2

1/2 ±

v := v, v = ‚ v dx (3.7)

k k

„¦ ± multi’index

|±|¤k

1/2 1/2

2 2

± ±

= ‚ v dx = ‚v .

0

„¦

± multi’index ± multi’index

|±|¤k |±|¤k

Greater ¬‚exibility with respect to the smoothness properties of the func-

tions that are contained in the de¬nition is obtained by requiring that v

and its weak derivatives should belong not to L2 („¦) but to Lp („¦). In the

norm denoted by · k,p the L2 („¦) and 2 norms (for the vector of the

derivative norms) have to be replaced by the Lp („¦) and p norms, respec-

tively (see Appendices A.3 and A.5). However, the resulting space, denoted

k

by Wp („¦), can no longer be equipped with a scalar product for p = 2. Al-

though these spaces o¬er greater ¬‚exibility, we will not use them except in

Sections 3.6, 6.2, and 9.3.

Besides the norms · k , there are seminorms | · |l for 0 ¤ l ¤ k in H k („¦),

de¬ned by

1/2

2

|v|l = ‚±v ,

0

± multi’index

|±|=l

such that

1/2

k

|v|2

v = ,

k l

l=0

In particular, these de¬nitions are compatible with those in (2.18),

vw + ∇v · ∇w dx ,

v, w :=

1

„¦

· for the L2 („¦) norm, giving a meaning to this

and with the notation 0

one.

3.1. Variational Equations and Sobolev Spaces 95

The above de¬nition contains some assertions that are formulated in the

following theorem:

Theorem 3.3 The bilinear form ·, · k is a scalar product on H k („¦); that

is, · k is a norm on H k („¦).

H k („¦) is complete with respect to · k , and is thus a Hilbert space.

2

Proof: See, for example, [37].

Obviously,

H k („¦) ‚ H l („¦) for k ≥ l ,

and the embedding is continuous, since

¤v for all v ∈ H k („¦) .

v (3.8)

l k

In the one-dimensional case (d = 1) v ∈ H 1 („¦) is necessarily continuous:

Lemma 3.4

H 1 (a, b) ‚ C[a, b] ,

and the embedding is continuous, where C[a, b] is equipped with the norm

· ∞ ; that is, there exists some constant C > 0 such that

¤C v for all v ∈ H 1 (a, b) .

v (3.9)

∞ 1

2

Proof: See Exercise 3.2.

Since the elements of H k („¦) are ¬rst of all only square integrable func-

tions, they are determined only up to points of a set of (d-dimensional)

measure zero. Therefore, a result as in Lemma 3.4 means that the func-

tion is allowed to have removable discontinuities at points of such a set of

measure zero that vanish by modifying the function values.

However, in general, H 1 („¦) ‚ C(„¦).

¯

As an example for this, we consider a circular domain in dimension d = 2:

„¦ = BR (0) = x ∈ R2 |x| < R , R < 1.

Then the function

1

v(x) := |log |x| |γ for some γ <

2

¯

is in H 1 („¦), but not in C(„¦) (see Exercise 3.3).

The following problem now arises: In general, one cannot speak of a

value v(x) for some x ∈ „¦ because a set of one point {x} has (Lebesgue)

measure zero. How do we then have to interpret the Dirichlet boundary

conditions? A way out is to consider the boundary (pieces of the boundary,

respectively) not as arbitrary points but as (d ’ 1)-dimensional “spaces”

(manifolds).

96 3. Finite Element Methods for Linear Elliptic Problems

The above question can therefore be reformulated as follows: Is it possible

to interpret v on ‚„¦ as a function of L2 (‚„¦) (‚„¦ “‚” Rd’1 ) ?

It is indeed possible if we have some minimal regularity of ‚„¦ in the

following sense: It has to be possible to choose locally, for some boundary

point x ∈ ‚„¦, a coordinate system in such a way that the boundary is

locally a hyperplane in this coordinate system and the domain lies on one

side. Depending on the smoothness of the parametrisation of the hyperplane

we then speak of Lipschitz, C k - (for k ∈ N), and C ∞ - domains (for an exact

de¬nition see Appendix A.5).

Examples:

(1) A circle „¦ = x ∈ Rd |x ’ x0 | < R is a C k -domain for all k ∈ N,

and hence a C ∞ -domain.

(2) A rectangle „¦ = x ∈ Rd 0 < xi < ai , i = 1, . . . , d is a Lipschitz

domain, but not a C 1 -domain.

(3) A circle with a cut „¦ = x ∈ Rd |x ’ x0 | < R, x = x0 + »e1 for 0 ¤

» < R is not a Lipschitz domain, since „¦ does not lie on one side of

‚„¦ (see Figure 3.1).

„¦ „¦ „¦

Circle Rectangle Circle with cut

Figure 3.1. Domains of di¬erent smoothness.

Hence, suppose „¦ is a Lipschitz domain. Since only a ¬nite number of

overlapping coordinate systems are su¬cient for the description of ‚„¦,

using these, it is possible to introduce a (d ’ 1)-dimensional measure on

‚„¦ and de¬ne the space L2 (‚„¦) of square integrable functions with respect

to this measure (see Appendix A.5 or [37] for an extensive description). In

the following, let ‚„¦ be equipped with this (d ’ 1)-dimensional measure

dσ, and integrals over the boundary are to be interpreted accordingly. This

also holds for Lipschitz subdomains of „¦, since they are given by the ¬nite

elements.

Theorem 3.5 (Trace Theorem) Suppose „¦ is a bounded Lipschitz do-

main. We de¬ne

C ∞ (Rd )|„¦ v : „¦ ’ R v can be extended to v : Rd ’ R and

:= ˜

∞

v ∈ C (R ) .

d

˜

Then, C ∞ (Rd )|„¦ is dense in H 1 („¦); that is, with respect to · 1 an arbitrary

w ∈ H 1 („¦) can be approximated arbitrarily well by some v ∈ C ∞ (Rd )|„¦ .

3.1. Variational Equations and Sobolev Spaces 97

The mapping that restricts v to ‚„¦,

γ0 : C ∞ (Rd )|„¦ , · ’ L2 (‚„¦), · ,

1 0

’ v|‚„¦ ,

v

is continuous.

Thus there exists a unique, linear, and continuous extension

γ0 : H 1 („¦), · ’ L2 (‚„¦), · .

1 0

2

Proof: See, for example, [37].

Therefore, in short form, γ0 (v) ∈ L2 (‚„¦), and there exists some constant

C > 0 such that

¤C v for all v ∈ H 1 („¦) .

γ0 (v) 0 1

Here γ0 (v) ∈ L2 (‚„¦) is called the trace of v ∈ H 1 („¦).

The mapping γ0 is not surjective; that is, γ0 (v) v ∈ H 1 („¦) is a real

subset of L2 (‚„¦). For all v ∈ C ∞ (Rd )|„¦ we have

γ0 (v) = v|‚„¦ .

In the following we will use again v|‚„¦ or “v on ‚„¦” for γ0 (v), but in

the sense of Theorem 3.5. According to this theorem, de¬nition (2.20) is

well-de¬ned with the interpretation of u on ‚„¦ as the trace:

De¬nition 3.6 H0 („¦) := v ∈ H 1 („¦) γ0 (v) = 0 (as a function on ‚„¦) .

1

Theorem 3.7 Suppose „¦ ‚ Rd is a bounded Lipschitz domain. Then

∞ 1

C0 („¦) is dense in H0 („¦).

2

Proof: See [37].

The assertion of Theorem 3.5, that C ∞ (Rd )|„¦ is dense in H 1 („¦), has

severe consequences for the treatment of functions in H 1 („¦) which are in

general not very smooth. It is possible to consider them as smooth functions

if at the end only relations involving continuous expressions in · 1 (and not

requiring something like ‚i v ∞ ) arise. Then, by some “density argument”

the result can be transferred to H 1 („¦) or, as for the trace term, new terms

can be de¬ned for functions in H 1 („¦). Thus, for the proof of Lemma 3.4

it is necessary simply to verify estimate (3.9), for example for v ∈ C 1 [a, b].

1

By virtue of Theorem 3.7, analogous results hold for H0 („¦).

Hence, for v ∈ H 1 („¦) integration by parts is possible:

Theorem 3.8 Suppose „¦ ‚ Rd is a bounded Lipschitz domain. The outer

unit normal vector ν = (νi )i=1,...,d : ‚„¦ ’ Rd is de¬ned almost everywhere

and νi ∈ L∞ (‚„¦).

98 3. Finite Element Methods for Linear Elliptic Problems

For v, w ∈ H 1 („¦) and i = 1, . . . , d,

‚i v w dx = ’ v ‚i w dx + v w νi dσ .

„¦ „¦ ‚„¦

2

Proof: See, for example, [14] or [37].

If v ∈ H 2 („¦), then due to the above theorem, v|‚„¦ := γ0 (v) ∈ L2 (‚„¦)

and ‚i v|‚„¦ := γ0 (‚i v) ∈ L2 (‚„¦), since also ‚i v ∈ H 1 („¦). Hence, the normal

derivative

d

‚ν v|‚„¦ := ‚i v|‚„¦ νi

i=1

is well-de¬ned and belongs to L2 (‚„¦).

Thus, the trace mapping

γ : H 2 („¦) ’ L2 (‚„¦) — L2 (‚„¦) ,

’ (v|‚„¦ , ‚ν v|‚„¦ ) ,

v

is well-de¬ned and continuous. The continuity of this mapping follows from

the fact that it is a composition of continuous mappings:

continuous continuous

v ∈ H 2 („¦) ’ ‚i v ∈ H 1 („¦) ’ ‚i v|‚„¦ ∈ L2 (‚„¦)

continuous

’ ‚i v|‚„¦ νi ∈ L2 (‚„¦) .

Corollary 3.9 Suppose „¦ ‚ Rd is a bounded Lipschitz domain.

(1) Let w ∈ H 1 („¦), qi ∈ H 1 („¦), i = 1, . . . , d. Then

q · ∇w dx = ’ ∇ · q w dx + q · ν w dσ . (3.10)

„¦ „¦ ‚„¦

(2) Let v ∈ H 2 („¦), w ∈ H 1 („¦). Then

∇v · ∇w dx = ’ ∆v w dx + ‚ν v w dσ .

„¦ „¦ ‚„¦

The integration by parts formulas also hold more generally if only it is

ensured that the function whose trace has to be formed belongs to H 1 („¦).

For example, if K = (kij )ij , where kij ∈ W∞ („¦) and v ∈ H 2 („¦), w ∈

1

H 1 („¦), it follows that

K∇v · ∇w dx = ’ ∇ · (K∇v) w dx + K∇v · ν w dσ (3.11)

„¦ „¦ ‚„¦

with conormal derivative (see (0.41))

d

‚νK v := K∇v · ν = ∇v · K ν = T

kij ‚j v νi .

i,j=1

Exercises 99

Here it is important that the components of K∇v belong to H 1 („¦), using

the fact that for v ∈ L2 („¦), k ∈ L∞ („¦),

kv ∈ L2 („¦) and ¤k

kv v .

∞

0 0

Theorem 3.10 Suppose „¦ ‚ Rd is a bounded Lipschitz domain.

If k > d/2, then

H k („¦) ‚ C(„¦) ,

¯

and the embedding is continuous.

2

Proof: See, for example, [37].

For dimension d = 2 this requires k > 1, and for dimension d = 3 we

need k > 3 . Therefore, in both cases k = 2 satis¬es the assumption of the

2

above theorem.

Exercises

3.1 Prove the Lax“Milgram Theorem in the following way:

(a) Show, by using the Riesz representation theorem, the equivalence of

(3.5) with the operator equation

A¯ = f

u

for A ∈ L[V, V ] and f ∈ V .

(b) Show, for Tµ ∈ L[V, V ], Tµ v := v ’ µ(Av ’ f ) and µ > 0, that for

some µ > 0, the operator Tµ is a contraction on V . Then conclude

the assertion by Banach™s ¬xed-point theorem (in the Banach space

setting, cf. Remark 8.5).

3.2 Prove estimate (3.9) by showing that even for v ∈ H 1 (a, b),

|v(x) ’ v(y)| ¤ |v|1 |x ’ y|1/2 for x, y ∈ (a, b) .

3.3 Suppose „¦ ‚ R2 is the open disk with radius 1 and centre 0. Prove

2

±

that for the function u(x) := ln |x| , x ∈ „¦ \ {0}, ± ∈ (0, 1 ) we have

2

u ∈ H 1 („¦), but u cannot be extended continuously to x = 0.

3.4 Suppose „¦ ‚ R2 is the open unit disk. Prove that each u ∈ H 1 („¦)

√

has a trace u|‚„¦ ∈ L2 (‚„¦) satisfying u 0,‚„¦ ¤ 4 8 u 1,„¦ .

100 3. Finite Element Methods for Linear Elliptic Problems

3.2 Elliptic Boundary Value Problems of Second

Order

In this section we integrate boundary value problems for the linear, sta-

tionary case of the di¬erential equation (0.33) into the general theory of

Section 3.1.

Concerning the domain we will assume that „¦ is a bounded Lipschitz

domain.

We consider the equation

(Lu)(x) := ’∇ · (K(x)∇u(x)) + c(x) · ∇u(x) + r(x)u(x) = f (x) for x ∈ „¦

(3.12)

with the data

K : „¦ ’ Rd,d , c : „¦ ’ Rd , r, f : „¦ ’ R.

Assumptions about the Coe¬cients and the Right-Hand Side

For an interpretation of (3.12) in the classical sense, we need

‚i kij , ci , r , f ∈ C(„¦) , i, j ∈ {1, . . . , d} ,

¯ (3.13)

and for an interpretation in the sense of L2 („¦) with weak derivatives, and

hence for a solution in H 2 („¦),

‚i kij , ci , r ∈ L∞ („¦) , f ∈ L2 („¦) , i, j ∈ {1, . . . , d} . (3.14)

Once we have obtained the variational formulation, weaker assumptions

about the smoothness of the coe¬cients will be su¬cient for the veri¬ca-

tion of the properties (3.2)“(3.4), which are required by the Lax“Milgram,

namely,

kij , ci , ∇ · c , r ∈ L∞ („¦) , f ∈ L2 („¦) , i, j ∈ {1, . . . , d} ,

(3.15)

ν · c ∈ L∞ (“1 ∪ “2 ) .

and if |“1 ∪ “2 |d’1 > 0 ,

Here we refer to a de¬nition of the boundary conditions as in (0.36)“(0.39)

(see also below). Furthermore, the uniform ellipticity of L is assumed: There

exists some constant k0 > 0 such that for (almost) every x ∈ „¦,

d

kij (x)ξi ξj ≥ k0 |ξ|2 for all ξ ∈ Rd (3.16)

i,j=1

(that is, the coe¬cient matrix K is positive de¬nite uniformly in x).

Moreover, K should be symmetric.

If K is a diagonal matrix, that is, kij (x) = ki (x)δij (this is in particular

the case if ki (x) = k(x) with k : „¦ ’ R, i ∈ {1, . . . , d}, where K∇u

becomes k∇u), this means that

” ki (x) ≥ k0 for (almost) every x ∈ „¦ , i ∈ {1, . . . , d} .

(3.16)

3.2. Elliptic Boundary Value Problems 101

Finally, there exists a constant r0 ≥ 0 such that

1

r(x) ’ ∇ · c(x) ≥ r0 for (almost) every x ∈ „¦ . (3.17)

2

Boundary Conditions

As in Section 0.5, suppose “1 , “2 , “3 is a disjoint decomposition of the

boundary ‚„¦ (cf. (0.39)):

‚„¦ = “1 ∪ “2 ∪ “3 ,

where “3 is a closed subset of the boundary. For given functions gj : “j ’

R , j = 1, 2, 3, and ± : “2 ’ R we assume on ‚„¦

• Neumann boundary condition (cf. (0.41) or (0.36))

K∇u · ν = ‚νK u = g1 on “1 , (3.18)

• mixed boundary condition (cf. (0.37))

K∇u · ν + ±u = ‚νK u + ±u = g2 on “2 , (3.19)

• Dirichlet boundary condition (cf. (0.38))

u = g3 on “3 . (3.20)

Concerning the boundary data the following is assumed: For the classical

approach we need

gj ∈ C(“j ) , ± ∈ C(“2 ) ,

j = 1, 2, 3 , (3.21)

whereas for the variational interpretation,

± ∈ L∞ (“2 )

gj ∈ L2 (“j ) , j = 1, 2, 3 , (3.22)

is su¬cient.

3.2.1 Variational Formulation of Special Cases

The basic strategy for the derivation of the variational formulation of

boundary value problems (3.12) has already been demonstrated in Sec-

tion 2.1. Assuming the existence of a classical solution of (3.12) the

following steps are performed in general:

Step 1: Multiplication of the di¬erential equation by test functions that

are chosen compatible with the type of boundary condition and

subsequent integration over the domain „¦.

Step 2: Integration by parts under incorporation of the boundary condi-

tions in order to derive a suitable bilinear form.

Step 3: Veri¬cation of the required properties like ellipticity and continuity.

102 3. Finite Element Methods for Linear Elliptic Problems

In the following the above steps will be described for some important special

cases.

(I) Homogeneous Dirichlet Boundary Condition

‚„¦ = “3 , g3 ≡ 0 , V := H0 („¦)

1