The LBB condition for the Taylor-Hood P2 − P1 and Scott-Vogelius
P2 − discP1 element pairs in 2-D
V.J. Ervin ∗
E.W. Jenkins ∗
Department of Mathematical Sciences
Clemson University
Clemson, SC, 29634-0975
USA.
Abstract
In this article we apply the Stenberg criteria to show that the Taylor-Hood P2 − P1 and the
Scott-Vogelius P2 − discP1 element pairs satisfy the LBB condition in IR2 . The Taylor-Hood
P2 − P1 pair is shown to be stable on a regular triangulation of the domain. For the ScottVogelius P2 − discP1 element pair the mesh is assumed to be a barycenter refinement of a regular
triangulation.
Key words. LBB condition; Stenberg criteria
AMS Mathematics subject classifications. 65N30
1
Introduction
In the approximation of fluid flow problem based on a weak formulation of the modeling equations,
specifically those modeling Navier-Stokes, Stokes or Darcy’s equations, an important component
in the approximation algorithm is ensuring that the velocity and pressure approximation spaces,
Xh ⊂ X and Qh ⊂ Q, respectively, satisfy an LBB condition, i.e.
inf sup
q∈Qh v∈Xh
b(q , v)
≥ β,
kqkQ kvkX
(1.1)
for some β ∈ IR+ , where
Z
b(q , v) =
q div(v) dx .
(1.2)
Ω
“Compatible pair” of velocity and pressure approximation spaces for fluid flow problems in Cartesian
coordinates are well documented in the literature, see for example [4, 2]. For the Stokes and NavierStokes equations Taylor-Hood Pk − Pk−1 approximation elements are a common choice. More
∗
email: [email protected] , [email protected].
1
recently there has been increasing interest in the use of Scott-Vogelius Pk − discPk−1 approximation
elements. In 2-D Scott-Vogelius approximation elements are known to satisfy (1.1) for triangulations
constructed as a barycentered refinement of a regular triangulation of the domain [1, 5].
In [6] Stenberg developed a sufficient condition based on checking a local criteria to determine if
approximation spaces Xh , Qh satisfied the LBB condition. In this paper we review the Stenberg
sufficient condition and show that both Taylor-Hood P2 − P1 and the Scott-Vogelius P2 − discP1
element pairs satisfy the local criteria.
The paper is organized as follows. In the following section we introduce the mathematical notation
used throughout and present the Stenberg sufficiency condition. Section 3 confirms that TaylorHood P2 −P1 element pairs are LBB stable on a regular triangulation of the domain. (In [6] Stenberg
showed that P2 − P1 element pairs were stable on a partition of the domain into quadrilaterals.)
In Section 4 we show that the Scott-Vogelius P2 − discP1 element are LBB stable on a barycenter
refinement of a regular triangulation.
2
Mathematical Preliminaries
In [6] Stenberg established a sufficiency condition on the partition, Th , and the approximation spaces,
Xh and Qh , for the LBB condition (1.1)(1.2) to be satisfied.
Next we summarize the Stenberg sufficiency condition. We discuss the case for a triangulation Th
of the domain Ω.
2.1
Stenberg sufficiency condition
We assume that Ω is a domain in IR2 with boundary Γ, and (Th )h denotes a family of regular
triangulations of Ω [3].
Let P2 (T ) denote the set of restriction to T of polynomials of degree less than or equal to 2. For
the velocity approximation space we consider
Xh = {w ∈ (C 0 (Ω))2 : w|Γ = 0 , w|T ∈ (P2 (T ))2 , ∀T ∈ Th } .
(2.1)
For the pressure space for Taylor-Hood P2 − P1 ,
Z
Qh = {q ∈ C 0 (Ω) :
q dx = 0 , q|T ∈ P1 (T ), ∀T ∈ Th } ⊂ Q .
(2.2)
Ω
For a regular triangulation Th the associated barycentered refined mesh, denoted ThB , is formed by
dividing each triangle T in Th into three triangles by connecting each vertex of T to its center. On
the family of triangulations (ThB )h the Scott-Vogelius P2 − discP1 approximation pair is defined by
Xh = {w ∈ (C 0 (Ω))2 : w|Γ = 0 , w|T ∈ (P2 (T ))2 , ∀T ∈ ThB } .
Z
Qh = {q :
q dx = 0 , q|T ∈ P1 (T ), ∀T ∈ Th } ⊂ Q .
Ω
2
(2.3)
(2.4)
Firstly we introduce the concept of a macroelement. A macroelement M is said to be equivalent to
c if there is a mapping FM : M
c → M satisfying the conditions:
a reference macroelement M
(i) FM is continuous and one-to-one.
c) = M .
(ii) FM (M
c = ∪m Tbj , where Tbj , j = 1, 2, . . . , m, are the triangles in M
c, then Tj = FM (Tbj ),
(iii) If M
j=1
j = 1, 2, . . . , m, are the triangles in M .
(iv) FM |Tb = FTj ◦ F b−1 , j = 1, 2, . . . , m, where FTbj and FTj are the affine mappings from the
Tj
j
reference triangle with vertices (0, 0), (1, 0) and (0, 1) onto Tbj and Tj , respectively.
c is denoted E c.
The family of macroelements equivalent with M
M
For a macroelement M define the spaces Xh,M , Qh,M , Nh,M and divXh,M as
Xh,M
Qh,M
Nh,M
= {w ∈ (C 0 (Ω))2 : w|Γ = 0 , w|Ω\M = 0, w|T ∈ (Pk (T ))2 , ∀T ∈ M } ,
Z
= {q :
q dx = 0 , q|T ∈ Pl (T ), ∀T ∈ M } ,
Ω
Z
=
q ∈ Qh,M :
q div(w) dx = 0 , ∀w ∈ Xh,M ,
(2.5)
(2.6)
(2.7)
Ω
divXh,M
= {q : q = div w , w ∈ Xh,M } .
(2.8)
Theorem 1 [6] [Stenberg Sufficiency Condition] If
(i) there exists a finite set of classes EM
ci , i = 1, . . . , n, n ≥ 1, such that for each M ∈ EM
ci , i =
1, . . . , n, the space Nh,M is one dimensional consisting of functions which are constant on M ,
(ii) for each Th ∈ (Th )h , the triangles can be grouped together to form macroelements Mj , j =
1, . . . , m, so that the so obtained macroelement partitioning of Ω, Mh satisfies that Mj belongs
to some EM
ci , for all Mj ∈ Mh ,
then (1.1)(1.2) is satisfied.
In the case linear elements are used for the velocity approximation there is one additional constraint
on Th .
(iii) If γ is the common part of two macroelements in (ii) then γ is connected and contains at least
two edges of triangles in Th .
The following two lemmas are used by Stenberg in [6] to establish Theorem 1.
R
Let Πh denote the projection, with respect to the innerproduct hq, pi := Ω q p dx, from Qh onto
the space
QC
(2.9)
h := {q ∈ Q : q|M is constant ∀M ∈ Mh } .
For Λ ⊂ Ω, denote by | · |s,Λ and k · ks,Λ the seminorm and norm of the Sobolev space (H s )α , where
s and α are integers. We omit the subscript T if T = Ω.
3
Lemma 1 [[6], Lemma 3.2] Under the conditions of Theorem 1, there is a constant C > 0 such
that for all qh ∈ Qh there is a vh ∈ Xh satisfying
Z
Z
(I − Πh )qh div(vh ) dx ≥ Ck(I − Πh )qh k20 , (2.10)
qh div(vh ) dx =
b(qh , vh ) =
Ω
Ω
and
|vh |1 ≤ k(I − Πh )qh k0 .
(2.11)
Lemma 2 [[6], Lemma 3.3] Under the conditions of Theorem 1, there is a constant C > 0 such
that for all qh ∈ Qh there is a vh ∈ Xh satisfying
Z
qh div(vh ) dx = kΠh qh k20 ,
(2.12)
b(qh , vh ) =
Ω
and
|vh |1 ≤ C kΠh qh k0 .
(2.13)
In order to establish Rthat Nh,M has dimension one, consisting of functions which are constant on M ,
we form the matrix Ω q div(w) dx = 0 , ∀w ∈ Xh,M . This matrix is constructed by performing
the computations on the reference triangle Tb. Next we summarize the affine transformation from
T ⊂ Ω → Tb, and give the quadratic and linear basis elements on Tb.
y
η
(0 , 1)
F
S3
S3
(x2 , y2)
T
S4
S5
T
S2
(0 , 0)
S1
(x3 , y3)
(1 , 0)
S6
S2
S1
ξ
(x1 , y1)
x
Figure 2.1: Mapping of the reference triangle Tb to triangle T .
We have
F :
Let
x = x1 + (x2 − x1 )ξ + (x3 − x1 )η
.
y = y1 + (y2 − y1 )ξ + (y3 − y1 )η
"
Jmat =
∂x
∂ξ
∂y
∂ξ
∂x
∂η
∂y
∂η
#
+
+
∂g
∂y
∂g
∂y
=
(x2 − x1 ) (x3 − x1 )
(y2 − y1 ) (y3 − y1 )
(2.14)
.
For ĝ(ξ, η) = g(x, y),
∂ĝ
∂ξ
∂ĝ
∂η
=
=
∂g
∂x
∂g
∂x
∂x
∂ξ
∂x
∂η
"
∂y
∂ξ
∂y
∂η
=
∂x
∂ξ
∂x
∂η
∂y
∂ξ
∂y
∂η
#"
∂g
∂x
∂g
∂y
#
T
⇐⇒ ∇ξ,η ĝ = Jmat
∇x,y g
−T
=⇒ ∇x,y g = Jmat
∇ξ,η ĝ .
4
(2.15)
Note that the quadrature rule
Z
1
1
1
fˆ(ξ, η) dξ dη ∼ fˆ(1/2, 0) + fˆ(1/2, 1/2) + fˆ(0, 1/2) ,
6
6
6
Tb
(2.16)
is exact for polynomials of degree ≤ 2.
A Lagrangian basis for P2 (Tb), with basis functions q̂i satisfying q̂i (Sbi ) = 1, q̂i (Sbj ) = 0, j 6= i,
i, j, = 1, 2, . . . , 6, is
q̂1 (ξ, η) = (1 − ξ − η)(1 − 2ξ − 2η) ,
q̂2 (ξ, η) = ξ (2ξ − 1) ,
q̂3 (ξ, η) = η (2η − 1) ,
q̂4 (ξ, η) = 4 ξ η ,
q̂5 (ξ, η) = 4 η (1 − ξ − η) ,
q̂6 (ξ, η) = 4 ξ (1 − ξ − η) .
A Lagrangian basis for P1 (Tb), with basis functions ˆli satisfying ˆli (Sbi ) = 1, ˆli (Sbj ) = 0, j 6= i,
i, j, = 1, 2, 3, is
ˆl1 (ξ, η) = (1 − ξ − η) ,
3
ˆl2 (ξ, η) = ξ ,
ˆl3 (ξ, η) = η .
The LBB condition for Taylor-Hood P2 −P1 elements on triangles
In this section we show that the Stenberg sufficiency criteria for satisfying the LBB condition is
satisfied for Taylor-Hood P2 − P1 elements on triangles. We begin by identifying an appropriate
macroelement, M , and then show that the corresponding vector space Nh,M has dimension one. For
notational convenience we suppress the h subscript, i.e. NM ≡ Nh,M .
Let M be given by the collection of three triangles in Figure 3.1.
y
R5
T2
R1
R3
R6
z
(x4 , y4)
R7 T3
(x5 , y5)
(x1 , y1)
R5
R4
(x3 , y3)
R1
R6
T2
T3
T1
T1
R2 (x2 , y2)
R3
R7
R8
R2 (R4)
x
r
Figure 3.1: Macroelement for Taylor-Hood
P2 − P1 .
Figure 3.2: Macroelement for Taylor-Hood
P2 − P1 .
We have that 1
0
1
0
XM = span v1 = q6 (x, y)
, v2 = q6 (x, y)
, v3 = q7 (x, y)
, v4 = q7 (x, y)
,
0
1
0
1
where q6 , q7 represent the (continuous) Lagrangian quadratic basis function which has value 1 at
node R6 , R7 , respectively, and vanish at all other nodes. QM = span{l1 (x, y), l2 (x, y), l3 (x, y), l4 (x, y),
5
l5 (x, y)}, where li , i = 1, . . . , 5, represents the (continuous) Lagrangian linear basis function which
has value 1 at node Ri and vanishes at nodes Rj , j = 1, 2, . . . , 5, j 6= i.
Note that the defining equation for NM , namely (div(v) , p) = 0, ∀v ∈ X0,M , generates four
equations for the five unknown constants p1 , p2 , p3 , p4 , p5 , where p(x, y) = p1 l1 (x, y) + p2 l2 (x, y) +
p3 l3 (x, y) + p4 l4 (x, y) + p5 l5 (x, y) .
Using Green’s theorem,
(div(v) , p)M = −(v , ∇p)M = −
3
X
(v , ∇p)Tj .
j=1
Also,
Z
Z
v · ∇p dA =
(v , ∇p)Tj =
T̂
Tj
−T
v̂ · Jmat
∇ξ,η p̂ |Jmat | dξ dη ,
(3.1)
where |Jmat | denotes the absolute value of the determinant of Jmat .
For v ∈ XM , p ∈ QM , v̂ is a (vector) quadratic function, ∇ξ,η p̂ is a constant vector, and Jmat is
a constant matrix. Hence the integrand in (3.1) is a polynomial of degree ≤ 2, and is therefore
evaluated exactly using (2.16).
3.1
Computation of (v , ∇p)T1
In terms of the mapping of T1 to the reference triangle, relative to (2.14), associate S1 ≡ R1 ,
S2 ≡ R2 , and S3 ≡ R3 .
We have that v3 |T1 and v4 |T1 are zero. Also,
1
−1
v1 (x, y)|T1 = q̂5 (F (x, y))
,
0
v2 (x, y)|T1 = q̂5 (F
−1
(x, y))
0
1
,
p(x, y) = p1 ˆl1 (F −1 (x, y)) + p2 ˆl2 (F −1 (x, y)) + p3 ˆl3 (F −1 (x, y)) ,
(x2 − x1 ) (x3 − x1 )
and Jmat =
.
(y2 − y1 ) (y3 − y1 )
Using (3.1) and (2.16) we have that
Z
1
v1 · ∇p dA = ((y2 − y3 )p1 + (y3 − y1 )p2 + (y1 − y2 )p3 ) ,
6
T1
and
3.2
(3.2)
Z
1
v2 · ∇p dA = − ((x2 − x3 )p1 + (x3 − x1 )p2 + (x1 − x2 )p3 ) .
6
T1
(3.3)
Computation of (v , ∇p)T2
In terms of the mapping of T2 to the reference triangle, relative to (2.14), associate S1 ≡ R1 ,
S2 ≡ R3 , and S3 ≡ R5 .
6
We have that
v1 (x, y)|T2 = q̂6 (F
−1
(x, y))
v3 (x, y)|T2 = q̂4 (F −1 (x, y))
1
0
1
0
v2 (x, y)|T2 = q̂6 (F
,
−1
(x, y))
v4 (x, y)|T2 = q̂4 (F −1 (x, y))
,
0
1
0
1
,
,
p(x, y) = p1 ˆl1 (F −1 (x, y)) + p3 ˆl3 (F −1 (x, y)) + p5 ˆl5 (F −1 (x, y)) ,
(x3 − x1 ) (x5 − x1 )
and Jmat =
.
(y3 − y1 ) (y5 − y1 )
Using (3.1) and (2.16) we have that
Z
1
v1 · ∇p dA = ((y3 − y5 )p1 + (y5 − y1 )p3 + (y1 − y3 )p5 ) ,
6
T2
Z
1
v2 · ∇p dA = − ((x3 − x5 )p1 + (x5 − x1 )p3 + (x1 − x3 )p5 ) ,
6
T2
Z
1
v3 · ∇p dA = ((y3 − y5 )p1 + (y5 − y1 )p3 + (y1 − y3 )p5 ) ,
6
T2
and
3.3
(3.4)
(3.5)
(3.6)
Z
1
v4 · ∇p dA = − ((x3 − x5 )p1 + (x5 − x1 )p3 + (x1 − x3 )p5 ) .
6
T2
(3.7)
Computation of (v , ∇p)T3
In terms of the mapping of T3 to the reference triangle, relative to (2.14), associate S1 ≡ R5 ,
S2 ≡ R3 , and S3 ≡ R4 .
We have that v1 |T3 and v2 |T3 are zero. Also,
1
−1
v3 (x, y)|T3 = q̂6 (F (x, y))
,
0
v4 (x, y)|T3 = q̂6 (F
−1
(x, y))
0
1
,
p(x, y) = p5 ˆl5 (F −1 (x, y)) + p3 ˆl3 (F −1 (x, y)) + p4 ˆl4 (F −1 (x, y)) ,
(x3 − x5 ) (x4 − x5 )
and Jmat =
.
(y3 − y5 ) (y4 − y5 )
Using (3.1) and (2.16) we have that
Z
1
v3 · ∇p dA = ((y4 − y5 )p3 + (y5 − y3 )p4 + (y3 − y4 )p5 ) ,
6
T3
and
(3.8)
Z
1
v4 · ∇p dA = − ((x4 − x5 )p3 + (x5 − x3 )p4 + (x3 − x4 )p5 ) .
6
T3
7
(3.9)
3.4
Dimension of NM
R
Using equations (3.2)-(3.9), corresponding to Ω q div(w) dx = 0 , ∀w
minor simplification) the following linear system of equations, Ap = 0, :


 p1
(y2 − y5 ) (−y1 + y3 ) (−y2 + y5 )
(y1 − y3 ) 
 (x2 − x5 ) (−x1 + x3 ) (−x2 + x5 )
  p2
(x
1 − x3 )  

p3
 (y3 − y5 )
(−y1 + y4 ) (−y3 + y5 ) (y1 − y4 )  
 p4
(x3 − x5 )
(−x1 + x4 ) (−x3 + x5 ) (x1 − x4 )
p5
∈ XM , we obtain (after


0


 0
 = 

 0

0


.

(3.10)
Note that p1 = p2 = p3 = p4 = p5 is a solution to (3.10), i.e. NM contains the constant functions.
To show that dim(NM ) = 1 it suffices to show that the matrix A has full rank, i.e. the rows of A
are linearly independent.
Lemma 3 The rows of the matrix A given in (3.10) are linearly independent.
Proof : Rows 1 and 2 are linearly dependent if and only if
y3 − y1
x3 − x1
=
y2 − y5
x2 − x5
That is, if and only if
and
y5 − y2
x5 − x2
=
y2 − y5
x2 − x5
and
y1 − y3
x1 − x3
=
.
y2 − y5
x2 − x5
y3 − y1
x3 − x1
y3 − y1
y2 − y5
=
⇔
=
.
y2 − y5
x2 − x5
x3 − x1
x2 − x5
(3.11)
Geometrically (3.11) would implies that the diagonals of the quadrilateral formed by R1 , R2 , R3 and
R5 are parallel! Hence we can conclude that rows 1 and 2 are linearly independent. (The degenerate
quadrilateral case, when R3 lies on the line segment between R2 and R5 also leads to the conclusion
that rows 1 and 2 are linearly independent.)
As R3 6= R5 , then either the (3, 4) entry or the (4, 4) entry of A is nonzero. Assume, without loss of
generality, that the (3, 4) entry is nonzero.
The fact that the (1 , 4) and (2 , 4) entries in A are zero, and the (3 , 4) entry is nonzero now
implies that rows 1, 2, and 3 are linearly independent. Thus, if the rows of A are linearly dependent
then row 4 must be expressible as a linear combination of rows 1, 2, and 3. That is, there must
exists constants C1 , C2 , C3 , not all zero, such that
(x3 − x5 ) = C1 (y2 − y5 ) + C2 (x2 − x5 ) + C3 (y3 − y5 )
(3.12)
0 = C1 (y3 − y1 ) + C2 (x3 − x1 ) + C3 (y3 − y5 )
(3.13)
(x4 − x1 ) = C1 (y5 − y2 ) + C2 (x5 − x2 ) + C3 (y4 − y1 )
(3.14)
(x5 − x3 ) = C3 (y5 − y3 )
(3.15)
(x1 − x4 ) = C1 (y1 − y3 ) + C2 (x1 − x3 ) + C3 (y1 − y4 )
(3.16)
From (3.15), if x3 = x5 then C3 = 0. The independence of rows 1 and 2 of A, (i.e. (3.12), (3.13))
then implies that C1 = C2 = 0. If x3 6= x5 , equations (3.15),(3.13),(3.16), imply that C3 6= 0 and
1
y5 − y3
y1 − y4
=
=
.
C3
x5 − x3
x1 − x4
8
(3.17)
Geometrically (3.17) implies that the diagonals of the quadrilateral formed by R1 , R3 , R4 and R5
are parallel! Hence we conclude that the rows of the matrix A are linearly independent.
By modifying the matrix in (3.10), it is straight forward to show that the three triangles depicted in
Figure 3.2 also form a macroelement for Taylor-Hood P2 − P1 approximation pair. Thus, we could
conclude that for any triangulation of the domain of Ω, which can be partitioned into groups of
three adjacent triangles, the Taylor-Hood P2 − P1 approximation pair is LBB stable. However often
the number of triangles in a triangulation is not exactly divisible by three. Next we demonstrate
that there are many choices of macroelements for the Taylor-Hood P2 − P1 approximation pair.
Lemma 4 Suppose M is a macroelement with NM = 1, consisting of functions which are constant
on M . Let M̃ be formed from M by adding an adjacent triangle (i.e. sharing an edge with M ).
Then M̃ is also a macroelement with the desired property that NM̃ = 1, consisting of functions
which are constant on M̃ .
Proof : We consider separately the two cases corresponding to M̃ being formed by adding a triangle
to M that: (i) shares two edges with M , and (ii) shares one edge with M .
Case (i): The added triangle shares two edges with M . (For example, see Figures 3.1, 3.3.)
y
R4
R5
R7 T3
T2
R1
R6
R4
R5
R9
R3
T1
y
R7 T3
T4
T2
R8
R1
R2
x
R6
R10
R3
T1 R8
T4
R2
x
Figure 3.3: New macroelement for TaylorHood P2 − P1 .
Figure 3.4: New macroelement for TaylorHood P2 − P1 .
Let A and à be the matrices associated with NM and NM̃ , respectively. For nM,Q the dimension
of QM , we have that rank(A) = nM,Q − 1, and that A is a m × nM,Q matrix with m = 2 ×
(number of interior edges in M ) ≥ nM,Q −1. Ã is therefore a m̃×nM,Q matrix with m̃ ≥ nM,Q +1,
as
R Ã must have at least one more interior edge that
R M . Note that every row in à comes from
M̃ vi · ∇p dA = 0, for vi ∈ XM̃ . As, ∀vi ∈ XM̃ , M̃ vi · ∇p dA = 0 is satisfied for p a constant
function, then p1 = p2 = . . . = pnM,Q satisfies Ãp = 0, and rank(Ã) ≤ nM,Q − 1. Since the
rank(A) = nM,Q − 1, this implies thatA hasnM,Q − 1 linearly independent rows. The fact that for
A
v ∈ XM , v|M̃ \M = 0; we have à =  · · · , which implies that à has at least nM,Q − 1 linearly
B
independent rows. Hence rank(Ã) = nM,Q − 1 and the dimension of NM̃ = 1.
9
Case (ii): The added triangle shares one edges with M . Let R2 and R3 denote the endpoints of the
shared triangle edge. (For example, see Figures 3.1, 3.4.)
In this case, along with two new triangle edges, an additional triangle vertex is added to M in
forming M̃ . Therefore, the dimension of QM̃ = dim(QM ) + 1, Rwith the increase in dimension
corresponding to the new added vertex. Again, as ∀vi ∈ XM̃ , M̃ vi · ∇p dA = 0 is satisfied
for p a constant function, then p1 = p2 = . . . = pnM,Q +1 satisfies Ãp = 0, which implies that


A 0
rank(Ã) ≤ nM,Q . Also, as for v ∈ XM , v|M̃ \M = 0; we have à =  · · · · , where the number
B b
of rows in the matrix B is two, corresponds to the velocity basis functions ṽ1 , ṽ2 , associated with
the shared triangle edge, added to QM to form QM̃ . As the added triangle lies in the support of ṽ1 ,
and ṽ2 , then from (3.1) (and corresponding minor simplifications) b = [−y2 + y3 − x2 + x3 ]T . As
R2 6= R3 , the number of independent rows in à must be greater than the number of independent
rows in A = nM,Q − 1. Hence rank(Ã) = nM,Q and the dimension of NM̃ = 1.
Corollary 1 The Taylor-Hood P2 − P1 approximation pair is LBB stable on a regular triangulation
of Ω.
4
The LBB condition for Scott-Vogelius P2 − discP1 elements on
triangles
In this section we show that the Stenberg sufficiency criteria for satisfying the LBB condition is
satisfied for Scott-Vogelius P2 −discP1 elements on a barycenter refinement of a regular triangulation.
We begin by identifying the macroelement, M , and then showing that the corresponding vector space
NM has dimension one. Again, we suppress the h subscript.
Let M be given by the three triangles in Figure 4.1, formed by a barycenter refinement of a triangle.
We have that
XM
=
1
0
0
1
1
0
0
1
span v1 = q5 (x, y)
, v2 = q5 (x, y)
, v3 = q6 (x, y)
, v4 = q6 (x, y)
,
1
0
1
0
v5 = q7 (x, y)
, v6 = q7 (x, y)
, v7 = q4 (x, y)
, v8 = q4 (x, y)
, where q4 , q5 , q6 ,
0
1
0
1
q7 represent the (continuous) Lagrangian quadratic basis function which has value 1 at node R4 ,
R5 , R6 , R7 , respectively, and vanish at all other nodes.
On T1 , let l1 (x, y), l2 (x, y), l3 (x, y) denote the Lagrangian linear basis functions associated with
nodes R1 , R2 , and R4 , respectively. For (x, y) ∈ T2 , T3 , define li (x, y) = 0, i = 1, 2, 3. On T2 ,
let l4 (x, y), l5 (x, y), l6 (x, y) denote the Lagrangian linear basis functions associated with nodes
R2 , R3 , and R4 , respectively. For (x, y) ∈ T1 , T3 , define li (x, y) = 0, i = 4, 5, 6. On T3 , let
l7 (x, y), l8 (x, y), l9 (x, y) denote the Lagrangian linear basis functions associated with nodes R3 , R1 ,
and R4 , respectively. For (x, y) ∈ T1 , T2 , define li (x, y) = 0, i = 7, 8, 9. (See Figure 4.1.) Then,
QM = span {l1 (x, y), l2 (x, y), . . . , l9 (x, y)}.
10
(x3 , y3) R3
p p
y
7
p
R1
8
R5
p
1
(x1 , y1)
R4
5
R7
T3
p
9
p
p
T1
T2
6
3
(x4 , y4)
R6
p
2
p
4
R2
(x2 , y2)
x
Figure 4.1: Macroelement for Scott-Vogelius P2 − discP1 .
Note that the defining equation for NM , namely (div(v) , p) = 0, ∀v ∈ X0,M , generates eight
equations for the nine unknown constants p1 , p2 , . . . , p9 , where p(x, y) = p1 l1 (x, y) + p2 l2 (x, y) +
. . . , + p9 l9 (x, y) .
Using Green’s theorem, and the fact that p is discontinuous across the edges of the triangles in M ,
!
Z
Z
3
3 Z
X
X
v · n p ds −
v · ∇p dA ,
(4.1)
div(v) p dA =
(div(v) , p)M =
j=1
Tj
j=1
Tj
∂Tj
where n denotes the outward pointing normal on ∂Tj , the boundary of Tj .
Consider now the evaluation of
Z
v · n p ds =
∂Tj
3 Z
X
k=1
v · n p ds ,
ek
where e1 , e2 , e3 , denote the edges of the triangle Tj .
Let R0 = (a0 , b0 ) and R1 = (a1 , b1 ) denote the beginning and end points of the edge ek , orientated in
a counter-clockwise direction with respect to Tj . Then the unit tangent t, and outer normal vector
n, on ek are given by
R1 − R0
1
1
a1 − a0
b1 − b0
t =
=
, and n =
,
|R1 − R0 |
|R1 − R0 | b1 − b0
|R1 − R0 | −(a1 − a0 )
where |R1 − R0 | denotes the length of ek .
As v is a quadratic function along ek , p a linear function along ek , then the line integral along ek is
evaluated exactly using Simpson’s quadrature rule:
Z
1
b1 − b0
b1 − b0
b1 − b0
v·n p ds =
v|R0 ·
p|R0 + 4 v|RM ·
p|RM + v|R1 ·
p|R1 ,
−(a1 − a0 )
−(a1 − a0 )
−(a1 − a0 )
6
ek
(4.2)
11
where RM = 1/2(R0 + R1 ).
Remark: Of interest is investigating the null space of (div(v) , p)M = 0. To be consistent
with the Taylor-Hood analysis in Section 3, we form the associated linear system corresponding
to −(div(v) , p)M = 0, i.e. from (4.1),
!
Z
Z
3
X
v · ∇p dA −
v · n p ds = 0 .
(4.3)
j=1
4.1
Computation of
R
T1
Tj
v · ∇p dA −
∂Tj
R
∂T1
v · n p ds
In terms of the mapping of T1 to the reference triangle, relative to (2.14), associate S1 ≡ R1 ,
S2 ≡ R2 , and S3 ≡ R4 .
We have that v5 |T1 and v6 |T1 are zero. Also,
1
−1
v1 (x, y)|T1 = q̂5 (F (x, y))
,
0
1
v3 (x, y)|T1 = q̂4 (F −1 (x, y))
,
0
1
−1
,
v7 (x, y)|T1 = q̂3 (F (x, y))
0
v2 (x, y)|T1 = q̂5 (F
−1
(x, y))
v4 (x, y)|T1 = q̂4 (F −1 (x, y))
v8 (x, y)|T1 = q̂3 (F
−1
(x, y))
0
1
0
1
0
1
,
,
,
p(x, y) = p1 ˆl1 (F −1 (x, y)) + p2 ˆl2 (F −1 (x, y)) + p3 ˆl3 (F −1 (x, y)) ,
(x2 − x1 ) (x4 − x1 )
.
and Jmat =
(y2 − y1 ) (y4 − y1 )
Using (3.1),(2.16), and (4.2) we have that
Z
Z
1
v1 · ∇p dA −
v1 · n p ds = ((y2 − y4 )p1 + (y4 − y1 )p2 +
6
T1
∂T1
4
1
1
p1 + p2
− (y1 − y4 )
6
2
2
1
= ((−2y1 + y2 + y4 )p1 + (−y1 + y4 )p2 +
6
Similarly,
Z
Z
v2 ·∇p dA −
(y1 − y2 )p3 )
(−y1 − y2 + 2y4 )p3 )(4.4)
.
1
v2 ·n p ds = − ((−2x1 + x2 + x4 )p1 + (−x1 + x4 )p2 + (−x1 − x2 + 2x4 )p3 )
6
T1
∂T1
(4.5)
Z
Z
1
v3 · ∇p dA −
v3 · n p ds = ((y2 − y4 )p1 + (−y1 + 2y2 − y4 )p2 + (y1 + y2 − 2y4 )p3 ) ,
6
T1
∂T1
(4.6)
Z
Z
1
v4 · ∇p dA −
v4 · n p ds = − ((x2 − x4 )p1 + (−x1 + 2x2 − x4 )p2 + (x1 + x2 − 2x4 )p3 ) .
6
T1
∂T1
(4.7)
12
For the velocity basis elements associated with the barycenter node R4 :
Z
Z
1
v7 · n p ds = ((−y1 + y2 )p1 + (−y1 + y2 )p2 + (−y1 + y2 )p3 ) ,
v7 · ∇p dA −
(4.8)
6
∂T1
T1
Z
Z
1
v8 · n p ds = − ((−x1 + x2 )p1 + (−x1 + x2 )p2 + (−x1 + x2 )p3 ) . (4.9)
v8 · ∇p dA −
6
∂T1
T1
4.2
Computation of
R
T2
v · ∇p dA −
R
∂T2
v · n p ds
In terms of the mapping of T2 to the reference triangle, relative to (2.14), associate S1 ≡ R2 ,
S2 ≡ R3 , and S3 ≡ R4 .
We have that v1 |T2 and v2 |T2 are zero. Also,
1
v3 (x, y)|T2 = q̂5 (F −1 (x, y))
,
0
1
−1
v5 (x, y)|T2 = q̂4 (F (x, y))
,
0
1
−1
,
v7 (x, y)|T2 = q̂3 (F (x, y))
0
v4 (x, y)|T2 = q̂5 (F −1 (x, y))
v6 (x, y)|T2 = q̂4 (F
−1
v8 (x, y)|T2 = q̂3 (F
−1
(x, y))
(x, y))
0
1
0
1
0
1
,
,
,
p(x, y) = p4 ˆl1 (F −1 (x, y)) + p5 ˆl2 (F −1 (x, y)) + p6 ˆl3 (F −1 (x, y)) ,
(x3 − x2 ) (x4 − x2 )
.
and Jmat =
(y3 − y2 ) (y4 − y2 )
Using (3.1),(2.16), and (4.2) we have that
Z
Z
1
v3 · ∇p dA −
v3 · n p ds = ((−2y2 + y3 + y4 )p4 + (−y2 + y4 )p5 + (−y2 − y3 + 2y4 )p6 ) ,
6
T2
∂T2
(4.10)
Z
Z
1
v4 ·∇p dA −
v4 ·n p ds = − ((−2x2 + x3 + x4 )p4 + (−x2 + x4 )p5 + (−x2 − x3 + 2x4 )p6 ) ,
6
T2
∂T2
(4.11)
Z
Z
1
v5 · ∇p dA −
v5 · n p ds = ((y3 − y4 )p4 + (−y2 + 2y3 − y4 )p5 + (y2 + y3 − 2y4 )p6 ) ,
6
T2
∂T2
(4.12)
Z
Z
1
v6 · ∇p dA −
v6 · n p ds = − ((x3 − x4 )p4 + (−x2 + 2x3 − x4 )p5 + (x2 + x3 − 2x4 )p6 ) ,
6
T2
∂T2
(4.13)
and for the velocity basis vectors associated with the barycenter node R4 :
Z
Z
1
v7 · ∇p dA −
v7 · n p ds = ((−y2 + y3 )p4 + (−y2 + y3 )p5 + (−y2 + y3 )p6 ) , (4.14)
6
T2
∂T2
Z
Z
1
v8 · ∇p dA −
v8 · n p ds = − ((−x2 + x3 )p4 + (−x2 + x3 )p5 + (−x2 + x3 )p6 ) . (4.15)
6
T2
∂T2
13
4.3
Computation of
R
v · ∇p dA −
T3
R
∂T3
v · n p ds
In terms of the mapping of T3 to the reference triangle, relative to (2.14), associate S1 ≡ R3 ,
S2 ≡ R1 , and S3 ≡ R4 .
We have that v3 |T3 and v4 |T3 are zero. Also,
1
−1
v5 (x, y)|T3 = q̂5 (F (x, y))
,
0
1
−1
,
v1 (x, y)|T3 = q̂4 (F (x, y))
0
1
,
v7 (x, y)|T3 = q̂3 (F −1 (x, y))
0
v6 (x, y)|T3 = q̂5 (F
−1
v2 (x, y)|T3 = q̂4 (F
−1
(x, y))
(x, y))
v8 (x, y)|T3 = q̂3 (F −1 (x, y))
0
1
0
1
0
1
,
,
,
p(x, y) = p7 ˆl1 (F −1 (x, y)) + p8 ˆl2 (F −1 (x, y)) + p9 ˆl3 (F −1 (x, y)) ,
(x1 − x3 ) (x4 − x3 )
.
and Jmat =
(y1 − y3 ) (y4 − y3 )
Using (3.1),(2.16), and (4.2) we have that
Z
Z
1
v5 · ∇p dA −
v5 · n p ds = ((y1 − 2y3 + y4 )p7 + (−y3 + y4 )p8 + (−y1 − y3 + 2y4 )p9 ) ,
6
T3
∂T3
(4.16)
Z
Z
1
v6 · ∇p dA −
v6 · n p ds = − ((x1 − 2x3 + x4 )p7 + (−x3 + x4 )p8 + (−x1 − x3 + 2x4 )p9 ) ,
6
T3
∂T3
(4.17)
Z
Z
1
v1 ·∇p dA −
v1 ·n p ds = ((y1 − y4 )p7 + (2y1 − y3 − y4 )p8 + (y1 + y3 − 2y4 )p9 ) , (4.18)
6
T3
∂T3
Z
Z
1
v2 · ∇p dA −
v2 · n p ds = − ((x1 − x4 )p7 + (2x1 − x3 − x4 )p8 + (x1 + x3 − 2x4 )p9 ) ,
6
T3
∂T3
(4.19)
and for the velocity basis vectors associated with the barycentered node R4 :
Z
Z
1
v7 · ∇p dA −
v7 · n p ds = ((y1 − y3 )p7 + (y1 − y3 )p8 + (y1 − y3 )p9 ) ,
(4.20)
6
T3
∂T3
Z
Z
1
v8 · ∇p dA −
v8 · n p ds = − ((x1 − x3 )p7 + (x1 − x3 )p8 + (x1 − x3 )p9 ) .
(4.21)
6
T3
∂T3
Next we form the linear system Ap = 0, corresponding to (div(v) , p) = 0 , ∀v ∈ X0,M , with
equation i, generated by basis vector vi , i = 1, . . . 8. That is, row 1 of A is generated by combining
equations (4.4) and (4.18), row 2 generated by combining (4.5) and (4.19), . . ., row 8 generated by
14
combining (4.9),(4.15) and (4.21).

−2y1 + y2 + y4
−y1 + y4
−y1 − y2 + 2y4


 −2x1 + x2 + x4
−x1 + x4
−x1 − x2 + 2x4


y2 − y4
−y1 + 2y2 − y4
y1 + y2 − 2y4
−2y2 + y3 + y4
−y2 + y4



x2 − x4
−x1 + 2x2 − x4 x1 + x2 − 2x4 −2x2 + x3 + x4
−x2 + x4

A =

y3 − y4
−y2 + 2y3 − y4



x3 − x4
−x2 + 2x3 − x4



−y1 + y2
−y1 + y2
−y1 + y2
−y2 + y3
−y2 + y3

−x1 + x2
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
−x1 + x2
−x1 + x2
−x2 + x3
−x2 + x3

..
.
y1 − y4
2y1 − y3 − y4
y1 + y3 − 2y4

..

.
x1 − x4
2x1 − x3 − x4 x1 + x3 − 2x4 

..

. −y2 − y3 + 2y4


..

. −x2 − x3 + 2x4


..
. y2 + y3 − 2y4
y1 − 2y3 + y4
−y3 + y4
−y1 − y3 + 2y4 


..
. x2 + x3 − 2x4 x1 − 2x3 + x4
−x3 + x4
−x1 − x3 + 2x4 


..

.
−y2 + y3
y1 − y3
y1 − y3
y1 − y3

..
.
−x2 + x3
x1 − x3
x1 − x3
x1 − x3
(4.22)
Observe that p1 = p2 = p3 = . . . = p9 satisfies Ap = 0, i.e., NM consisting of functions which are
constant on M .
Lemma 5 The rows of the matrix A given in (4.22) are linearly independent, i.e. rank(A) = 8.
Proof : To see that rank(A) = 8, consider a change of basis for NM corresponding to a (scaled)
Lagrangian basis on each triangle with node located at the midpoint of each edge. Introduce:
b 1 (ξ, η) =
dl
b 3 (ξ, η) =
dl
1 ˆ
−l1 (ξ, η) + ˆl2 (ξ, η) + ˆl3 (ξ, η) ,
2
1 ˆ
l1 (ξ, η) + ˆl2 (ξ, η) − ˆl3 (ξ, η) ,
2
b 2 (ξ, η) = 1 ˆl1 (ξ, η) − ˆl2 (ξ, η) + ˆl3 (ξ,(4.23)
dl
η) ,
2
(4.24)
and p(x, y) = p̃1 dl1 (x, y) + p̃2 dl2 (x, y) + . . . , + p̃9 dl9 (x, y) . Under this change of basis the linear
system of equations Ap = 0 transforms to Ãp̃ = 0, where the first column of Ã, ÷,1 , corresponding
to the coefficients of p̃1 , is giving by (see (4.23)) ÷,1 = 1/2(−A·,1 + A·,2 + A·,3 ), the second column
of Ã, ÷,2 = 1/2(A·,1 − A·,2 + A·,3 ), . . ., the ninth column of Ã, ÷,9 = 1/2(A·,7 + A·,8 − A·,9 ).
15
Then we have

−y2 + y4
−x2 + x4
y2 − y4
x2 − x4





à = 




 −y1 + y2
−x1 + x2
−y1 + y4
−x1 + x4
y1 − y4
x1 − x4
−y1 + y2
−x1 + x2
−y1 + y2
−x1 + x2
−y1 + y2
−x1 + x2
−y1 + y2
−x1 + x2
−y3 + y4
−x3 + x4
y3 − y4
x3 − x4
−y2 + y3
−x2 + x3
−y2 + y4
−x2 + x4
y2 − y4
x2 − x4
−y2 + y3
−x2 + x3
−y2 + y3
−x2 + x3
−y2 + y3
−x2 + x3
−y2 + y3
−x2 + x3
y1 − y4
x1 − x4
y3 − y4
x3 − x4
−y1 + y4
−x1 + x4
y1 − y3
x1 − x3
−y3 + y4
−x3 + x4
y1 − y3
x1 − x3

y1 − y3
x1 − x3 





y1 − y3 

x1 − x3 

y1 − y3 
x1 − x3
(4.25)
Note that, as the transformation is invertible, rank(A) = rank(Ã).
Next we obtain Ã2 from à by applying the following elementary row operations:
Ã21,· = 1/2(Ã1,· + Ã3,· + Ã5,· ) ,
Ã22,· = 1/2(Ã2,· + Ã4,· + Ã6,· ) ,
Ã23,· = 1/2(−Ã1,· + Ã3,· + Ã5,· ) ,
Ã24,· = 1/2(−Ã2,· + Ã4,· + Ã6,· ) ,
Ã25,· = 1/2(Ã1,· + Ã3,· − Ã5,· ) ,
Ã26,· = 1/2(Ã2,· + Ã4,· − Ã6,· ) ,
Ã27,· = Ã7,· − 1/2(Ã1,· + Ã3,· + Ã5,· ) ,
Ã28,· = Ã8,· − 1/2(Ã2,· + Ã4,· + Ã6,· ) ,
i.e.,

−y1 + y2
−x1 + x2


 y2 − y4

 x2 − x4
Ã2 = 




 −y1 + y2
−x1 + x2
−y2 + y3
−x2 + x3
−y2 + y3
−x2 + x3
y1 − y4
x1 − x4
−y1 + y2
−x1 + x2
−y1 + y2
−x1 + x2
−y3 + y4
−x3 + x4
−y2 + y3
−x2 + x3
−y2 + y4
−x2 + x4
−y2 + y3
−x2 + x3
−y1 + y4
−x1 + x4
y1 − y4
x1 − x4
y1 − y3
x1 − x3
−y3 + y4
−x3 + x4
y3 − y4
x3 − x4
y1 − y3
x1 − x3

y1 − y3
x1 − x3 




.





(4.26)
Again, as the transformation is invertible, rank(A) =
rank(Ã2 ).
To further simplify the determination of the rank(A) we apply one more transformation, this time
to the columns of Ã2 . Let Ã3 be obtained from Ã2 by:
Ã3·,2 = Ã2·,2 − Ã2·,1 ,
Ã3·,5 = Ã2·,5 − Ã2·,4 ,
Ã3·,8 = Ã2·,8 − Ã2·,7 .
i.e.,



 y2 − y4

 x2 − x4
3
à = 




 −y1 + y2
−x1 + x2
−y1 + y2
−x1 + x2
−y2 + y3
−x2 + x3
−y2 + y3
−x2 + x3
y1 − y2
x1 − x2
−y1 + y2
−x1 + x2
−y3 + y4
−x3 + x4
−y2 + y3
−x2 + x3
−y2 + y3
−x2 + x3
−y1 + y4
−x1 + x4
y1 − y4
x1 − x4
y1 − y3
x1 − x3
y1 − y3
x1 − x3
−y1 + y3
−x1 + x3

y1 − y3
x1 − x3 




.





(4.27)
As the rank(Ã3 ) = rank((Ã3 )T ), we consider the linear system (Ã3 )T c = 0, where c = [ci ], i =
1, . . . , 8. Introduce
yi − yj
,
si,j :=
xi − xj
16
i.e., si,j represents the slope of the line segment from (xi , yi ) to (xj , yj ). Note that si,j = sj,i .
From rows 2, 5, and 9, respectively, of (Ã3 )T c = 0, we obtain
(y1 − y2 )c3 + (x1 − x2 )c4 = 0 ,
⇒ c4 = −s1,2 c3 ,
(assuming x1 6= x2 )
(4.28)
(y3 − y2 )c5 + (x3 − x2 )c6 = 0 ,
⇒ c6 = −s2,3 c5 ,
(assuming x2 6= x3 )
(4.29)
(y1 − y3 )c1 + (x1 − x3 )c2 = 0 ,
⇒ c2 = −s1,3 c1 ,
(assuming x1 6= x3 )
(4.30)
Combining (4.28)-(4.30) with row 3 of (Ã3 )T c = 0, we obtain
(−y1 + y2 )c1 + (−x1 + x2 )c2 + (−y1 + y2 )c5 + (−x1 + x2 )c6 = 0
(−y1 + y2 )c1 − (−x1 + x2 )s1,3 c1 + (−y1 + y2 )c5 − (−x1 + x2 )s2,3 c5 = 0,
s1,2 − s1,3
c1 .
on simplifying becomes c5 = −
s1,2 − s2,3
Combining (4.29) and (4.31)
c6 = s23
s1,2 − s1,3
c1 .
s1,2 − s2,3
(4.31)
(4.32)
Note that s1,2 − s2,3 6= 0, as this would imply that the points P1 , P2 , P3 , lie on a line.
Consider next row 6 of (Ã3 )T c = 0.
(−y2 + y3 )c1 + (−x2 + x3 )c2 + (−y2 + y3 )c3 + (−x2 + x3 )c4 = 0
⇒
s2,3 c1 + c2 + s2,3 c3 + c4 = 0
⇒
s2,3 c1 − s1,3 c1 + s2,3 c3 − s1,2 c3 = 0
s2,3 − s1,3
⇒ c3 = −
c1 ,
s2,3 − s1,2
s2,3 − s1,3
and then with (4.28) ⇒ c4 = s1,2
c1 .
s2,3 − s1,2
(4.33)
(4.34)
Row 8 of (Ã3 )T c = 0 implies
(y1 − y3 )c3 + (x1 − x3 )c4 + (−y1 + y3 )c5 + (−x1 + x3 )c6 = 0
⇒
⇒
s1,3 c3 + c4 − s1,3 c5 − c6 = 0
⇒ (s1,3 − s1,2 )c3 + (s2,3 − s1,3 )c5 = 0
(s1,2 − s1,3 )(s2,3 − s1,3 )
(s1,3 − s1,2 )(s2,3 − s1,3 )
+
c1 = 0
s2,3 − s1,2
s1,2 − s2,3
(s1,3 − s1,2 )(s2,3 − s1,3 )
⇒ 2
c1 = 0 , ⇒ c1 = 0 ,
s2,3 − s1,2
(4.35)
as, s1,3 − s1,2 6= 0, and s2,3 − s1,3 6= 0.
Thus from (4.30)-(4.34), c1 = 0 , ⇒
c2 = c3 = c4 = c5 = c6 = 0.
If x1 = x2 (⇒ y1 6= y2 , as P1 6= P2 ), then from (4.28) c3 = 0, and in place of (4.31), (4.33), (4.35),
we would obtain, respectively:
c5 = −c1
c4 = (s1,3 − s2,3 )c1
2(s1,3 − s2,3 )c1 = 0 , ⇒
17
c1 = 0 ,
and c2 = c3 = c4 = c5 = c6 = 0. Similarly, the other exceptional cases also lead to the same
conclusion that ci = 0 , i = 1, 2, . . . , 6.
Rows 1 and 4 of (Ã3 )T c = 0 imply, respectively,
Hence
(−y1 + y2 )c7 + (−x1 + x2 )c8 = 0 ,
⇒
c8 = −s1,2 c7 ,
(−y2 + y3 )c7 + (−x2 + x3 )c8 = 0 ,
⇒
c8 = −s2,3 c7 .
⇒
c8 = 0 .
(s2,3 − s1,2 )c7 = 0 ,
⇒
c7 = 0 ,
(4.36)
As (Ã3 )T c = 0 has only the trivial solution, c = 0, we conclude that rank((Ã3 )T ) = rank(A) = 8,
which then implies the dimension of NM = 1.
Corollary 2 The Scott-Vogelius P2 − discP1 approximation pair is LBB stable on a barycenter
refinement of a regular triangulation of Ω.
Corollary 3 The space QM can be decomposed into QM = NM ⊕ divXM . In addition, NM and
divXM are orthogonal with respect to the L2 innerproduct on QM .
R
Proof : Let hq1 , q2 iM := M q1 q2 dA denote the L2 innerproduct on QM . We have that NM is a
⊥ . Then
closed subspace of QM . Denote its orthogonal complement with respect to h· , ·iM by NM
⊥
⊥
QM = NM ⊕ NM . Note that as dim(QM ) = 9, dim(NM ) = 1, then dim(NM ) = 8.
R
⊥.
We have that for v ∈ XM , M div(v) q dA = 0 , ∀q ∈ NM . Therefore it follows that divXM ⊂ NM
⊥
The fact that rank(A) = 8 implies that dim(divXM ) ≥ 8. Hence dim(divXM ) = NM .
Corollary 4 There exists matrices Q1 ∈ IR8×8 , Q2 ∈ IR9×9 , and a diagonal matrix Σ ∈ IR8×8 , such
that
 2

σ1
0

σ22
0 


(4.37)

..  = Σ = Q1 A Q2 .
..

.
. 
σ82 0
1, (x, y) ∈ M
0, (x, y) 6∈ M
denote an orthogonal basis for divXM .
Proof : Let φ9 (x, y) :=
. Note that φ9 is a basis for NM . Let {φ1 , φ2 , . . . φ8 }
Let Q1 , Q2 , be defined by
Q1 (i, ·)[div(v1 ), div(v2 ), . . . , div(v8 )]T
Then, σi2 =
R
M
= φi , i = 1, 2, . . . , 8,
(4.38)
[l1 , l2 , . . . , l9 ]Q2 (·, j) = φj , j = 1, 2, . . . , 9.
(4.39)
φ2i dA, i = 1, 2, . . . , 8.
18
References
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[2] F. Brezzi and M. Fortin. Mixed and Hybrid Finite Element Methods. Springer-Verlag, New York,
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[3] V. Girault and P.A. Raviart. Finite element methods for Navier-Stokes equations. SpringerVerlag, Berlin, 1986.
[4] M.D. Gunzburger. Finite element methods for viscous incompressible flows. Computer Science
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[5] J. Qin. On the convergence of some low order mixed finite elements for incompressible fluids.
PhD thesis, Pennsylvania State University, 1994.
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