk of Pk such that:

ϕj(σi) = δij , 1 ≤ i, j ≤ Nk .

Proof.

At first, we notice that the cardinal of Σk and the dimension of the vector space Pk

coincide:

card(Σk) = dim(Pk) =(d + k)!

d!k!.

Indeed, we can write the elements of Σk as follows:

Σk =

(dX

j=1

αj

kai +

`1−

dX

j=1

αj

k

´α0 , 0 ≤ α1 + · · · + αd ≤ k

),



where the αj ∈ N. We know that the mapping associating to every polynomial Pk

its values on the lattice Σk is a linear mapping. Hence, it is sufficient to show thatit is an injection to have the bijective property. We will prove by recurrence on thedimension d that if p ∈ Pk is such that p(x) = 0 for all x ∈ Σk the p = 0 on Rd.At first, notice that a polynomial of degree k that vanishes in k + 1 points of R isidentically null. Suppose this is also true for the dimension d−1. We use a recurrenceon the degree k. For k = 1, an affine function that vanishes at the vertices of a non-degenerated simplex K is identically null according to the relation (5.13). Supposethis property is true for all polynoials of degree k − 1 and let consider a degree kpolynomial p that vanishes on Σk. We observe that Σk contains the subset

Σ′k = x ∈ Σk , λ0(x) = 0 ,

that corresponds to the principal lattice of order k of the d − 1-simplex of vertices(a1, . . . , ad). Since the restriction of p to the hyperplane generated by (a1, . . . , ad) isa polynomial of degree k in d− 1 variables, then p = 0 on this hyperplan, tnaks tothe recurrence hypothesis. If we introduce a system of coordinates (x1, . . . , xd) suchthat the hyperplane 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 Σk −Σ′k since xd is

non null on this set. The set Σk −Σ′k is a principal lattice of order k − 1 and thus

the recurrence hypothesis leads to conclude that q = 0 and consquently that p = 0.

The results follows. !

In practice, we consider only polynomials of degree 1 or 2. Equation (5.13)provides the characterization of a polynomial of degree one. Given a d-simplexK of vertices (aj)1≤j≤d+1, we define the edge midpoints (ajj′)1≤j<j′≤d+1 bytheir barycentric coordinates:

λj(ajj′) = λ′j(ajj′) =12

, λl(ajj′) = 0 l 1= j, j′ , .

The principal lattice Σ2 is exactly composed of the vertices and the edgemidpoints and every polynomial p ∈ P2 can be written as:

p(x) =d+1∑

j=1

p(aj)λj(x)(2λj(x)− 1) +∑

1≤j<j′≤d+1

4p(ajj′)λj(x)λj′(x) , (5.16)

where the (λj(x))1≤j≤d+1 are the barycentric coordinates of x ∈ Rd.

5.3.2 Triangular Lagrange Pk finite elements

Suppose the domain Ω is covered by a simplicial mesh Th. The finite ele-ment method for triangular Lagrange Pk elements involves the discrete finitedimensional functional space:

V kh =

v ∈ C0(Ω) , vh|Kj ∈ Pk , Kj ∈ Th

,



and its subspace:

V k0,h =

vh ∈ V k

h , vh = 0 on ∂Ω

.

Definition 5.3.6. A triangular Lagrange Pk finite element is locally definedby a triad (K, Pk, Σk), 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)Σk is the principal lattice of order k of the simplex K ∈ Th.

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

We consider the set of points (ai)1≤i≤Ndof of the principal lattices of orderk of each of the simplices Ki ∈ Th, where Ndof is the number of degreesof freedom of the Pk finite element method. We call degrees of freedom of afunction vh ∈ V k

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

Remark 5.3.3. We observe that the nodes of the degrees of freedom coincideexactly with the vertices of the simplices Ki ∈ Th, when k = 1. The nodesof the degrees of freedom are composed by the mesh vertices and the edgemidpoints, when k = 2.

Lemma 5.3.2. The space V kh is a subspace of the space H1(Ω) of finite di-

mension corresponding to the number of degrees of freedom. Furthermore,there exists a basis (ϕj)1≤j≤Ndof of V k

h defined by:

ϕi(aj) = δij , 1 ≤ i, j ≤ Ndof ,

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

vh(x) =Ndof∑

i=1

vh(ai)ϕi(x) .

Proof.

It is easy to see that the elements of V kh belong to H1(Ω). The Lemma 5.3.1 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 (ϕj)1≤j≤Ndof is defined.

!Corollary 5.3.1. The subspace V k

0,h is a subspace of H10 (Ω) of finite dimen-

sion corresponding to the number of internal degrees of freedom, i.e.] not takingthe nodes on ∂Ω into account.

Definition 5.3.7. The triad (K, Pk, Σk) is said to be unisolvent if and onlyif the mapping v ∈ Pk -→ (ϕ1(v), . . . ,ϕNdof (v)) ∈ RNdof is an isomorphism.

The unisolvency property is equivalent to say that every function in the poly-nomial space Pk is entirely determined by its node values.



5.3.3 Finite element approximation of a boundary-value problem

We return to the numerical approximation of the solution of the homogeneousDirichlet boundary-value problem (5.12). The variational formulation of theinternal approximation reads:

find uh ∈ V k0,h such that

∫

Ω(∇uh ·∇vh)(x)+

∫


∫

Ωf(x)vh(x) dx , ∀vh ∈ V k

0,h ,

(5.17)By decomposing uh on the canonical basis (ϕj)1≤j≤Ndof and considering astest functions vh = ϕi, we obtain:

Ndof∑

j=1

uh(aj)(∫

Ω(∇ϕj ·∇ϕi)(x) dx +

∫


)=

∫

Ωf(x)ϕi(x) dx .

This formulation leads to solving a linear system in RNdof :

AhUh = Fh ,

where we have introduced the notations Uh = (uh(aj))1≤j≤Ndof and Fh =(∫

Ω fϕi dx)1≤i≤Ndof and

Ah =(∫

Ω(∇ϕj ·∇ϕi)(x) dx +

∫


)

1≤i,j≤Ndof

.

It is easy to see that the matrix Ah can be decomposed as a sum of a stiffnessmatrix Kh and a mass matrix Mh. Actually, this result is independent of thedimension.

Coefficients of Ah

Since the shape functions ϕj have a small support around a node ai, theintersection of the supports of ϕj and ϕi if often the emptyset and thus theresulting matrix Ah will contains lot of zero coefficients. It is a sparse matrix.The coefficients of Ah can be computed via an exact integration formula. Letdenote (λi(x))1≤i≤d+1 the barycentric coordinates of the point x in a simplexK ∈ Th. For every (αi)1≤i≤d+1 we have:

∫

Kλ1(x)α1 . . . λd+1(x)αd+1 dx = |K| α1 + . . . αd+1! d!( ∑

1≤j≤d+1

αj + d)!,

where |K| = meas(K) denotes the ”volume” of the simplex K. The integralsof the right-hand side term Fh are approximated using quadrature formulas in



a1

a12

a2

a23

3a

a13

P = P1 P2 Σ = p(ai), 1 ≤ i ≤ 3 Σ = p(ai), 1 ≤i ≤ 3; p(aij), 1 ≤ i < j ≤ 3

P1

K λi(m) 1 ≤ i ≤ 3

λi(m) = 1 −aim · ni

aiaj · ni, j #= i,

m = (x, y) ni Kai λi(m) j

i j1 j2 i aj1aj2 · ni = 0λi(m) ai aj

ak (ajak)P2

λi(2λi − 1), 1 ≤ i ≤ 3, 4λiλj , 1 ≤ i < j ≤ 3.

Ω ⊂ R2

k = 1Th N V k

h (ϕ1, . . . , ϕN )A ∈ RN,N

Aij =

∫

Ω∇ϕi ·∇ϕj =

∑

K∈Th

∫

K∇ϕi ·∇ϕj , 1 ≤ i, j ≤ N.

NeK(m) 1 ≤ m ≤ Ne (θm,1, . . . , θm,nf )

K(m) nf = 3 k = 1 nf = 6 k = 2(ϕ1, . . . , ϕN )

(K1, . . . , KNe)

(θm,1, . . . , θm,nf )

noeud(1 : nf , 1 : Ne),

noeud(ni, m) = ϕK(m) θm,ni.

Fig. 5.7. Nodes of the degrees of freedom on a finite element (K, P2, Σ2).

each simplex K ∈ Th. For instance the generalization of the trapeze formulafor a simplex K is:

∫

Kϕj(x) dx ≈ |K|

d + 1

d+1∑

i=1

ϕj(ai) ,

where (ai)≤i+. 1 represent the vertices of K. These formulas are exact for affinefunctions. For instance, considering a triangle K ∈ R2 of vertices (ai)1≤i≤3

and denoting by (aij)1≤i,j≤3 the midpoints of the edges aiaj (cf. Figure 5.7),the following quadrature formula is exact for ϕk ∈ P2:

∫

Kϕk(x) ≈ |K|

3

∑

1≤i,j≤3

ϕk(aij) ,

where |K| represents the area of simplex K, while the following formula isexact for ϕk ∈ P3:

∫

Kϕk(x) dx ≈ |K|

60

∑

1≤i≤3

ϕk(ai) + 8∑

1≤i<j≤3

ϕk(aij) + 27ϕk(a0)

,

where the point a0 = 1d+1

∑1≤i≤d+1 ai denotes here the barycenter of K.

Remark 5.3.4. As we mentioned in one dimension of space, the analysis canbe simplified by considering an affine transformation allowing to consider anyd-simplex K ∈ Th as the image of a reference element K. Hence, all compu-tations can be performed on this reference simplex.

5.3.4 The reference finite element

By convention, the vertices of the reference simplex K are given by the origina0 = (0, . . . , 0) and the points ai = (0, . . . , 1, . . . , 0) , for which all coordi-nates are equal to zero except the ith coordinate that is equal to one. For anon-degenerated simplex K, we denote by FK : Rd → Rd the unique affinetransformation that maps ai on ai for all i = 0, . . . , d. Hence, we write;



FK(x) = a0 + BK x ,

where BK is a d×d matrix such that the column i is given by the coordinatesof ai − a0. Since the simplex K is non-degenerated, BK is invertible and FK

is a one-to-one mapping that maps K on K and we have:

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

d!.

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

x = F−1K (x) = B−1

K (x− a0) ⇔ x = FK(x) .

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

v(x) = v(x) ⇔ v = v FK = v(FK(x)) ,

and we remove the · notation when dealing with F−1K , for every function v

defined on K, we write:

v(x) = v F−1K (x) = v(F−1

K (x)) .

Similarly, if ψ is a linear form acting on the functions defined on K, we definethe transported linear form ψ acting on the functions defined on K as:

ψ(v) = ψ(v) .

Notice that the barycentric coordinates are preserved by the affine trasnfor-mation FK :

λi(x) = λi(x) .

This leads to conclude that the principal lattice of order k of the simplex Kis the image by FK of the principal lattice of order k of K, denoted by Σk.Finally, we observe that the space of polynoials of degree k is left invariantby the transformation FK .

Proposition 5.3.2. We denote by ‖ ·‖ the Euclidean norm of Rd and itssubordinate norm. Hence, we have:

‖BK‖ ≤hK

ρ bK, ‖B−1

K ‖ ≤h bKρK

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

.

Proof.

by definition,

‖BK‖ = supv∈Rd

‖BKv‖‖v‖ = sup

‖v‖=ρcK

‖BKv‖ρ bK

,



a1

FKa2

a1a0

K

K

a0

a2

Fig. 5.8. Affine transformation of the reference linear triangle in two dimensions.

and the sup is attained. Let v ∈ Rd be a vector of norm ρ bK that correspond to thesup. Hence, there exists two points y and z on the boundary of the inscribed sphereof bK 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 andbK. The third equality is obtained by noticing that |det(BK)| is the Jacobian of FK .Hence,

|K| =

Z

K

dx =

Z

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

!As a consequence, we deduce that there exists two constants C1 and C2,

independent of K such that:

C2ρdK ≤ |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) , (5.18)

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

∇v(x) = BtK∇v(x) ,

and we obtain the identity:

|v|2H1(K) =∫

K‖∇v(x)‖2 dx =

∫

bK‖(Bt

K)‘−1∇v(x)‖2 |det(BK)| dx

= C2 d!|K|ρ2

K

∫

bK‖∇v(x)‖2 dx

,

and then:|v|H1(K) ≤ C

hK

|K|1/2ρK |v|H1( bK) , (5.19)

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

|v|H1( bK) ≤ C|K|1/2

ρK|v|H1(K) . (5.20)

We give the following generalization result.

Theorem 5.3.1. If v ∈ Hm(K), then v ∈ Hm(K) and there exists a constantC1, independent of K, such that:

∀v ∈ Hm(K) , |v|Hm( bK) ≤ C1‖BK‖m|det(BK)|−1/2|v|Hm(K) . (5.21)

If v ∈ Hm(K), then v ∈ Hm(K) and there exists a constant C2, independentof K, such that:

∀v ∈ Hm(K) , |v|Hm(K) ≤ C2‖B−1K ‖m|det(BK)|1/2|v|Hm( bK) . (5.22)

Proof.

Result admitted here. !

Remark 5.3.5. Now, we are able to construct finite elements from a unisolventtriad (K, Pk, Σk) by defining directly the finite element (K, Pk, Σk) as thetransport of these quantities by the transformation FK . It is easy to see thatthe new finite element is unisolvent.



5.3.5 Convergence of the finite element method

We introduce a preliminary result.

Proposition 5.3.3. Let consider (Th)h>0 a sequence of regular triangulationsof Ω ⊂ Rd (d ≤ 3). Then, for every v ∈ Hk+1(Ω), the interpolate rhv isdefined and there exists a constant C, independent of h such that:

‖v − rhv‖H1(Ω) ≤ C hk ‖v‖Hk+1(Ω) .

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

Theorem 5.3.2. Let consider (Th)h>0 a sequence of regular triangulations ofΩ ⊂ Rd (d ≤ 3) where all elements in all triangulations are affine equivalentto a same reference element (K, Pk, Σk) of class C0 for a given k ≥ 1 suchthat Pk ⊂ P ⊂ H1(K). Then, the Pk finite element method converges, i.e. theapproximate solution uh of the problem (5.17) converges toward the solutionu of problem (5.12) in H1(Ω):

limh→0

‖u− uh‖H1(Ω) = 0 .

Furthermore, if u ∈ Hk+1(Ω), then there exists a constant C independent ofh such that:

‖u− uh‖H1(Ω) ≤ C hk ‖u‖Hk+1(Ω) ,

Proof.

To show the convergence, we use Theorem (??) where we consider V = C∞c (Ω) ⊂

Hk+1(Ω) dense in H1(Ω). The estimate in the previous proposition allows to verifythe assumption of Theorem (??). Then, we use Cea’s lemma to write:

‖u− uh‖H1(Ω) ≤ C infvh∈V k

0,h

‖u− vh‖H1(Ω) ≤ C‖u− rhu‖H1(Ω) ,

if rhu ∈ H1(Ω). The previous proposition allows to conclude. !

Remark 5.3.6. This result involves the exact solution uh of the internal ap-proximation problem in V k

0,h. This requires computing all integrals in Ah andFh exactly. Since numerical integration formulas are used in practice, this re-sult may not be valid. Nonetheless, if these integration formulas are exact orat least very accurate, the order of convergence of the Lagrange finite elementmethod is preserved,

The Theorem 5.3.2 shows that the regularity of the exact solution has adirect impact on the order of convergence of the finite element approximation.Hence, it is often important to have an a priori knowledge of its regularity.The following result gives an indication for a convex domain.



Theorem 5.3.3. Let Ω be a convex polygon and let f ∈ R2. Then, the solu-tion u of the homogeneous Dirichlet problem:find u such that:

∫

Ω∇u.∇v =

∫

Ωfv , ∀v ∈ H1

0 (Ω) ,

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

∀f ∈ L2(Ω) , ‖u‖H2(Ω) ≤ ‖f‖L2(Ω) .

5.3.6 Numerical resolution of the linear system

We remind here that the finite element approximation of the boundary-valueproblem (5.12) implies the numerical solving of a linear system in RNdof :

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

where Uh = (uh(aj))1≤j≤Ndof and Fh = (∫

Ω fϕi dx)1≤j≤Ndof . The genericterm of the stiffness matrix Kh is of the form:

kij = a(ϕj , ϕi) , 1 ≤ i, j ≤ Ndof .

where a(·, ·) is the bilinear form defined on the functional space H10 and

(ϕj)1≤j≤Ndof is the canonical basis of the approximation space. We have men-tioned already that the stiffness matrix Kh inherits of the properties of thebilinear form a. More precisely, if a is V-elliptic on X, then Kh is positivedefinite and if a is symmetric then Kh is symmetric. We have also indicatedthat 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 methods are preferred tosolve this linear system (cf. Chapter 3).

Let denote 0 < λ1(Kh) < · · · < λNdof (Kh) the eigenvalues of Kh and0 < λ1(Mh) < · · · < λNdof (Mh) the eigenvalues of Mh. We introduce thenotations:

κ(Kh) =λNdof (Kh)

λ1(Kh), and κ(Mh) =

λNdof (Mh)λ1(Mh)

,

and we have the following result.

Proposition 5.3.4. Suppose (Th)h>0 is a quasi-uniform sequence. Then,there exists two constants C1 > 0, C2 > 0 and a constant C3 > 0 suchthat:

C1 ≤ h2κ(Kh) ≤ C2 , and κ(Mh) ≤ C3 .

We observe that the mass matrix Mh is always well-conditioned since its con-dition number κ(Mh) is independent of the mesh size h, under the conditionthat the refinement is quasi-uniform. On the other hand, the stiffness matrixKh becomes ill -conditioned, i.e. κ(Kh) 4 1, when the mesh size h is small.



Remark 5.3.7. We have shown that the approximation error between the exactsolution u and the Galerkin finite element solution uh is bounded by above bya constant times the ”distance” between he spaces V and Vh. The smaller thedistance, the better the approximation error will be. Is seems then natural toincrease the dimension of the approximation space Vh in order to improve theaccuracy of the solution. To this end, we can either:

1. increase the number of elements and thus increase the global numberof degrees of freedom while preserving the local reference element. Thisstrategy is known as h-refinement,

2. increase the degree of the polynomials and thus modifying the local ref-erence element while preserving the total number of elements. Obviously,this choice can be only be applied if the exact solution u is suffcientlysmooth. This strategy is known as p-refinement.

We will discuss these adaptation options and others in the chapter 6.

5.4 The finite element method for the Stokes problem

To illustrate the application of the finite element method to solve a system ofpartial differential equations, we consider the Stokes problem in two dimen-sions of space.

We consider the flow of a fluid inside a bounded domain Ω ⊂ R2 andsubjected to an external force field f . The flow is considered as stationaryand inertial forces are supposed negligible here, hence the Stokes equationsintroduced in Chapter ?? describe this viscous flow problem:Find (u, p) in appropriate Hilbert spaces such that:

−ν∆u +∇p = f in Ω

div u = g in Ω

u = 0 on ∂Ω

(5.23)

where the unknowns u and p represent the velocity and the pressure of thefluid, respectively and ν > 0 is the viscosity. Usually, in the applications, theflow is incompressible and the function g is equal to zero.

5.4.1 Mixed formulation

We have already seen the mixed formulation of this problem in Chapter ??.We introduce the functional space:

L20(Ω) = q ∈ L2(Ω) ,

∫

Ωq dx = 0 ,

and the mixed formulation reads:Given f ∈ L2(Ω) and g ∈ L2

0(Ω), find u ∈ H10 (Ω)2 and p ∈ L2

0(Ω) such that:


5.4 The finite element method for the Stokes problem 247

a(u, v) + b(v, p) = f(v) , ∀ v ∈ H1

0 (Ω)2 ,

b(u, q) = g(q) , ∀ q ∈ L20(Ω)

(5.24)

where we posed a(u, v) = ν

∫

Ω∇u : ∇v, b(v, p) = −

∫

Ωp div v, f(v) =

∫

Ωfv

and g(q) = −∫

Ωgq. The functional spaces H1

0 (Ω)2 and L2(Ω)2 are endowed

with the canonical norms:

‖u‖H1(Ω) =

∑

i=1,2

‖ui‖2H1(Ω)

1/2

, ‖f‖L2(Ω) =

∑

i=1,2

‖fi‖2L2(Ω)

1/2

,

for u = (ui)1≤i≤2 ∈ H10 (Ω)2 and f = (fi)1≤i≤2 ∈ L2(Ω)2.

Remark 5.4.1. If f ∈ C0(Ω)2 and g ∈ C0(Ω), if Ω is of class C2 and if thesolution (u, p) of the problem (5.24) is such that u ∈ C2(Ω)2 and p ∈ C1(Ω),then (u, p) is the classical solution of the Stokes problem.

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

V0 = v ∈ H10 (Ω)2 , div v = 0 .

Since the divergence operator is surjective, there exists a function ug ∈H1

0 (Ω)2 such that for every g ∈ L20(Ω), we have:

div ug = g , and ‖ug‖H1(Ω)2 ≤ c‖g‖L2(Ω) ,

where c > 0 is a constant (see [?] for more details). Introducing the changeof variables u′ = u− ug leads to the condition div u′ = 0 and to consider thefollowing weak formulation:find u′ ∈ V0 such that:

a(u′, v) = 〈f, v〉 − a(ug, v) , ∀v ∈ V0 . (5.25)

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

Proposition 5.4.1. The problem (5.25) is a well-posed problem.

Proof.

It is a direct application of Lax-Milgam theorem. ! Although thisnew formulation has some theoretical advantages, many practical drawbacksprevent its application for solving a discrete problem. It is indeed difficult toconstruct divergence-free finite elements. The reader interested is referred tothe book [?] for more details on this topic.



5.4.2 Numerical approximation

We will now investigate the condition required for the approximation spacesin velocity and pressure to define a well-posed discrete problem.

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

the space L20(Ω). We introduce the discrete Stokes problem:

Find uh ∈ Xh and ph ∈ Mh such that:

ν

∫

Ω∇uh : ∇vh −

∫

Ωph div vh =

∫

Ωf · vh , ∀vh ∈ Xh ,

∫

Ωqh div uh =

∫

Ωgqh , ∀qh ∈ Mh .

(5.26)

We introduce the space

Ker(Bh) = vh ∈ Xh ,

∫

Ωqh div vh = 0 , ∀qh ∈ Mh ,

and we enounce the fundamental compatibility result.

Theorem 5.4.1. The discrete problem (5.26) is a well-posed problem if andonly if the spaces Xh and Mh satisfy the following compatibility condition:

∃βh > 0 , infqh∈Mh

supvh∈Xh

∫

Ωqh div vh

‖qh‖L2(Ω)‖vh‖H1(Ω)≥ βh . (5.27)

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

‖u− uh‖H1(Ω) ≤ c1h infvh∈Xh

‖u− vh‖H1(Ω) + c2h infqh∈Mh

‖p− qh‖L2(Ω) (5.28)

‖p− ph‖L2(Ω) ≤ c3h infvh∈Xh

‖u− vh‖H1(Ω) + c4h infqh∈Mh

‖p− qh‖L2(Ω) (5.29)

where the constants (cih)1≤i≤4 are given by Theorem ??.

Proof.

We know that the bilinear form a(·, ·) is V -elliptic on H10 (Ω)2 × H1

0 (Ω)2 and the

result is then a direct consequence of Theorem ??. !The following result provides a practical method for checking whether the

compatibility condition is satisfied or not.

Lemma 5.4.1 (Fortin). Let consider the spaces X = H10 (Ω)2 and M =

L20(Ω) and let b ∈ L(X,M). Suppose there exists β > 0 such that:

infq∈M

supv∈X

b(v, q)‖v‖X‖q‖M

≥ β .

Then, there exists βh > 0 such that:



infqh∈Mh

supvh∈Xh

b(vh, qh)‖vh‖Xh‖qh‖Mh

≥ βh ,

if and only if there exists γh > 0 such that for every v ∈ X, there exists arestriction 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 ≤ γh‖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

≥ 1γh

supv∈X

b(v, qh)‖v‖X

≥ βγh‖qh‖M .

Conversely, if the discrete condition is satisfied, then the operator Bh : Xh → Mh

defined by b(vh, qh) = 〈Bhvh, qh〉 defines 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 ≤ 1β‖Bhrh(v)‖M′

≤ 1β

supqh∈Mh

b(rh(v), qh)‖qh‖M

≤ 1β

supah∈Mh

b(v, qh)‖qh‖M

≤ γh

β‖v‖X .

The operator rh has the desired properties. !This proof provides a technique to verify the discrete compatibility con-

dition; it consists in constructing the operator rh using the solution of thesystem:

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, it is possible to construct a simpler operator for specificapproximation spaces and finite elements and to check if the compatibilitycondition is satisfied.

It is easy to construct an internal variational approximation of the Stokesproblem, as we have already seen:find uh ∈ Xh and ph ∈ Mh such that:



a(uh, vh) + b(vh, ph) = f(vh) , ∀vh ∈ Xh ,

b(uh, qh) = g(qh) , ∀qh ∈ Mh ,(5.30)

If we denote by nu (resp. np) the dimension of the space Xh (resp. Mh), weintroduce the basis (ϕj)1≤j≤nu of Xh and the basis (ψj)1≤j≤np of Mh definedusing the basis shape functions of the finite elements and we decompose uh

and ph on these basis:

uh =nu∑

j=1

uh(aj)ϕj(x) , and ph(x) =np∑

j=1

ph(a′j)ψj(x) .

We obtain the following linear system to solve:(

Ah Bth

Bh 0

) (Uh

Ph

)=

(Fh

Gh

). (5.31)

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 = ν

∫

Ω∇ϕi : ∇ϕj dx , and bki = −

∫

Ωψi div ϕk dx .

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

fi =∫

Ωf · ϕi dx , and gk =

∫

Ωg ψk dx .

Lemma 5.4.2. 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 =

Z

Ω

rh div wh dx .

Since rh = 1 is in Mh and since

∀wh ∈ Xh ,

Z

Ω

div wh dx =

Z

∂Ω

wh · n ds = 0 .

we conclude that Rh = 1 ∈ Ker(Bth). !

Finding a pair of approximation spaces (Xh, Mh) for the Stokes problemhas been an active research field in numerical analysis. We present here a fewexamples of such pairs.



The P1/P1 finite element

This seems the obvious first choice for implementing the finite element methodfor the Stokes problem. We consider the unit domain Ω =]0, 1[2 and we sup-pose a regular Cartesian mesh is defined on Ω (Figure 5.9). We introduce theapproximation spaces:

Xh = uh ∈ C0(Ω)2 , ∀K ∈ Th , uh|K ∈ P21 , uh|∂Ω = 0 ,

Mh = ph ∈ L20(Ω) ∩ C0(Ω) , ∀K ∈ Th , ph|K ∈ P1 .

For a given element K ∈ Th, we denote (aiK)1≤i≤3 its vertices. Then, wehave:

∀vh ∈ Xh ,

∫

Ωph 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 byits values 0, −1, 1 at the three 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 ,

∫

Ωph div vh = 0 .

This is clearly in contradiction with the requirement of the inf-sup condi-tion. Hence, the P1/P1 finite element shall be discarded for solving the Stokesproblem.

0 1

-1

0

-1

1

-1

0

1

1

0-1

1

1

0

-1 0

-1

1

1

-1

0

1

-10

Fig. 5.9. A regular Cartesian mesh of the unit domain Ω =]0, 1[2 and the unstablepressure mode for the P1/P1 finite element.



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 aniterative method will result in oscillations in the pressure values. On the otherhand, if dim(Ker(Bt

h)) = 1, the discrete pressure can be uniquely determinedby fixing its value at a mesh node or by setting its average value on the domainΩ.

The P1/P0 finite element

This finite element presents the a priori advantage of being very simple toimplement. Given a triangulation of the domain Ω, we define a piecewisecontinuous approximation of the velocity and a piecewise constant (and thusdiscontinuous) approximation for the pressure. Since the velocity is piecewiseaffine, its divergence is piecewise constant and thus by testing the divergenceusing constant functions, the divergence of the discrete field is strongly forcedto be null. Hence, this finite element allows to obtain exactly a divergencefree velocity field. Nonetheless, the P1/P0 finite element does not fulfill theinf -sup condition and Fortin’s criterion.

Again, we provide a counter-example in two dimensions. Suppose the tian-gulation Th of the simply connected polygonal domain Ω contains t triangles,vi internal vertices and vb boundary vertices. We shall have 2vi degrees offreedom 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-freeconstraints. The Euler formulas 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 theoremallows 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, the space Mh is too rich to impose the Bhuh = 0 constraint. Afunction uh ∈ Xh is thus overconstrained and a 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 spaceof the degrees of freedom of the velocity is not large enough. To overcome this



244 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). !

Fig. 5.10. The P1-bubble/P1 (mini) element for velocity (left) and pressure (right)approximation.

problem, it seems natural to attempt enriching the velocity space rather thanimpoverishing the pressure space. However, it is not necessary to consider aspace of polynomials of degree two for approximating the velocity. Crouzeixand Raviart suggested (in 1973) to add one extra degree of freedom to thebarycenter of every simplex of the triangulation Th of the domain Ω. Hence,we define in two dimensions:

P1 = (P1(K)⊕ spanbK , K ∈ Th)2 ,

with the bubble function bK defined in two dimensions as:

bK = 273∏

i=1

λi , and generalized in any dimension as bK = (d+1)d+1d+1∏

i=1

λi ,

where (λi)1≤i≤3 denote the barycentric coordinates on K. The degrees offreedom associated with P1 are defined as:

Σk = v -→ v(ai) , 1 ≤ i ≤ 3 ∪ v -→∫

ei

v · ni dσ .

Then, we consider the approximation spaces:

Xh = vh ∈ C0(Ω)2 , ∀K ∈ Th , vh|K ∈ P1 , uh|∂Ω = 0 ,

Mh = ph ∈ L20(Ω) ∩ C0(Ω) , ∀K ∈ Th , ph|K ∈ P1 .

(5.32)It is easy to see that Xh ⊂ (H1

0 (Ω))2.Before giving a compatibility result for this finite element, we need to

introduce an interpolation operator rh. Since v ∈ H10 (Ω)2) is not necessarily

continuous, its Lagrange interpolate might not be defined. Therefore, in eachtriangle, we define rh as:

rh(v) = Ch(v) + αK bK ,

where Ch(v) = (Ch(v1), Ch(v2)) is the Clement interpolate defined hereafterand αK ∈ R2 is chosen so as to satisfy the equation:



∫

K(rh(v)− v) = 0 .

To this end, we define:

αK =(∫

KbK

)−1 ∫

K(v − Ch(v)) .

For every v ∈ Xh and qh ∈ Mh, we write:∫

Ωqh div(v − rh(v)) =

∫

Ω∇qh(rh(v)− v) =

∑

K∈Th

∫

K∇qh(rh(v)− v) = 0 ,

since ∇qh is piecewise constant on every K ∈ Th.

Definition 5.4.1. Let Th be a triangulation of Ω, Γh be the set of the internalvertices of Th. We introduce Θh the space of continuous piecewise linear func-tions on Th which are zero on the boundary and we denote (ϕγ)γ∈Γh the basisof P1 finite elements with homogeneous boundary conditions. We introduce theoperator Ch ∈ L(H1

0 (Ω), Θh), proposed by Clement in 1975, and defined asfollows:

Ch(v) =∑

γ∈Γh

(1

|ωγ |

∫

ωγ

v

)ϕγ ,

where the set ωγ is defined as the union of all triangles K ∈ Th for which γis a vertex.

In this definition, the nodal value v(γ) is replaced in the definition of theinterpolant rh by the local average value around γ. The Clement operator canbe applied to any function in L2(Ω). We have the following result.

Lemma 5.4.3. If the family of triangulations (Th)h>0 is quasi-uniform, thenthe operator Ch satisfies the local estimates:

∀v ∈ H10 (Ω) , ‖v − Ch(v)‖L2(K) ≤ C hK |v|H1(ωK) ,

∀v ∈ H10 (Ω) , |Ch(v)|H1(K) ≤ C |v|H1(ωK) ,

where ωK is the union of all triangles sharing at least one of the vertices ofK.

Theorem 5.4.2. If the family of triangulations (Th)h>0 is quasi-uniform, theapproximation spaces Xh and Mh are uniformely compatible with respect to h,i.e. there exists a constant βh, independent f h, satisfying the condition (5.27).



Proof.

Let consider v ∈ H10 (Ω)2. We construct the operator rh(v) ∈ Xh such that

∀qh ∈ Mh ,

Z

Ω

qh div(rh(v)) =

Z

Ω

qh div(v) .

Since Mh ⊂ H1(Ω), the previous identity is equivalent to writing:

∀qh ∈ Mh ,X

K∈Th

Z

K

rh(v) ·∇qh =X

K∈Th

Z

K

v ·∇qh .

Since ∇qh ∈ Pd0, the previous identity leads to conclude that:

∀K ∈ Th ,

Z

K

rh(v) =

Z

K

v .

Since the operator rh is composed of two terms, we analyze them separately. Re-garding the Clement interpolate Ch, we invoke the previous estimate:

∀v ∈ H10 (Ω) , |Ch(v)|H1(K) ≤ C |v|H1(ωK) .

For the term αKbK , we write:

|αKbK |H1(K) ≤ C|αK | ,

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

|αK | =

„Z

K

bK

«−1

|Z

K

(v − Ch(v))|

≤ C1

|K| |Z

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(ωK) , to obtain:

|αK | ≤ C |v|H1(ωK) ,

and finally:|rh(v)|H1(K) ≤ C |v|H1(ωK) ,

By taking the square of this expression and summing over all K ∈ Th we concludethat:

‖rh(v)‖H1(Ω) ≤ C‖v‖H1(Ω) .

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

Proposition 5.4.2. Suppose the solution (u, p) of the problem (5.24) is sif-ficiently smoth, i.e. u ∈ H2(Ω)2 ∩ H1

0 (Ω)2 and p ∈ H1(Ω) ∩ L20(Ω). Then,

the solution (uh, ph) of the problem (5.26) with the spaces defined by (5.32) issuch that:

‖u− uh‖H1(Ω) + ‖p− ph‖L2(Ω) ≤ C h (‖u‖H2(Ω) + ‖p‖H1(Ω)) . (5.33)



5.4.3 Numerical resolution

We have seen that the discrete problem (5.30) leads to solve the linear system(5.31). We assume here the natural hypothesis that the bilinear form a(·, cdot)is V -elliptic on X with a constant α and that the form b(·, ·) satisfies onxh ×Mh a inf -sup condition uniformely in h, and we denote the constant β.

The resolution of the linear system using a direct method is likely to becostly because of the number of degrees of freedom and thus the size of thelinear system. Direct methods are appropriate up to Ndof = 104. For thisreason, iterative methods are favored. However, the matrix of the system(5.31) is neither positive nor definite. It is indeed an indefinite matrix (butsymmetric) with that zero block on the diagonal. It will have positive and alsonegative eigenvalues. It is possible to make it positive by solving the system:

AhU + Bt

hP = Fh

−BhU = −Gh. (5.34)

However, the new matrix corresponding to this system is still indefinite andis no longer symmetric. We know from Chapter 3 that gradient methods arenot efficient on this type of matrix. We present two commonly used methodsfor 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 problems in general) because of its versatility and facilityto be implemented. It consists in replacing the problem (5.31) by the followingproblem:

AhUε + BthPε = Fh

−BhUε + εShPε = −Gh, (5.35)

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

(ShP,Q) = 〈ph, qh〉M ,

where (·, ·) and 〈·, ·〉M denote the standard Euclidean scalar product in Rnp

and the inner product in M , respectively. In the system (5.35), ε is a penal-ization coefficient sufficiently small.

The main idea is simple. Suppose Ah is a positive definite symmetric ma-trix, the system (5.31) is then equivalent to:

infBhVh=Gh

12(AhVh, Vh)− (Fh, Vh)

.

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



infVh

12(AhVh, Vh) +

12ε

(S−1h (BhVh −Gh, BhVh −Gh)− (Fh, Vh)

,

or equivalently, by writing Pε = (1/ε)S−1h (BhVh −Gh):

AhUε + Bt

hPε = Fh

BhUε −Gh = εShPε. (5.36)

By replacing the quantity Pε in the first equation, we obtain the linear system:(

Ah +1εBt

hS−1h Bh

)Uε = Fh +

1εBt

hS−1h Gh . (5.37)

If the matrix S−1h is a sparse matrix, this provides an efficient technique to

solve our problem. This system can be solved by an iterative technique like theconjugate gradient, since the matrix is now positive definite and symmetric.Notice however that the term 1/εBt

hS−1h Bh has a strong negative impact on

the condition number of the linear system (5.4.3).We now investigate the effect of changing the constrained problem to a

penalized problem. We have the following error estimate.

Proposition 5.4.3. Let consider ε > 0 and let (U, P ) and (Uε, Pε) denote thesolutions of system (5.34) and (5.36), respectively. Then, we have the errorestimate:

αβ‖U − Uε‖X + αβ2‖P − Pε‖M ≤ Cε‖P‖M . (5.38)

Proof.

By difference, we obtain the system:(

Ah(U − Uε) + Bth(P − Pε) = 0

−Bh(U − Uε)− εShPε = 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 , α‖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≥ β ,

or equivalently:∀P ∈ Rnp , β‖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 − Pε‖M ≤ 1β‖Bt

h(P − Pε)‖∗

=1β‖Ah(U − Uε)‖∗ ≤

Cβ‖U − Uε‖X .

By multiplying the first equation by (U−Uε) and using the V -ellipticity of the forma we obtain for the second equation:

α‖U − Uε‖2X ≤ (Ah(U − Uε), U − Uε) = (Bth(Pε − P ), U − Uε) = (Pε − P, Bh(U − Uε)) = −ε(Pε − P, ShPε)

= −ε(Pε − P, Sh(Pε − P ))− ε(Pε − P, ShP )

≤ −ε(Pε − P, ShP )

≤ ε‖Pε − P‖M ‖P‖M .

The error estimate is obtained by combining the two previous inequalities. !Remark 5.4.2. Notice that discretizing a penalized problem is not necessarilyequivalent to penalize a discrete problem [?]. The penalty method is consideredin this last case as a procedural (algorithmic) technique, since a choice ofspaces Xh ⊂ X and Mh ⊂ M has been done. On the other hand, discetizingthe penalized problem consists in choosing Qh = BhVh that is in general apoor choice.

Uzawa’s operator

To conclude this section, we introduce an iterative, although very efficient,algorithm to solve such problem when th operator Ah associated with thebilinear form a is invertible. We know since Chapter ?? that the Stokes equa-tions are equivalent to solving a problem of energy minimization. More pre-cisely, the resolution of the linear system (5.31) is equivalent to the followingminimization:

J(Uh) = minVh∈KerBh

J(Vh) with J(Vh) =12(AhVh, Vh)− (Fh, Vh) .

(5.39)In the optimization terminology, the pressure is the Lagrange multiplier thatimposes the constraint div u = g when the energy is minimized.

First, we eliminate the variable Uh from the linear system (5.31), if thematrix Ah is invertible, thus leading to:

Uh = A−1h (Fh −Bt

hP )

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

BhA−1h Bt

h Ph = BhA−1h Fh −Gh . (5.40)

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

Proposition 5.4.4. If matrix Ah is positive definite, this matrix is also pos-itive definite.



Proof.

One has:(BhA−1

h BthP, P ) = (A−1

h BthP, Bt

hP ) ≥ α‖BthP‖ .

It is positive definite if KerBth = 0. !

Hence, problem (5.40) is more easy to solve than the original problem (5.31) asefficient methods exist for positive definite systems. Unfortunately, even if thematrix Ah is a sparse matrix (i.e. contains lot of zero values), its inverse A−1

h islikely to be a full matrix. The resolution of (5.40) must be performed using aniterative method since inverting the matrix A−1

h is extremely computationallycostly. The convergence rate of iterative methods is strongly related to thecondition number of the matrix.

Proposition 5.4.5. If the matrix U is symmetric, we have the estimate:

κ(U) ≤ C

α2β2κ(Sh) .

Remark 5.4.3. If the family of mesh (Th)h>0 is quasi-uniform, the conditionnumber κ(Sh) is a constant independent of h. Hence, the previous estimateshows that the conditioning of Uzawa’s operator is of order one (if α and βare of order one, which is a reasonable hypothesis). The iterative techniquesconverge fast without any preconditioning.


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5 The ﬁnite element approximation · 2017. 5. 9. · 210 5 The ﬁnite element approximation (iii) If in addition the bilinear form a(·,·) is symmetric the previous estimate becomes: