v ∈ V has been chosen arbitrarily. 2

For applications e.g. in structural mechanics as above, the minimization

problem is called the principle of minimal potential energy.

Remark 2.4 Lemma 2.3 holds for general vector spaces V if a is a sym-

metric, positive bilinear form and the right-hand side f, v 0 is replaced by

b(v), where b : V ’ R is a linear mapping, a linear functional. Then the

variational equation reads as

¬nd u ∈ V for all v ∈ V ,

with a(u, v) = b(v) (2.13)

and the minimization problem as

¬nd u ∈ V F (u) = min F (v) v ∈ V

with , (2.14)

2.1. Variational Formulation 51

1

a(v, v) ’ b(v) .

where F (v) :=

2

Lemma 2.5 The weak solution according to (2.10) (or (2.11)) is unique.

Proof: Let u1 , u2 be two weak solutions, i.e.,

a(u1 , v) = f, v ,

0

for all v ∈ V .

a(u2 , v) = f, v ,

0

By subtraction, it follows that

a(u1 ’ u2 , v) = 0 for all v ∈ V.

Choosing v = u1 ’ u2 implies a(u1 ’ u2 , u1 ’ u2 ) = 0 and consequently

2

u1 = u2 , because a is de¬nite.

Remark 2.6 Lemma 2.5 is generally valid if a is a de¬nite bilinear form

and b is a linear form.

So far, we have de¬ned two di¬erent norms on V : · a and · 0 . The

di¬erence between these norms is essential because they are not equivalent

on the vector space V de¬ned by (2.7), and consequently, they generate

di¬erent convergence concepts, as will be shown by the following example:

Example 2.7 Let „¦ = (0, 1), i.e.

1

a(u, v) := u v dx ,

0

and let vn : „¦ ’ R for n ≥ 2 be de¬ned by (cf. Figure 2.1)

±

for 0 ¤ x ¤ n ,

1

nx ,

for n ¤ x ¤ 1 ’ n ,

1 1

1,

vn (x) =

n ’ nx , for 1 ’ n ¤ x ¤ 1 .

1

vn

1

1 n-1

n1

n

Figure 2.1. The function vn .

Then

1/2

1

¤

vn 1 dx = 1,

0

0

52 2. Finite Element Method for Poisson Equation

1/2

1

√

1

n

2n ’ ∞ for n ’ ∞ .

n2 dx + n2 dx

vn = =

a

1

0 1’ n

¤C v

Therefore, there exists no constant C > 0 such that v for all

a 0

v ∈V.

However, as we will show in Theorem 2.18, there exists a constant C > 0

such that the estimate

¤C v for all v ∈ V

v 0 a

holds; i.e., · a is the stronger norm.

It is possible to enlarge the basic space V without violating the previous

statements. The enlargement is also necessary because, for instance, the

proof of the existence of a solution of the variational equation (2.13) or

the minimization problem (2.14) requires in general the completeness of V.

However, the actual de¬nition of V does not imply the completeness, as

the following example shows:

Example 2.8 Let „¦ = (0, 1) again and therefore

1

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

0

For u(x) := x± (1’x)± with ± ∈ 1 , 1 we consider the sequence of functions

2

±

for x ∈ n , 1 ’ n ,

1 1

u(x)

for x ∈ 0, n ,

1 1

un (x) := n u( n ) x

n u(1 ’ n ) (1 ’ x) for x ∈ 1 ’ n , 1 .

1 1

Then

un ’ um ’0 for n, m ’ ∞ ,

a

un ’ u ’0 for n ’ ∞ ,

a

but u ∈ V , where V is de¬ned analogously to (2.7) with d = 1.

/

In Section 3.1 we will see that a vector space V normed with · a exists

˜

such that u ∈ V and V ‚ V . Therefore, V is not complete with respect

˜ ˜

to · a ; otherwise, u ∈ V must be valid. In fact, there exists a (unique)

completion of V with respect to · a (see Appendix A.4, especially (A4.26)),

but we have to describe the new “functions” added by this process. Besides,

integration by parts must be valid such that a classical solution continues to

be also a weak solution (compare with Lemma 2.1). Therefore, the following

idea is unsuitable.

Attempt of a correct de¬nition of V :

Let V be the set of all u with the property that ‚i u exists for all x ∈ „¦

without any requirements on ‚i u in the sense of a function.

2.1. Variational Formulation 53

For instance, there exists Cantor™s function with the following properties:

f : [0, 1] ’ R, f ∈ C([0, 1]), f = 0, f is not constant, f (x) exists with

f (x) = 0 for all x ∈ [0, 1].

x

Here the fundamental theorem of calculus, f (x) = 0 f (s) ds+f (0), and

thus the principle of integration by parts, are no longer valid.

Consequently, additional conditions for ‚i u are necessary.

To prepare an adequate de¬nition of the space V, we extend the de¬nition

of derivatives by means of their action on averaging procedures. In order

to do this, we introduce the multi-index notation.

A vector ± = (±1 , . . . , ±d ) of nonnegative integers ±i ∈ {0, 1, 2, . . .} is

d

called a multi-index. The number |±| := i=1 ±i denotes the order (or

length) of ±.

For x ∈ Rd let

x± := x±1 · · · x±d . (2.15)

1 d

A shorthand notation for the di¬erential operations can be adopted by this:

For an appropriately di¬erentiable function u let

‚ ± u := ‚1 1 · · · ‚d d u .

± ±

(2.16)

We can obtain this de¬nition from (2.15) by replacing x by the symbolic

vector

T

∇ := (‚1 , . . . , ‚d )

of the ¬rst partial derivatives.

For example, if d = 2 and ± = (1, 2), then |±| = 3 and

‚3u

‚ ± u = ‚1 ‚2 u =

2

.

‚x1 ‚x2

2

Now let ± be a multi-index of length k and let u ∈ C k („¦). We then

∞

obtain for arbitrary test functions • ∈ C0 („¦) by integration by parts

‚ ± u • dx = (’1)k u ‚ ± • dx .

„¦ „¦

The boundary integrals vanish because ‚ β • = 0 on ‚„¦ for all multi-indices

β.

Therefore, we make the following de¬nition:

De¬nition 2.9 v ∈ L2 („¦) is called the weak (or generalized) derivative

∞

‚ ± u of u ∈ L2 („¦) for the multi-index ± if for all • ∈ C0 („¦),

v • dx = (’1)|±| u ‚ ± • dx .

„¦ „¦

54 2. Finite Element Method for Poisson Equation

The weak derivative is well-de¬ned because it is unique: Let v1 , v2 ∈

2

L („¦) be two weak derivatives of u. It follows that

∞

(v1 ’ v2 ) • dx = 0 for all • ∈ C0 („¦) .

„¦

∞

Since C0 („¦) is dense in L2 („¦), we can furthermore conclude that

(v1 ’ v2 ) • dx = 0 for all • ∈ L2 („¦) .

„¦

If we now choose speci¬cally • = v1 ’ v2 , we obtain

v1 ’ v2 (v1 ’ v2 ) (v1 ’ v2 ) dx = 0 ,

2

=

0

„¦

and v1 = v2 (a.e.) follows immediately. In particular, u ∈ C k („¦) has weak

¯

derivatives ‚ ± u for ± with |±| ¤ k, and the weak derivatives are identical

to the classical (pointwise) derivatives.

Also the di¬erential operators of vector calculus can be given a weak

de¬nition analogous to De¬nition 2.9. For example, for a vector ¬eld q

with components in L2 („¦), v ∈ L2 („¦) is the weak divergence v = ∇ · q if

∞

for all • ∈ C0 („¦)

v• dx = ’ q · ∇• dx .

„¦ „¦

1

The correct choice of the space V is the space H0 („¦), which will be

de¬ned below. First we de¬ne

u : „¦ ’ R u ∈ L2 („¦) , u has weak derivatives

H 1 („¦) :=

(2.17)

‚i u ∈ L2 („¦) for all i = 1, . . . , d .

A scalar product on H 1 („¦) is de¬ned by

∇u(x) · ∇v(x) dx

u, v := u(x)v(x) dx + (2.18)

1

„¦ „¦

with the norm

1/2

|u(x)| dx + |∇u(x)| dx

2 2

u := u, u = (2.19)

1 1

„¦ „¦

induced by this scalar product.

The above “temporary” de¬nition (2.7) of V takes care of the boundary

condition u = 0 on ‚„¦ by conditions for the functions. I.e. we want to

choose the basic space V analogously as:

H0 („¦) := u ∈ H 1 („¦) u = 0

1

on ‚„¦ . (2.20)

Here H 1 („¦) and H0 („¦) are special cases of so-called Sobolev spaces.

1

For „¦ ‚ Rd , d ≥ 2, H 1 („¦) may contain unbounded functions. In par-

ticular, we have to examine carefully the meaning of u|‚„¦ (‚„¦ has the

2.2. The Finite Element Method with Linear Elements 55

d-dimensional measure 0) and, in particular, u = 0 on ‚„¦. This will be

described in Section 3.1.

Exercises

2.1

(a) Consider the interval (’1, 1); prove that the function u(x) = |x| has

the generalized derivative u (x) = sign(x).

(b) Does sign(x) have a generalized derivative?

N

2.2 Let „¦ = l=1 „¦l , N ∈ N, where the bounded subdomains „¦l ‚ R2

are pairwise disjoint and possess piecewise smooth boundaries. Show that

a function u ∈ C(„¦) with u|„¦l ∈ C 1 („¦l ), 1 ¤ l ¤ N, has a weak derivative

‚i u ∈ L2 („¦), i = 1, 2, that coincides in N „¦l with the classical one.

l=1

2.3 Let V be the set of functions that are continuous and piecewise con-

tinuously di¬erentiable on [0, 1] and that satisfy the additional conditions

u(0) = u(1) = 0. Show that there exist in¬nitely many elements in V that

minimize the functional

1

2

1 ’ [u (x)]2

F (u) := dx.

0

2.2 The Finite Element Method

with Linear Elements

The weak formulation of the boundary value problem (2.1), (2.2) leads to

particular cases of the following general, here equivalent, problems:

Let V be a vector space, let a : V — V ’ R be a bilinear form, and let

b : V ’ R be a linear form.

Variational equation:

Find u ∈ V a(u, v) = b(v) for all v ∈ V .

with (2.21)

Minimization problem:

Find u ∈ V with F (u) = min F (v) v ∈ V ,

(2.22)

1

where F (v) = a(v, v) ’ b(v) .

2

The discretization approach consists in the following procedure: Replace

V by a ¬nite-dimensional subspace Vh ; i.e., solve instead of (2.21) the ¬nite-

dimensional variational equation,

¬nd uh ∈ Vh for all v ∈ Vh .

with a(uh , v) = b(v) (2.23)

56 2. Finite Element Method for Poisson Equation

This approach is called the Galerkin method. Or solve instead of (2.22) the

¬nite-dimensional minimization problem,

¬nd uh ∈ Vh F (uh ) = min F (v) v ∈ Vh .

with (2.24)

This approach is called the Ritz method.

It is clear from Lemma 2.3 and Remark 2.4 that the Galerkin method

and the Ritz method are equivalent for a positive and symmetric bilinear

form. The ¬nite-dimensional subspace Vh is called an ansatz space.

The ¬nite element method can be interpreted as a Galerkin method (and

in our example as a Ritz method, too) for an ansatz space with special

properties. In the following, these properties will be extracted by means of

the simplest example.

1

Let V be de¬ned by (2.7) or let V = H0 („¦).

The weak formulation of the boundary value problem (2.1), (2.2)

corresponds to the choice

∇u · ∇v dx ,

a(u, v) := b(v) := f v dx .

„¦ „¦

Let „¦ ‚ R2 be a domain with a polygonal boundary; i.e., the boundary

“ of „¦ consists of a ¬nite number of straight-line segments as shown in

Figure 2.2.

„¦

Figure 2.2. Domain with a polygonal boundary.

Let Th be a partition of „¦ into closed triangles K (i.e., including the

boundary ‚K) with the following properties:

(1) „¦ = ∪K∈Th K;

(2) For K, K ∈ Th , K = K ,

int (K) © int (K ) = … , (2.25)

where int (K) denotes the open triangle (without the boundary ‚K).

(3) If K = K but K ©K = …, then K ©K is either a point or a common

edge of K and K (cf. Figure 2.3).

A partition of „¦ with the properties (1), (2) is called a triangulation

of „¦. If, in addition, a partition of „¦ satis¬es property (3), it is called a

conforming triangulation (cf. Figure 2.4).

2.2. Linear Elements 57

r ¨ r

¨

rr rr d

¨

rr ¨¨ rr d

not

r

¨

¨rr d

¨

allowed: ¨

allowed:

¨ ¨

rr d

¨¨ ¨

¨

r d

¨ r ¨ d

Figure 2.3. Triangulations.

The triangles of a triangulation will be numbered K1 , . . . , KN . The

subscript h indicates the ¬neness of the triangulation, e.g.,

h := max diam (K) K ∈ Th ,

where diam (K) := sup |x ’ y| x, y ∈ K denotes the diameter of K.

Thus here h is the maximum length of the edges of all the triangles.

Sometimes, K ∈ Th is also called a (geometric) element of the partition.

The vertices of the triangles are called the nodes, and they will be

numbered

a1 , a2 , . . . , aM ,

i.e., ai = (xi , yi ), i = 1, . . . , M , where M = M1 + M2 and

∈

a1 , . . . , aM1 „¦,

(2.26)