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in¬‚ow boundary; that is, the conditions
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).

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