A note on the optimal degree of the weak gradient of the stabilizer free weak Galerkin finite element method
AA NOTE ON THE OPTIMAL DEGREE OF THE WEAK GRADIENTOF THE STABILIZER FREE WEAK GALERKIN FINITE ELEMENTMETHOD
AHMED AL-TAWEEL ∗ AND
XIAOSHEN WANG † Abstract.
Recently, a new stabilizer free weak Galerkin method (SFWG) is proposed, which iseasier to implement. The idea is to raise the degree of polynomials j for computing weak gradient. Itis shown that if j ≥ j for some j , then SFWG achieves the optimal rate of convergence. However,large j will cause some numerical difficulties. To improve the efficiency of SFWG and avoid numericallocking, in this note, we provide the optimal j with rigorous mathematical proof. Key words. weak Galerkin finite element methods, weak gradient, Error estimates.
AMS subject classifications.
Primary: 65N15, 65N30; Secondary: 35J50
1. Introduction.
In this note, similar to [1], we will consider the following Pois-son equation − ∆ u = f in Ω , (1.1) u = 0 on ∂ Ω , (1.2)as the model problem, where Ω is a polygonal domain in R . The variational formu-lation of the Poisson problem (1.1)-(1.2) is to seek u ∈ H (Ω) such that( ∇ u, ∇ v ) = ( f, v ) , ∀ v ∈ H (Ω) . (1.3)A stabilizer free weak Galerkin finite element method is proposed by Ye and Zhangin [1] as a new method for the solution of Poisson equation on polytopal meshes in2D or 3D. Let j be the degree of polynomials for the calculation of weak gradient. Ithas been proved in [1] that there is a j > j ≥ j . However, when j is too large, themagnitude of the weak gradient can be extremely large, causing numerical instability.In this note, we provide the optimal j to improve the efficiency and avoid unnecessarynumerical difficulties, which has mathematical and practical interests.
2. Weak Galerkin Finite Element Schemes.
Let T h be a partition of thedomain Ω consisting of polygons in 2D. Suppose that T h is shape regular in the sensedefined by (2.7)-(2.8) and each T is convex. Let E h be the set of all edges in T h , andlet E h = E h \ ∂ Ω be the set of all interior edges. For each element T ∈ T h , denote by h T the diameter of T , and h = max T ∈T h h T the mesh size of T h . ∗ Department of Mathematics, University of Arkansas at Little Rock, Little Rock, AR 72204(email: [email protected]) † Department of Mathematics, University of Arkansas at Little Rock, Little Rock, AR 72204(email:[email protected] ) 1 a r X i v : . [ m a t h . NA ] J u l On T , let P k ( T ) be the space of all polynomials with degree no greater than k .Let V h be the weak Galerkin finite element space associated with T ∈ T h defined asfollows: V h = { v = { v , v b } : v | T ∈ P k ( T ) , v b | e ∈ P k ( e ) , e ∈ ∂T, T ∈ T h } , (2.1)where k ≥ v symbolizes theinterior value of v , and the component v b symbolizes the edge value of v on each T and e , respectively. Let V h be the subspace of V h defined as: V h = { v : v ∈ V h , v b = 0 on ∂ Ω } . (2.2)For any v = { v , v b } ∈ V h + H (Ω) , the weak gradient ∇ w v ∈ [ P j ( T )] is definedon T as the unique polynomial satisfying( ∇ w v, q ) T = − ( v , ∇ · q ) T + (cid:104) v b , q · n (cid:105) ∂T , ∀ q ∈ [ P j ( T )] , j ≥ k. (2.3)where n is the unit outward normal vector of ∂T .For simplicity, we adopt the following notations,( v, w ) T h = (cid:88) T ∈T h ( v, w ) T = (cid:88) T ∈T h (cid:90) T vwdx, (cid:104) v, w (cid:105) ∂ T h = (cid:88) T ∈T h (cid:104) v, w (cid:105) ∂T = (cid:88) T ∈T h (cid:90) ∂T vwdx. For any v = { v , v b } and w = { w , w b } in V h , we define the bilinear forms asfollows: A ( v, w ) = (cid:88) T ∈T h ( ∇ w v, ∇ w w ) T . (2.4) Stabilizer free Weak Galerkin Algorithm 1.
A numerical solution for(1.1)-(1.2) can be obtain by finding u h = { v , v b } ∈ V h , such that the followingequation holds A ( u h , v ) = ( f, v ) , ∀ v = { v , v b } ∈ V h . (2.5) where A ( · , · ) is defined in (2.4). Accordingly, for any v ∈ V h , we define an energy norm ||| · ||| on V h as:(2.6) ||| v ||| = (cid:88) T ∈T h ( ∇ w v, ∇ w v ) T = (cid:88) T ∈T h (cid:107)∇ w v (cid:107) T . An H norm on V h is defined as: (cid:107) v (cid:107) ,h = (cid:88) T ∈T h (cid:0) (cid:107)∇ v (cid:107) T + h − T (cid:107) v − v b (cid:107) ∂T (cid:1) . The following well known lemma and corollary will be needed in proving our mainresult.
Lemma 2.1.
Suppose that D ⊂ D are convex regions in R d such that D canbe obtain by scaling D with a factor r > . Then for any p ∈ P n ( D ) , (cid:107) p (cid:107) D ≤ (cid:107) p (cid:107) D ≤ Cr n (cid:107) p (cid:107) D , where C depends only on d and n . Corollary 2.1.
Suppose that D and D satisfy the conditions of Lemma 2.1and D ⊆ D ⊆ D . Then (cid:107) p (cid:107) D ≤ (cid:107) p (cid:107) D ≤ Cr n (cid:107) p (cid:107) D , ∀ p ∈ P n ( D ) , where C depends only on d and n . The following Theorem shows that for certain j >
0, whenever j ≥ j , (cid:107) · (cid:107) ,h isequivalent to the ||| · ||| defined in (2.6), which is crucial in establishing the feasibilityof SFWG. Later on, we will show that j is optimal. Theorem 2.2. (cf.[1], Lemma 3.1) Suppose that ∀ T ∈ T h , T is convex with atmost m edges and satisfies the following regularity conditions: for all edges e t and e s of T | e s | < α | e t | ;(2.7) for any two adjacent edges e t and e s the angle θ between them satisfies θ < θ < π − θ , (2.8) where ≤ α and θ > are independent of T and h . Let j = k + m − or j = k + m − when each edge of T is parallel to another edge of T . When deg ∇ w v = j ≥ j ,then there exist two constants C , C > , such that for each v = { v , v b } ∈ V h , thefollowing hold true C (cid:107) v (cid:107) ,h ≤ ||| v ||| ≤ C (cid:107) v (cid:107) ,h , where C and C depend only on α and θ .Proof . For simplicity, from now on, all constant are independent of T and h ,unless otherwise mentioned. They may depend on k, α , and θ . Suppose ∇ w v ∈ [ P j ( T )] , v ∈ P k ( T ). We know that( ∇ w v, (cid:126)q ) T = ( ∇ v, (cid:126)q ) T + (cid:104) v b − v , (cid:126)q · (cid:126)n (cid:105) ∂T ∀ (cid:126)q ∈ [ P j ( T )] . (2.9)Suppose ∂T = ( ∪ m − i =0 e i ), and we want to construct a (cid:126)q ∈ [ P j ( T )] ⊆ [ P j ( T )] , where j ≤ j , so that ( (cid:126)q, p ) T = 0 , ∀ p ∈ [ P k − ( T )] , (2.10) (cid:126)q · (cid:126)n i = 0 on e i , i = 1 , . . . , m − , (2.11) (cid:104) v b − v − (cid:126)q · n , p (cid:105) e = 0 , p ∈ P k ( e ) , (2.12)where (cid:126)n i is the unit outer normal to e i . Without loss of generality, we may assumethat e = { ( x, | x ∈ (0 , } , ∀ ( x, y ) ∈ T, y >
0, and one of the vertices of e is(0 , (cid:96) i ∈ P ( T ) , i = 2 , . . . , m − (cid:96) i ( e i ) = 0 , (cid:96) i ( p ) ≥ , ∀ p ∈ T, and max p ∈ T (cid:96) i ( p ) = 1. Let (cid:126)q = (cid:96) . . . (cid:96) m − Q (cid:126)t = LQ (cid:126)t = Q(cid:126)t, (2.13)where (cid:126)t is a unit tangent vector to e and Q ∈ P j − m +2 ( T ). It is easy to see that (cid:126)q · (cid:126)n i | e i = 0 , i = 1 , . . . , m −
1. We want to find Q so that (cid:10) v b − v − (cid:96) . . . (cid:96) m Q ( (cid:126)t · (cid:126)n ) , p (cid:11) e = 0 , ∀ p ∈ P j − m +2 ( e ) , (2.14) ( (cid:96) . . . (cid:96) m Q , p ) T = 0 , ∀ p ∈ P k − ( T ) . (2.15)Let j − m + 2 = k or j = k + m −
2. It is easy to see that if Q satisfies (2.14)-(2.15), then (cid:126)q satisfies (2.10)-(2.12). In addition, (2.14)-(2.15) has ( k + 1)( k + 2) / Q ∈ P k ( T ) we set v b − v = 0. Setting p = Q in (2.14) yields (cid:104) (cid:96) . . . (cid:96) m − Q , Q (cid:105) e = 0 . Since L = (cid:96) . . . (cid:96) m − > e , Q ( x, ≡
0. Thus Q = yQ , where Q ∈ P k − ( T ). Similarly ( y(cid:96) . . . (cid:96) m − Q , p ) T = 0 ∀ p ∈ P k − ( T ) , implies Q = 0 and thus Q = 0. Note that for any j ≥ j , since (cid:126)q ∈ [ P j ( T )] ⊂ [ P j ( T )] , there exists at least one such (cid:126)q in [ P j ( T )] . Note also that when every edgeof ∂T is parallel to another one, we can lower the number of (cid:96) i in (2.13) by one. Thusfor such a T , we have j = m + k −
3. Next, we want to show that the unique solutionof (2.10)-(2.12) satisfies (cid:107) (cid:126)q (cid:107) T ≤ γ (cid:107) (cid:126)q · (cid:126)n (cid:107) e , (2.16)for some γ >
0. Note that( (cid:126)q · (cid:126)n ) = Q sin θ = (cid:126)q · (cid:126)q sin θ ≥ (cid:126)q · (cid:126)q sin θ , where θ is the angle between e and e . Thus, (2.16) is equivalent to (cid:107) Q (cid:107) T ≤ γ (cid:107) Q (cid:107) e , (2.17)for some γ > e = { (0 , y ) | ≤ y ≤ } , because of the assumption (2.7). Now let T r be the triangle spanned by e and e and label the third edge of T r as e t . Let ˆ T r ⊂ T r be the triangle with ver-tices (0 , . , (0 , .
75) and (0 . , . T r connecting (0 , .
75) and(0 . , .
25) as ˆ e t . For i = 2 , . . . , m −
1, let d i = dist ( ˆ T r , L i ) ,D i = max p ∈ T dist ( p, L i ) , where L i is the line containing e i . Then d i ≥ dist (ˆ e t , e t ) = √ / ,D i ≤ ( | e | + | e | + · · · + | e m − | ) / ≤ mα . (2.18)Thus ∀ p ∈ ˆ T r , (cid:96) i ( p ) ≥ √ mα = ε, i = 2 , . . . , m − . (2.19)Now we will prove (2.17) in 3 steps: I : β (cid:107) Q (cid:107) e ≤ (cid:107) Q (cid:107) e ,II : β (cid:107) Q (cid:107) T r ≤ (cid:107) Q (cid:107) e ,III : β (cid:107) Q (cid:107) T ≤ (cid:107) Q (cid:107) T r . I. Let ˆ e = { ( x, | ≤ x ≤ . } . Then similar to (2.19), ∀ p ∈ ˆ e, (cid:96) i ( p ) ≥ ε, i =2 . . . . , m −
1. Thus ε m − (cid:107) Q (cid:107) ˆ e ≤ (cid:107) Q (cid:107) ˆ e ≤ (cid:107) Q (cid:107) e . It follows from Corollary 2.1 that (cid:107) Q (cid:107) e ≥ ε m − (cid:107) Q (cid:107) ˆ e ≥ Cε m − (cid:107) Q (cid:107) e = β (cid:107) Q (cid:107) e . (2.20)II. It is easy to see that (cid:107) Q (cid:107) T r = (cid:90) (cid:90) − y L Q dxdy = (cid:90) (cid:90) − y L ( Q − Q ( x,
0) + Q ( x, dxdy ≤ (cid:90) (cid:90) − y (cid:104) L ( Q − Q ( x, + L ( Q ( x, (cid:105) dxdy. It follows from (2.19) and Corollary 2.1 that (cid:107) LQ ( x, (cid:107) T r = (cid:90) (cid:90) − y L ( Q ( x, dxdy ≤ (cid:90) (cid:90) ( Q ( x, dxdy = (cid:107) Q (cid:107) e (2.21)Note that Q − Q ( x,
0) = y ¯ Q, ¯ Q ∈ P k − , and (cid:90) T (cid:2) LQ ( x,
0) + Ly ¯ Q (cid:3) P dA = 0 , ∀ p ∈ P k − . (2.22) Since Ly ≥ ε for ( x, y ) ∈ ˆ T r and Ly ≥ , ∀ ( x, y ) ∈ T , (cid:90) T Ly ¯ Q dA ≥ ε (cid:90) ˆ T r ¯ Q dA = δ (cid:107) ¯ Q (cid:107) T r . (2.23)Since (cid:107) · (cid:107) ˆ T r and (cid:107) · (cid:107) T r are equivalent norms on P k − ( T r ), δ (cid:107) ¯ Q (cid:107) ˆ T r ≥ β (cid:107) ¯ Q (cid:107) T r . (2.24)Let T ce be the circumscribed isosceles-right triangle obtained by scaling T r . Itis easy to see that diam ( T ce ) ≤ mα . (2.25)Since T r ⊆ T ⊆ T ce , it follows from (2.23), (2.24), and Corollary 2.1 that (cid:90) T Ly ¯ Q dA ≥ β (cid:107) ¯ Q − (cid:107) T r ≥ δ (cid:107) ¯ Q (cid:107) T , where δ depend on (2.25) and k . Letting p = ¯ Q in (2.22) yields (cid:90) T Ly ¯ Q dA = − (cid:90) T LQ ( x,
0) ¯
QdA ≤ (cid:107) Q ( x, (cid:107) T (cid:107) ¯ Q (cid:107) T , and thus (cid:107) ¯ Q (cid:107) T ≤ δ (cid:107) Q ( x, (cid:107) T . Then (cid:107) ¯ Q (cid:107) T ≤ δ (cid:90) N (cid:90) M Q ( x, dxdy = CN (cid:90) M Q ( x, dx = C (cid:107) Q (cid:107) ,M ] , where N = max ( x,y ) ∈ T y ≤ mα and M = max ( x,y ) ∈ T xε ≤ mα . It followsfrom Corollary 2.1 and (2.23) (cid:107) Ly ¯ Q (cid:107) ˆ T r ≤ (cid:107) ¯ Q (cid:107) T ≤ C (cid:107) Q (cid:107) e . (2.26)Combining (2.26) and (2.21) yields (cid:107) Q (cid:107) T r ≤ C (cid:107) Q (cid:107) e . (2.27)III. Applying Corollary 2.1 again yields C (cid:107) Q (cid:107) T ≤ (cid:107) Q (cid:107) T r . (2.28)Now we have completed the 3-step argument.By a scaling argument, we have (cid:107) Q (cid:107) T ≤ γ h / (cid:107) Q (cid:107) e . (2.29)Plugging (cid:126)q into (2.9) yields (cid:107) Q (cid:107) T (cid:107)∇ w v (cid:107) T ≥ ( ∇ w v, (cid:126)q(cid:126)n ) ˆ T = (cid:104) v b − v , (cid:126)q(cid:126)n · (cid:126)n (cid:105) e = (cid:107) v b − v (cid:107) e = (cid:107) v b − v (cid:107) e (cid:107) Q (cid:107) e . (2.30)It follows from (2.29) that (cid:107)∇ w v (cid:107) T ≥ Ch / (cid:107) v b − v (cid:107) e . Similarly, (cid:107) v b − v (cid:107) e i ≤ Ch / T (cid:107)∇ w v (cid:107) T , i = 1 , . . . , m − . Thus (cid:107) v b − v (cid:107) ∂T ≤ Ch / T (cid:107)∇ w v (cid:107) T . By letting q = ∇ w v in (2.9), we arrive at( ∇ w v, ∇ w v ) T = ( ∇ v , ∇ w v ) T + (cid:104) v b − v , ∇ w v · n (cid:105) ∂T . It follows from the trace inequality and the inverse inequality that (cid:107)∇ w v (cid:107) T ≤ (cid:107)∇ v (cid:107) T (cid:107)∇ w v (cid:107) T + (cid:107) v − v b (cid:107) ∂T (cid:107)∇ w v (cid:107) ∂T ≤ (cid:107)∇ v (cid:107) T (cid:107)∇ w v (cid:107) T + Ch / T (cid:107) v − v b (cid:107) ∂T (cid:107)∇ w v (cid:107) T . Thus (cid:107)∇ w v (cid:107) T ≤ C (cid:16) (cid:107)∇ v (cid:107) T + h / T (cid:107) v − v b (cid:107) ∂T (cid:17) . To show that j = k + m − j = k + m − Example 2.1.
Suppose
Ω = (0 , × (0 , and is partitioned as shown in Figure2.1. Fig. 2.1 . Let j = k = 1 = j − . To see if (2.5) has a unique solution in V h , we set f = 0 and v = u h . Then ∇ w u h = 0 . By definition, ∇ w u h , (cid:126)q ) T h = − ( u h , ∇ (cid:126)q ) T h + (cid:104) u b , (cid:126)q · (cid:126)n (cid:105) ∂ T h , ∀ (cid:126)q ∈ [ P ( T h )] . Note that the degree of freedom of [ P ( T h )] is while the degree of freedom of V h is . Thus, there exist u h (cid:54) = 0 so that ∇ w u h = 0 . Thus Algorithm will not work. Example 2.2.
Let
Ω = (0 , × (0 , and is partitioned as shown in Figure 2.2.Let j = k = 1 = j − . Note that the degree of freedom of [ P ( T h )] is whilethe degree of freedom of V h is . A similar argument as in example 2.1 shows thatAlgorithm will not work. Fig. 2.2 . For the completeness, we list the following from [1].
Theorem 2.3.
Let u h ∈ V h be the weak Galerkin finite element solution of (2.5).Assume that the exact solution u ∈ H k +1 (Ω) and H -regularity holds true. Then,there exists a constant C such that ||| u − u h ||| ≤ Ch k | u | k +1 , (2.31) (cid:107) u − u (cid:107) ≤ Ch k +1 | u | k +1 . (2.32)
3. Numerical Experiments.
We are devoting this section to verify our the-oretical results in previous sections by two numerical examples. The domain in allexamples is Ω = (0 , × (0 , N = 2 , , , , , , P k ( T ) , P k ( e ) , [ P k +1 ( T )] ) , k = 1 , u h = { u , u b } in the computation. Fig. 3.1 . The first two triangle grids in computation
Example 3.1.
In this example, we solve the Poisson problem (1.1)-(1.2) posedon the unit square Ω , and the analytic solution is u ( x, y ) = sin( πx ) sin( πy ) . (3.1) The boundary conditions and the source term f ( x, y ) are computed accordingly. Table3.1 lists errors and convergence rates in H -norm and L -norm. Table 3.1
Error profiles and convergence rates for P k elements with P k +1 weak gradient, ( k = 1 , . k h ||| u h − Q h u ||| Rate (cid:107) u − Q u (cid:107) Rate
Example 3.2.
In this example, we consider the problem − ∆ u = f with boundarycondition and the exact solution is u ( x, y ) = x (1 − x ) y (1 − y ) . (3.2) The results obtained in Table 3.2 show the errors of SFWG scheme and the con-vergence rates in the ||| · ||| norm and (cid:107) · (cid:107) norm. The simulations are conducted ontriangular meshes and polynomials of order k = 1 , . The SFWG scheme with P k el-ement has optimal convergence rate of O ( h k ) in H -norm and O ( h k +1 ) in L -norm. Table 3.2
Error profiles and convergence rates for P k elements with P k +1 weak gradient, ( k = 1 , . k N ||| e h ||| Rate (cid:107) e (cid:107) Rate
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