Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Galerkin Methods
A Project
submitted to the African University of Science and Technology (AUST)
Abuja-Nigeria
in partial fulfilment of the requirement for

MASTER DEGREE IN PURE AND APPLIED MATHEMATICS

By
Edith Laure Ngouanfo Fopa
Supervisor:
Doctor Guy Degla

African University of Science and Technology
www
...
edu
...
M
...
C
...


/home/engouanfo/Desktop/FinalMsC/logo
...
it is certain that the real function of art is to increase our
self-consciousness; to make us more aware of what we are, and therefore of
what the universe in which we live really is
...
It is an art, and a great art
...
N
...
The gratefulness goes also to Prof C
...
I would like to thank also my supervisor
Dr Guy Degla for having introduced me to the study of Galerkin Method
and for all the time, the great patience, and the attention he devoted to my
work
...

Many thanks goes furthermore to Prof
...
Leonard
Todjihounde, to Dr Ngalla Djitte, to Prof
...
Mbianda Gilbert from my home institute
...
Among them I
would like to thank my AUST classmates, especially the dear ones Johnson,
and Usman
...

But my first thought and my warmest thank goes to Bolaji : “let me dedicate this to you
...
It deals with Galerkin Methods that convert, by approximation,
a continuous operator problem (in an infinite dimensional separable Hilbert
space) to a sequence of discrete problems (in Euclidean spaces)
...
We also recall the Lax-Milgram Theorem (in
its general form) which is fundamental here and proves the well-posedness of
variational problems of the form

Find u ∈ H such that,
(P)
A(u, v) = L(v), ∀v ∈ H
where H is a Hilbert space, L : H −→ R a bounded linear functional and
A : H × H −→ R is a coercive and continuous bilinear form
...
Furthermore, the motivation for solving (P)
numerically comes from the fact that it models an important class of Boundary Value Problems and its symmetric case (i
...
, when A is symmetric) is
equivalent to an optimization problem; more precisely, the minimization of
the quadratic functional u 7→ A(u, u)/2 − L(u)
...
On the other hand we also
give an abstract proof of the Galerkin method (cf
...
5
...

Moreover we emphasize on the variants of Galerkin Methods and their relation
with the Finite Element Method and finally provide some examples
...

In this section, we review some basic notions of Functional Analysis and Variational Equations
...
1

Banach spaces

Definition 1
...
1
Let X be a linear space over K
...


ii) ||λx|| = |λ| ||x|| ,

∀ x ∈ X and ∀ λ ∈ K
...


A linear space X endowed with a norm || · || is called a normed linear space
and is denoted by (X, || · ||)
...

Definition 1
...
2
Let X be a K-linear space
...


Definition 1
...
3
Let (X, || · ||) be a normed linear space
...


n→+∞

In this case a is unique and we also say that (xn )n converges to a with
respect to the norm || · || and then write
lim xn = a
...
1
...

1
...
That
is:
(an )n ⊂ A converges in X =⇒
lim an ∈ A
...
A subset U of X is said to be open if its complement; U c = X \ U , is
closed
...
The collection = of all open sets of the normed linear space (X, || · ||)
define a topology on X
...

4
...
The closure of A is in fact the intersection of all closed sets
of X containing A
...

A subset D ⊆ X of which closure is equal to X, is said to be dense in
X
...
The interior of a subset A of X is the largest open set (of X) which is
contained in A
...

˚ or int(A)
...
A subset A ⊂ X is said to be bounded if there exists a positive constant
M such that
x ∈ A =⇒ ||x|| < M
...

Definition 1
...
5
A normed linear space (X, || · ||) is said to be separable if it contains a dense
subset D which is at most countable
...
1
...
Then a sequence of elements of X converges with
respect to || · ||1 if and only if it does with respect to || · ||2
...

Vice-versa if two norms define the same topology, then they are equivalent
...
1
...
A sequence (xn )n of elements of X is
said to be a Cauchy sequence if
lim

m,n→+∞

||xn − xm || = 0 ;

that is,
∀ε > 0, ∃ N : ||xn − xm || < ε for all n ≥ N and all m ≥ N
...
1
...

In a normed linear space, every convergent sequence is a (bounded) Cauchy
sequence
...
1
...

Definition 1
...
10
Given a normed linear space (X, || · ||X ), any Banach space (E, || · ||E ) such
that there exists an isometry j : X → E with dense range in E; i
...
,
(i) ||j(x)||E = ||x||X for all x ∈ X, and
(ii) j(X) = E,
is called a completion of X
...

Theorem 1
...
11 (Hausdorff )
Every normed linear space has a completion
...
1
...
A map
f : X → Y is said to be continuous at a point a ∈ X if for every sequence


(xn )n of X converging to a with respect to || · ||X , the sequence f (xn ) n converges to f (a) in Y with respect to || · ||Y
...

Equivalently, f is continuous if and only if the pre-image of every open set in
Y is an open set in X
...

Equivalently, T : X −→ Y is linear if for all x1 , x2 ∈ X and for all α ∈ K,
we have
T (αx1 + x2 ) = αT (x1 ) + T (x2 )
...
1
...
Then a linear map
T : X → Y is continuous if and only if T is a bounded linear map in the sense
that there exists a constant real number α ≥ 0 such that
||T (x)||Y ≤ α ||x||X

∀x ∈ X
...
1
...

The set of all bounded linear maps (i
...
continuous linear maps) from X into
Y is a linear space that will be denoted by B(X, Y )
...

4

We denote by X ∗ := B(X, K) the topological dual of X; that is, the set of all
continuous linear functionals of X
...
1
...
Then for every T ∈ B(X, Y ), we have
||T (x)||Y ≤ ||T || ||x||X

∀x ∈ X ,

and
||T || =

sup ||T (x)||Y =
||x||X ≤1

sup ||T (x)||Y =
||x||X =1

||T (x)||Y

...
1
...
Then


1
...

2
...


B(X, Y ), || · ||B(X,Y )




is

Corollary 1
...
17
The dual X ∗ of any normed linear space X is (always) a Banach space
...
1
...

Moreover there exists a canonical injection J : X ,→ X ∗∗ defined by
J : X −→ X ∗∗
x 7−→ J(x) ,
where J(x) the continuous form on X ∗ defined by
hJ(x), f i := hf, xi := f (x) ; ∀ f ∈ X ∗
...
1
...


5

Theorem 1
...
20 (Hahn-Banach)
Let (X, ||
...

If f : E −→ R is a continuous linear functional, then there exists F ∈ X ∗ that
extends f such that
||F ||X ∗ = ||f ||E ∗
...
1
...

Theorem 1
...
22 (Uniform Boundedness Principle)
Let (X, || · ||X ) and (Y, || · ||Y ) be two Banach spaces and let {Ti }i∈I be a
family (not necessarily countable) of continuous linear operators from X into
Y
...

i∈I

Then
sup ||Ti ||B(X,Y ) < ∞
...

Theorem 1
...
23 (Open Mapping Theorem)
Let (X, || · ||X ) and (Y, || · ||Y ) be two Banach spaces and let T be a continuous
linear operator from X into Y that is surjective
...

Corollary 1
...
24
Let (X, || · ||X ) and (Y, || · ||Y ) be two Banach spaces and let T be a continuous
linear operator from X into Y that is bijective
...
1
...
Assume that the graph of T , G(T ), is closed in E × F
...

Remark 1
...
26
The converse is obviously true, since the graph of any continuous map (linear
or not) is closed
...
2

Hilbert spaces

Definition 1
...
1 (Inner product)
An inner product on a K-linear space E is any functional h , i defined on E × E
which is a positive Hermitian and nondegenerate form; that is,
h , i : E × E −→ K
(x, y) 7→ hx, yi
and satisfies the following conditions :
1
...

2
...


3
...


∀ x1 , x 2 , y ∈ E

A linear space endowed with ann inner product is called an inner product
space or a pre-Hilbert space
...
2
...
Then
|hx, yi|2 ≤ hx, xi hy, yi

∀ x, y ∈ E
...

Theorem 1
...
3 (Norm induced by an inner product)
Let (E, h , i) be an inner product space
...

Definition 1
...
4 (Hilbert space)
The inner product space E is called be Hilbert space, and denote by H, if
(E, ||
...

Theorem 1
...
5 (Parallelogram law and Polarization identity)
Let (E, h , i) be an inner product space
...
the parallelogram law :
||x + y||2 + ||x − y||2 = 2(||x||2 + ||y||2 )
7

∀ x, y ∈ E ,

2
...


Definition 1
...
6 (Orthogonality of vectors, Orthogonal of a set)
Let (E, h , i) be an inner product space
...

If M is a non empty subset of E, we write x⊥M (and read x orthogonal to
M ) if x is orthogonal to every element of M
...
That is,
n
o
M ⊥ = x ∈ E , hx, yi = 0 ∀ y ∈ M
...

Theorem 1
...
7 (Projection Theorem)
Let H be a Hilbert space and M a closed subspace of H
...

m∈M

Furthermore, x ∈ M is such a vector if and only if (x − x)⊥M
...
2
...
Then E is said to be the
direct sum of X and Y if
E = X + Y and X ∩ Y = {0}
...

in this case we write E = X ⊕ Y
...
2
...
Then,
H = M ⊕ M⊥
...
2
...
Then,
8

(i) There exists a unique vector yo in H such that
f (x) = hx, yo i,

for every x ∈ H
...

Corollary 1
...
11
If H be a Hilbert space, then

H ' H∗

via the canonical isometric bijection
ϕ : H −→ H ∗
a 7−→ ϕa ,
with

ϕa : H −→
R
x 7−→ hx, ai
...


h , i is used to denote a duality pair
...


J is linear, injective, and is an isometry (i
...

If J is onto, then E is said to be reflexive
...
2
...

Definition 1
...
14
Let E be a real Banach space
...

We denote the family of all such maps from E into R by {φf (x)}f ∈E ∗
...

9

Proposition 1
...
15
Let (E, ω) be a real Banach space endowed with the weak topology ω
...

Consequently if a sequence {xn }n∈N in (E, ω) converges weakly to some x
(i
...

Proposition 1
...
16
A sequence {xn }n∈N in a real Banach space E converges weakly to some x in
E if and only if f (xn ) −→ f (x) for each f ∈ E ∗
...
2
...

Definition 1
...
18
Let E be a real reflexive space
...


(ii) A bilinear form A : E × E → R ∪ {+∞} is said to be coercive (or elliptic)
if
∃β > 0 such that A(u, u) ≥ β||u||E 2 for all u ∈ E
...
3

Basic terminologies on classical Partial Differential Equations (PDEs)

Let Ω be a nonempty open subset of Rn , n ∈ N
...
3
...

We can write a typical PDE in a form as follows:


F Dk u(x), Dk−1 u(x),
...
1)
10

where k ≥ 1 is an integer and
k

F : Rn × Rn

k−1

× · · · × Rn × R × Ω −→ R,

f : Ω −→ R,

are given functions,
u : Ω −→ R,
is the unknown function, and


∂ku
k
D u =

...
∂xik i1 ,i2 ,
...
,n}
Definition 1
...
2
1
...


|α|≤k

for given functions aα (| α | ≤ k) and f such that that the functions aα
are not all equal to zero
...

2
...
, Du, u, x) = 0
...
A partial differential equation is said to be quasilinear if it has the form

X

aα (Dk−1 u,
...
, Du, u, x) = 0
...
A partial differential equation is said to be fully nonlinear if it depends
nonlinearly upon the highest order derivatives
...

The classification of second order linear PDEs
n
X
∂u
∂ 2u
+
bi (x)
+ c(x)u = f ,
aij (x)
∂xi ∂xj
∂xi
i=1
i,j=1
n
X

(1
...
, n}, is done according to the signature of
the symmetric matrix
 
A = aij

...
3
...
2) is said to be


1
...
hyperbolic if all the eigenvalues of A = aij 1≤i,j≤n are nonzero and have
the same sign except one,


3
...

Example 1
...
4
Consider the linear second-order PDE in two variables with
a11

∂ 2u
∂ 2u
∂ 2u
∂u
∂u
+
2a
+
a
+
b
+
b
+ cu = f ,
12
22
1
2
∂x2
∂x∂y
∂y 2
∂x
∂y

(1
...

Then at the point (xo , yo ), the equation (1
...

(ii) parabolic if d = 0
...

Definition 1
...
5 (Boundary conditions)
In general, in mathematical models, additional information on the boundary
of the domain Ω (according to the time or the spatial variable) are supplied
with the PDE
...

The PDE is not generally sufficient to ensure the unicity of the solution u
...
Such data is also called a boundary condition
...

Three main types of boundary conditions exist: the Dirichlet Condition, the
Neumann Condition and the Robin Condition
...
3
...
Dirichlet condition
...

Numerically, the Dirichlet condition guarantees existence and uniqueness
of second order elliptic PDEs
...
For example, a 1D wave equation
with a constant boundary condition describes the motion of a string with
fixed ends
...
Neumann Condition
The Neumann condition specifies the values of the normal derivative on
the boundary of the domain
...


In this condition, the values of u on the boundary are unknown
...
When
the Neumann condition is constant, it implies that the rate of variation of
u across the boundary does not change
...

3
...
One can define the Robin condition with the following
prescribtion :
∂u
c1 u + c2 ∂n
= g on ∂Ω

13

1
...

(i) A multi-index is a vector of the form α = (α1 ,
...
, k}
...

(a) We write
x = (x1 ,
...


u(x) = u(x1 ,
...

(c) If u is continuous, then the support of u is defined and denoted by
n
o
supp(u) = x ∈ Ω; u(x) 6= 0
...

(e) We denote by


∇u =

where ei = (δik )1≤k≤n

∂u
∂u
,
...

(f) Let α be a multi-index and u ∈ C ∞
...

α
α
∂x1 1 ∂x2 2
...
e
...


(iv) Cc ∞ (Ω) denote the space of infinitely many times differentiable functions
u : Ω −→ R with compact support in Ω
...

¯ is the space of all functions v such that v is the restriction on Ω
(b) D(Ω)
of a function of D(Rn )
...

(vii)
(a)
n
o
k
C (Ω) = u : Ω −→ R, u is k−times continuously differentiable
¯ =
C k (Ω)

n
o
u ∈ C k (Ω), Dα (u) is uniformly continuous for all |α| ≤ k
...

C k (Ω)

k=0

¯ then D u continuously extends to Ω
¯ for each multi-index
Thus if u ∈ C (Ω)
α, |α| ≤ k
...
4
...

We say that v is the αth -weak derivative of u and write Dα u = v if
Z
Z
α
|α|
v φdx,
u D φdx = (−1)
Ω

Ω

for all φ ∈ D(Ω)
...

Theorem 1
...
2 (Dubois Raymond)
If f ∈ L1 loc (Ω) and,

Z
f (x)φ(x)dx = 0
Ω

for each φ ∈ D(Ω) then f = 0 a
...

Lemma 1
...
3 (Uniqueness of weak derivatives)
A weak αth -partial derivative of u, if it exists, is uniquely defined up to a set
of measure zero
...
4
...

Definition 1
...
5
Let k be a non negative integer and 1 ≤ p ≤ ∞
...
4
...

Definition 1
...
7
We denote by W0k,p (Ω) the closure of D(Ω) with respect to the norm ||
...

Theorem 1
...
8
W k,p (Ω) are Banach spaces for 1 ≤ p ≤ ∞, separable for 1 ≤ p < ∞ (i
...


16

Remark 1
...
9
(a) In the case p = 2, we write H k (Ω) = W k,2 (Ω)
...

|α|≤k

Ω

In our work, the sobolev space H 1 (Ω) will be of particular interest
...
, xn−1 )
...


Definition 1
...
10
(a) We say that Ω is of class C k , k ∈ N or equivalently has a boundary of class
C k if for every x ∈ ∂Ω, there exists an open neighbourhood U of x in Rn and
a map
φ : Q −→ U
such that :
(i) φ is a bijection,
(ii) φ ∈ C k (Q, UT
) , φ−1 ∈ C k (U, Q),T
(iii) φ(Q+ ) = Ω U , φ(Q0 ) = ∂Ω U
...

(b) Ω is said to be Lipschitz or equivalently, has a Lipschitz boundary if φ is
Lipschitz
...

(c) If ∂Ω is C 1 , then along ∂Ω is defined the outward pointing unit normal
vector field ν = (ν 1 ,
...
The unit normal at any point xo ∈ ∂Ω is ν(xo ) =
ν = (ν1 ,
...

¯ We call
(d) Let u ∈ C 1 (Ω)
...

17

Definition 1
...
11
Ω is called a domain if it is open and connected
...

Theorem 1
...
12 (Existence of the trace operator)
Suppose Ω be bounded with C 1 boundary and Ω lies on one side of its boundary
∂Ω
...

γ0 : W k,p (Ω) −→ Lp (∂Ω)
¯ γ0 (u) is the restriction of u to ∂Ω
...
4
...
||Lp (Ω) is Lp (Ω); that is,
Im(γ0 ) = Lp (Ω)
...


Definition 1
...
13
Let X and Y be real normed vector spaces
...

The embedding X ⊂ Y is said to be continuous if ı is continuous
...

The embedding X ⊂ Y is said to be compact if ı is a compact linear operator
...


18

Theorem 1
...
14 (Rellich-Kondrachov)
Let Ω be a bounded domain of class C 1 , then the following are satisfied with
compact embeddings
...

(a) If 1 ≤ p < n, then W 1,p (Ω) ,→,→ Lq (Ω), for all q ∈ [1, p? [ , where p? = n−p
1,p
q
(b) If p = n, then W (Ω) ,→,→ L (Ω), for all q ∈ [1, ∞[
...

Note that for 1 ≤ p < n, we also have the continuous embedding
?

where p∗ =

W 1,p (Ω) ,→ Lp (Ω)

np

...
4
...
It is not the case in dimension two
...

Let Ω be the open ball B(O; 1/2)
number and
u(x, y) =

⊂ R2 with O = (0, 0), 0 < α <
 p
α


ln x2 + y 2 
...
In fact, we have
lim

u(x, y) =

(x,y)−→(0,0)

lim | ln r|α = +∞

r−→0+

by using polar coordinates
...
So
R
|u|2 dx < ∞
...

Theorem 1
...
16
If n ≥ 1 then
¯ for all m ≥ 0 and k ≥ 1 such that m < k −
H k (Ω) ⊂ C m (Ω)

n
2

with continuous embedding; that is,
∀ u ∈ H k (Ω) ,

∃ C > 0,

sup |u(x)| ≤ C ||u||H k (Ω)
...

Theorem 1
...
17 (Poincaré Inequality)
Let Ω be a domain, then
∃CΩ > 0 such that, ||u||L2 (Ω) ≤ CΩ ||∇u||L2 (Ω) , ∀u ∈ H01 (Ω)
Therefore the semi-norm of H 1 (Ω) defined by
||u|| = ||∇u||L2 (Ω)
is a norm on H01 (Ω)
...

Remark 1
...
18
(a) We often write
||u||2H 1 (Ω) = ||∇u||L2 (Ω)

for every u ∈ H01 (Ω)
...


and there exists some α1 , α2 > 0 such that,
α1 ||u||H 1 (Ω) ≤ ||u||H01 (Ω) ≤ α2 ||u||H 1 (Ω)
(b) Obviously Poincaré inequality is not true in H 1 (Ω)
...
Then u ∈ H 1 (Ω) but we
have ||uo ||L2 (Ω) > C ||∇uo ||L2 (Ω) for all C > 0
...
4
...
Then if u, v ∈ H 1 (Ω) ,



Z
Z
Z 
∂u
∂v
v dx = − u
dx +
(γ0 u) (γ0 v) νi dσ,
∂xi
∂xi
Ω
Ω
∂Ω
for each i ∈ {1,
...

Theorem 1
...
20 (Second Green’s Formula)
Let Ω be a nonempty bounded domain of class C 1 , u ∈ H 2 (Ω) and v ∈ H 1 (Ω)
...

(∆u)vdx = − ∇u · ∇v dx +
∂ν
Ω
Ω
∂Ω

21

CHAPTER

2

Galerkin Method

2
...


Partial Differential Equations (PDE’s) are fundamental in many areas of Mathematics such as Differential Geometry and Stochastic Processes
...

As there is no general theory known for solving all PDE’s, and given the variety of phenomena modelled by such equations, research focuses on particular
PDE’s that are important for theory or applications
...

Thus in our dissertation, we would like to present a constructive method for
solving Boundary Value Problems (i
...
, PDE’s subjected to Boundary Conditions) of variational type that have the variational formulation :

Find u ∈ H such that,
(P )
A(u, v) = L(v), ∀v ∈ H
where H is an infinite dimensional Hilbert space, L : H −→ R is a bounded
linear form and A : H × H −→ R is coercive and continuous bilinear form
...


2
...

Theorem 2
...
1 (Generalized Lax Milgram)
Let V and W be two real Hilbert spaces
...


sup

(2
...
2)

and
sup A(w, v) > 0 ∀v ∈ V \ {0}
...
3)

w∈W

Then for any L ∈ V ∗ , there exists a unique solution u ∈ W of (P)
...

α

(2
...
2
...

Proof of Theorem 2
...
1
For each w fixed in W , A(w,
...


By setting Aw = v1 , we have a well-defined mapping
A : W −→ V
...

Linearity: Let α1 , α2 ∈ R and let w1 , w2 ∈ W
...

So A(α1 w1 + α2 w2 ) = α1 Aw1 + α2 Aw2
...

Continuity: Let w ∈ W
...

Therefore A is continuous
...
Then
α||w|| ≤

A(w, v)
=
v∈V \{0} ||v||V
sup

and so
||w|| ≤

hAw, vi
≤ γ||Aw|| ,
v∈V \{0} ||v||V

||Aw||
α

sup

∀w ∈ W
...
5)

The injectivity clearly follows
...

Indeed let (vn )n be a sequence of elements of ImA converging to some v in

24

V
...

Moreover for all m, n, we have
||wn − wm || ≤

||A(wn − wm )||
||vn − vm )||
=
α
α

by the inequality (2
...
Therefore (wn )n is a Cauchy
sequence and so converges in the Hilbert W to some element w
...
This completes the proof of the closedness
of ImA
...
Then hvo , viV = 0 for all v ∈ ImA which means also
that
hvo , AwiV = hAw, vo iV = 0 ,
∀ w ∈ W
...


Therefore vo = 0 according to (2
...

Hence A is a continuous linear bijection with a continuous inverse (cf
...
5))
...


By setting v¯ = B(L), it is clear that B defines a linear continuous and isometric map from V ∗ onto V
...

(2
...



Let us now consider the particular case in which V = W in problem (P)
...
2
...

Assume moreover that A is continuous i
...
e,
∃α > 0 :

A(v, v) ≥ α||v||2V ,

∀v ∈ V
...
7)

Then there exists a unique u ∈ V solution to the problem (P ) and
||L||V ∗

...
2
...

Obviously from (2
...
2) and (2
...
So we are done
...
2
...
2
...
2
...

Next we link the symmetric case of the Lax-Milgram theorem to variational
calculus
...
2
...

Suppose A is symmetric and positive i
...
,
A(u, v) = A(v, u)

∀ u, v ∈ V

and
A(u, u) ≥ 0

∀u ∈ V
...
Suppose u is solution of problem (P )
...

So
J(u)

= J(v) − A(u, v − u) + L(v − u) − 21 A(v − u, v − u)
= J(v) − 12 A(v − u, v − u) since A(u, v) = L(v) ∀v ∈ V
≤ J(v) since A is positive
...

Conversely, suppose that u minimizes J
...

J(w) − J(u)

= 12 A(u + λv, u + λv) − L(u + λv) − 12 A(u, u) + L(u)
2
= λ(A(u, v) − L(v)) + λ2 A(v, v) ≥ 0 since J(v) ≥ J(u) ∀v ∈ V
...


Substituting v by −v yields to
−A(u, v) ≥ −L(v) , ∀v ∈ V

⇐⇒

A(u, v) ≤ L(v) ∀v ∈ V


...

Remark 2
...
6
(i) Consequently when A is symmetric and positive, we have

Find u ∈ V such that
(P ) ⇐⇒
(b)
J(u) ≤ J(v) ∀v ∈ V
(ii) In many applications, the functional J corresponds to an energy of the
deformation of an elastic membrane
...

Theorem 2
...
7 (Existence)
Let X be a real Banach space
...
Let L ∈ X ∗ , M be a nonempty closed
convex subset of X then, the abstract minimization problem

Find x ∈ M such that
J(x) = inf u∈M J(u)
27

where

J : X −→ R

...

Proof:
Symmetry of A implies that A is an inner product on X
...
There exists α, γ > 0 such that
α||u||2X ≤ A(u, u) ≤ γ||u||2X from coercivity and continuity of A
...
Therefore ||
...
|| define two equivalent norms on X
...
Then
by Riez representation theorem for each m in H there exists a unique uo ∈ X
such that L(u) = A(uo , u)
...

(1
...

inf J(u) =

u∈M

So we have
0 ≤ ||x − uo || ≤ ||u − uo || ∀u ∈ M =⇒ 0 ≤ ||x − uo ||2 ≤ ||u − uo ||2 ∀u ∈ M
which yield the fact that
∃! x ∈ X ; ||x − uo ||2 = inf u∈M ||u − uo ||2
Then from (1
...

Let us go back to problem (P ) with H = V
...
However when dimH = ∞
and a total orthonormal system S is prescribed for H, the determination of u
is equivalent to the computation of its components with respect to S which
represent infinitely many unknown parameters which is hard to do in general
...

For this end we shall consider Galerkin Methods that we shall introduce in two
cases
...

Proposition 2
...
7
Problem (PN ) has a unique solution
...
So by applying the Lax-Milgram Theorem and (b) we have
the result
...
Let us denote it by {φi }i∈{1,
...
The
application
f :

VN −→ RN
P

...
, uN ) = U

is a bijection
...

So find a solution uN of (PN ), is equivalent to find U solution of

Find U ∈ RN such that
(P 0 )
j(U ) ≤ j(R) ∀R ∈ RN
where j(R) = 21 Rt KR − Rt G
...
The matrix K is also definite positive i
...

In fact Let η ∈ VN such that
η=

N
X

ηj φj 6= 0,

j=1

then
hKη, ηi

PN PN
=
i=1 Kij ηi ηj
Pj=1
N
=
i,j=1 ηi A(φi , φj )ηj
= A(η, η)
...


Remark 2
...
8
It is also easy to check directly that the stiffness matrix K is invertible without
using the proof of Theorem 2
...
1
...
Let η ∈ R , η 6= 0, such that η = N
j=1 ηj φj ,
then,
N
X
Kη = 0 =⇒
A(φi , φj )ηj = 0
...
e A(η, η) = 0, which yields to η = 0 by coercivity of A
...
3

The Ritz-Galerkin method

i) Construct VN ⊂ V
ii) Find a basis of VN
iii) Compute the matrix K generally called stiffness matrix, then the vector
G generally called the load vector
...

v) Compute the Galerkin approximation solution
u(N ) =

N
X

ui φi
...

Remark 2
...
1
They are several methods to solve real algebraic linear systems of the form
Ax = b
where A is an invertible square matrix of dimension N such as: Gaussian
elimination method Jacobi method, etc
...

The choice of the approximation space VN , is very fundamental for the development of the approximation method
...
However, the choice of the basis is
(N )
capital, even if u
depends on the choice of VN not on the choice of the basis
...
e
S
{basis of VN +1 } = {basis of VN } {φN +1 }
basis are then overlapping each other
...
3
...
In fact the stiffness matrix
we get from this kind of basis is not sparse i
...
We need to avoid this kind of matrix because it is very cost in time
and memory (computer)
...
, λN ) and explicitly
u(N ) =

N
N
X
X
L(φi )
L(φi )
φi =
φi
...

Unfortunately, it is rare to know explicitly basis functions φi
...

The technique of finite element that we will see later is an example of this
second choice
...


2
...
e if ,
u(N ) −→ u as N −→ ∞
...

Definition 2
...
1 (Consistency)
We say that the Galerkin approximation (in this symmetric case) defined by
the space VN ⊂ V is consistent if for every u ∈ V ,
d(u, VN ) −→ 0 when N −→ ∞
or equivalently
inf v∈VN ||u − v||V −→ 0 when N −→ ∞
Now let us state the following fundamental theorem
...
4
...
Let u be a unique solution of (P ) and u(N ) the
unique solution of (PN )
...
We first prove that u(N ) is a projection of u on VN with the inner
product (· , ·) induced by p
A (since it is symmetric) on V ×V defined by (u, v) =
A(u, v)
...

As we have seen earlier ||·|| is equivalent to ||
...
||) is a Hilbert space
...
By the characterization of the orthogonal projection on a closed linear
subspace of a Hilbert space, we have
v − u ∈ (VN )⊥
...
We deduce from (P ) that
A(v, w) = A(u, w) = L(w) ,

∀w ∈ VN
...
By uniqueness
of the solution of problem (QN ) , v = uN
...
We have to establish an estimation of the V -norm of the difference
between u and uN
...
Therefore by the definition of a projection,
||u − u(N ) || ≤ ||u − v|| ,
=⇒
=⇒

||u − u(N ) ||2 ≤ ||u − v||2 ,

∀v ∈ VN
∀v ∈ VN

A(u − u(N ) , u − u(N ) ) ≤ A(u − v, u − v)

∀v ∈ VN

=⇒ α||u − u(N ) ||2V ≤ γ||u − v||2V ,
∀v ∈ VN
by coercivity and continuity of A
...

α
||u − u

(N )

||V ≤

33

Case 2 (General Case) : A not necessarily symmetric
The general principle of Galerkin method is similar to that of the symmetric
case
...
4
...

Proof : It is obvious by Lax-Milgram Theorem
...
,N } , of VN such that for each
v ∈ VN ,
N
X
v=
vi φi ,
i=1

where (v1 ,
...
By writing down that w
satisfies (Q0N ) for all v ∈ VN , and so for all v = φj with j ∈ {1,
...
, N }
and by developing w on the basis {φi }i∈{1,
...
, N }

i=1

which corresponds to
Kw = G
...
8)

Observe that, finding a solution of (Q0N ) is equivalent to find a solution of
(2
...

Indeed, if w is a solution of (Q0N ), then it is obviously a solution of (2
...

Conversely, if w a solution of (2
...

Thus w solves (Q0N ), since {vj }, j ∈ {1,
...

Remark 2
...
4
the stiffness matrix K is still invertible because A is nondegenerate
...
5

An Abstract on Galerkin method when A is
not necessarily symmetric

i) Construct VN ⊂ V
ii) Find a basis of VN
iii) Compute the matrix K (stiffness matrix), then the vector G (load vector)
iv) Solve KU = G
v) Compute the Galerkin approximation solution
(N )

u

=

N
X

ui φi
...

Remarks 2
...
1
(i) The method described in CASE I is generally called Ritz-Galerkin method
and the second CASE is Galerkin method
...

(iii) In Galerkin method we have the same steps as in Ritz method
...

(iv) Problems (PN ), (PN0 ), (QN ), and (Q0N ) are discrete problems
...
Concerning the choice of the basis, possibilities are the same as in
Ritz method
...

The convergence theorem in this case is the following
...
5
...
(A not necessarily symmetric)
...


Remark 2
...
3
The estimation obtained here with Galerkin method is less good than the
estimation obtained from Ritz method This is normal because Ritz method
is a particular case

of Galerkin method
...
This is another property of consistance we
need
...


36

Proposition 2
...
4 (Characterization of consistency)
Let W be a dense subspace of V
...

Proof : Let w ∈ W
...
(g)

Since u ∈ V and W is dense in V we have
∀ > 0, ∃ w¯ ∈ W ; ||u − w||
¯ V ≤


2

We choose w which verified the latter inequality
...


So if N ≥ N , we have d(u, VN ) ≤  from (g), which implies that d(u, VN )
tends to 0 as N tends to ∞
...
In each finite dimensional space VN , (PN ) is solved
exactly as we have shown it previously
...
5
...
Then there exists
{Vk }k∈N an increasing sequence of finite dimensional subspaces of V such that
dim Vk = k for each k ∈ N and
[
Vk = V
k∈N

Lemma 2
...
6
Let u be the solution of (P ) and u(N ) be the solution of (Q0N ), where V is
furthermore a separable space
...
9)
N ∈N

Then
37

||u − u(N ) || −→ 0 as N −→ ∞
Under the assumptions stated in the previous lemma, the following Proposition
shows that Galerkin method can be used to provide another proof of the
existence and uniqueness of a solution of (P)
...
5
...
The approximate
solutions {u(N ) }N ∈N where each u(N ) ∈ VN , converges to the exact solution u
of (P)
...
Therefore we can bear in mind that
L 6≡ 0
...
We show that {u(N ) }N ∈N is bounded and thus it has a subsequence
{u(Ni ) }i∈N which converges weakly to some element u in V
...
With the fact that weak
topology is Hausdorff, u is unique
...
We show that for any fixed integer N ≥ 1 and any v ∈ VN ,
A(u, v) = L(v)
...

Since {u(Ni ) }i∈N converges weakly to u in V and A(· , v) ∈ V ∗ , we have
A(u(Ni ) , v) −→ A(u, v) as i → ∞ ,
which yields
L(v) = A(u, v)
since each u(Ni ) solves (Q0Ni ) for all i ∈ N, and v ∈ VNi for all Ni ≥ N
...
We show that (P) is valid (on the whole V ) using the density of
S
N ∈N VN in V
...
9), there exists a sequence
[
{v (k) }k∈N ⊆
VN
N ∈N

such that v (k) −→ v

as

k −→ ∞
...

From step 2,

L(v (k) ) = A(u, v (k) )
...

Step 4
...

Suppose by contradiction that {u(N ) }N ∈N does not converges to u
...


(2
...
The sequence {u(Ni ) }i∈N has also in its turn a subsequence {u(Nik ) }k∈N
which converges weakly to some w in V (from step 1) such that
A(w, v) = L(v), ∀v ∈ V

(from step 3)

For each k ∈ N, we have:
α||u(Nik ) − w||2 V

≤
=
=
=

A(w − u(Nik ) , w − u(Nik ) )
A(w, w) − A(u(Nik ) , w) − A(w, u(Nik ) ) + A(u(Nik ) , u(Nik ) )
A(w, w) − A(u(Nik ) , w) − L(u(Nik ) ) + L(u(Nik ) )
A(w, w) − A(u(Nik ) , w)
...
But
A(w, v) = L(v) and

A(u, v) = L(v),

∀v ∈ V

so that
A(w − u, w − u) = 0 and then w − u = 0 from coercivity of A
...
10); the fact that the sequence {u(Ni ) }i∈N has no subsequence that converges to u
...
5
...
5
...
5
...

Note that problem (P ) generally comes from a weak formulation of a
given boundary value problem
...

39

2
...
In particular, it does not indicate whether the
solution will be analytic or infinitely many times differentiable
...
It would be wiser
to require for a solution of a PDE of order k, to be at least k times continuously
differentiable
...
Such a solution is called a classical solution of the boundary value
problem
...
For example, in general, conservation laws do not admit classical
solutions but are well-posed
...
Definition
1
...
1) and look for a different type of solutions called weak solutions
...

Furthermore, the weak formulation of a given boundary value problem gives
often access to existence and uniqueness results for the solution and that is well
suited for the numerical approximation of such problems
...
Thus we can say that, a given solution is refereed to as a weak
solution, because it has less requirement on the solution’s differentiability
...
6
...


2
...


Consider the following boundary value problem in one dimension with f ∈
L2 (Ω) = L2 (0, 1):

−u00 = f in
Ω = (0, 1)
(S)
u = 0 on ∂Ω = {0, 1}
1
...
By multiplying the first equation of (S) by v and by
integrating by part, we have,
40

Z

1

−

Z

00

1

fv

u v=

0

−[u

=⇒

v]10

Z

1
0 0

0 0

0

Z

1

fv

uv =

+

0

1

Z

f v + u0 (1)v(1) − u0 (0)v(0)

uv =

=⇒

1

0

0

0

Z

(i1 )

0

To find a weak formulation of (S) we start by finding a Hilbert space of functions which give sense to (i1 ) and which will be compatible with boundary
conditions
...
Likewise, the left hand side of (i1 ), is well defined
whenever u0 , v 0 ∈ L2 (0, 1)
...

Recall that in dimension 1
H01 (0, 1) ,→ C[0, 1]
So u(0) and u(1) are well defined
...

A is a symmetric, continuous, coercive, and continuous bilinear form; L is
linear and continuous, so we can use Galerkin method as we can use Ritz
method
...
Moreover we have:
Continuity of L(v):
|L(v)| ≤ ||f ||L2 (0,1) ||v||L2 (0,1)
=⇒

=⇒

|L(v)| ≤ ||f ||L2 (0,1) ||v||H 1 (0,1)

|L(v)| ≤ c||f ||L2 (0,1) ||v||H01 (0,1)

from Poincaré inequality
Continuity of A: By Cauchy-Schwarz inequality
41

for some c > 0

|A(u, v)| ≤ ||u0 ||L2 (0,1) ||v 0 ||L2 (0,1) ,

(i2 )
1

||u||H 1 (0,1) = (||u||2L2 (0,1) + ||u0 ||2L2 (0,1) ) 2
implies that
||u0 ||L2 (0,1) ≤ ||u||H 1 (0,1) , (i3 )
Then from (i2 ) and (i3 )
A(u, v) ≤ ||u||H 1 (0,1) ||v||H 1 (0,1)
...

Thus, by applying the Lax-Milgram Theorem, we see that, the weak problem has a unique solution
...
Notice that a function in V must vanish on
{0,1}
...
, N
...

(ii) A basis of VN is
φi defined by φi (x) = xi (1 − x)

i = {1,
...


(iii) Let us compute the stiffness N × N matrix K
...

Observe here that {basis of VN } ⊆ {basis of VN +1 }
...
, N
...
, N }

ψi = sin(iπx)

Note that in this example it is possible to construct an increasing sequence
{VN }N ∈N of finite dimensional subspaces of V
...

The coefficients of the Galerkin approximation u(N ) are then,
Z 1
2
ui = 2 2
f (t) sin(iπt)dt
...

Note that in this example it is possible to construct an increasing sequence
{VN }N ∈N of finite dimensional subspaces of V since V is a separable Hilbert
space so the Galerkin approximation converges to the exact solution of (S)
from Lemma 2
...
6
Instead of gambling, finding a way to have a sparse matrix we shall introduce the notion of Finite Element Method which is a special case of
Galerkin method
...
The important issue is the
sparsity of A for the linear system to be solved efficiently
...

43

When the Hilbert space V in which we are looking for the solutions, consists
of functions defined in Ω and the Galerkin subspaces VN comprise piecewisepolynomial functions, the Galerkin method is called the finite element method
...
While the Galerkin Method assumes an arbitrary basis
of the subspaces VN , the FEM (up to rare exceptions) prefers basis functions
with small supports so that as many of them as possible are disjoints
...


44

CHAPTER

3

Finite Element Method

3
...
A mesh is a tessellation of a space by very simple
elementary volumes : segments when Ω ⊆ R, triangles and rectangles when
Ω ⊆ R2 , tetrahedron and hexahedron when Ω ⊆ R3
...

In this context, the parameter h, correspond to the maximum size of the mesh
...
This will have two important consequences on the
one hand Vh −→ V as h −→ 0 , and on the other hand , the stiffness matrix
K will be sparse
...
1
...
Some examples are triangles,
rectangles, pentagons,
...
1
...


45

Definition 3
...
3
A polyhedron is a solid with flat faces (from Greek poly- meaning "many" and
-edron meaning faces ) where each faces is a polygon
...

Remark 3
...
4
(a) To apply FEM in R, Ω must be a bounded interval
...
If not, it can not be partitioned into straight sided polyhedron
...

In that case it is Ωh that will be partitioned
...
This deserves another survey of FEM, because it can leads us to a
situation like Vh * V even if Ωh ⊆ Ω which is not our concern
...

Discretization of the closure Ω of Ω
...
We partitioned Ω into N intervals
Ki = [xi−1 , xi ], called elements such that
Ω=

N
[

Ki ,

i=1

xi are called nodes and xo = a < x1 <
...
We set
hi = xi − xi−1 ,

h = max hi
...
e K is non degenerate
...

(iv) diam(K) ≤ h for each K ∈ τh
...
τh is said to be regular by property (ii)
...
g in R2 we can choose to construct a
triangulation which contains both triangles and quadrilaterals
...
We will focus only on a triangulation
of Ω in R2 which contains only triangles
...


3
...
Xn ] the space of polynomials of degree
less than or equal to m in the variables x1 , x2 ,
...
, xn , i
...
,in xi11
...
, xn ) ∀p ∈ Pnm
i1 ,
...
+ in ≤ m

p(x) =

X

αi1 ,
...
xinn

avec

x = (x1 ,
...


...

≤ in ≤ m

The most commonly used space Vh is defined by
(a)
n
Vhm = vh ∈ C(Ω); vh|K ∈ Pnm , ∀ K ∈ τh }
with m ≥ 1, and where the elements K can be
(i) intervals in R and Vhm is then called, the space of continuous finite
elements since functions in the space H 1 (Ω) in one spatial dimension are continuous
...

or
(b)
Vhm

n
= vh ∈ C(Ω),

o
vh|K oTK ∈ Qm , ∀K ∈ τh ,
47

(m ≥ 1)

which is called the space of parallelepipedal finite elements because the elements
K here are parallelograms in R2 or parallelepipeds in R3
...
e TK (x) = BK x + bK , BK being a non singular matrix
...
But we will see
that there is equivalence of (a) and (b) with C(Ω) on one hand, and of (a) and
(b) with V
...
This is
in fact a consequence of the following result
...
2
...
A function v : Ω −→ R belongs to H 1 (Ω) if and only if
(a) v|K ∈ H 1 (K) for each K ∈ τh T
(b) For each common face F = K1 K2 , K1 , K2 ∈ τh , the traces on F of v|K1
and v|K2 are the same
...
Assume (a) and (b) true
...
Using (a), we define the functions wj ∈ L2 (Ω) as
wj |K = Dj (v|K ) ∀K ∈ τh
...


Using integration by part formula, for every φ ∈ D(Ω)
R
R
P
wφ
wφ =
K∈τ
Ω j
R
P
P h KR j
= − K∈τh K (v|K )Dj φ + K∈τh ∂K (v|K )φνj,K
where νj,K is the j th component of the outward normal vector to ∂K
...
In other
words, the support cannot touch the boundary ∂Ω and consequently, there
must be a belt along ∂Ω where φ vanishes
...

Conversely, if we assume that, v ∈ H 1 (Ω), it follows immediately that (a)
holds
...

48

Furthermore, we have the following proposition
...
2
...
, N }
V
(ii) Degree of polynomials are chosen such that v is entirely determined by its
values at nodals
...
Before, observe that when m = 1
we have the following results
...
2
...

(a) If the restriction vh on each element K is smooth, a necessary and sufficient
condition for having vh ∈ H 1 (Ω), is that vh ∈ C(Ω)
...

T
denote by v1 = vh |K1 , v2 = vh |K2 and w = v1 − v2
...
THence, w = 0, i
...
Conversely, let K1 , K2 ∈ S
τh , K1 K2 6= ∅ such that vh be a
¯
piecewise linear function, then, vh ∈ C(K1 K2 )
...


3
...
Then a basis function corresponding to the subspace
n
oT
Vh = vh ∈ C([a, b]) ; vh|Ki ∈ P1 , ∀i ∈ {1,
...
, N − 1}
(1)
φi (x) =
, x ∈ [xi , xi+1 ],
hi+1






0
otherwise,
49

Let us show that
Vh =

n
o\
vh ∈ C([a, b]); vh|Ki ∈ P1 , ∀i ∈ {1,
...
It suffices to show that the
functions defined above constitute a basis for Vh
...
Observe also that φi (xj ) = δij
...
Therefore
f (x1 ) = α1 = 0,
...
Furthermore, any vh ∈ Vh is uniquely
written (because a polynomial of degree 1 on
interval [c, d] is uniquely
P an
−1
determined by its values on c and d) by vh = N
i=1 βi φi
...
, vh (xN −1 ) = βN −1
...
Therefore
n
o
Vh = span φi , 1 ≤ i ≤ N − 1
=⇒ dim Vh = N − 1

Note that the support of φi is
suppφi = [xi−1 , xi+1 ] ,

i ∈ [1, N − 1]
...
So let Ke be a triangle
with vertices (xi , yi ),(xj , yj ) and (xk , yk ) taken in the anti-clockwise direction
...
At the nodes we get
u(Ke ) (xi , yi ) = ui = a1 + a2 xi + a3 yi ,
u(Ke ) (xj , yj ) = uj = a1 + a2 xj + a3 yj ,
u(Ke ) (xk , yk ) = uk = a1 + a2 xk + a3 yk
...

1 xk u k 

Observe that 12 ∆ is the area of the triangle Ke
...

By straight computations we have
∆ = (xj yk − xk yj ) − (xi yk − xk yi ) + (xi yj − xj yi )
∆1 = (xj yk − yj xk )ui − (xi yk − xk yi )uj + (xi yj − yi xj )uk ,
∆2 = −(yk − yj )ui + (yk − yi )uj − (yj − yi )uk ,
∆3 = (xk − xj )ui − (xk − xi )uj + (xj − xi )uk
...
Nr e are called
shape functions
...
, N } ,
Vh = vh ∈ C(Ω)
Ki
in two dimensional space
...

Theorem 3
...
2 (Caratheodory Theorem)
Let K be a non degenerate element of the regular triangulation τh defined
above, with vertices ti , i = 1,
...
, n, the barycentric coordinates of x = (x1 ,
...
, n + 1 such that
Pn+1
Pn+1
i=1 λi ti,j
i=1 λi = 1 and xj =
Where {ti,j }nj=1 are cartesian coordinates of the vector ti
...
, n + 1 are furthermore affine
...
3
...
Then
for all r = i, j, k, there exists a unique polynomial λr ∈ P1 (K) satisfying
∀s = i, j, k

λr (ts ) = δrs
Proof :
From above we just need to take
(K)

∀r = i, j, k
...

λr are called barycentric coordinates
...
3
...
Then
for all fr , r = i, j, k, there exists a unique polynomial p ∈ P1 (K) such that,
p(tr ) = fr ,

∀r = i, j, k
52

Proof : Just take
p=

X

fr λr
...
3
...
,M be the set of all vertices of the triangulation, i
...
Then for all cr ∈ R, r = 1,
...
, M

Proof : From the previous theorem, for each K ∈ τh , there exists a unique
pK ∈ P1 such that,
pK (Sri ) = cri ,

∀ i = 1, 2, 3,

with Sr1 , Sr2 , Sr3 ∈ K
...


Then v is a well-defined function according to Proposition 3
...
3 and is morover
continuous by the Pasting Lemma
...
3
...

Theorem 3
...
6
For each s = 1,
...
, M
then the set {ws }s=1,
...

Proof : From the previous theorem, given v ∈ Vh we have equivalently v(Sr ),
r = 1,
...
Therefore dimVh = M and furthermore it is clear that
X
v=
v(Ss )ws
s=1,
...
,M generates Vh
...

Remark 3
...
7
For every proper subspace of H 1 (Ω) we can still take λr (ts ) = δrs for all vertices
ts of the triangulation, as a basis for the subspace
n
o\
¯
Vh = vh ∈ C(Ω); vh|Ki ∈ P1 , ∀i ∈ {1,
...

Example 3
...
8
n
o
2
Q1 = p : R −→ R; p(x, y) = a + bx + cy + dxy / a, b, c, d ∈ R
n
o
P1 = p : R2 −→ R; p(x, y) = a + by + cx / a, b, c ∈ R
...

Conditions to choose nodals (refer to [?, page 57 to page 60]
...
Then
m
(H0 ) In each element K the number of nodals must be given by dimPm = Cn+m
(resp dimQm = (m + 1)n )
...

(H2 ) on each interface of the triangulation, there must be sufficient nodals to
determine completely a function in Pm (resp Qm ) of dimension n − 1
...

These results extend to R3
...
3
...
Let p ∈ Pm (resp p ∈ Qm )
...
, Cn+m
(resp
n
(m + 1) )
...


54

Theorem 3
...
10
Let τh be a regular triangulation of Ω containing only triangles if n = 2 or
tetrahedrals if n = 3
...
, Nh
the set of nodals in the mesh satisfying hypothesis (H0 ), (H1 ), (H2 )
...
, Nh defined by

φi |K ∈ Pm ,
∀ϕ ∈ τh ,
φi (cj ) = δij , ∀j = 1,
...
It is particularly suitable
for higher dimensional problems
...
For each element Ki
in the mesh we define then an affine reference map FKi : Ka −→ Ki and use
it to transfer the shape functions from Ka to Ki
...

Example 3
...
11
Suppose that the triangulation τh contains only triangles
...
Shape functions φi,r i = 1, 2, 3, are given by the following barycentric
coordinates functions λi i = 1, 2, 3
φ1,r (x, y) = 1 − x − y = λ1 (x, y),
φ2,r (x, y) = x = λ2 (x, y),
φ3,r (x, y) = y = λ3 (x, y),
for all (x, y) ∈ Tr
...
3
...
On the other hand by assuming
p(x, y) = a + by + cx, a, b, c ∈ R, ∀x, y ∈ Tr we obtain a unique solution
c = α2 − α1 , b = α3 − α1 , and a = α1 from p(ti ) = αi
...
Thus for (α1 , α2 , α3 ) = (1, 0, 0),
(0, 1, 0), and (0, 0, 1) we have respectively by unicity of p
55

p1 (x, y) = 1 − x − y = λ1 (x, y)
p2 (x, y) = x = λ2 (x, y)
p3 (x, y) = y = λ3 (x, y)
...
Which constitute a basis on Tr from Theorem 3
...
4 called shape
functions
...


where F is fixed such that F (0, 0) = t1 , F (1, 0) = t2 , F (0, 1) = t3 ,
In general, the stiffness matrix and the load vector are not easy to compute
exactly
...
Among the wide
scale of existing numerical quadrature methods, the Gaussian quadrature rules
are of high efficiency
...
4

Gaussian Quadrature Rules

One dimensional case
Definition 3
...
1
The (m+1)-point Gaussian quadrature rule in the interval Ka = (−1, 1) has
the form
Z 1
m
X
g(ξ)dξ ≈
wm+1,i g(ξm+1,i ),
−1

i=0

where g is a real bounded continuous function on [−1, 1], ξm+1,i ∈ (−1, 1),
i = 0,
...

i=0

Definition 3
...
2 (Legendre polynomial):
Let the integer m ≥ 0
...

Remark 3
...
3
Legendre polynomials can be defined through the following recursive relation:

L0 (x) = 1,


L1 (x) = x, 


1
 Lm+1 (x) =
(2m + 1)xLm (x) − mLm−1 (x) , m ≥ 2
m+1

Gauss-Legendre quadrature formula:
Let x0 < x2 <
...
, wm be the solution of the linear system
m
X

xji wi

Z

1

xj dx,

=

0≤j≤m

(a)
...
, m and
Z

1

p(x)dx =
−1

m
X

p(xm+1,i )wm+1,i

∀p ∈ P2m+1 (−1, 1)
...
, m, such that (a)
holds for all p ∈ P2m+2 (−1, 1)
...
Thus the
inclusion of such points provides the following two quadratures formulas
...
We construct it
after the following polynomial:
q(x) = Lm+1 (x) + aLm (x),

(b)

where a is such that q(−1) = 0, that is,
(−1)
a = − LLm+1
m (−1)

Let x0 < x1 <
...
, wm ) be the solution
of the linear system
m
X
i=0

xji wi

Z

1

=

xj dx,

−1

57

0≤j≤m


...


i=0

In order to have the Gauss-Radau formula to include the point x = 1, the
variable a is taken in such a way that q(1) = 0 and a similar result as the
previously presented is valid
...

Let x0 < x1 <
...
, wm be the solution
of the linear system
m
X

xji wi

Z
=

Z

1

p(x)dx =
−1

xj dx,

0≤j≤m

−1

i=0

Then

1

m
X

p(xm+1,i )wm+1,i

∀p ∈ P2m−1 (−1, 1)
...
4
...
Furthermore we have the
relation
wm+1,i =

2
,
2
(1−ξm+1,i
)L0m+1 (ξ)2

i = 0,
...


Quadrature in arbitrary intervals
Let K = (xi−1 , xi ) ⊂ R be an arbitrary interval
...

Therefore, the new integration points ξ˜m+1,i ∈ K are then defined as
ξ˜m+1,i = FK (ξm+1,i ), i = 0,
...


58

and

Z

Z
|JK (ξ)|(goFK )(ξ)dξ,

g(x)dx =
K

Ka

which yields the new integration weights w˜m+1,i = JK wm+1,i = c2 wm+1,i , where
JK is the constant Jacobian of the affine map FK which is positive
...
, ξm1 +1,m1 and wm1 +1,0 ,
...
, ηm2 +1,m2 and wm2 +1,0 ,
...

Remark 3
...
5
To compute

Z
gdΩ,
Ω

where g is a real bounded continuous function on Ω ⊂ Rn , it suffices to make a
change of variables on an integral on [−1, 1]n to apply the Gaussian quadrature
formula
...

How to improve the approximation ?
To have d(V, VN ) very small, it seems reasonable to increase the dimension of
the space VN
...

- By increasing the polynomials degree of the basis functions
...

Remark 3
...
6
The second choice can be made only if the solution is sufficiently smoooth
...
5

Example of an application of Finite Element
Method in one dimensional case

Consider the following Neumann boundary value problem
...

We will just compute the stiffness matrix
...
, N , such
S
that [0, 1] = N
K
i , and x0 = 0, xN +1 = 1, xi < xi+1 for all i ∈ {0,
...

i=0
A weak formulation of (S) using the Integration by Parts (ou Green’s formula
for the higher dimensional case) is

Find u ∈ H 1 (0, 1) such that,
(S1 )
A(u, v) = L(v), ∀v ∈ H 1 (0, 1)
where

Z

1

Z
∇u · ∇v dx +

A(u, v) =

1

uv,
0

0

and

Z
L(v) =

1

f v
...

2/ We choose a finite dimensional subspace of H 1 (0, 1) to be
n
o
Vh = vh ∈ C([0, 1]); vh|Ki ∈ P1 , ∀i ∈ {0,
...


Vh is a subspace of H 1 (0, 1) by proposition 3
...
1
...


φN +1 (x) =

x−xN
,
hN

∀i ∈ {1,
...

60


φ0 (x) =

x1 −x
,
h0

x ∈ [x0 , x1 ],
0, otherwise
...
5
...
, N }
Suppφi = [xN , xN +1 ] for i ∈ {0, N + 1}
...
, N

i = 2,
...
, N }
...
, N }
, x ∈ [xi , xi+1 ],
h2i


0,
otherwise,
(
(x1 −x)2
, x ∈ [x0 , x1 ],
2
h20
φ0 (x) =
0,
otherwise,
(
(x−xN )2
, x ∈ [xN , xN +1 ],
2
h2N
φN +1 (x) =
0,
otherwise,
61

φ0i φ0i+1 (x)


=

φ0i φ0i−1 (x) =

0,




φ00 φ01 (x)

φ02
i (x)

A(φi , φj ) =











−1
,
h2i


=

x ∈ (xi , xi+1 ),
otherwise,

−1
,
h2i−1

x ∈ (xi−1 , xi )
0,
otherwise,

−1
,
h20

x ∈ [x0 , x1 ],
0, otherwise,

1
,
h2i−1
1
,
h2i

x ∈ (xi−1 , xi ],
∀i ∈ {1,
...
, N }
i ∈ {2,
...
, N }
i ∈ {0, N + 1}
i, j ∈ {0,
...


if

j=i

Thus, K is the following matrix :
 h0
3

+ h1

0

 h0 − 1
 6 h0
 0
 0

 0
0

h0
6

− h1

0

0

h0 +h1
+ h1 + h1
3
0
1
−h1
− h1
3
1

0

h1 +h2
3

+

−h1
− h1
3
1
1 + 1 −h2
h1
h2
3
−h2
− h1
3
2

···

0

···

0

− h1

2

0


...



...


...


63

−h3
− h1
3
3
hN −1 +hN
1
+
3
hN −1
h
− 3N − h1
N

0
+ h1
N

−hN
3
hN +1
3

− h1
N
+ h 1
N +1









Bibliography

[1] Ern A
...
L
...
Springer, 2002
...
C
...
American Mathematical Society, 1997
...
and Mishra, G
...
Department of Mathematics,
Motilal Nehru National Institute of Technology, Allahabad, India, 2011
...
D
...
SIAM 2000
...
and Valli, A
...
Springer, Berlin-Heidelberg-New York,
1997
...
, Galerkin Finite Element methods for parabolic problems,
Springer-Verlag, Berlin Heidelberg New York Tokyo 1984
...
H
...
Approximation des Équations aux Dérivées Partielles,
2011-2012

---