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h2
= ,
K
K∈Th i:
ai ∈K

since due to the regularity of (Th )h there is a uniform lower bound for
the angles of K ∈ Th (see (3.93)) and thus a uniform upper bound on the
number of K ∈ Th such that K © „¦i = ….
2
Corollary 6.17 Under the assumptions of Theorem 6.15 and for a regular
family of triangulations (Th )h there exists a constant ± > 0 independent of
h such that
ah (vh , vh ) ≥ ± vh for all vh ∈ Vh .
2
1


Proof: By Remark 6.16 and Theorem 6.15,
2
ah (vh , vh ) ≥ k |vh |1 + r0 C1 vh
2 2
,
0

i.e., we can take ± := min{k; r0 C1 } . 2
2


Theorem 6.15 (or Corollary 6.17) asserts the stability of the method. It
is the fundamental result for the proof of an error estimate.
Theorem 6.18 Let {Th }h∈(0,h] be a regular family of conforming triangu-
¯
lations, where in the case of the Voronoi diagram it is additionally required
that all triangles be nonobtuse. Furthermore, suppose in (6.5) that k > 0,
c1 , c2 , ∇ · c, r ∈ C(„¦), r + 1 ∇ · c ≥ r0 = const > 0 on „¦, f ∈ C 1 („¦), and
2
that the approximations γij , respectively ri , are chosen according to (6.7),
6.2. Finite Volume Method on Triangular Grids 277

respectively (6.8). Let the rij be de¬ned by (6.12) with R satisfying (P1)
and (P2). If the exact solution u of (6.5) belongs to H 2 („¦) and uh ∈ Vh
denotes the solution of (6.11), then

u ’ uh ¤ C h[ u + |f |1,∞ ] ,
2
1

where the constant C > 0 is independent of h.

Proof: The proof rests on a similar idea to those in the proof and the
application of Strang™s ¬rst lemma (Theorem 3.38) in Section 3.6.
Denoting by Ih : C „¦ ’ Vh the interpolation operator de¬ned in (3.71)
¯
and setting vh := uh ’ Ih (u), we have

ah (vh , vh ) = ah (uh , vh ) ’ ah (Ih (u), vh )
= f, vh 0,h ’ ah (Ih (u), vh )

’ f dx ’ ah (Ih (u), vh ).
= f, vh vi f dx + vi
0,h
„¦i „¦i
i∈Λ i∈Λ

By the de¬nition of the discrete form f, vh 0,h and by the di¬erential
equation (6.5), considered as an equation in L2 („¦), we get

(fi ’ f ) dx + Lu dx ’ ah (Ih (u), vh ) ,
ah (vh , vh ) = vi vi
„¦i „¦i
i∈Λ i∈Λ

where Lu = ’∇ · (k ∇u ’ c u) + r u .
For f ∈ C 1 („¦) and the choice fi := f (ai ), it is easy to see that

|fi ’ f (x)| ¤ |f |1,∞ max hK ¤ C h|f |1,∞ for all x ∈ „¦i .
K:ai ∈K

So it follows that

(fi ’ f ) dx ¤ C h|f |1,∞ |vi |mi
vi
„¦i
i∈Λ i∈Λ
1/2 1/2

¤ C h|f |1,∞ 2
vi mi mi
i∈Λ i∈Λ

¤ |„¦|

¤ C h|f |1,∞ vh 0,h .

For the other choice of fi (see (6.8)), the same estimate is trivially satis¬ed.
The di¬cult part of the proof is to get an estimate of the consistency error

Lu dx ’ ah (Ih (u), vh ) .
vi
„¦i
i∈Λ
278 6. Finite Volume Method

This is very extensive, and so we will omit the details. A complete proof of
the following result is given in the paper [40]:

1/2
Lu dx ’ ah (Ih (u), vh ) ¤ C h u |vh |2 + vh 2
vi . (6.22)
2 1 0,h
„¦i
i∈Λ

Putting both estimates together and taking into consideration Remark 6.16,
we arrive at
1/2
ah (vh , vh ) ¤ C h [ u + |f |1,∞ ] |vh |2 + vh 2
2 1 0,h
¤ C h[ u + |f |1,∞ ] vh .
2 1

By Corollary 6.17, we conclude from this that
¤ C h[ u + |f |1,∞ ] .
vh 1 2

It remains to apply the triangle inequality and the standard interpolation
error estimate (cf. Theorem 3.29 with k = 1 or Theorem 3.35)
u ’ uh ¤ u ’ Ih (u) ¤ C h[ u + |f |1,∞ ] .
+ vh
1 1 1 2

2

We point out that the error measured in the H 1 -seminorm is of the same
order as for the ¬nite element method with linear ¬nite elements.
Now we will turn to the investigation of some interesting properties of
the method.

Global Conservativity
Here we consider the boundary value problem
’∇ · (k ∇u ’ c u) = f in „¦ ,
ν · (k ∇u ’ c u) = g on ‚„¦ .
Integrating the di¬erential equation over „¦, we conclude from Gauss™s
divergence theorem that

’ ∇ · (k ∇u ’ c u) dx = ’ ν · (k ∇u ’ c u) dσ = ’ g dσ ,
„¦ ‚„¦ ‚„¦

and hence

g dσ + f dx = 0 .
‚„¦ „¦

This is a necessary compatibility condition for the data describing the bal-
ance between the total ¬‚ow over the boundary and the distributed sources.
We will demonstrate that the discretization requires a discretized version of
this compatibility condition, which is called discrete global conservativity.
Therefore, we ¬rst have to de¬ne the discretization for the above type of
boundary conditions. Obviously, for inner control volumes „¦i (i ∈ Λ), there
6.2. Finite Volume Method on Triangular Grids 279

is no need for any modi¬cations. So we have to consider only the boundary
control volumes „¦i (i ∈ ‚Λ := Λ \ Λ).
In the case of the Voronoi diagram we have

’ ∇ · (k ∇u ’ c u) dx = ’ ν · (k ∇u ’ c u) dσ
„¦i ‚„¦i

’ ν · (k ∇u ’ c u) dσ ’ ν · (k ∇u ’ c u) dσ (6.23)
=
‚„¦i ©‚„¦
“ij
j∈Λi

’ ν · (k ∇u ’ c u) dσ ’
= g dσ .
‚„¦i ©‚„¦
“ij
j∈Λi

Since the line integrals over “ij can be approximated in the standard way,
we get the following equation:
ui ’ uj
+ γij [rij ui + (1 ’ rij ) uj ]
vi µij mij (6.24)
dij
j∈Λi
i∈Λ

’ vi g dσ = fi vi mi ,
‚„¦i ©‚„¦
i∈Λ i∈Λ

where the ansatz and test space Vh consists of all continuous functions over
¯
„¦ that are piecewise linear with respect to the underlying triangulation
(that is, in the boundary nodes no function values are prescribed).
It is again assumed that the rij are de¬ned by (6.12) with a function R
satisfying (P1) and γji = ’γij .
Obviously, the particular function ih :≡ 1 belongs to Vh . So we are
allowed to set vh = ih in the discretization. Then, repeating the above
symmetry argument (cf. the proof of Lemma 6.12), we get
mij mij
µij (ui ’ uj ) =’ µij (ui ’ uj ) ,
dij dij
i∈Λ j∈Λi i∈Λ j∈Λi

that is,
mij
µij (ui ’ uj ) = 0.
dij
i∈Λ j∈Λi

On the other hand, using the same argument, we have

[rij ui + (1 ’ rij ) uj ] γij mij
i∈Λ j∈Λi

[rji uj + (1 ’ rji ) ui ] γji mji
=
i∈Λ j∈Λi

=’ [(1 ’ rij ) uj + rij ui ] γij mij . (6.25)
i∈Λ j∈Λi
280 6. Finite Volume Method

Consequently, this term vanishes, too. Because of

vi g dσ = g dσ ,
‚„¦i ©‚„¦ ‚„¦
i∈Λ

it follows that

’ ≈
g dσ = fi vi mi = f i mi f dx . (6.26)
‚„¦ „¦
i∈Λ i∈Λ

This is the mentioned compatibility condition. It ensures the solvability of
the discrete system (6.24).
In the case of the Donald diagram, we obviously have
k∇uh , ∇vh = 0.
0

Since the proof of (6.25) does not depend on the particular type of the
control volumes, the property of discrete global conservativity in the sense
of (6.26) is satis¬ed for the Donald diagram, too.

Inverse Monotonicity
The so-called inverse monotonicity is a further important property of the
boundary value problem (6.5) that is inherited by the ¬nite volume dis-
cretization without any additional restrictive assumptions. Namely, it is
well known that under appropriate assumptions on the coe¬cients, the
solution u is nonnegative if the (continuous) right-hand side f in (6.5) is
nonnegative in „¦.
We will demonstrate that this remains true for the approximative solu-
tion uh . Only at this place is the property (P3) of the weighting function
R used; the preceding results are also valid for the simple case R(z) ≡ 1 .
2
There is a close relation to the maximum principles investigated in Sec-
tions 1.4 and 3.9. However, the result given here is weaker, and the proof
is based on a di¬erent technique.
Theorem 6.19 Let the assumptions of Theorem 6.15 be satis¬ed, but R
in (6.12) has to satisfy (P1)“(P3). Further, suppose that f ∈ C(„¦) and
f (x) ≥ 0 for all x ∈ „¦. Moreover, in the case of the Donald diagram, only
the weighting function R(z) = 1 [sign (z) + 1] is permitted.
2
Then
uh (x) ≥ 0 for all x ∈ „¦ .

Proof: We start with the case of the Voronoi diagram. Let uh be the
solution of (6.11) with f (x) ≥ 0 for all x ∈ „¦. Then we have the following
additive decomposition of uh :
uh = u+ ’ u’ , where u+ := max {0, uh } .
h h h

In general, u+ , u’ do not belong to Vh . So we interpolate them in Vh and
h h
set in (6.11) vh := Ih (u’ ), where Ih : C „¦ ’ Vh is the interpolation
¯
h
6.2. Finite Volume Method on Triangular Grids 281

operator (3.71). It follows that
= ah (uh , vh ) = ah Ih (u+ ), Ih (u’ ) ’ ah Ih (u’ ), Ih (u’ ) .
0 ¤ f, vh 0,h h h h h

By Theorem 6.15, we have
2
k Ih (u’ ) ¤ ah Ih (u’ ), Ih (u’ ) ¤ ah Ih (u+ ), Ih (u’ ) .
h h h h h
1

If we were able to show that ah Ih (u+ ), Ih (u’ ) ¤ 0, then the theorem
h h
would be proven, because this relation implies Ih (u’ ) 1 = 0, and from this
h

we immediately get uh = 0, and so uh = uh ≥ 0.
+

Since u+ u’ = 0 for all i ∈ Λ, it follows from (6.19) in the proof of
i i
Lemma 6.12 that
bh Ih (u+ ), Ih (u’ ) = (1 ’ rij ) u+ u’ γij mij . (6.27)
j i
h h
i∈Λ j∈Λi

Furthermore, obviously dh Ih (u+ ), Ih (u’ ) = 0 holds. Thus
h h

µij +
ah Ih (u+ ), Ih (u’ ) uj + γij (1 ’ rij ) u+ u’ mij

= j i
h h
dij
i∈Λ j∈Λi

µij γij dij
(1 ’ rij ) u+ u’ mij .
’ 1’
= j i
dij µij
i∈Λ j∈Λi

Due to 1 ’ [1 ’ R(z)] z ≥ 0 for all z ∈ R (cf. property (P3) in (6.13)) and
u+ u’ ≥ 0, it follows that
j i

ah Ih (u+ ), Ih (u’ ) ¤ 0 .
h h

So it remains to investigate the case of the Donald diagram. The function
R(z) = 1 [sign (z) + 1] has the property
2
1
[1 ’ R(z)] z = [1 ’ sign (z)] z ¤ 0 for all z ∈ R ,
2
that is (cf. (6.27)),
bh Ih (u+ ), Ih (u’ ) ¤ 0 .
h h

Taking u+ u’ = 0 into consideration, we get
i i

ah Ih (u+ ), Ih (u’ ) k ∇Ih (u+ ), ∇Ih (u’ )
¤
h h h h 0

u+ u’ ∇•j , ∇•i
=k .
j i 0
i∈Λ j∈Λi

Now Lemma 3.47 implies that
k
ah Ih (u+ ), Ih (u’ ) ¤ ’ u+ u’ cot ±K + cot ±K ,
ij ij
j i
h h
2
i∈Λ j∈Λi

where K and K are a pair of triangles sharing a common edge with vertices
ai , aj .
282 6. Finite Volume Method

Since all triangles are nonobtuse, we have cot ±K ≥ 0, cot ±K ≥ 0, and
ij ij
hence
ah Ih (u+ ), Ih (u’ ) ¤ 0 .
h h

2


Exercises
6.2 Suppose that the domain „¦ ‚ R2 can be triangulated by means of
equilateral triangles with edge length h > 0 in an admissible way.
(a) Give the shape of the control domains in the case of the Voronoi and
the Donald diagrams.
(b) Using the control domains from subproblem (a), discretize the Poisson
equation with homogeneous Dirichlet boundary conditions by means
of the ¬nite volume method.
1
6.3 Formulate an existence result for the weak solution in H0 („¦) of the
boundary value problem (6.5) similar to Theorem 3.12. In particular, what
form will condition (3.17) take?

6.4 Verify Remark 6.7; i.e., show that ·, · possesses the properties of
0,h
a scalar product on Vh .

6.5 Prove Remark 6.16 in detail.

6.6 Verify or disprove the properties (P1)“(P3) for the three weighting
functions given before (6.13) and for R ≡ 1 .
2

6.7 Let K be a nonobtuse triangle with the vertices a1 , a2 , a3 . The length
of the segments “K := “ij © K is denoted by mK , and dij is the length
ij ij
K
of the edge connecting ai with aj . Finally, ±ij is the interior angle of K
opposite that edge.
Demonstrate the following relation: 2mK = dij cot ±K .

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