# Nonconforming finite element Stokes complexes in three dimensions

Two nonconforming finite element Stokes complexes ended with the nonconforming P_1-P_0 element for the Stokes equation in three dimensions are constructed. And commutative diagrams are also shown by combining nonconforming finite element Stokes complexes and interpolation operators. The lower order H(grad curl)-nonconforming finite element only has 14 degrees of freedom, whose basis functions are explicitly given in terms of the barycentric coordinates. The H(grad curl)-nonconforming elements are applied to solve the quad-curl problem, and optimal convergence is derived. By the nonconforming finite element Stokes complexes, the mixed finite element methods of the quad-curl problem is decoupled into two mixed methods of the Maxwell equation and the nonconforming P_1-P_0 element method for the Stokes equation, based on which a fast solver is developed.

## Authors

• 11 publications
• ### A family of finite element Stokes complexes in three dimensions

We construct finite element Stokes complexes on tetrahedral meshes in th...
08/09/2020 ∙ by Kaibo Hu, et al. ∙ 0

• ### A weak Galerkin-mixed finite element method for the Stokes-Darcy problem

In this paper, we propose a new numerical scheme for the coupled Stokes-...
02/28/2021 ∙ by Hui Peng, et al. ∙ 0

• ### New stabilized P_1× P_0 finite element methods for nearly inviscid and incompressible flows

This work proposes a new stabilized P_1× P_0 finite element method for s...
09/09/2021 ∙ by Yuwen Li, et al. ∙ 0

• ### Global and local pointwise error estimates for finite element approximations to the Stokes problem on convex polyhedra

The main goal of the paper is to show new stability and localization res...
07/16/2019 ∙ by Niklas Behringer, et al. ∙ 0

• ### A geometric multigrid method for space-time finite element discretizations of the Navier-Stokes equations and its application to 3d flow simulation

We present a parallelized geometric multigrid (GMG) method, based on the...
07/22/2021 ∙ by Mathias Anselmann, et al. ∙ 0

• ### Preconditioning immersed isogeometric finite element methods with application to flow problems

Immersed finite element methods generally suffer from conditioning probl...
08/11/2017 ∙ by Frits de Prenter, et al. ∙ 0

• ### Adaptive geometric multigrid for the mixed finite cell formulation of Stokes and Navier-Stokes equations

Unfitted finite element methods have emerged as a popular alternative to...
03/18/2021 ∙ by S. Saberi, et al. ∙ 0

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## 1. Introduction

In this paper we shall consider nonconforming finite element discretization of the following Stokes complex in three dimensions

 (1.1) R\autorightarrow$⊂$H1(Ω)\autorightarrow$∇$H(gradcurl,Ω)\autorightarrow$curl$H1(Ω;R3)\autorightarrow$div$L2(Ω)\autorightarrow0,

where . Conforming finite element Stokes complexes on triangles and rectangles in two dimensions are devised in [22, 33]. And a conforming virtual element discretization of the Stokes complex (1.1) in three dimensions is advanced in [4]. To the best of our knowledge, there is no finite element discretization of the Stokes complex (1.1) in three dimensions in literature. Recently -conforming finite elements in three dimensions are constructed with in [34], whose space of shape functions includes all the polynomials with degree no more than . The number of the degrees of freedom for the lowest-order element in [34] is . Due to the large dimension of the conforming element spaces, nonconforming elements to discretize in three dimensions are preferred. The -nonconforming Zheng-Hu-Xu element in [37] only has degrees of freedom.

We will construct an -nonconforming finite element possessing fewer degrees of freedom than those of the Zheng-Hu-Xu element, but preserving the same approximation error in energy norm. Then we build up the nonconforming finite element Stokes complexes from the new -nonconforming finite element and the Zheng-Hu-Xu element. The finite element discretization of in the Stokes complex (1.1) should be a stable divergence-free pair for the Stokes equation, which suggests us to use the nonconforming linear element and piecewise constant to discretize and respectively. On the other hand, the direct sum decomposition  [1, 2] implies the curl operator is injective. This motivates us to take the space of shape functions with . Note that is exactly the space of shape functions of the Zheng-Hu-Xu element. The dimension of is , which is six fewer than the dimension of . The degrees of freedom for are given by

 ∫ev⋅teds on each e∈E(K), ∫F(curlv)×nds on each F∈F(K).

By comparing the degrees of freedom, the lower order nonconforming element for is very similar as the Morley-Wang-Xu element [32] for . The explicit expressions of the basis functions of are shown in terms of the barycentric coordinates. Furthermore, we construct the commutative diagram

 (1.2)

The -nonconforming element together with the Lagrange element is applied to solve the quad-curl problem. The discrete Poincaré inequality is established for the -nonconforming element space , as a result the coercivity on the weak divergence-free space follows. Then we acquire the discrete stability of the bilinear form from the evident discrete inf-sup condition, and derive the optimal convergence of the nonconforming mixed finite methods. Since the interpolation operator is not well-defined on , in the error analysis we exploit a quasi-interpolation operator defined on , which is constructed by combining a regular decomposition for the space , and the Scott-Zhang interpolation operator [30].

By the nonconforming finite element Stokes complex in the bottom line of the commutative diagram (1.2), we equivalently decouple the mixed finite element methods of the quad-curl problem into two mixed methods of the Maxwell equation and the nonconforming - element method for the Stokes equation, as the decoupling of the quad-curl problem in the continuous level [9, 36]. A fast solver based on this equivalent decoupling is developed for the mixed finite element methods of the quad-curl problem.

In addition to the Stokes complex (1.1), another kind of Stokes complex is  [24]

 (1.3) R\autorightarrow⊂H2(Ω)\autorightarrowcurlH1(Ω;R2)\autorightarrowdivL2(Ω)\autorightarrow0

in two dimensions, and

 (1.4) R\autorightarrow$⊂$H2(Ω)\autorightarrow$∇$H1(curl,Ω)\autorightarrow$curl$H1(Ω;R3)\autorightarrow$div$L2(Ω)\autorightarrow0

in three dimensions, where . We refer to [16, 26, 3, 25, 17, 19, 20, 35, 18] for some finite element discretizations of the Stokes complex (1.3) in two dimensions, and [31, 19, 29] for some finite element discretizations of the Stokes complex (1.4) in three dimensions. While the finite elements corresponding to the Stokes complexes (1.3)-(1.4) are not suitable to discretize the quad-curl problem, since is not a subspace of .

The rest of this paper is organized as follows. In Section 2, we devise a lower order -nonconforming finite element. Nonconforming finite element Stokes complexes are developed in Section 3. In Section 4, we propose the nonconforming mixed finite element methods for the quad- problem. And the decoupling of the mixed finite element methods and a fast solver are discussed in Section 5.

## 2. The H(gradcurl)-nonconforming finite elements

In this section we will present -nonconforming finite elements.

### 2.1. Notation

Given a bounded domain and a nonnegative integer , let be the usual Sobolev space of functions on , and

the vector version of

. The corresponding norm and semi-norm are denoted, respectively, by and . Let be the standard inner product on or . If is , we abbreviate , and by , and , respectively. Denote by the closure of with respect to the norm . Let stand for the set of all polynomials in with the total degree no more than , and be the vector version of . Let be the -orthogonal projector, and its vector version is denoted by . Set . The gradient operator, curl operator and divergence operator are denoted by , and respectively. And define Sobolev spaces , , , and in the standard way.

Assume is a contractible polyhedron. Let be a regular family of tetrahedral meshes of . For each element , denote by the unit outward normal vector to , which will be abbreviated as for simplicity. Let , , and be the union of all faces, interior faces, edges and vertices of the partition , respectively. We fix a unit normal vector for each face , and a unit tangent vector for each edge . For any , denote by , and the set of all faces, edges and vertices of , respectively. For any , let be the set of all edges of . And for each , denote by the unit vector being parallel to and outward normal to . Set , where is the exterior product. For elementwise smooth function , define

 ∥v∥21,h:=∑K∈Th∥v∥21,K,|v|21,h:=∑K∈Th|v|21,K.

Let , and be the elementwise version of , and with respect to .

### 2.2. Nonconforming finite elements

We focus on constructing nonconforming finite elements for the space in this subsection. To this end, recall the direct sum of the polynomial space [1, 2]

 (2.1) Pk(K;R3)=∇Pk+1(K)⊕x×Pk−1(K;R3)∀ K∈Th.

The decomposition (2.1) implies that is injective. We intend to use the nonconforming linear element to discretize , then the decomposition (2.1) and the complex (1.1) motivate us that the space of shape functions to discrete should include . The direct sum in (2.1) also suggests to enrich with for some positive integer to get the space of shape functions. Hence for each , define the space of shape functions as

 Wk(K):=∇Pk+1(K)⊕x×P1(K;R3) for k=0,1.

By the decomposition (2.1), we have , and

 dimWk(K)={14,k=0,20,k=1.

Then choose the following local degrees of freedom

 (2.2) ∫ev⋅teqds ∀ q∈Pk(e) on each e∈E(K), (2.3) ∫F(curlv)×nds on each F∈F(K).

The degrees of freedom (2.2)-(2.3) are inspired by the degrees of freedom of nonconforming linear element and the Nédélec element [27, 28]. Note that the triple is exactly the nonconforming finite element in [37]. Here we embed this nonconforming finite element into the discrete Stokes complex. And we also construct the lowest order triple .

###### Lemma 2.1.

The degrees of freedom (2.2)-(2.3) are unisolvent for the shape function space .

###### Proof.

Notice that the number of the degrees of freedom (2.2)-(2.3) is same as the dimension of . It is sufficient to show that for any with vanishing degrees of freedom (2.2)-(2.3).

For each , applying the integration by parts on face , we get from the vanishing degrees of freedom (2.2) that

 ∫F(curlv)⋅nFds =∫Fdiv(v×nF)ds=∑e∈E(F)∫e(v×nF)|F⋅nF,eds =∑e∈E(F)∫ev⋅(nF×nF,e)ds=∑e∈E(F)∫ev⋅tF,eds=0,

which together with the vanishing degrees of freedom (2.3) implies

 ∫Fcurlvds=0.

Since , we acquire from the unisolvence of the nonconforming linear element that . Thus there exists such that . By the vanishing degrees of freedom (2.2), it holds , which implies that we can choose such that for each . Noting that , we acquire and . ∎

By comparing the degrees of freedom, the lower order nonconforming element for is very similar as the Morley-Wang-Xu element [32] for .

### 2.3. Basis functions

We will figure out the basis functions of in the subsection. We refer to [37] for the basis functions of . Let , , and be the barycentric coordinates of point with respect to the vertices and of the tetrahedron respectively. Let be the face of opposite to . And the vertices of denoted by , and with . Set , which is a tangential vector to the edge with vertices and , and similarly define other tangential vectors with different subscripts. For ease of presentation, let

 Meij(v):=1|eij|∫eijv⋅tijds,MFl(v):=∫Fl(curlv)×nds,
 MFl,1(v):=1|Fl|(tl1l2×tl1l3)⋅nFl∫Fl(curlv)⋅(tl1l3×nFl)ds,
 MFl,2(v):=1|Fl|(tl1l3×tl1l2)⋅nFl∫Fl(curlv)⋅(tl1l2×nFl)ds.

The degrees of freedom and are equivalent to , i.e. (2.3).

#### 2.3.1. Basis functions corresponding to the face degrees of freedom

Define

 φFl,1 :=12(2λl+cl−1)x×tl1l2−12nl3⋅tl1l3(cl−1)xl1⋅(tl1l2×tl1l3)nl3 −12nl⋅tl1lclxl1⋅(tl1l2×tl1l)nl,
 φFl,2 :=12(2λl+cl−1)x×tl1l3−12nl2⋅tl1l2(cl−1)xl1⋅(tl1l3×tl1l2)nl2 −12nl⋅tl1lclxl1⋅(tl1l3×tl1l)nl,

where constant . We will show that and are the basis functions being dual to and .

###### Lemma 2.2.

It holds

 (2.4) (clxl+(1−cl)xl1)⋅nl=0.
###### Proof.

Since is constant, we have

 cl=−xl1⋅∇λl=1−xl⋅∇λl.

Hence it follows

 (clxl+(1−cl)xl1)⋅∇λl =clxl⋅∇λl+(1−cl)xl1⋅∇λl =cl(1−cl)−(1−cl)cl=0.

Therefore the identity (2.4) follows from the fact is parallel to . ∎

###### Lemma 2.3.

Functions and are the basis functions of being dual to and , respecrtively. That is

 (2.5) Me(φFl,1)=Me(φFl,2)=0∀ e∈E(K),
 (2.6) MF(φFl,1)=MF(φFl,2)=0∀ F∈F(K)∖{Fl},
 (2.7) MFl,2(φFl,1)=MFl,1(φFl,2)=0,MFl,1(φFl,1)=MFl,2(φFl,2)=1.
###### Proof.

Apparently . Since , , , and is constant on , we get

 Mel2l3(φFl,1)=12∫el2l3((2λl+cl−1)(xl1×tl1l2)⋅tl2l3−(cl−1)xl1⋅(tl1l2×tl1l3))ds=0.

Again, from , and (2.4), it follows

 Mel3l(φFl,1) =12∫el3l((2λl+cl−1)(xl×tl1l2)⋅tl3l+(cl−1)xl1⋅(tl1l2×tl1l3))ds −12∫el3lclxl⋅(tl1l2×tl1l)ds =12∫el3l(cl(xl×tl1l2)⋅tl3l1+(cl−1)xl1⋅(tl1l2×tl1l3))ds =12∫el3l(clxl+(1−cl)xl1)⋅(tl1l2×tl3l1)ds=0.

Similarly we can show that for other edges and . Hence (2.5) holds.

On the other hand, by the identity , we have

 curlφFl,1=12curl((2λl+cl−1)x×tl1l2)=(1−3λl)tl1l2,
 curlφFl,2=12curl((2λl+cl−1)x×tl1l3)=(1−3λl)tl1l3.

We conclude (2.6)-(2.7) by the fact that is the basis function of the nonconforming element. ∎

#### 2.3.2. Basis functions corresponding to the edge degrees of freedom

Next we construct the basis function corresponding to the degree of freedom . Recall the basis function of the lowest order Nédélec element of the first kind . Thanks to (2.5)-(2.6), function can be modified by and to derive the basis function of corresponding to the degree of freedom .

###### Lemma 2.4.

Let

 φeij:=λi∇λj−λj∇λi+4∑l=1(cijl,1φFl,1+cijl,2φFl,2)

with constants

 cijl,1:=2(tl1l2×tl1l3)⋅nFl(∇λj×∇λi)⋅(tl1l3×nFl),
 cijl,2:=2(tl1l3×tl1l2)⋅nFl(∇λj×∇λi)⋅(tl1l2×nFl).

Then

 Meij(φeij)=1,Me(φeij)=0,MF(φeij)=0.

for each and .

###### Proof.

The identities and follow from (2.5) and the fact

 Meij(λi∇λj−λj∇λi)=1,Me(λi∇λj−λj∇λi)=0∀ e∈E(K)∖{eij}.

On the other hand, we get from that

 MFl,r(φeij)=MFl,r(λi∇λj−λj∇λi)+cijl,r=0

for . ∎

In summary, we arrive at the basis functions being dual to the degrees of freedom , and :

1. Two basis functions on each face ()

 φFl,1 =12(2λl+cl−1)x×tl1l2−12nl3⋅tl1l3(cl−1)xl1⋅(tl1l2×tl1l3)nl3 −12nl⋅tl1lclxl1⋅(tl1l2×tl1l)nl,
 φFl,2 =12(2λl+cl−1)x×tl1l3−12nl2⋅tl1l2(cl−1)xl1⋅(tl1l3×tl1l2)nl2 −12nl⋅tl1lclxl1⋅(tl1l3×tl1l)nl,

where constant .

2. One basis function on each edge ()

 φeij=λi∇λj−λj∇λi+4∑l=1(cijl,1φFl,1+cijl,2φFl,2)

with constants

 cijl,1=2(tl1l2×tl1l3)⋅nFl(∇λj×∇λi)⋅(tl1l3×nFl),
 cijl,2=2(tl1l3×tl1l2)⋅nFl(∇λj×∇λi)⋅(tl1l2×nFl).

## 3. Nonconforming finite element Stokes complexes

We will consider the nonconforming finite element discretization of the Stokes complex (1.1) in this section. The homogeneous version of the Stokes complex (1.1) is

where . Since , it holds , where

 H0(curl2,Ω):={v∈H0(curl,Ω):curlv∈H0(curl,Ω)}.

We can use the Lagrange element, the nonconforming linear element and piecewise constant to discretize , and in the Stokes complex (1.1), respectively. Take the Lagrange element space

 Vgh:={vh∈H1(Ω):vh|K∈Pk+1(K) % for each K∈Th}

with , the nonconforming linear element space

 Vsh:={vh∈L2(Ω;R3): vh|K∈P1(K;R3) for each K∈Th, and ∫F⟦vh⟧ds=0 for each F∈Fih},

and the piecewise constant space

 Qh:={qh∈L2(Ω):qh|K∈P0(K) for each K∈Th}.

Here is the jump of across . Define the global -nonconforming element space

 Wh:={vh∈L2(Ω;R3): vh|K∈Wk(K) for % each K∈Th, and all the degrees of freedom (???)-(???) are single-valued}.

According to the proof of Lemma 2.1, it holds

 (3.1) ∫F⟦curlvh⟧ds=0∀ vh∈Wh,F∈Fih.

To prove the exactness of the nonconforming discrete Stokes complexes, we need the help of the Nédélec element spaces [27, 28]

 Vch:={vh∈H(curl,Ω):vh|K∈Vck(K) for % each K∈Th},

where with . The degrees of freedom for are

 (3.2) ∫ev⋅teqds ∀ q∈Pk(e) on each e∈E(K).

It is observed that the degrees of freedom (3.2) are exactly same as (2.2). By the finite element de Rham complexes [1, 2], we have

 (3.3) Vch∩ker(curl)=∇Vgh.

The notation means the kernel space of the operator .

###### Lemma 3.1.

It holds

 Wh∩ker(curlh)=∇Vgh.
###### Proof.

Apparently we have