A P4 BUBBLE ENRICHED P3 DIVERGENCE-FREE

FINITE ELEMENT ON TRIANGULAR GRIDS

SHANGYOU ZHANG

DEDICATED TO PROFESSOR PETER MONK ON THE OCCASION OF HIS 60THBIRTHDAY

Abstract. On triangular grids, the continuous Pk plus discontinuousPk−1 mixed finite element is stable for polynomial degree k ≥ 4. Whenk = 3, the inf-sup condition fails and the mixed finite element convergesat an order that is two orders lower than the optimal order. We enrichthe continuous P3 by adding some P4 divergence-free bubble functions,to be exact, one P4 divergence-free bubble function each componenteach edge. We show that such an enriched P3-P2 mixed element is inf-sup stable, and converges at the optimal order. Numerical tests arepresented, comparing the new element with the P4-P3 element and theunstable P3-P2 element.

AMS subject classifications. 65M60, 65N30, 76M10, 76D07.

Keywords. finite element, divergence-free element, Stokes equations,triangular grid.

1. Introduction

It is a challenge to construct stable H1 conforming mixed finite elementssatisfying the incompressible condition exactly, in computing the Stokes orNavier-Stokes equations. That is, the velocity is approximated by the con-tinuous piecewise polynomials of degree k and the pressure is approximatedby the discontinuous piecewise polynomials of one degree less. Here themethod is truly conforming in the sense that the finite element velocity isthe H1 projection of the true solution in a polynomial subspace. A break-through on the method was done by Scott and Vogelius in 1985 [18, 19] thatthe method is stable and consequently of the optimal order of convergenceon 2D triangular grids, for the Pk-Pk−1 element, if k ≥ 4, a magic number.What is this magic number k in 3D? Or if there is such a magic number in3D? The problem remains open, after it was posted explicitly for so manyyears [18].

Scott and Vogelius showed that the Pk-Pk−1 element is not stable ongeneral triangular grids, if k < 4. However, on special triangular grids,

1

low order elements may be stable. On Hsieh-Clough-Tocher macro-elementgrids, where each base triangle is split into 3 triangles by connecting thebarycenter with three vertices, the P2-P1 and the P3-P2 mixed elements arestable, cf. [1, 17, 22]. On Powell-Sabin macro-element grids, where eachtriangle is split into 6 sub-triangles, even the P1-P0(subspace) element isstable [26]. When enriching the continuous Pk velocity space by some ra-tional functions, Guzman and Neilan showed the enriched Pk-Pk−1 elementis table for all k ≥ 1, [9, 10]. With additional continuity constraints, Falkand Neilan showed that the Pk-Pk−1 element is stable if the continuous Pkvelocity is also C1 at vertices and the discontinuous Pk−1 pressure is C0 atvertices, cf. [8].

In 3D, the Pk-Pk−1 mixed element is stable for all k ≥ 3 on the Hsieh-Clough-Tocher macro-element tetrahedral grids (where each base tetrahe-dron is split into 4 sub-tetrahedra by connecting the barycenter with fourvertices), cf. [25]. If splitting further a base tetrahedron into 12 sub-tetrahedra(connecting the barycenter with 4 vertices and 4 face-triangle barycenters),the Pk-Pk−1 is stable for all k ≥ 2, cf. [29]. On the uniform tetrahedralgrids, i.e., each cube is subdivided into 6 tetrahedra, the continuous Pk withdiscontinuous Pk−1 mixed finite element is stable for all k ≥ 6, cf. [28]. Withadditional constraints on the finite element spaces, Neilan showed that thePk-Pk−1 element is stable for k ≥ 6 on general tetrahedral grids if the con-tinuous velocity finite element is C2 continuous at all vertices and also C1continuous on all edges, and the discontinuous pressure finite element func-tion is C1 at all vertices and C0 on all edges, cf. [16]. But the Scott-Vogeliusproblem is still open, on general tetrahedral grids. On rectangular grids,this problem is simple that Qk,k−1 ×Qk−1,k-Qk−1 element (and its nD ver-sion) is stable for all k ≥ 2, where Qk,k−1 denotes the continuous piecewisepolynomials of separated degrees k and k− 1 in its first variable and secondvarialbe, respectively, cf. [14, 15, 27].

The mixed finite element of continuous P3 velocity and discotinuous P2pressure is not stable on general triangular grids in 2D, cf. [18, 19]. Inthis work, we enrich the continuous P3 space by some divergence-free P4bubble functions. Such P4 bubble functions do not provide additional ap-proximation power, but do provide additional degrees of freedom to relaxthe locking problem of the divergence-free constraint. Such a finite elementenrichment technique is used before, many times. For example, mentionedabove, Guzman and Neilan enrich the continuous P3 velocity by some ratio-nal bubble functions to obtain an inf-sup stable mixed finite element [9, 10].Here, instead of rational functions (whose numerical integration formula areunknown) we use the P4 bubble polynomials in this work. In the low ordermixed finite element methods for the linear elasticity equation, the H(div)Pk finite element space must be enriched by Pd+1 divergence-free bubblefunctions in d-dimensional space, cf. [2, 3, 11, 12, 13].

2

2. The enriched P3 divergence-free element

In this section, we define the P4-enriched P3 divergence-free finite element.Its uni-solvence is shown.

We consider a model stationary Stokes problem: Find the velocity u andthe pressure p on a 2D polygonal domain Ω, such that

(2.1)

−∆u+∇p = f in Ω,divu = 0 in Ω,

u = 0 on ∂Ω.

The standard variational form for (2.1) is: Find u ∈ H10 (Ω)2 and p ∈L20(Ω) := L

2(Ω)/C = {p ∈ L2 |∫Ω p = 0} such that

(2.2)a(u,v) + b(v, p) = (f ,v) ∀v ∈ H10 (Ω)2,

b(u, q) = 0 ∀q ∈ L20(Ω).

Here H10 (Ω)2 is the subspace of the Sobolev space H1(Ω)2 (cf. [7]) with zero

boundary trace, and the blinear forms are defined by

a(u,v) =

∫Ω∇u · ∇v dx,

b(v, p) = −∫Ωdivv p dx,

(f ,v) =

∫Ωf v dx.

Let Th be an initial triangulation of Ω. We refine each triangle into fourcongruent triangles by connecting the three mid-edge points. This way, onegrid is refined to the next level grid. We denote each triangulation in thesequence of grids also by Th, where h is the grid size. Here we introducethe multigrids [23, 24], instead of general quasi-uniform grids, to avoid thetechnical details of nearly-singular points. For an initial triangulation, wemay have a few singular points [4, 18]. Here a singular point is a point atwhich all edges of an triangulation fall into two crossing lines at the point.There are exactly four types of singular points, three boundary ones and oneinternal one, shown in Figure 2.1. There is a minor constraint for the discretepressure functions at the singular points. However, all singular points of theinitial grid will stay singular and no new singular points appear, after themultigrid refinement.

Let the P4-enriched P3 velocity space be, for K ∈ Th,

VK = {v ∈ P4(K)2 | divv ∈ P2}.(2.3)

Here Pk stands for the space of polynomials of degree k or less. We notethat, for a P 24 vector, the divergence is a P3 polynomial, not a P2 polynomial.That is, under the constraint, the x3, x2y, xy2 and y3 coefficients of thepolynomial of the divergence must be zero. With these four constraints, we

3

s

AAA

�����

�������s

AA

A

""

"""

�������������s

AAA

"""""

BBBB

AAAAAAAAAA

AAAAAAAAAAAA

s

BBBB

"""""

BBBB"

"""

"

Figure 2.1. There boundary singular points (left three)and an internal singular point.

would expect the dimension of VK be

2 dimP4 − 4 = 26 = 2dimP3 + 6.This is to be proved in Lemma 2.1. But we would like to give another(equivalent) definition of VK . As the range of the divergence operator onP 23 space is P2 already, the newly added P

24 functions would be divergence-

free. That is, the above 6 dimension space is spanned by the following 6divergence-free P 24 polynomials:

curlx5, curlx4y, curlx3y2 =

(2x3y

−3x2y2), curlx2y3, curlxy4, curly5.

Also there, in Lemma 2.1, the 26 degrees of freedom in VK are given byv(xi), three vertex values of two components,∫eiv · xj ds, 0-th, 1st, 2nd moments on three edges,∫

K v dx, 0-th moment on the element.

(2.4)

The mixed element spaces are defined by, also equivalently by (2.3) and(2.12),

Vh ={vh ∈ C(Ω)2 | vh|K ∈ VK ∀K ∈ Th, and vh|∂Ω = 0

},(2.5)

Ph = {divvh | vh ∈ Vh} .(2.6)

Since∫Ω ph =

∫Ω divvh =

∫∂Ω uh = 0 for any ph ∈ Ph, we conclude that

Vh ⊂ H10 (Ω)2, Ph ⊂ L20(Ω),i.e., the mixed-finite element pair is conforming.

The resulting system of finite element equations for (2.2) is: Find uh ∈ Vhand ph ∈ Ph such that

(2.7)a(uh,v) + b(v, ph) = (f ,v) ∀v ∈ Vh,

b(uh, q) = 0 ∀q ∈ Ph.Traditional mixed-finite elements require the inf-sup condition to guar-

antee the existence of discrete solutions. As (2.6) provides a compatibilitybetween the discrete velocity and discrete pressure spaces, the linear systemof equations (2.7) always has a unique solution, independent of the inf-supcondition.

4

Proposition 2.1. There is a unique solution in the discrete linear system(2.7).

Proof. As (2.7) is a square system, the uniqueness implies existence. Let(uh, ph) be a solution for the homogeneous equations (2.7). Let v = uh andq = ph in (2.7). We have

a(uh,uh) + b(uh, ph) = 0,

b(uh, ph) = 0.

Subtracting the second equation from the first one, we get

a(uh,uh) = 0 ⇒ uh = 0.

The first equation of (2.7) is now

b(v, ph) = 0, forall v ∈ Vh.(2.8)

By (2.6), we have some wh ∈ Vh such that ph = divwh. Let v = wh in(2.8). It is then ∥divwh∥2L2 = 0. So ph = divwh = 0.

Further, by the second equation in (2.7) and the definition of Ph in (2.6),we conclude that

(2.9) b(uh, q) = b(uh,−divuh) = ∥divuh∥2L2(Ω)2 = 0

and that

divuh = 0,

i.e. uh is divergence-free. In this case, we call the mixed finite element adivergence-free element. It is apparent that the discrete velocity solution isdivergence-free if and only if the discrete pressure finite element space is thedivergence of the discrete velocity finite element space, i.e., (2.6). In fact,it is trivial to show ([5, 6, 26]), in the next theorem, that uh is the uniquea(·, ·) orthogonal projection from the divergence-free space Z to its subspaceZh, defined by

Z :={v ∈ H10 (Ω)2 | divv = 0

},(2.10)

Zh := {v ∈ Vh | divv = 0} .(2.11)

That is,

uh ∈ Zh, a(u− uh,vh) = 0 ∀vh ∈ Zh.

We note that the definition of Ph is abstract, which may not be goodenough for computation (the basis of functions of Ph is unknown.) But asthe pressure space is defined implicitly by the velocity space, the unknownpressure solution can be implicitly defined by a function in the velocity space,by an iterative method, cf. [27]. Indeed, such a computation saves half ofcoding work on the pressure finite element and one-third unknowns in theresulting linear system of equations. But we do give another, traditionaldefinition of the pressure space Ph below. By (2.6), Ph may not be the full

5

space of discontinuous P2 polynomials, but a proper subspace if singularvertices are present. The next definition describes Ph precisely.

Ph = {ph ∈ L20(Ω) | ph|K ∈ P2 ∀K ∈ Th,(2.12)i0∑i=1

(−1)ivh|Ki(s) = 0 at a singular vertex s },

where s is one of the four types of vertexes depicted in Figure 2.1, and Kiare the i0 (= 1, 2, 3, 4) triangles around the singular vertex s.

Lemma 2.1. The dimension of VK in (2.3) is 26. The 26 degrees of freedomare listed in (2.4).

Proof. dimP 24 = 30. Adding the four constraints of P3 coefficients of thedivergence to the 26 degrees of freedom, we have a square system of linearequations. The existence of the solution is implied by the uniqueness, whichwill be proved next.

The proof is done in two steps, on the reference triangle K̂ and on thegeneral triangle K. Let the reference triangle be

K̂ = {x̂ ≥ 0, ŷ ≥ 0, x̂+ ŷ ≤ 1}.(2.13)

With all 26 dof in (2.4) of v̂h be zero, v̂h = 0 on the boundary of K̂, as eachcomponent of v̂|êi is a degree 4 polynomial with 5 zeros. Thus

v̂h = x̂ŷ(1− x̂− ŷ)(c1 + c2x̂+ c3ŷc4 + c5x̂+ c6ŷ

),(2.14)

for some constants c1, ... c6. We show these constants are all zero. The x̂3

and ŷ3 coefficients of d̂ivv̂h are −c5 and −c3, respectively. Thus c3 = c5 = 0.The x̂2y coefficient of d̂ivv̂h is−3c2−2c5−2c6 = −3c2−2c6 = 0. Similarly, bychecking the x̂ŷ2 coefficient of d̂ivv̂h, we have −2c2−2c3−3c6 = −2c2−3c6 =0. Together, we get c2 = c6 = 0. Thus

v̂h = x̂ŷ(1− x̂− ŷ)(c1c4

).

Because the bubble function x̂ŷ(1 − x̂ − ŷ) is positive inside K̂, by the 0thmoment of v̂h on K̂ in (2.4), c1 = c4 = 0. Thus v̂h = 0.

Now, we show the uniqueness on a general triangle K ∈ Th. Let FK bean affine mapping from K̂ to K, cf. Figure 2.2, such that

FK(x̂, ŷ) = x0 +Bx̂ =

(x0y0

)+(

⃗x0x1 ⃗x0x2)(x̂

ŷ

).(2.15)

Now, if the 26 dof’s of function vh have value 0, then vh is identically zeroon the three edges of K:

vh = λ1λ2λ3

(p(1)1

p(2)1

),

6

K̂ :

@@@

@@@@

@x̂0x̂1

x̂2-FK = Bx̂+ x0

K :

ccc

cc�

������������

��

��

��

�x2

x0

x1

Figure 2.2. An affine mapping FK from the reference tri-angle K̂ to a general triangle K.

where λi are three area-coordinates on K, and p(i)1 are two P1 polynomials.

We define a Piola transform by

v̂h(x̂) = B−1vh(FK(x̂)),(2.16)

where B is defined in (2.15). Because vh is identically zero on the boundaryof K, so is v̂h. On the other side, if vh ̸= 0 on the boundary of K, we cannotuse the Piola transformation as it would destroy the tangential continuityof H1 functions. The v̂h defined in (2.16) can also be expressed as (2.14).Since the Piola transform preserves the divergence,

divvh = traceB−T (∇̂Bv̂h(x̂))T = trace (∇̂v̂h(x̂))T = d̂ivv̂h,

d̂ivv̂h is also a P2 function. Further, as B is invertible, two linear combina-tions of 2 zero 0-moment component of vh have also zero 0-moment. Thus,by the analysis on v̂h above, v̂h = 0. So is vh = Bv̂h(F

−1K (x)) = 0.

3. Stability and convergence

In this section, we will prove the on-to mapping property of the divergenceoperator, from the discrete velocity space to the finite element pressurespace. Consequently we prove the inf-sup stability condition and the optimalorder of convergence for the finite element solution.

Remark 3.1. The inf-sup condition (3.11), i,e, Theorem 3.1, can be provedas a corollary of Soctt-Vogelius’ Theorem 5.1 [18]. But we give an inde-pendent proof. The difference between the proof here and the Scott-Vogelius’proof is in the construction of uh in next lemma, Lemma 3.1. We use bubble-enriched P3 polynomials (equivalently P4 polynomials) to construct uh whileScott and Vogelius used only P3 polynomials, cf. [21, Lemma 2.3] and [18,Lemma 4.1]. Thus the Scott-Vogelius result requires the grid size h suffi-ciently small, [18, Remark 5.1, Lemma 5.1, and (5.6)]. But the theory heredoes not have any restriction on h.

7

Lemma 3.1. There is a uh ∈ Vh, supported on two triangles K,K ′, cf. Fig-ure 3.1, such that divuh has a nodal value 1 at x1 on the K side, and nodalvalue 0 at the rest 11 P2 Lagrange nodes on the two triangles, as long as thetwo edges x0x1 and x1x0,K′ do not fall into a same line.

K̂

@@

@@

@@

@@

b bbb

bbbb

bbb K̂ ′

�

K

cc

ccc�

����������

������

��

AAAAAAAA

��

��

��

��

�x0

x1 = x2,K′

x2 = x1,K′

x0,K′

K ′bb

bb

b-

divuh = 1b

bb

bb

b

Figure 3.1. C−1-P2 Lagrange nodes on two neighboringtriangles and the reference mapping.

Proof. On the reference triangle K̂ in Figure 3.1, we find all such vectorsû01, which vanish on the boundary of (K̂ ∪ K̂ ′) and whose divergence is theP2 polynomial having value 1 at vertex x̂1 only.

d̂ivû01(x̂, ŷ) = 2x̂2 − x̂, on K̂.

Note that there are precisely three divergence-zero vectors. They are thecurl of functions

x̂3ŷ2, x̂2ŷ2 and x̂2ŷ3.

So, on K̂,

û01(x̂, ŷ) = x̂ŷ

((0

2x̂− 1

)+ c1

(2x̂2

−3x̂ŷ

)+c2

(2x̂−2ŷ

)+ c3

(3x̂ŷ−2ŷ2

)).

(3.1)

On the other reference triangle K̂ ′ in Figure 3.1, we need d̂ivû01 = 0. Sothe vector must be a linear combination of the curls of, on K̂ ′,

(1− x̂)2(1− ŷ)3, (1− x̂)2(1− ŷ)2 and (1− x̂)3(1− ŷ)2.

û01(x̂, ŷ) = (1− x̂)(1− ŷ)(c4

(−3(1− x̂)(1− ŷ)

2(1− ŷ)2)

+c5

(−2(1− x̂)2(1− ŷ)

)+ c6

(−2(1− x̂)2

3(1− x̂)(1− ŷ)

)).

(3.2)

8

We map these two vector functions to triangles K and K ′, preserving thedivergence. Because û01 = 0 on the four outside edges of triangles K̂ andK̂ ′, after the Piola transformation, the piecewise P4 vector function remainszero on the four outside edges of triangles K and K ′. Then we only matchthe interface values of two P4 vector functions at the common edge x1x2.Each P4 function has 3 internal Lagrange nodal-values on an edge. We endup with 6 equations for the matching on x1x2. Though we have 6 degreesof freedom in (3.1) and (3.2), the system does not have a unique solution,as the curl of the C1-P5 Argyris normal derivative basis at the mid-point ofx1x2 is a solution of the homogeneous system. We have either no solution,if x0x1 and x1x0,K′ are colinear, or infinitely many solutions, if the two linesare different. We will show the detail next.

The reference mapping from K̂ to K is defined in (2.15). On K, u01 isdefined by the Piola transformation,

u01(x) = Bû01(F−1K (x)).(3.3)

The reference mapping from K̂ ′ to K ′ is(xy

)= FK′(x̂, ŷ) = x0,K′ +B

′(1− x̂1− ŷ

),

where B′ =(

⃗x0,K′x1,K′ ⃗x0,K′x2,K′). In order to keep the divergence, sim-

ilar to the Piola transformation (3.3), we let, on K ′,

u01(x, y) = −B′û0,1(F−1K′ (x)).(3.4)

When restricted on the common edge x1x2, i.e., ŷ = 1− x̂, we have

B

(2x̂2c1 + 2x̂c2 + 3(x̂− x̂2)c3

2x̂− 1 + 3(−x̂+ x̂2)c1 + 2(−1 + x̂)c2 + (−2 + 4x̂− 2x̂2)c3

)= −B′

(1

1

)(2x̂2c4 + 2x̂c5 + 3(x̂− x̂2)c6

3(−x̂+ x̂2)c4 + 2(−1 + x̂)c5 + (−2 + 4x̂− 2x̂2)c6

).

Matching coefficients of 1, x̂ and x̂2 of the two components, we have thefollowing 6× 6 system of equations, in block matrix form,

(0 0 00 −2 −2

)A

(0 0 00 −2 −2

)(

0 2 3−3 2 4

)A

(0 2 3−3 2 4

)(2 0 −33 0 −2

)A

(2 0 −33 0 −2

)

c1c2c3c4c5c6

=

010−200

,9

where the common matrix A is defined in (3.5).

A = B−1B′(

11

)=(

⃗x0x1 ⃗x0x2)−1 (

⃗x0,K′x1,K′ ⃗x0,K′x2,K′)( 1

1

)=(

⃗x0x1 ⃗x0x2)−1 (

⃗x0,K′x2 ⃗x0,K′x1)( 1

1

)=(

⃗x0x1 ⃗x0x2)−1 ((

⃗x0x1 ⃗x0x2)+(

⃗x0,K′x0 ⃗x0,K′x0))

=

(1

1

)+(B−1 ⃗x0,K′x0 B

−1 ⃗x0,K′x0)

=

(1 + z1 z1z2 1 + z2

).

(3.5)

Here

(z1z2

)= B−1 ⃗x0,K′x0. By adding the first block and the third block to

the second block in the linear system, we have a simplified linear system

0 0 0 0 −2z1 −2z10 −2 −2 0 −2− 2z2 −2− 2z22 2 0 2 + 2z1 2 + 2z1 00 0 0 2z2 2z2 02 0 −3 2 + 5z1 0 −3− 5z13 0 −2 3 + 5z2 0 −2− 5z2

c1c2c3c4c5c6

=

010−100

.(3.6)

There are two cases, z2 = 0 or z2 ̸= 0. z2 = 0 if and only if ⃗x0,K′x0 = c ⃗x0x1(i.e. x0,K′ is on the straight line x0x1.) If z2 = 0, by the fourth equation in(3.6), there is no solution.

When z2 ̸= 0, we let c6 = 0 (or a new constant). By the first equation,if z1 ̸= 0, c5 = 0. But if z1 = 0, i.e., x0,K′ is on the straight line x0x2, wehave another degree of freedom and we also let it be zero, i.e., c5 = 0. Bythe fourth equation in (3.6), c4 = −1/(2z2). By this time, the system (3.6)

10

is reduced to 0 −2 −2 | 12 2 0 | (1 + z1)/z22 0 −3 | (2 + 5z1)/(2z2)3 0 −2 | (3 + 5z2)/(2z2)

R1+R2−→

0 −2 −2 | 12 0 −2 | (1 + z1 + z2)/z22 0 −3 | (2 + 5z1)/(2z2)3 0 −2 | (3 + 5z2)/(2z2)

−R2+R3,(−3/2)R2+R4−→

0 −2 −2 | 12 0 −2 | (1 + z1 + z2)/z20 0 −1 | (3z1 − 2z2)/(2z2)0 0 1 | (−3z1 + 2z2)/(2z2)

R4+R3,2R4+R2,2R4+R1−→

0 −2 0 | (−3z1 + 3z2)/z22 0 0 | (1− 2z1 + 3z2)/z20 0 0 | 00 0 1 | (−3z1 + 2z2)/(2z2)

.Thus, we find a solution

c3 =−3z12z2

+ 1

c2 =3z12z2

− 32

c1 =1− 2z12z2

+3

2.

Lemma 3.2. There is a uh ∈ Vh, supported on two triangles K,K ′, cf. Fig-ure 3.2, such that divuh has a nodal value 1 at x1 on both K and K

′, andnodal value 0 at the rest 10 P2 Lagrange nodes on the two triangles whenthe two edges x0x1 and x1x0,K′ fall into a same line.

Proof. The proof repeats that for Lemma 3.1. We still give the details.On the reference triangle K̂ in Figure 3.2, we find all such vectors ûh,

which vanish on the boundary of (K̂ ∪ K̂ ′) and whose divergence is the P2polynomial having value 1 at vertex x̂1 only.

d̂ivûh(x̂, ŷ) = 2x̂2 − x̂, on K̂.

Note that there are precisely three divergence-zero vectors. They are thecurl of functions

x̂3ŷ2, x̂2ŷ2 and x̂2ŷ3.

11

K̂

@@

@@

@@

@@

b bbb

bbb

bbb K̂ ′ -

FK , FK′

K

����

����

��������CCCCCCCCx0

x1 = x2,K′

x2 = x1,K′

x0,K′

K ′b bb b b b

b b

�������

divuh = 1

Figure 3.2. C−1-P2 Lagrange nodes on two neighboringtriangles and the reference mapping.

So, on K̂,

ûh(x̂, ŷ) = x̂ŷ

((0

2x̂− 1

)+ c1

(2x̂2

−3x̂ŷ

)+c2

(2x̂−2ŷ

)+ c3

(3x̂ŷ−2ŷ2

)).

(3.7)

On the other reference triangle K̂ ′ in Figure 3.2, similarly, we have

ûh(x̂, ŷ) = (1− x̂)(1− ŷ)((

2ŷ − 10

)+ c4

(−3(1− x̂)(1− ŷ)

2(1− ŷ)2)

+c5

(−2(1− x̂)2(1− ŷ)

)+ c6

(−2(1− x̂)2

3(1− x̂)(1− ŷ)

)).

(3.8)

We map these two vector functions to triangles K and K ′, preserving thedivergence, noting that they vanish on the outside of two edges of the trian-gles where they are defined. Then we match the interface values of two P4vector functions at the common edge x1x2. The reference mapping from K̂to K is defined in (2.15). In order to keep the divergence, we define the twoPiola transformations in (3.3) and (3.4). When restricted on the commonedge x1x2, i.e., ŷ = 1− x̂, we have

B

(2x̂2c1 + 2x̂c2 + 3(x̂− x̂2)c3

2x̂− 1 + 3(−x̂+ x̂2)c1 + 2(−1 + x̂)c2 + (−2 + 4x̂− 2x̂2)c3

)= −B′

(1

1

)(2x̂2c4 + 2x̂c5 + 3(x̂− x̂2)c6

1− 2x̂+ 3(−x̂+ x̂2)c4 + 2(−1 + x̂)c5 + (−2 + 4x̂− 2x̂2)c6

).

12

Matching coefficients of 1, x̂ and x̂2 of the two components, we have thefollowing 6× 6 system of equations, in block matrix form,

(0 0 00 −2 −2

)A

(0 0 00 −2 −2

)(

0 2 3−3 2 4

)A

(0 2 3−3 2 4

)(2 0 −33 0 −2

)A

(2 0 −33 0 −2

)

c1c2c3c4c5c6

=

(01

)−A

(01

)(

0−2

)−A

(0−2

)(00

)

,

where the common matrix A is defined in (3.5). As x0,K′ is on the straightline x0x1, z2 = 0 in (3.5), and z1 ̸= 0 (unless the two triangles degenerateto a line segment).

A =

(1 + z1 z1

0 1

).

By the row operations of (3.6), it follows that0 0 0 0 −2z1 −2z10 −2 −2 0 −2 −22 2 0 2 + 2z1 2 + 2z1 00 0 0 0 0 02 0 −3 2 + 5z1 0 −3− 5z13 0 −2 3 0 −2

c1c2c3c4c5c6

=−z10z1000

.

From the first equation, we choose c6 = 0 and c5 = 1/2. Then adding thesecond equation to the third equation, we can let c1 = c3 = c4 = 0 in thenew third equation to satisfy it. By the second equation, c2 = −c5 = −1/2.Thus, we find a solution uh which also satisfies, cf. Figure 3.2,∫

Kdivuhdx =

∫K̂(2x̂2 − x̂)dx̂ = 0,(3.9) ∫

K′divuhdx =

∫K̂′(1− ŷ)(1− 2ŷ)dx̂ = 0.(3.10)

Theorem 3.1. For any qh ∈ Ph (2.6), there is a vh ∈ Vh (2.5), such that

divvh = qh and ∥vh∥H1 ≤ C∥qh∥L2 ,(3.11)

where C is independent of h, but dependent on the first level grid Th1.

Proof. As qh ∈ Ph ⊂ L20(Ω), there is an H1 vector v ∈ H10 (Ω)2 such that,by [4, 6],

divv = qh and ∥v∥H1 ≤ C∥qh∥L2 .

Let v1 = Ihv be the Scott-Zhang [20] interpolation of v, by the 26 degreesof freedom in (2.4) (so that the edge flux is preserved.) Then ∥v1∥H1 ≤

13

C∥v∥H1 , and ∫Kdivv1dx =

∫∂K

v1 · nds =∫∂K

v · nds

=

∫Kdivvdx =

∫Kqhdx,

on every triangle K ∈ Th.Let q1 = qh−divv1 ∈ Ph. There are two types of vertices in triangulation

Th, singular points and non-singular points. If s is an internal singular point,shown in Figure 3.3, the other end point of an edge having s as an end point(points s1, s2, s3, s4 in Figure 3.3) cannot be a singular point, as the sum oftwo angles at s is already π. At s, the Ph function q1 has three degrees offreedom, not four, i.e.,

(q1|K1 − q1|K2 + q1|K3 − q1|K4)(s) = 0,(3.12)

cf. Figure 3.3.

s

BBBBBBB

""

""""

""""

BBBBBBB"

"""

""""

""

s1

K1

s2

s3

K2

s1

K4

s4

K3

Figure 3.3. An internal singular point s and its four neigh-boring triangles.

Let v1,2 ∈ (P3(K1)× P3(K2)) ∩H10 (K1 ∪K2) be such that

∇v12(s1) = ∇12v(s2) = ∇12v(s3) = 0, ∂ss2v12 = 1,

where ∂ss2 denotes the directional derivative along the direction from s tos2. That is, on K1,

v1,2 = cλss1λ2s1s2 , c = (∂ss2λss1)

−1λs1s2(s)−2,

where λss1 is the linear function assuming 0 on edge ss1 and 1 at the oppositevertex s2. On K2, v12 is defined symmetrically. Let

v12 = [q1|K1(s)v12]tss2 ,14

where tss2 is the unit (tangent) vector along the direction from s to s2,cf. Figure 3.3. It follows that

divv12|K1(s) = q1|K1(s) [(tss2)1∂x1v12(s) + (tss2)2∂x2v12(s)]= q1|K1(s)∂ss2v12(s) = q1|K1(s).

But, on K2, divv12(s) = q1|K1(s), not matching q1|K2(s) yet. At the restnodes and the node s of (other) triangles, divv12|Kj (si) = 0. Also, due tothe tangent vector in the definition, via integration by parts, we have∫

K1

divv12dx =

∫∂K1

v12 · nds =∫∂K1

0ds = 0.

The integral is also zero on K2. Similarly, we define

v23 = (q1|K2(s)− q1|K1(s))v23tss3 ,v34 = (q1|K3(s)− q1|K2(s) + q1|K1(s))v34tss4 .

Together, we let

vs = v12 + v23 + v34.

By the construction, divvs(s) = q1(s), on K1, K2 and K3. By the constraint(3.12), on K4,

divvs(s) = q1|K3(s)− q1|K2(s) + q1|K1(s) = q1|K4(s).

So the divergence of the constructed vs matches the four values of q1 at thesingular point s. By the scaling argument, we have

∥vs∥H1(Ω) ≤ Ch max1≤i≤4

|q1|Ki(s)| ≤ C∥q1∥L2(∪4i=1Ki).

In the same way, we can define a discrete velocity at each singular point,including the boundary singular points depicted in Figure 2.1. Summingthese velocity functions, we name it v2 such that

divv2(s) =

{q1(s) at all singular vertices on all triangles,

0 at rest vertices on all triangles,∫Kdivv2dx = 0 on all triangles,

∥v2∥H1(Ω) ≤ C∥q1∥L2(Ω).

Let q2,1 = q1 − divv2 ∈ Ph. The rest vertices are non-singular. But thereis a special type of nonsingular vertex, shown in Figure 3.4, that the triangleK and both its neighboring triangles at vertex x, K ′ and K ′′, form a straightline passing through x. The matching at x on K (must be done before thematching on the rest triangles at x) can be done by the above method fortreating the singular vertex. But we give another construction in Lemma3.2. Let x be a non-singular vertex on the middle triangle K, cf. Figure 3.4.

15

x

BBBBBBB

""

""""

""""

BBBBBBB"

"""

""""

""

x1

K

x2

x3

K ′

x1

K ′′

x4

BBBB

Figure 3.4. A non-singular point x in K with two“straight-line” neighboring triangles, K ′ and K ′′.

By lemma 3.2, (3.9) and (3.10), there is a vx,K,K′ ∈ Vh∩H10 (K∪K ′)2, suchthat

divvx,K,K′ |K(x) = divvx,K,K′ |K′(x) = q2,1|K(x),divvx,K,K′ |Ki(xj) = 0 at rest vertices xj on triangles Ki = K,K ′,∫

Ki

divvx,K,K′dx = 0 on all triangles Ki = K,K′,

∥vx,K,K′∥H1(Ω) ≤ Ch∣∣q2,1|K(x)∣∣ ≤ C∥q2,1∥L2(K∪K′).

There is likely no such “half-singular” vertex. But if there are, we sum suchvelocity functions vx,K,K′ and name it v2,1.

Let q2 = q2,1 − divv2,1 ∈ Ph. q2 keeps all properties of q2,1 except hav-ing different nodal values at “half-singular” vertex that q2|K(x) = 0 andq2|K′(x) = q2,1|K′(x)− q2,1|K(x).

For each x of the rest vertices on one of its associated triangle K (itmust have at least one neighboring triangle K ′ not forming a straight linewith K), by Lemma 3.1, there is a vector vx,K ∈ Vh, supported on K andanother neighboring triangle K ′, such that

divvx,K |K(x) = q2|K(x),divvx,K |K1(x1) = 0 at rest vertices x1 and triangles K1,∫

K1

divvx,Kdx = 0 on all triangles K1,(3.13)

∥vx,K∥H1(Ω) ≤ Ch∣∣q2|K(x)∣∣ ≤ C∥q2∥L2(K∪K′).

16

Here (3.13) holds because, in the construction of vx,K in Lemma 3.1, wehave vx,K ∈ C0(K ∪K ′)2 and divvx,K |K′ = 0 so that∫

Kdivvx,Kdx =

∫K∪K′

divvx,Kdx−∫K′

divvx,Kdx

=

∫∂(K∪K′)

vx,K · nds−∫K′

0dx

=

∫∂(K∪K′)

0ds− 0 = 0.

Once more, summing all these vx,K , we get a vector function v3 ∈ Ph, suchthat divv3 matches all non-zero vertex values of q2, and does not destroythe properties by the constructions v1 and v2.

Let q3 = q2 − divv3 ∈ Ph. On each triangle, q3 is P2 polynomialvanishing at three vertices and having mean value

∫K q3dx = 0. On a

triangle K, the dimension of q3 functions is 2, not 3. We construct aP3 vector (bubble) on each triangle so that its divergence matches twoof the mid-edge values, q3(m1) and q3(m2), of q3 on the triangle. LetvK = BK(q3(m1)t1 + q3(m2)t2) where

BK = λ1λ2λ3,

A =(∇BK(m1) ∇BK(m2)

),(

t1 t2)= A−T ,

and λi is the linear function vanishing on edge ei and assuming value 1on the opposite vertex of K. Note that, the matrix A is invertible as thetriangle K is non-singular. By this construction, we have

divvK(m1) = ∂x1BK(m1)(q3(m1)t1 + q3(m2)t2)1

+ ∂x2BK(m1)(q3(m1)t1 + q3(m2)t2)2

= q3(m1)∇BK(m1) · t1 + q3(m2)∇BK(m1) · t2= q3(m1) · 1 + q3(m2) · 0 = q3(m1).

Also divvK(m2) = q3(m2). As BK has two directional derivatives 0 atthe vertices, divvK(xi) = 0 at all three vertices. Also as BK |∂K = 0,∫K divvK =

∫∂K ∇vK · n = 0. That is, the P2 polynomial q3 − divvK ,

on K, has 5 zero values at 5 Lagrange nodes and zero mean value. Thusq3 = divvK on K. Summing all such vK of all triangles K ∈ Th, we let itbe v4 that

divv4 = q3, ∥v4∥H1(Ω) ≤ Ch|q3|L∞(Ω) ≤ C∥q3∥L2(Ω).

Combining the four constructed vectors, we let

vh = v1 + v2 + v2,1 + v3 + v4.

17

Then divvh = ph, and, due to the finite over-lapping,

∥vh∥H1 ≤ ∥v1∥H1 + ∥v2∥H1 + ∥v2,1∥H1 + ∥v3∥H1 + ∥v4∥H1≤ ∥v1∥H1 + ∥v2∥H1 + ∥v2,1∥H1 + ∥v3∥H1 + C∥q3∥L2≤ ∥v1∥H1 + ∥v2∥H1 + ∥v2,1∥H1 + C∥v3∥H1 + C∥q2∥L2≤ C∥v1∥H1 + C∥q1∥L2≤ C∥qh∥L2 .

Because of the homogeneous boundary condition, the divergence operatormay not have a bounded right inverse in an high order Sobolev norm, evenif the solution p is smooth, [4, Theorem 3.1]. So we assume, by [4, Theorem3.1], for the solution p of (2.1), there is a w ∈ H10 (Ω)2 ∩H4(Ω)2 such that

divw = p and ∥w∥H4 ≤ C∥p∥H3 .(3.14)

Theorem 3.2. Let the grids be defined by the multigrid refinement of aninitial grid Th1. The finite element solution (uh, ph) of (2.7) is quasi-optimalin approximating the exact solution (u, p) of the Stokes equation (2.1), as-suming (3.14) holds,

∥u− uh∥H1 + ∥p− ph∥L2 ≤ C infvh∈Vh,qh∈Ph

(∥u− vh∥H1 + ∥p− qh∥L2)

≤ Ch3(∥u∥H4 + ∥p∥H3).

Proof. By (3.11), the following inf-sup condition holds,

infqh∈Ph

supvh∈Vh

(divvh, qh)

∥vh∥H1∥qh∥L2≥ C

with C independent of the grid size h. By the standard theory on saddle-point approximation [5, 6], the quasi-optimality inequality holds. By theScott-Zhang interpolation operator [20], we have

infvh∈Vh

∥u− vh∥H1 ≤ C∥u− Ihu∥H1 ≤ Ch3∥u∥H4 ,

infqh∈Ph

∥p− qh∥L2 ≤ C∥divw − div Ihw∥L2 ≤ Ch3∥w∥H4 ≤ Ch3∥p∥H3 .

4. Numerical tests

We solve the Stokes (2.1) on the unit square Ω = (0, 1)2, where the exactsolution is

u = curl g, p = ∆g, where g = 28x2(1− x)2y2(1− y)2.(4.1)

The first grid is the northwest-southeast cut of the domain, shown in Figure4.1. Then the standard multigrid refinement is applied to generate higherlevels of grids, shown in Figure 4.1. For such grids, there are only two

18

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@

@@@

@@

@@

@

@@@

@@

@@

@

@@@

@@

@@

@

@@@

@@

@@

@

Figure 4.1. The level 1, 2 and 3 uniform grids.

Table 4.1. The errors, eu = uI −uh and ep = pI −ph, andthe order of convergence, by the P3 + B4/P2 finite element(2.5) and (2.6), for the problem (4.1).

∥eu∥L2 hn |eu|H1 hn ∥ep∥L2 hn dimVh1 1.7347540 0.0 14.82305 0.0 168.5918 0.0 432 0.4304411 2.0 6.21139 1.3 5.8976 4.8 1733 0.0328937 3.7 0.98969 2.6 1.1996 2.3 6234 0.0021759 3.9 0.13048 2.9 0.1674 2.8 22895 0.0001381 4.0 0.01654 3.0 0.0200 3.1 8691

singular-points, the northeast corner and the southwest corner, i.e., (1, 1)and (0, 0).

We first apply the newly proposed P+3 -P2 mixed finite element method(2.5). The output data are listed in Table 4.1. As proved in Theorem 3.2,the finite element solution converges at the optimal order.

Table 4.2. The errors, eu = uI −uh and ep = pI −ph, andthe order of convergence, by the P4/P3 Scott-Vogelius finiteelement, for the problem (4.1).

∥eu∥L2 hn |eu|H1 hn ∥ep∥L2 hn dimVh1 2.161128 0.0 16.855690 0.0 93.254046 0.0 502 0.156839 3.8 2.598032 2.7 9.471127 3.3 1623 0.006195 4.7 0.224538 3.5 1.065502 3.2 5784 0.000166 5.2 0.013377 4.1 0.055986 4.3 21785 0.000004 5.3 0.000760 4.1 0.003264 4.1 8450

In the second numerical test, we compute the solution (4.1) again by theScott-Vogelius P4-P3 element, the lowest order stable element of [18]. Theerror and the order of convergence are listed in Table 4.2. The optimal orderof convergence is obtained there.

In the third numerical test, we use an unstable mixed finite, the contin-uous P3 velocity with discontinuous P2 pressure (subset, Ph = div(C

0-P3)),to solve the Stokes equation (4.1). The resulting linear system of equationsis solved by the iterated-penalty method, cf. [26, 27]. The error and the

19

Table 4.3. The errors, eu = uI −uh and ep = pI −ph, andthe order of convergence, by the P3-P2 finite element, for theproblem (4.1).

∥eu∥L2 hn |eu|H1 hn ∥ep∥L2 hn dimVh1 1.424589 0.0 10.660649 0.0 12.860813 0.0 282 0.325989 2.1 5.254361 1.0 6.636293 1.0 703 0.062660 2.4 2.033926 1.4 3.308077 1.0 2144 0.006759 3.2 0.408822 2.3 1.374595 1.3 7425 0.000787 3.1 0.090568 2.2 0.625108 1.1 27586 0.000096 3.0 0.021744 2.1 0.297753 1.1 10630

order of convergence are listed in Table 4.3. In this case, the pressure con-verges two orders below the optimal order, to the true solution. The finiteelement velocity is one order sub-optimal. But as the inf-sup condition fails,we do not have a theory to cover the above observation, i.e., the sub-optimalconvergence is not proved.

Acknowledgment

We thank an anonymous referee for pointing out Remark 3.1, and Figure3.4 (which was not discussed in an earlier version of the paper.)

This research is partially supported by the NSFC project 11571023.

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Department of Mathematical Sciences, University of Delaware, Newark,DE 19716, USA. szhang@udel.edu.

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