c · ν ≥ 0 on “1 and c · ν ¤ 0 on “2

hold. Frequently prescribed boundary conditions are

’(c u ’ K∇u) · ν = ’ν · c u on “1 ,

’(c u ’ K∇u) · ν = g2 on “2 .

They are based on the following assumptions: On the in¬‚ow boundary “2

the normal component of the total (mass) ¬‚ux is prescribed but on the

out¬‚ow boundary “1 , on which in the extreme case K = 0 the boundary

conditions would drop out, only the following is required:

• the normal component of the total (mass) ¬‚ux is continuous over “1 ,

• the ambient mass ¬‚ux that is outside „¦ consists only of a convective

part,

• the extensive variable (for example, the concentration) is continuous

over “1 , that is, the ambient concentration in x is also equal to u(x).

Therefore, after an obvious reformulation we get, in accordance with the

de¬nitions of “1 and “2 due to (3.18), (3.19), the Neumann boundary

3.2. Elliptic Boundary Value Problems 109

condition (3.18), and the mixed boundary condition (3.19),

K∇u · ν =0 on “1 ,

K∇u · ν + ± u = g2 on “2 ,

where ± := ’ν · c.

Now the conditions of Theorem 3.15 can be checked:

We have r’ 1 ∇·c = r + 1 ∇·c; therefore, for the latter term the inequality

˜2

2

in (1) and (4)(b) must be satis¬ed. Further, the condition ν · c ≥ 0 on

“1 holds due to the characterization of the out¬‚ow boundary. Because of

± + 1 ν · c = ’ 2 ν · c, the condition (3) is satis¬ed due to the de¬nition of

1

2

the in¬‚ow boundary.

Now we address the case of inhomogeneous Dirichlet boundary

conditions (|“3 |d’1 > 0).

This situation can be reduced to the case of homogeneous Dirich-

let boundary conditions, if we are able to choose some (¬xed) element

w ∈ H 1 („¦) in such a way that (in the sense of trace) we have

γ0 (w) = g3 on “3 . (3.34)

The existence of such an element w is a necessary assumption for the exis-

tence of a solution u ∈ H 1 („¦). On the other hand, such an element w can

˜

exist only if g3 belongs to the range of the mapping

v ’ γ0 (v)|“3 ∈ L2 (“3 ).

H 1 („¦)

However, this is not valid for all g3 ∈ L2 (“3 ), since the range of the trace

operator of H 1 („¦) is a proper subset of L2 (‚„¦).

Therefore, we assume the existence of such an element w. Since only

the homogeneity of the Dirichlet boundary conditions of the test functions

plays a role in derivation (3.31) of the bilinear form a and the linear form

b, we ¬rst obtain with the space V , de¬ned in (3.30), and

V := v ∈ H 1 („¦) : γ0 (v) = g3 on “3 = v ∈ H 1 („¦) : v ’ w ∈ V

˜

the following variational formulation:

Find u ∈ V such that

˜˜

for all v ∈ V .

a(˜, v) = b(v)

u

However, this formulation does not ¬t into the theoretical concept of

˜

Section 3.1 since the space V is not a linear one.

If we put u := u + w, then this is equivalent to the following:

˜

Find u ∈ V such that

a(u, v) = b(v) ’ a(w, v) =: ˜

b(v) for all v ∈ V . (3.35)

Now we have a variational formulation for the case of inhomogeneous

Dirichlet boundary conditions that has the form required in the theory.

110 3. Finite Element Methods for Linear Elliptic Problems

Remark 3.17 In the existence result of Theorem 3.1, the only assumption

is that b has to be a continuous linear form in V .

For d = 1 and „¦ = (a, b) this is also satis¬ed, for instance, for the special

linear form

δγ (v) := v(γ) for v ∈ H 1 (a, b),

where γ ∈ (a, b) is arbitrary but ¬xed, since by Lemma 3.4 the space

H 1 (a, b) is continuously embedded in the space C[a, b]. Thus, for d = 1

point sources (b = δγ ) are also allowed. However, for d ≥ 2 this does not

hold since H 1 („¦) ‚ C(„¦).

¯

Finally, we will once again state the general assumptions under which the

variational formulation of the boundary value problem (3.12), (3.18)“(3.20)

in the space (3.30),

V = v ∈ H 1 („¦) : γ0 (v) = 0 on “3 ,

has properties that satisfy the conditions of the Lax“Milgram Theorem

(Theorem 3.1):

• „¦ ‚ Rd is a bounded Lipschitz domain.

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

|“1 ∪ “2 |d’1 > 0, ν · c ∈ L∞ (“1 ∪ “2 ) (i.e., (3.15)).

• There exists some constant k0 > 0 such that in „¦, we have ξ ·K(x)ξ ≥

k0 |ξ|2 for all ξ ∈ Rd (i.e., (3.16)),

• gj ∈ L2 (“j ) , j = 1, 2, 3, ± ∈ L∞ (“2 ) (i.e., (3.22)).

• The following hold:

r ’ 1 ∇ · c ≥ 0 in „¦ .

(1) 2

ν · c ≥ 0 on “1 .

(2)

± + 1 ν · c ≥ 0 on “2 .

(3) 2

(4) Additionally, one of the following conditions is satis¬ed:

(a) |“3 |d’1 > 0 .

(b) There exists some „¦ ‚ „¦ with |„¦|d > 0 and r0 > 0 such

˜ ˜

that r ’ 1 ∇ · c ≥ r0 on „¦.

˜

2

(c) There exists some “1 ‚ “1 with |“1 |d’1 > 0 and c0 > 0

˜ ˜

such that ν · c ≥ c0 on “1 .

˜

(d) There exists some “2 ‚ “2 with |“2 |d’1 > 0 and ±0 > 0

˜ ˜

such that ± + 1 ν · c ≥ ±0 on “2 .

˜

2

• If |“3 |d’1 > 0 , then there exists some w ∈ H 1 („¦) with γ0 (w) = g3

on “3 (i.e., (3.34)).

3.2. Elliptic Boundary Value Problems 111

3.2.2 An Example of a Boundary Value Problem of Fourth

Order

The Dirichlet problem for the biharmonic equation reads as follows:

Find u ∈ C 4 („¦) © C 1 („¦) such that

¯

∆2 u = f in „¦ ,

(3.36)

‚ν u = u = 0 on ‚„¦ ,

where

d

2 2 2

∆ u := ∆ (∆u) = ‚i ‚j u .

i,j=1

In the case d = 1 this collapses to ∆2 u = u(4) .

For u, v ∈ H 2 („¦) it follows from Corollary 3.9 that

(u ∆v ’ ∆u v) dx = {u ‚ν v ’ ‚ν u v}dσ

„¦ ‚„¦

and hence for u ∈ H 4 („¦), v ∈ H 2 („¦) (by replacing u with ∆u in the above

equation),

∆2 u v dx ’

∆u ∆v dx = ‚ν ∆u v dσ + ∆u ‚ν v dσ .

„¦ „¦ ‚„¦ ‚„¦

For a Lipschitz domain „¦ we de¬ne

H0 („¦) := v ∈ H 2 („¦) v = ‚ν v = 0 on ‚„¦

2

2

and obtain the variational formulation of (3.36) in the space V := H0 („¦):

Find u ∈ V , such that

for all v ∈ V .

a(u, v) := ∆u ∆v dx = b(v) := f v dx

„¦ „¦

More general, for a boundary value problem of order 2m in conservative

form, we obtain a variational formulation in H m („¦) or H0 („¦).

m

3.2.3 Regularity of Boundary Value Problems

In Section 3.2.1 we stated conditions under which the linear elliptic bound-

ary value problem admits a unique solution u (˜, respectively) in some

u

1

subspace V of H („¦). In many cases, for instance for the interpolation of

the solution or in the context of error estimates (also in norms other than

the · V norm) it is not su¬cient that u (˜, respectively) have only ¬rst

u

2

weak derivatives in L („¦).

Therefore, within the framework of the so-called regularity theory, the

question of the assumptions under which the weak solution belongs to

H 2 („¦), for instance, has to be answered. These additional conditions

contain conditions about

112 3. Finite Element Methods for Linear Elliptic Problems

• the smoothness of the boundary of the domain,

• the shape of the domain,

• the smoothness of the coe¬cients and the right-hand side of the

di¬erential equation and the boundary conditions,

• the kind of the transition of boundary conditions in those points,

where the type is changing,

which can be quite restrictive as a whole. Therefore, in what follows we

often assume only the required smoothness. Here we cite as an example

one regularity result ([13, Theorem 8.12]).

Theorem 3.18 Suppose „¦ is a bounded C 2 -domain and “3 = ‚„¦. Further,

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

¯

as well as (3.16). Suppose there exists some function w ∈ H 2 („¦) with

γ0 (w) = g3 on “3 . Let u = u + w and let u be a solution of (3.35). Then

˜

u ∈ H („¦) and

2

˜

¤ C{ u + w 2}

u

˜ +f

2 0 0

with a constant C > 0 independent of u, f , and w.

One drawback of the above result is that it excludes polyhedral domains.

If the convexity of „¦ is additionally assumed, then it can be transferred

to this case. Simple examples of boundary value problems in domains with

reentrant corners show that one cannot avoid such additional assumptions

(see Exercise 3.5).

Exercises

3.5 Consider the boundary value problem (1.1), (1.2) for f = 0 in the

sector „¦ := (x, y) ∈ R2 x = r cos •, y = r sin • with 0 < r < 1, 0 < • <

± for some 0 < ± < 2π, thus with the interior angle ±. Derive as in (1.23),

by using the ansatz w(z) := z 1/± , a solution u(x, y) = w(x + iy) for an

appropriate boundary function g. Then check the regularity of u, that is,

u ∈ H k („¦), in dependence of ±.

3.6 Consider the problem (1.29) with the transmission condition (1.30)

and, for example, Dirichlet boundary conditions and derive a variational

formulation for this.

3.7 Consider the variational formulation:

Find u ∈ H 1 („¦) such that

∇u · ∇v dx = for all v ∈ H 1 („¦) ,

f v dx + gv dσ (3.37)

„¦ „¦ ‚„¦

Exercises 113

where „¦ is a bounded Lipschitz domain, f ∈ L2 („¦) and g ∈ L2 (‚„¦).

(a) Let u ∈ H 1 („¦) be a solution of this problem. Show that ’∆u exists

in the weak sense in L2 („¦) and

’∆u = f .

(b) If additionally u ∈ H 2 („¦), then ‚ν u|‚„¦ exists in the sense of trace in

L2 (‚„¦) and

‚ν u = g

where this equality is to be understood as

(‚ν u ’ g)v dσ = 0 for all v ∈ H 1 („¦) .

‚„¦

3.8 Consider the variational equation (3.37) for the Neumann problem

for the Poisson equation as in Exercise 3.7.

(a) If a solution u ∈ H 1 („¦) exists, then the compatibility condition

f dx + g dσ = 0 (3.38)

„¦ ‚„¦

has to be ful¬lled.

(b) Consider the following bilinear form on H 1 („¦) :

∇u · ∇v dx +

a(u, v) :=

˜ u dx v dx .

„¦ „¦ „¦

1

Show that a is V -elliptic on H („¦).

˜

Hint: Do it by contradiction using the fact that a bounded sequence in

H 1 („¦) possesses a subsequence converging in L2 („¦) (see, e.g., [37]).

(c) Consider the unique solution u ∈ H 1 („¦) of

˜

for all v ∈ H 1 („¦) .

a(u, v) =

˜ f v dx + gv dσ

„¦ ‚„¦

Then:

|„¦| u dx =

˜ f dx + g dσ .

„¦ „¦ ‚„¦

Furthermore, if (3.38) is valid, then u is a solution of (3.37) (with

˜

„¦ u dx = 0).

˜

3.9 Show analogously to Exercise 3.7: A weak solution u ∈ V ‚ H 1 („¦)

of (3.31), where V is de¬ned in (3.30), with data satisfying (3.14) and

(3.22), ful¬lls a di¬erential equation in L2 („¦). The boundary conditions

are ful¬lled in the following sense:

g2 v dσ for all v ∈ V .

‚νK u v dσ+ (‚νK u+± u)v dσ = g1 v dσ+

“1 “2 “1 “2

114 3. Finite Element Methods for Linear Elliptic Problems

3.3 Element Types and A¬ne Equivalent

Triangulations

In order to be able to exploit the theory developed in Sections 3.1 and 3.2

we make the assumption that „¦ is a Lipschitz domain.

The ¬nite element discretization of the boundary value problem (3.12)

with the boundary conditions (3.18)“(3.20) corresponds to performing a

Galerkin approximation (cf. (2.23)) of the variational equation (3.35) with

the bilinear form a and the linear form b, supposed to be de¬ned as in

(3.31), and some w ∈ H 1 („¦) with the property w = g3 on “3 . The solution

of the weak formulation of the boundary value problem is then given by

u := u + w, if u denotes the solution of the variational equation (3.35).

˜

Since the bilinear form a is in general not symmetric, (2.21) and (2.23),

respectively (the variational equation), are no longer equivalent to (2.22)

and (2.24), respectively (the minimization problem), so that in the following

we pursue only the ¬rst, more general, ansatz.

The Galerkin approximation of the variational equation (3.35) reads as

follows: Find some u ∈ Vh such that

a(uh , v) = b(v) ’ a(w, v) = ˜ for all v ∈ Vh .

b(v) (3.39)

The space Vh that is to be de¬ned has to satisfy Vh ‚ V . Therefore, we

speak of a conforming ¬nite element discretization, whereas for a non-

conforming discretization this property, for instance, can be violated. The

ansatz space is de¬ned piecewise with respect to a triangulation Th of „¦

with the goal of getting small supports for the basis functions. A trian-

gulation in two space dimensions consisting of triangles has already been

de¬ned in de¬nition (2.25). The generalization in d space dimensions reads

as follows:

De¬nition 3.19 A triangulation Th of a set „¦ ‚ Rd consists of a ¬nite

number of subsets K of „¦ with the following properties:

Every K ∈ Th is closed.

(T1)

For every K ∈ Th its nonempty interior int (K) is a Lipschitz domain.

(T2)

„¦ = ∪K∈Th K.

(T3)

For di¬erent K1 and K2 of Th the intersection of int (K1 ) and int (K2 )

(T4)

is empty.

The sets K ∈ Th , which are called somewhat inaccurately elements in the

following, form a nonoverlapping decomposition of „¦. Here the formulation

is chosen in such a general way, since in Section 3.8 elements with curved

boundaries will also be considered. In De¬nition 3.19 some condition, which

corresponds to the property (3) of de¬nition (2.25), is still missing. In the

following this will be formulated speci¬cally for each element type. The

3.3. Element Types and A¬ne Equivalent Triangulations 115

parameter h is a measure for the size of all elements and mostly chosen as

h = max diam (K) K ∈ Th ;

that is, for instance, for triangles h is the length of the triangle™s largest

edge.

For a given vector space Vh let

PK := {v|K | v ∈ Vh } for K ∈ Th , (3.40)

that is,

Vh ‚ v : „¦ ’ R v|K ∈ PK for all K ∈ Th .

In the example of “linear triangles” in (2.27) we have PK = P1 , the poly-

nomials of ¬rst order. In the following de¬nitions the space PK will always

consist of polynomials or of smooth “polynomial-like” functions, such that

we can assume PK ‚ H 1 (K) © C(K). Here, H 1 (K) is an abbreviation for

H 1 (int (K)). The same holds for similar notation.

As the following theorem shows, elements v ∈ Vh of a conforming ansatz

space Vh ‚ V have therefore to be continuous :

Theorem 3.20 Suppose PK ‚ H 1 (K) © C(K) for all K ∈ Th . Then

Vh ‚ C(„¦) ⇐’ Vh ‚ H 1 („¦)

¯

and, respectively, for V0h := v ∈ Vh v = 0 on ‚„¦ ,

V0h ‚ C(„¦) ⇐’ V0h ‚ H0 („¦) .

¯ 1

Proof: See, for example, [9, Theorem 5.1 (p. 62)] or also Exercise 3.10. 2

If Vh ‚ C(„¦), then we also speak of C 0 -elements. Hence with this notion

¯

we do not mean only the K ∈ Th , but these provided with the local ansatz

space PK (and the degrees of freedom still to be introduced). For a bound-

ary value problem of fourth order, Vh ‚ H 2 („¦) and hence the requirement

Vh ‚ C 1 („¦) are necessary for a conforming ¬nite element ansatz. There-

¯

fore, this requires, analogously to Theorem 3.20, so-called C 1 -elements. By

degrees of freedom we denote a ¬nite number of values that are obtained

for some v ∈ PK from evaluating linear functionals on PK . The set of

these functionals is denoted by ΣK . In the following, these will basically

be the function values in ¬xed points of the element K, as in the example

of (2.27). We refer to these points as nodes. (Sometimes, this term is used

only for the vertices of the elements, which at least in our examples are

always nodes.) If the degrees of freedom are only function values, then we

speak of Lagrange elements and specify Σ by the corresponding nodes of

the element. Other possible degrees of freedom are values of derivatives in

¬xed nodes or also integrals. Values of derivatives are necessary if we want

to obtain C 1 -elements.

116 3. Finite Element Methods for Linear Elliptic Problems

As in the example of (2.27) (cf. Lemma 2.10), Vh is de¬ned by specifying

PK and the degrees of freedom on K for K ∈ Th . These have to be chosen

such that, on the one hand, they enforce the continuity of v ∈ Vh and,

on the other hand, the satisfaction of the homogeneous Dirichlet bound-

ary conditions at the nodes. For this purpose, compatibility between the

Dirichlet boundary condition and the triangulation is necessary, since it

will be required in (T6).

As can be seen from the proof of Lemma 2.10, it is essential

that the interpolation problem, locally de¬ned on K ∈

(F1) (3.41)

Th by the degrees of freedom, is uniquely solvable in PK ,

that this also holds on the (d ’ 1)-dimensional boundary

surfaces F of K ∈ Th for the degrees of freedom from F

and the functions v|F where v ∈ PK ; this then ensures

(F2) (3.42)

the continuity of v ∈ Vh , if PK and PK match in the

sense of PK |F = PK |F for K, K ∈ Th intersecting in F

(see Figure 3.2).

. .