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for the existence of a minimum at µ = 0. Thus u solves (2.10), because
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)

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