PDE & FEM TERMINOLOGY.

BASIC PRINCIPLES OF FEM.

Sergey Korotov

Basque Center for Applied Mathematics / IKERBASQUE

http://www.bcamath.org & http://www.ikerbasque.net

1

Introduction

The analytical solution of problems described by partial differential

equations (PDEs) is known only in a few cases on special domains

(like balls, cubes, half-spaces, etc). It is often necessary to use some

numerical methods to get an approximation of this solution.

One of the most powerful numerical methods for solving PDEs is

the Galerkin method.

The standard finite element method (FEM) is, roughly speaking,

the Galerkin method with a special choice of basis functions.

2

The FEM has been developed during last 70 years. The discovery of

FEM is usually attributed to Richard Courant [Courant, 1943]. However,

in [Ciarlet, Lions] we can find some older references to finite element-like

methods. The first monograph on FEM is probably that of Synge

[Synge] of 1957.

The notion element was introduced in the 1950-th by aerospace engineers

performing elasticity computations as they devided a continuum into

small pieces called elements. The notion finite element was introduced

by mathematicians later, in the 1960-th. From that time the theory of

FEM has also been rigorously investigated.

R. Courant. Variational methods for problems of equilibrium and vibration.

Bull. Amer. Math. Soc. 49 (1943), 1–23.

P. G. Ciarlet, J. L. Lions (eds.). Handbook of Numerical Analysis. Vol. II

Finite element methods. North-Holland, Amsterdam, 1991.

J. L. Synge. The Hypercircle in Mathematical Physics. Cambridge Univ.

Press, Cambridge, 1957.

3

Nowadays FEM seems to be one of the most efficient numerical

methods for solving problems of mathematical physics which are

based on variational principles. One may solve by FEM some

variational problems which do not correspond to any PDEs (e.g.

the obstacle problem).

Moreover, FEM enables perfect description of the examined

domain, which was not possible by classical numerical methods

(such as the collocation method, finite difference method, etc.).

4

The flow-chart of numerical simulation:

to be solved

A physical problemSet up mathematical Is a computer

solution

No

to be considered ?

of solution

Continue with analytical

(mathematical problem) problem

description of the

(non−computer) methods

model into form suitable

Establish mathematical

for numerical solution discrete problem

and solve the resulting

Perform discretizationCheck correctness of

numerical resultsYes

There are many ways how to set up a mathematical model and then design it

into a form suitable for numerical solution. In our case it will be so that we

give a variational formulation of the problem in an appropriate function space:

We want to find a function minimizing a convex functional over a closed set of

admissible functions. The basic idea of the discretization will consist then in

transforming the problem formulated in function spaces with infinite dimension

into appropriate problems in finite-dimensional spaces.

5

First we shall remind a variational (weak) formulation of problems

described by second order PDEs of elliptic type with some

boundary conditions.

This formulation is useful in explaining the mathematical

background of FEM and is based on the theory of Sobolev spaces.

Sobolev spaces are extensions of spaces of differentiable functions in

the classical sense. Such extensions are natural as the solution of a

variational problem usually does not have the classical derivatives.

6

Model Problem

In a bounded domain Ω with a Lipschitz boundary ∂Ω, consider the

following second order PDE for the unknown function u ∈ C2(Ω):

−d∑

i,j=1

∂

∂xi

(

aij∂u

∂xj

)

+ cu = f .

Here c, f ∈ C(Ω) and the matrix A = (aij) ∈ (C1(Ω))d×d are given. We

further assume that

aij = aji , i, j = 1, . . . , d , c(x) ≥ 0 , x ∈ Ω ,

and that there is a constant M > 0 such that

d∑

i,j=1

aij(x)ξiξj ≥ M

d∑

i=1

ξ2i ∀(ξ1, . . . , ξd)T ∈ Rd, x ∈ Ω .

The above assumptions ensure that our model problem is elliptic.

7

Three (standard) types of boundary conditions characterizing the

(homogeneous) Dirichlet, Neumann and Newton classical problem:

u = 0 on ∂Ω ,

∂

∂nAu = 0 on ∂Ω ,

αu+∂

∂nAu = 0 on ∂Ω .

Here α = α(s) ≥ 0 on ∂Ω is continuous,

∂

∂nAu =

d∑

i,j=1

aij∂u

∂xj

ni = nTA gradu

denotes the conormal derivative, ni are the components of the unit

outward normal to ∂Ω and nA = An is the conormal. The vector

function A gradu on Ω is said to be the cogradient and the scalar

function nTA gradu|∂Ω the boundary flux.

8

The model PDE with one of the above boundary conditions represents

the so-called boundary value problem (BVP) (classically formulated).

The function u ∈ C2(Ω) satisfying the PDE and the associated boundary

condition is called the classical solution.

Such BVP may describe a stationary magnetic, electric, or temperature

fields, etc. The coefficients aij describe physical properties of the

medium Ω. If aij are constants (i.e., independent of x), we call the

medium homogeneous (otherwise nonhomogeneous). If aii(x) = a11(x)

for i = 2, . . . , d, and aij(x) = 0 for i 6= j, the medium is said to be

isotropic (otherwise anisotropic). If A is only diagonal, the medium is

orthotropic.

In real-life problems the coefficients aij , c, α are often nonsmooth (they

are, e.g., piecewise constant) and we cannot use the equation as it

stands, since the classical derivatives in it need not exist.

This is why we introduce a weak formulation of the classical problems,

which enables us to consider also nonsmooth coefficients and/or

nonsmooth f .

9

Consider e.g. the Newton classical problem and introduce the space of

test functions V := H1(Ω). Multiplying the PDE by an arbitrary test

function v ∈ V and integrating over Ω we get

−

∫

Ω

∑

i,j

∂

∂xi

(

aij∂u

∂xj

)

v dx+

∫

Ω

cuv dx =

∫

Ω

fv dx .

Employing now Green’s formula, we find that

∫

Ω

∑

i,j

aij∂u

∂xj

∂v

∂xi

dx−

∫

∂Ω

∑

i,j

aij∂u

∂xj

niv ds+

∫

Ω

cuv dx =

∫

Ω

fv dx .

Using further the boundary condition, we obtain that

∫

Ω

∑

i,j

aij∂u

∂xj

∂v

∂xi

dx+

∫

Ω

cuv dx+

∫

∂Ω

αuv ds =

∫

Ω

fv dx ∀v ∈ V .

10

We see that any classical solution of the Newton problem (if it exists)

satisfies the equation

∫

Ω

∑

i,j

aij∂u

∂xj

∂v

∂xi

dx+

∫

Ω

cuv dx+

∫

∂Ω

αuv ds =

∫

Ω

fv dx ∀v ∈ V . (∗)

Let us observe that the integrals in above are well defined even when aij ,

c ∈ L∞(Ω), α ∈ L∞(∂Ω), f ∈ L2(Ω), and when prescribed conditions on

the coefficients hold almost everywhere (a.e.) in Ω only.

When introducing the space of test functions V we started to use

implicitly the concept of completion. Instead of looking for u ∈ C2(Ω)

satisfying the PDE and the boundary condition, we shall solve the

integral form equation (∗). Its solution will be searched in the Hilbert

space H1(Ω), which is, of course, complete and bigger than C2(Ω). This

allows us to employ some useful results from functional analysis (e.g.

Lax-Milgram lemma) and apply the finite element method.

11

Weak Solution

The function u ∈ H1(Ω) is called the weak (or generalized) solution of

the Newton problem if

∫

Ω

∑

i,j

aij∂u

∂xj

∂v

∂xi

dx+

∫

Ω

cuv dx+

∫

∂Ω

αuv ds =

∫

Ω

fv dx ∀v ∈ V . (∗)

For convenience the weak solution is denoted by u as the classical

solution. Note that the classical solution from C2(Ω) need not exist.

However, if it does exist, it is also the weak solution.

Notice further that the weak formulation (∗) contains only the first

(generalized) derivatives of u which may be advantageous for

approximation.

12

In order to establish under what conditions the problem posed in a weak

form has a unique solution in the Sobolev space H1(Ω), we recall some

definitions and present two abstract theorems.

Let V be a linear space. A mapping a(·, ·) : V × V → R1 is called a

bilinear form, if for any fixed v ∈ V the mappings a(v, ·) : V → R1 and

a(·, v) : V → R1 are linear.

If, in addition,

a(v, w) = a(w, v) ∀v, w ∈ V ,

the bilinear form a(·, ·) is said to be symmetric.

13

Theorem 1: Let V be a Banach space equipped with the norm ‖ · ‖V

and let F : V → R1 be a continuous linear form. Let a(·, ·) : V × V → R1

be a symmetric and continuous bilinear form, i.e., there exists C1 > 0

such that

|a(v, w)| ≤ C1‖v‖V ‖w‖V

∀v, w ∈ V.

Moreover, we suppose that there exists C2 > 0 such that

a(v, v) ≥ C2‖v‖2V ∀v ∈ V (V -ellipticity condition) .

Then the problem: Find u ∈ V such that

a(u, v) = F (v) ∀v ∈ V (variational equality)

has one and only one solution.

14

P r o o f : Due to the symmetry of a(·, ·) and V -ellipticity, the bilinear

form a(·, ·) is a scalar product on V since a(v, v) = 0 implies v = 0. The

corresponding norm√

a(·, ·) is, obviously, equivalent to the given norm

‖ · ‖V . Thus V equipped with the scalar product a(·, ·) is a Hilbert space.

Since F (·) is a continuous linear form, there exists by the Riesz

representation theorem a unique element u = uF ∈ V such that

F (v) = a(uF , v) ∀v ∈ V .

15

Theorem 2: Let all the assumptions of the previous theorem be

fulfilled. Then the problem: Find u ∈ V such that

a(u, v) = F (v) ∀v ∈ V ,

is equivalent to the problem: Find u ∈ V such that

J(u) = infv∈V

J(v) ,

where

J(v) =1

2a(v, v)− F (v) .

16

Functional J(v) = 12a(v, v)− F (v) is called the energy functional.

Norm√

a(·, ·) (equivalent to ‖ · ‖V -norm) is called the energy norm.

To apply Theorem 1 to the weak formulation (∗) we set

V = H1(Ω) ,

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx+

∫

∂Ω

αvw ds , v, w ∈ V ,

F (v) =

∫

Ω

fv dx, v ∈ V

and verify whether all the assumptions of the theorem are fulfilled.

17

Useful Inequalities

Let Ω be a bounded domain with a Lipschitz boundary ∂Ω and let

v, w ∈ H1(Ω). Then

v|∂Ω ∈ L2(∂Ω) ,

‖v‖L2(∂Ω) ≤ C‖v‖H1(Ω) (trace theorem) ,

∫

Ω

∂v

∂xj

w dx+

∫

Ω

v∂w

∂xj

dx =

∫

∂Ω

vwnj ds (Green’s formula) ,

18

‖v‖H1(Ω) ≤ C

(

∫

Ω

d∑

j=1

(

∂v

∂xj

)2

dx+

∫

∂Ω0

v2 ds

)1

2

(Friedrichs’ inequality) ,

where ∂Ω0 ⊂ ∂Ω is such that meas∂Ω0 > 0,

‖v‖H1(Ω) ≤ C

(

∫

Ω

d∑

j=1

(

∂v

∂xj

)2

dx+

∫

B

v2 dx

)1

2

(X) ,

where B ⊂ Ω is a ball,

‖v‖H1(Ω) ≤ C

(

∫

Ω

d∑

j=1

(

∂v

∂xj

)2

dx+

(∫

Ω

v dx

)2)

1

2

(Poincare’s inequality) .

For the proof, see [Dautray, Lions, vol. 2], [Necas, 1967].

19

• V is a Hilbert space.

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx+

∫

∂Ω

αvw ds, v, w ∈ V ,

F (v) =

∫

Ω

fv dx, v ∈ V

• The linearity of F (·) and bilinearity of a(·, ·) are obvious.

• The continuity of F (·) follows immediately from the Cauchy-Schwarz

inequality

|F (v)| =

∣

∣

∣

∣

∫

Ω

fv dx

∣

∣

∣

∣

≤ ‖f‖0,Ω ‖v‖0,Ω ≤ ‖f‖0,Ω ‖v‖1,Ω

as ‖ · ‖0,Ω ≤ ‖ · ‖1,Ω.

20

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx+

∫

∂Ω

αvw ds, v, w ∈ V ,

• The bilinear form is symmetric as aij = aji and continuous as:

|a(v, w)| ≤d∑

i,j=1

‖aij‖L∞(Ω)

∥

∥

∥

∥

∂v

∂xi

∥

∥

∥

∥

0,Ω

∥

∥

∥

∥

∂w

∂xj

∥

∥

∥

∥

0,Ω

+ ‖c‖L∞(Ω)‖v‖0,Ω‖w‖0,Ω + ‖α‖L∞(∂Ω)‖v‖L2(∂Ω)‖w‖L2(∂Ω)

≤C‖v‖1,Ω‖w‖1,Ω ,

where we used the trace theorem and the relation

‖v‖21,Ω = ‖v‖20,Ω +

d∑

j=1

∥

∥

∥

∥

∂v

∂xj

∥

∥

∥

∥

2

0,Ω

.

21

• What remains is to prove the V -ellipticity condition for

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx+

∫

∂Ω

αvw ds,

Suppose that α(x) ≥ α0 > 0 on some part ∂Ω0 ⊂ ∂Ω which has a

positive measure. Then we can use Friedrichs’ inequality, and get

a(v, v) ≥ M

∫

Ω

d∑

i=1

(

∂v

∂xi

)2

dx+ α0

∫

∂Ω0

v2 ds ≥ C‖v‖21,Ω .

So, in this case the variational problem (∗) has a unique solution by

virtue of Theorem 1.

22

• Another situation.

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx+

∫

∂Ω

αvw ds,

Let now c(x) ≥ c0 > 0 on a ball B ⊂ Ω. Then we get the V -ellipticity in

a view of inequality (X):

a(v, v) ≥ M

∫

Ω

d∑

i=1

(

∂v

∂xi

)2

dx+ c0

∫

B

v2 dx ≥ C‖v‖21,Ω ,

which gives the existence of just one solution u ∈ V .

23

• We can show that the weak solution of (∗) belonging to C2(Ω) is also

the classical solution of the Newton problem.

• If c = 0 in Ω and simultaneously α = 0 on ∂Ω then we get the so-called

Neumann problem, which requires a special treatment (not to be

considered now).

• We can similarly handle the Dirichlet problem. For the space of test

functions, we choose

V = H10 (Ω) =

v ∈ H1(Ω) | v = 0 on ∂Ω

.

Now from Friedrichs’ inequality we immediately find that the associated

symmetric bilinear form

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx , v, w ∈ V ,

fulfils the V -ellipticity condition.

24

Consider now the nonhomogeneous mixed boundary conditions:

u = u on Γ1 ,

∂

∂nAu = g on Γ2 ,

αu+∂

∂nAu = g on Γ3 ,

where u ∈ H1(Ω), g ∈ L2(Γ2 ∪ Γ3), α ∈ L∞(Γ3), α ≥ 0, Γ1,Γ2,Γ3 are

mutually disjoint and are open sets in ∂Ω (one or two of them may be

empty),

Γ0 ∪ Γ1 ∪ Γ2 ∪ Γ3 = ∂Ω ,

and Γ0 (meas Γ0 = 0) is a set of those points where one type of

boundary condition changes into another.

25

In order to apply the FEM later, we shall assume from now on that each

set Γi (i = 1, 2, 3) has a finite number of components.

Let the space of test functions be chosen as follows:

V = v ∈ H1(Ω) | v = 0 on Γ1 .

26

If the problem with the above mixed boundary conditions is not purely

Neumann with c = 0 then the corresponding variational formulation

consists in finding u = u0 + u, where u0 ∈ V satisfies

a(u0, v) = F (v) ∀v ∈ V ,

and where

a(v, w) =

∫

Ω

∑

i,j

aij∂v

∂xj

∂w

∂xi

dx+

∫

Ω

cvw dx

+

∫

Γ3

αvw ds, v, w ∈ V ,

F (v) =

∫

Ω

fv dx+

∫

Γ2∪Γ3

gv ds− a(u, v), v ∈ V .

The desired V -ellipticity condition follows again from Friedrichs’

inequality or (X).

Note that it may be sometimes difficult to find u ∈ H1(Ω) with

prescribed values on Γ1.

27

Lax-Milgram Lemma

Sometimes we need to solve an elliptic problem whose associated bilinear

form a(·, ·) is not symmetric. Then, instead of Theorem 1, the following

theorem (often called Lax-Milgram lemma in the numerical analysis) is

applied.

Theorem 3: Let V be a Hilbert space, let a(·, ·) : V × V → R1 be a

continuous V -elliptic bilinear form and let F : V → R1 be a continuous

linear form. Then the problem: Find u ∈ V such that

a(u, v) = F (v) ∀v ∈ V ,

has one and only one solution.

If a(·, ·) is not symmetric then there is no associated minimization

problem as in Theorem 2.

28

Basic Idea of FEM

Now we shall show how to construct finite element subspaces Vh of V .

The FEM in its simplest setting is a Galerkin method characterized by

the following basic aspects in the construction of Vh:

(i) a triangulation Th is established over the domain Ω,

(ii) the functions vh ∈ Vh are piecewise polynomials,

(iii) there exists a basis in Vh whose functions have small supports.

• The so-called discretization parameter h is “associated” with a

characteristic size of triangulations used - it will be defined more

precisely later.

29

Let now Vh be an arbitrary finite-dimensional subspace of V . Then the

Galerkin method for approximating the solution of the problem: Find

u ∈ V such that

a(u, v) = F (v) ∀v ∈ V ,

consists of finding uh ∈ Vh such that

a(uh, vh) = F (vh) ∀vh ∈ Vh . (†)

Applying Lax-Milgram lemma, we observe that this problem has one and

only one solution uh, which we shall call the discrete solution.

30

When a(·, ·) is symmetric, the discrete solution is also characterized by

the property (see Theorem 2)

J(uh) = infvh∈Vh

J(vh) ,

where

J(v) =1

2a(v, v)− F (v) .

This definition of discrete solution is known as the Ritz method.

Note that

J(u) ≤ J(uh) ,

since Vh ⊂ V .

31

Obviously, we have the following orthogonality relation

a(u− uh, vh) = 0 ∀vh ∈ Vh ,

which means that the error u− uh is orthogonal to Vh with respect to

the scalar product a(·, ·).

This orthogonality relation implies that uh is the projection of u on Vh

with respect to a(·, ·) and that

a(u− uh, u− uh) = infvh∈Vh

a(u− vh, u− vh) ,

i.e. the Ritz method yields the best approximation with respect to the

energy norm.

32

Let a(·, ·) be not necessarily symmetric. And let vim

i=1 be basis in Vh.

We shall look for the discrete solution uh as a linear combination

uh =

m∑

j=1

cjvj .

Then by (†) it is true that

a

( m∑

j=1

cjvj , vi

)

= F (vi), i = 1, . . . ,m .

This finally leads to the following system of algebraic equations:

m∑

j=1

a(

vj , vi)

cj = F (vi), i = 1, . . . ,m , (‡)

for the unknowns c1, . . . , cm.

33

The matrix A =(

a(vj , vi))m

i,j=1and the vector

(

F (vi))m

i=1are often

called (by reference to elasticity problems) the stiffness matrix and the

load vector, respectively.

• We prove that A is nonsingular.

From the bilinearity of a(·, ·) and the V -ellipticity condition we have

(Aξ, ξ) =∑

i,j

a(vj , vi)ξiξj = a

(

∑

j

ξjvj ,∑

i

ξivi

)

= a(v, v) ≥

≥ C2||v||2V > 0 ∀ξ = (ξ1, . . . , ξm)T ∈ Rm, ξ 6= 0 ,

where v =∑

i

ξivi, and v 6= 0 since vi is a basis and ξ 6= 0. Hence, the

homogeneous equation Aξ = 0 cannot have a nontrivial solution ξ 6= 0.

34

The discrete solution uh is obviously independent of basis functions

v1, . . . , vm ∈ Vh, whereas the structure of A depends considerably on

v1, . . . , vm. Thus, concerning the choice of the basis vi, it is very

important from the numerical point of view that the matrix A possesses

as many zero entries as possible.

For instance, if the intersection of the supports of basis functions vp and

vq has zero measure then a(vp, vq) = 0. That is why the condition (iii) is

required. It may even happen that a(vp, vq) = 0, p 6= q, though the

intersection of supports has a positive measure. Thus according to (iii),

only a few entries (more precisely O(m) entries) of the m×m stiffness

matrix A remain different from zero, i.e., A is sparse.

Hence, we need less computer memory and fewer arithmetic operations

to solve (‡) than for full matrices which arise, in general, from the

classical Galerkin method.

35

In order to introduce some typical Vh, we first establish a triangulation

Th over domain Ω, i.e., we subdivide the set Ω into a finite number of

subsets K (called elements) in such a way that the following properties

hold:

(1) Ω =⋃

K∈Th

K ,

(2) for each K ∈ Th, the set K is closed and its interior K0 is non-empty,

(3) for each distinct K1,K2 ∈ Th, one has K01 ∩K0

2 = ∅,

(4) for each K ∈ Th, the boundary ∂K is the Lipschitz one.

Later we shall give further assumptions on Th. The discretization

parameter h is the maximum diameter of all K ∈ Th. Note that there is

a certain ambiguity in the meaning of Th as for a given sufficiently small

h there may exist many different triangulations Th. Nevertheless, we will

keep this often used notation. Sometimes Th is called a partition or a

decomposition of Ω into elements.

• From now on we shall write Hk(K) instead of Hk(K0) for simplicity.

36

In what follows we shall construct finite element spaces Vh consisting of

piecewise polynomial functions over Th. An important property of such

spaces is given in the following theorem.

Theorem 4: Let Th be a triangulation (decomposition) of Ω formed by

convex elements. Let Vh be a subspace of L2(Ω) such that the space

PK = vh|K | vh ∈ Vh

consists of polynomial functions for any K ∈ Th. Then Vh ⊂ H1(Ω) iff

Vh ⊂ C(Ω), i.e. a piecewise polynomial function is from H1(Ω) iff it is

continuous.

37

P r o o f : So let Vh ⊂ H1(Ω) and let there exist a function vh ∈ Vh

which is not continuous. Then there exist two adjacent elements

K1, K2 ∈ Th and an open ball B ⊂ K1 ∪K2 such that

B ∩ S 6= ∅ and vh|K1> vh|K2

on B ∩ S ,

where S = K1 ∩K2. Let w ∈ C∞0 (Ω) be a “hill” function such that

w(x) > 0 ∀x ∈ B and w(x) = 0 ∀x ∈ Ω\B .

Denote by nj = (nj1, . . . , n

jd)

T the outward unit normal to ∂Kj , j = 1, 2.

We see that S is contained in a hyperplane of Rd as K1 and K2 are

convex. Therefore nj is constant on S and there exists i ∈ 1, . . . , d

such that n1i 6= 0.

38

Now, referring to Green’s formula , we arrive at

0 =

∫

Ω

∂vh∂xi

w dx+

∫

Ω

vh∂w

∂xi

dx =2∑

j=1

(∫

Kj

∂vh∂xi

w dx+

∫

Kj

vh∂w

∂xi

dx

)

=2∑

j=1

(

−

∫

Kj

vh∂w

∂xi

dx+

∫

∂Kj

vh|Kjwnji ds+

∫

Kj

vh∂w

∂xi

dx

)

=

∫

S

(

vh|K1− vh|K2

)

wn1i ds = n1

i

∫

B∩S

(vh|K1− vh|K2

)w ds .

But this is a contradiction, since n1i 6= 0 and the last integral is positive

due to our assumptions upon w and the piecewise polynomial function

vh.

39

Conversely, let vh ∈ Vh be continuous. Then evidently vh ∈ L2(Ω) and

we show that it has also the first generalized derivatives in L2(Ω). Since

any K has a Lipschitz boundary and PK ⊂ H1(K) for all K ∈ Th, we

may apply Green’s formula for i ∈ 1, . . . , d to get∫

K

∂vh∂xi

w dx+

∫

K

vh∂w

∂xi

dx =

∫

∂K

vh|KwnKi ds ∀w ∈ C∞

0 (Ω) , (•)

where nKi is the ith component of the unit outward normal to ∂K. Let

zi ∈ L2(Ω) be defined through the relation

zi|K =∂vh∂xi

, K ∈ Th .

40

Summing (•) over all the elements K ∈ Th, we obtain∫

Ω

ziw dx+

∫

Ω

vh∂w

∂xi

dx =∑

K∈Th

∫

∂K

vh|KwnKi ds .

We see, however, that the sum vanishes. Either a portion of ∂K is a

portion of the boundary ∂Ω, where w = 0, or the contribution of any two

adjacent elements K,K′ ∈ Th is zero as nK + nK′

= 0. Hence,∫

Ω

ziw dx = −

∫

Ω

vh∂w

∂xi

dx ∀w ∈ C∞0 (Ω)

and the functions zi, i = 1, . . . , d, are the first generalized derivatives of

vh. Thus we may write zi = ∂vh/∂xi in the whole Ω.

41

The theorem can be also modified to the case when PK contains

generally nonpolynomial functions from C(K) ∩H1(K) and for

nonconvex elements. This may be useful to construct some more

complicated spaces of finite elements.

Suppose further that in the case d ≥ 2 elements of Th are convex

polygons or polyhedra, which is fulfilled quite often. Then Ω is, of course,

a polygon or polyhedron. We add two more assumptions upon Th:

(5) any face of any K ∈ Th is either a subset of the boundary ∂Ω, or a

face of another element K′ ∈ Th,

(6) the interior of any face of any K ∈ Th is disjoint with Γ0, where Γ0

is a set of those points where one type of boundary condition

changes into another.

42

Remark: For any generally nonconvex polygon (polyhedron) Ω there

exists a triangulation Th satisfying (1)–(5) such that all the elements

K ∈ Th are triangles (tetrahedra). Construction can be done as follows.

Let S1, . . . , Sq be faces of Ω and let L1, . . . , Lq be straight lines (planes)

such that Sj ⊂ Lj . First show that components of the set Ω \⋃q

j=1 Lj

are convex. Then cut these components into simplexes so that (5) holds.

43

Finite Elements

The following general definition of the finite element will be used to

construct finite element spaces Vh.

Definition: The finite element is a triple (K,P,Σ), where:

(I) K is a closed subset of Rd with a non-empty interior and a Lipschitz

boundary,

(II) P is a space of real-valued functions defined over the set K,

(III) Σ is a finite set of linearly independent linear forms Φi, 1 ≤ i ≤ N ,

defined over the space P (or over a space which contains P ).

44

The set Σ is said to be P -unisolvent if for any real scalars αi, 1 ≤ i ≤ N ,

there exists a unique function p ∈ P which satisfies

Φi(p) = αi , 1 ≤ i ≤ N .

Consequently, if Σ is P -unisolvent then there exist functions pi ∈ P,

1 ≤ i ≤ N , which satisfy

Φj(pi) = δij , 1 ≤ j ≤ N ,

where δij is Kronecker’s symbol. Since

∀p ∈ P p =

N∑

i=1

Φi(p)pi ,

we have dimP = N .

The linear forms Φi, 1 ≤ i ≤ N , are called the degrees of freedom of the

finite element, and the functions pi, 1 ≤ i ≤ N , are called the basis

functions of the finite element.

45

In what follows we introduce several examples of the most used finite

elements, where K will be a convex set of a simple form and P a space of

polynomials.

Functions from P are sometimes called ansatz-functions.

The space of all polynomials of degree at most k defined on K is denoted

by Pk(K).

We introduce only some lower order elements, since the use of higher

order elements requires certain higher smoothness of the true solution,

which is usually not present in practical problems.

46

Linear simplicial element

The set K is a d-simplex, the space P = PK is the space P1(K) of linear

functions

p(x1, . . . , xd) = γ0 + γ1x1 + . . .+ γdxd , γi ∈ R1 ,

dimP1(K) = d+ 1 .

AA A AA

A

A

1 2 1

A

21

2

3

3

A4

47

The set of degrees of freedom Σ = ΣK consists of the forms

Φi(p) = p(Ai), 1 ≤ i ≤ d+ 1, p ∈ P1(K) ,

where A1, . . . , Ad+1 are the vertices of K. For the sake of simplicity we

shall write only symbolically

ΣK = p(Ai), 1 ≤ i ≤ d+ 1 .

A A1

A

2

3

A A 2

11p

1 p1

1

In the case d = 2, this element is also known as Courant’s triangle

(element). The first basis function p1 is sketched above for d = 1, 2.

48

Quadratic simplicial element

K is again a d-simplex, PK = P2(K) is the space of quadratic functions

p(x1, . . . , xd) = γ0 +

d∑

i=1

γixi +∑

1≤i≤j≤d

γijxixj , γi, γij ∈ R1 ,

and

ΣK = p(Ai), 1 ≤ i ≤ d+ 1; p(Aij), 1 ≤ i < j ≤ d+ 1 ,

where Ai are the vertices of K and

Aij =1

2(Ai + Aj), 1 ≤ i < j ≤ d+ 1 ,

are the midpoints of the edges of the d-simplex K:

49

AA A AA

A

A

1 2 1

A

21

2

3

3

A4

A

A

A

A

AA A

A

A12 12

23

12

13

A13

34

14

23

24

We may easily check that

dimP2(K) =1

2(d+ 1)(d+ 2) .

50

Two figures in below show two qualitatively different types of basis

functions of the quadratic element for d = 1 and d = 2, respectively:

51

Bilinear and trilinear rectangular elements

Here K is a d-rectangle, PK = Q1(K) is the space of d-linear functions

p(x1, . . . , xd) =∑

0≤qi≤1, 1≤i≤d

γq1...qd xq11 . . . x

qdd , γq1...qd ∈ R1

(for d = 2 and d = 3 these functions are called bilinear and trilinear

polynomials, respectively), dimQ1(K) = 2d and

ΣK =

p(Ai), 1 ≤ i ≤ 2d

,

where Ai are the vertices of K.

A1 A2

A4 A3

A A

A

A

A

A

A

A

1 2

34

5

84 7

6

52

Prismatic element

The set K is a prismatic domain. The space PK is the tensor product of

the space P1 in variables x1, x2 by the space Q1 in the variable x3, i.e.

dim PK = 6 and PK consists of polynomials of the form

p(x1, x2, x3) = γ0 + γ1x1 + γ2x2 + γ3x3 + γ4x1x3 + γ5x2x3 , γi ∈ R1 ,

and

ΣK = p(Ai) , 1 ≤ i ≤ 6 ,

where Ai are again the vertices of K.

A 1A 3

A

A

A2

A4

5

6

x

xx

1

3

2

53

All the previous elements are called Lagrange finite elements since the

degrees of freedom are of the form p 7→ p(A), A ∈ K. The point A is

then called the node. These elements are PK-unisolvent, i.e, given any

(α1, . . . , αN )T ∈ RN there exists just one polynomial p ∈ PK such that,

for all the nodes A1, . . . , AN , p(Ai) = αi holds.

If at least one directional derivative occurs as a degree of freedom, the

associated finite element is said to be a Hermite finite element.

54

As an example we introduce the Hermite cubic element which is also

PK-unisolvent. Here K is a triangle, PK = P3(K) is the space of cubic

polynomials, and

ΣK =

p(Ai),∂p

∂x1(Ai),

∂p

∂x2(Ai), i = 1, 2, 3 ; p(G)

,

where Ai are the vertices of K and G is its centre of gravity.

AA1 2

A3

G

55

Some other types of degrees of freedom often used in FEM:

Φ(p) =∂p

∂n(A) – normal derivative

Φ(p) = (Dip)(A), |i| ≥ 1 – higher order derivative

Φ(p) =

∫

K

p(x) dx – integral over the element K

Φ(v) = n1v1(A)+· · ·+ndvd(A) – normal component of vector-function (v1, . . . , vd)T

Note that, in engineering literature, the degrees of freedom of a finite

element are sometimes called parameters.

56

Finite Element Spaces

We shall give now a description of a finite element space generated by

Lagrange elements (K,PK ,ΣK), K ∈ Th.

The sets of degrees of freedom of adjacent finite elements will be related

as follows: Whenever (Kℓ, PKℓ,ΣKℓ

) with ΣKℓ= p(Aℓ

i), 1 ≤ i ≤ Nℓ,

ℓ = 1, 2, are two adjacent finite elements, then

(

N1⋃

i=1

A1i

)

∩K2 =

(

N2⋃

i=1

A2i

)

∩K1 .

Let us denote this set by NS , where S = K1 ∩K2.

57

Note that the situation in below cannot occur due to property (5).

K1

K2

When we want to joint adjacent finite elements, we moreover need that

p1|S | p1 ∈ PK1 = p2|S | p2 ∈ PK2

.

We denote this space by PS . Finally suppose that if p ∈ PS and p(A) = 0

for all A ∈ NS , then p = 0 on S.

58

Let us define the set

Nh =⋃

K∈Th

NK ,

where, for each finite element (K,PK ,ΣK), the symbol NK denotes the

set of nodes. The associated finite element space Xh is then given by

Xh = vh ∈ C(Ω) | vh|K ∈ PK ∀K ∈ Th .

Therefore a function in the space Xh is uniquely determined by the set

Σh = v(A), A ∈ Nh ,

which is called the set of degrees of freedom of the finite element space

Xh. Given values at nodal points, we always get a continuous function.

59

An example of function from Xh defined by the linear finite elements for

d = 2.

60

According to Theorem 4 and the consistency condition (6) we define

finite element subspaces of the space

V = v ∈ H1(Ω) | v = 0 on Γ1 .

by

Vh = vh ∈ Xh | vh(A) = 0 for A ∈ Nh ∩ Γ1 ⊂ V .

It is really a subspace of V , as vh = 0 on Γ1, by our assumptions.

61

To guarantee the requirement (iii) we choose the basis vimi=1 ⊂ Vh

vi(Aj) = δij ,

where m = dimVh, i, j = 1, . . . ,m, Ai ∈ Nh, and Ai 6∈ Γ1.

0

1

1

62

Let for example d = 1, Ω = (0, 1),

a(v, w) =

(

∂v

∂x,∂w

∂x

)

0,Ω

, v, w ∈ H10 (Ω),

and 0 < A1 < A2 < · · · < Am < 1. Then

a(vi, vj) =

(

∂vi

∂x,∂vj

∂x

)

0, suppvi∩suppvj

and the standard Courant basis functions yield a(vi, vj) = 0 whenever

|i− j| > 1. Thus we see that the matrix A = (a(vi, vj))mi,j=1 is only

tridiagonal.

For d > 1 the nodes can be numbered so that A is a band matrix.

63

When ∂Ω is piecewise curved, there are several ways of constructing

finite element spaces. One way is to generate them by the so-called

isoparametric (curved) elements. Note that a curved element, K need

not be convex and PK may be formed e.g. by rational functions.

K

Ω

Another way is to approximate Ω by a polygonal (polyhedral) domain

Ωh ⊂ Ω and then to extend finite element functions from Ωh to the

whole Ω in an appropriate manner.

64

Consider for instance a bounded plane domain with a Lipschitz boundary

which consists of a finite number of smooth convex and concave arcs.

Ω

Ω h

The sides of ∂Ωh are chords or tangents of convex or concave arcs,

respectively. The points of inflexion of ∂Ω coincide with some vertices of

∂Ωh.

65

For such a polygon Ωh there is a triangulation Th consisting of triangles

such that the maximum length of all their sides equals h. Let us define

the space of continuous piecewise linear functions as follows

Vh = vh ∈ C(Ω) | vh|K ∈ P1(K) ∀K ∈ Th , vh|Ω\Ωh= 0 .

We see now that Vh is entirely contained in the space of test functions

V = H10 (Ω) for the homogeneous Dirichlet problem.

Note that in the case of the mixed boundary conditions, the parts Γ2

and Γ3 need not be approximated by “polygonal” curves and we still

have Vh ⊂ V .

66

When Vh ⊂ V , and when the bilinear and linear form of the discrete

problem are identical to the original ones, the finite element method is

said to be conforming. That was the case before.

Ω

Ω h

A nonconforming method arises when Vh 6⊂ V or when e.g. some

numerical integration is used. Figure in above shows another manner of

boundary approximation which is often used in practice and which also

yields Vh 6⊂ V .

A nonconforming method is also obtained when the function u ∈ H1(Ω)

from the Dirichlet boundary condition is approximized by a piecewise

polynomial continuous function.

67