The polynomial-preserving recovery for higher order finite ... - Math

DISCRETE AND CONTINUOUS DYNAMICAL SYSTEMS–SERIES B Volume 5, Number 30, August 2005

Website: http://AIMsciences.org pp. 769–798

THE POLYNOMIAL-PRESERVING RECOVERY FOR HIGHER ORDER FINITE ELEMENT METHODS IN 2D AND 3D

A. Naga and Z. Zhang

1

Department of Mathematics Wayne State University Detroit, MI 48202, USA

(Communicated by Jie Shen) Abstract. The Polynomial-Preserving Recovery (PPR) technique is extended to recover continuous gradients from C 0 finite element solutions of an arbitrary order in 2D and 3D problems. The stability of the PPR is theoretically investigated in a general framework. In 2D, the stability is established under a simple geometric condition. The numerical experiments demonstrated that the PPR-recovered gradient enjoys superconvergence, and the Zienkiewicz-Zhu error estimator based on the PPR-recovered gradient is asymptotically exact.

1. Introduction. Adaptive C 0 Finite Element Methods (FEMs) are very efficient tools in approximating solutions of partial differential equations. A crucial part of this kind of adaptation is the error estimation. Since the pioneering work by Babuˇska and Rheinboldt [4], a posteriori error estimators have been the focus of intensive research. For the history and the advances in this field, the reader is referred to [1, 5, 8, 15, 23]. In 1987, Zienkiewicz and Zhu [28] introduced their simple error estimator, the ZZ error estimator. This estimator relies on constructing a continuous gradient by a postprocessing operation. To complete their work, they [29] developed a gradient recovery technique known as the Superconvergence Patch Recovery (SPR). According to [30], using the SPR-recovered gradient in the ZZ error estimator produced a robust estimator; namely the ZZ-SPR. Later, Babuˇska et al. established the computer-based theory that enabled them to study and compare the known error estimators for a wide range of problems. One of their primary results [6, 7] is that the ZZ-SPR is the most robust error estimator among the tested ones. Basically, the SPR uses the gradient of the finite element solution and discrete least-squares to recover the gradient at mesh nodes. Naturally, one asks: is it possible to modify the SPR to use the finite element solution itself? If yes, is the ZZ error estimator based upon the new recovery technique as robust as the ZZ-SPR? Fortunately, these questions have positive answers. In [26], the authors introduced the Polynomial-Preserving Recovery (PPR) that uses the finite element solution. In [20], it was shown that the PPR-recovered gradient enjoys superconvergence. This result was established for C 0 linear FEM in 2D meshes that meet some mild 1991 Mathematics Subject Classification. 65N30, 65N15, 65N12, 65D10,74S05, 41A10, 41A25. Key words and phrases. finite element method, SPR, PPR, discrete least-squares fitting, superconvergence, a posteriori error estimator. 1 This research was partially supported by the National Science Foundation grants DMS0074301 and DMS-0311807.

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conditions. The ZZ-PPR (the ZZ error estimator based on the PPR) was validated in [27] and it was shown to be as good as or better than the ZZ-SPR. The goal of this paper is to extend the PPR to higher order C 0 FEMs in 2D and in 3D problems. The stability of the PPR recovery is studied in Section 3. The main result of that section shows that the PPR recovery is stable in regular meshes satisfying a certain requirement. In Section 4, we show that this requirement is guaranteed in 2D regular meshes satisfying a simple geometric condition. Although the stability of the PPR is studied in the context of regular meshes, we conjecture that the PPR is also stable in anisotropic meshes. This conjecture is verified through an example at the end of Section 4. Finally, Section 6 gives some numerical experiments that illustrate the efficiency of the PPR and the ZZ-PPR. 1.1. Notations. Let s ∈ [0, ∞], p ∈ [1, ∞], m, k, l ∈ Z+ (the set of positive integers), and A ⊂ Rd with d = 2 or 3. The spaces Wps (A) and H s (A) are the classical Sobolev spaces equipped with the norms k · ks,p,A and k · ks,A , respectively, and the seminorms | · |s,p,A and | · |s,A , respectively. The measure of A in Rd will be denoted by |A|. For vectors in Rm , k · k will denote the Euclidean length. If m = d, we may use | · | instead of k.k. If p is a polynomial over A, then deg(p) will denote the total degree of p. The vector space {p : p is a polynomial over A and deg(p) ≤ m} will be denoted by Pm (A) and nm d 1 Y will denote the dimension of Pm (A) where nm = (m+i). Let P∗m (A) ⊂ Pm (A) d! i=1 denote the subset {p : p ∈ Pm (A), p is homogeneous, and deg(p) = m}. The space of real matrices of order k ×l is denoted by Rk×l . The identity of Rk×k is denoted by Ik , while 1k (0k ) denotes the k × 1 column vector whose entries are all ones (zeros). Let σ1 (K) ≥ σ2 (K) ≥ . . . ≥ σmax(k,l) (K) ≥ 0 denote the singular 2 values of a matrix K ∈ Rk×l . Recall that σm (K) = σm (K T K) = σm (KK T ) for all 1 ≤ m ≤ min(k, l) and σm (K) = 0 for all min(k, l) < m ≤ max(k, l). Also, if K is symmetric positive semi-definite, then the set of its singular values is exactly the same as the set of its eigenvalues. The null space of K and the range of K are denoted by Null(K) and Range(K), respectively. To avoid difficulties in naming the constants that will appear in the course of this work, we adopt the following notations. Let E1 and E2 be two mathematical quantities and let C > 0 be some constant independent of any “essential” ingredients of E1 and E2 . Instead of E1 ≤ CE2 , E1 ≥ CE2 , and E2 /C ≤ E1 ≤ CE2 , we will use the notations E1 . E2 , E1 & E2 , and E1 ∼ = E2 , respectively. 1.2. Preliminaries. Consider the boundary value problem   −∇(D∇u) + αu = f in Ω ν · (D∇u) = gN on ΓN  u = gD on ΓD

(1.1)

where Ω ⊂ Rd is a bounded domain with Lipschitz boundary ∂Ω = ΓN ∪ ΓD , the boundary segments ΓN and ΓD are disjoint, ν is the unit outward normal vector to ∂Ω, and D is a d × d symmetric positive definite matrix. If ΓN = ∂Ω and α = 0, R R the compatibility condition f + g = 0 must be satisfied and the condition Ω ∂Ω R u = 0 is used to ensure the uniqueness. For simplicity, Ω is assumed to be a Ω polygon when d = 2 or a polyhedron when d = 3. In the variational form of (1.1), we seek u ∈ V such that B(u, v) = L(v) ∀ v ∈ V0

(1.2)

THE POLYNOMIAL-PRESERVING RECOVERY

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where V = {v ∈ H 1 (Ω) : v|ΓD = gD }, V0 = {v ∈ H 1 (Ω) : v|ΓD = 0}, Z Z Z B(u, v) = [(D∇u)∇v + αuv], and L(v) = fv + gN v. Ω

Ω

ΓN

If ΓD is empty, we take V = H 1 (Ω). Under suitable assumptions [13, pp. 35-53], the bilinear operator B is continuous and is V −elliptic, the linear operator L is bounded, and (1.2) has a unique solution u ∈ V . Let Th be a conforming partition of Ω. The elements in Th are triangles when d = 2 and tetrahedrons when d = 3. To simplify notations, we assume the elements in Th to be closed, i.e., T = T ∀ T ∈ Th . For T ∈ Th , let hT denote the diameter of T and set h = max{hT : T ∈ Th }. Also, let %T denote the diameter of the largest ball (circle when d=2 or sphere when d=3) inscribed in T . The regularity of T hT is measured with the aspect ratio . The mesh Th is said to be regular if there %T exists a finite positive constant γ, independent of h, such that hT ≤ γ ∀ T ∈ Th . (1.3) %T For r ∈ Z+ , define the finite element space Sh = {v ∈ C(Ω) : v|T ∈ Pr (T ) ∀ T ∈ Th }. Let Nh denote the set of the mesh nodes. A mesh node z ∈ Nh will be called internal (boundary) node if z ∈ Ω (z ∈ ∂Ω), and z will be called a mesh vertex if it is a vertex of an element T ∈ Th . The basis for Sh is the standard Lagrange basis {φz : z ∈ Nh } where P φz (´ z ) = δzz´ ∀ z, z´ ∈ Nh . Using this basis, a function v ∈ Sh 0 takes the form v = z∈Nh v(z)φz . Let Ih : C (Ω) → Sh denote the Lagrange interpolation operator where X Ih w = w(z)φz ∀ w ∈ C 0 (Ω). z∈Nh

The finite element approximation of u is the solution uh ∈ Sh of (1.2) when v varies over Sh ∩ V , i.e., B(uh , v) = L(v) ∀ v ∈ Sh ∩ V.

(1.4)

Qd 2. The PPR and the ZZ-PPR. Let Gh : Sh → i=1 Sh denote the PPR operator. As in the SPR, the structure of Gh uh , the PPR-recovered gradient of uh , relies on the fact that every function in Sh is uniquely defined by its values at the mesh nodes. If the values {(Gh uh )(z) : z ∈ Nh } are well-defined, then X (Gh uh )(z)φz . Gh uh , z∈Nh

Before exploring the definitions of Gh uh at the mesh nodes, we need the following notations. If A is a union of mesh elements in Th and v ∈ Sh , let N (A) denote the number of mesh nodes in A, M (A) denote the number of mesh triangles in A, and v A denote the column vector whose entries are the values of v at the mesh nodes in A. If z is a mesh vertex and n ∈ Z+ , let L(z, n) denote the union of mesh elements in the first n layers around z, i.e., [ L(z, n) = {T : T ∈ Th , T ∩ L(z, n − 1) 6= ∅} where L(z, 0) , {z}.

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2.1. Definition of the PPR. Let z ∈ Nh be a mesh vertex and let Kz denote a patch of mesh elements around z. Let pz ∈ Pr+1 (Kz ) be the polynomial that best fits uh at the mesh nodes in Kz in discrete least-squares sense, i.e., X X |(uh − pz )(˜ z )|2 = min |(uh − p)(˜ z )|2 . z˜∈Nh ∩Kz

p∈Pr+1 (Kz )

z˜∈Nh ∩Kz

For easy referencing, pz will be called the least-squares polynomial approximation, LSPA, of uh at z. Then, (Gh uh )(z) , ∇pz (z). To complete the definition of the PPR, we need to define Kz . Note that besides the dependence of its definition on the location of z, Kz must have at least nr+1 nodes distributed around z in a way that leads to a unique pz . For an internal mesh vertex z, we first define an initial patch Kz,0 such that N (Kz,0 ) ≥ nr+1 . This is achieved with the following definition:   L(z, 1) if N (L(z, 1)) ≥ nr+1 L(z, 2) if d = 3, r = 1, and M (L(z, 1)) = 4 Kz,0 , .  L(z, 1) ∪ {T ∈ Th : T ∩ L(z, 1) is a (d − 1)-simplex} otherwise The first part of this definition is illustrated in Fig. 1(b) while the third part is illustrated in Fig. 1(a). The formula in the third part means that we extend L(z, 1) by adding mesh elements having common (d − 1)-simplices (edges if d = 2 and faces when d = 3) with L(z, 1). Fig. 2 depicts this idea. Unfortunately, if d = 3 and r = 1, extending L(z, 1) out does not work as the number of nodes after extension is 9 < 10 = n2 . The second part of the Kz,0 definition takes care of this situation. Although N (Kz,0 ) ≥ nr+1 , this does not ensure the uniqueness of pz . If Kz,0 does not lead to a unique pz and L(z, n) ⊆ Kz,0 ( L(z, n + 1), set Kz,0 to L(z, n + 1) and recompute pz . The patch Kz is defined to be the smallest Kz,0 that leads to a unique pz . Next, we define Kz at a boundary mesh vertex z. Let n0 be the smallest positive integer such that L(z, n0 ) has at least one internal mesh vertex. Then, Kz , L(z, n0 ) ∪ {Kz˜ : z˜ ∈ L(z, n0 ) and z˜ is an internal vertex}. This definition ensures the uniqueness of pz as shown in Lemma 3.6. Examples for patches corresponding to boundary nodes are shown in Fig. 1. If r = 1, then all mesh nodes are vertices and Gh uh is completely defined. If r > 1, let z ∈ Nh be a non-vertex node. Then, z may lie on an edge between two vertices, inside a mesh element, or inside a face of a tetrahedron in 3D. Formally, ◦ z ∈τ = τ \ ∂τ where τ ⊆ T for some T ∈ Th and τ is a d1 -simplex whose vertices are in Nh and d1 ≤ d. Let z0 , . . . , zd1 be the vertices of τ , then d

(Gh uh )(z) ,

1 1 X ∇pzi (z) d1 + 1 i=0

where pzi is the LSPA of uh at zi . This completes the definition of Gh uh for r > 1.


(a) d=2, r=1

(b) d=2, r=2

Fig. 1. Examples for patches used in the PPR

Extend Out

(a) d=2, r=1

Extend Out

(b) d=3, r=1

Fig. 2. Examples for extending L(z, 1) out, where z is an internal mesh vertex

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2.2. The computational aspects of the PPR. Consider a mesh vertex z whose patch is Kz , let v ∈ Sh , and let pz be the LSPA of v at z. We now outline the details of computing pz . Let z1 , z2 , . . . , zN (Kz ) denote the mesh nodes in Kz . Without loss of generality, let z = z1 and set hz = max{|zi − z1 | : 2 ≤ i ≤ N (Kz )}. To avoid the computational instability resulting from small hz , the computations will be carried out on the patch ωz where x − z1 ωz , Fz (Kz ) and Fz : x ∈ Kz → x ˆ= . (2.5) hz The patch ωz will be called the reference patch associated with z. For 1 ≤ i ≤ N (Kz ), let zî = F (zi ) and set vi = v(zi ). Let pˆz ∈ Pr+1 (ωz ) be such that pz = pˆz ◦ Fz . For x ˆ ∈ ωz , pˆz (ˆ x) can be written in the form pˆz (ˆ x) = p(ˆ x)T cz (2.6) £ ¤T £ ¤T where cz = cz,1 cz,2 cz,3 · · · cz,nr+1 , p(ˆ x) = p1 (ˆ x) p2 (ˆ x) · · · pnr+1 (ˆ x) , and d the set {pl : 1 ≤ l ≤ nr+1 } is the monomials-basis of Pr+1 (R ). In other words, if Qd α y = (y1 , . . . , yd ) ∈ Rd and 1 ≤ l ≤ nr+1 , then pl (y) = y αl = j=1 yj l,j , where αl = (αl,1 , . . . , αl,d ) is a multi-index. For technical reasons (see the proof Lemma 3.6), an ordering of these monomials must satisfy the condition |αl | ≤ |αm | and αl,1 ≤ αm,1 ∀ 1 ≤ l ≤ m ≤ nr+1 . From the definition of pz , one can show that cz is the solution of the linear system ATz Az cz = ATz v Kz where

   Az ,  

p(ˆ z1 )T p(ˆ z2 )T .. .



Setting Bz =



     and v Kz ,   

p(ˆ zN (Kz ) )T ATz Az ,

(2.7) v1 v2 .. .

   . 

(2.8)

vN (Kz )

we have cz = Bz−1 ATz v Kz .

(2.9)

Note that Bz is a real symmetric positive-definite matrix of order nr+1 × nr+1 . The following lemma lists the basic facts about discrete least-squares. For the proof of this lemma, see [22, pp. 162-164]. Lemma 2.1. Let z ∈ Th be a mesh vertex whose patch is Kz . Then, the following are equivalent. 1. pz is unique. 2. Rank(Az ) = nr+1 . 3. The matrix Bz is invertible. 4. There exists no q ∈ Pr+1 (Kz ) such that q(zi ) = 0 for i = 1, 2, . . . , N (Kz ). The equivalence of the first three claims in Lemma 2.1 is somewhat obvious, at least from the definitions of pz , Az , and Bz . The equivalence of the first and the fourth claims says that fitting, by discrete least-squares, the values of some function at the nodes in Kz by a polynomial of degree r + 1 leads to a unique polynomial pz if and only if the nodes in Kz do not lie on a curve (in 2D), or surface (in 3D), corresponding to a polynomial q of degree r + 1. This equivalence is our main tool in analyzing the uniqueness of discrete least-squares best fitting by polynomials.


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2.3. The ZZ-PPR error estimator. Following the standard notations, let ηh denote the ZZ-PPR error estimator of eh , u − uh . As usual, ηh is obtained from the error indicators {ηT,h : T ∈ Th } where sZ sX 2 . ηT,h , | Gh uh − ∇uh |2 and ηh , ηT,h T

T ∈Th

Note that the ZZ-PPR, like the ZZ-SPR, is obtained from the ZZ error estimator by using the PPR-recovered gradient. By definition, ηT,h is a good approximation of |eh |1,T if and only if Gh uh is a good approximation of of ∇u in T . 2.4. Properties of the PPR. The PPR has the following properties. 1. By definition, Gh is linear. 2. Gh satisfies the consistency condition, i.e., Gh (Ih p) = ∇p ∀ p ∈ Pr+1 (Ω).

(2.10)

This fact is proved in [26]. Consequently, Gh is a polynomial-preserving operator and enjoys the approximation property k∇u − Gh (Ih u)kL2 (Ω) . hr+1 |u|r+2,Ω ∀ u ∈ H r+2 (Ω). For more details about proving this property, see [13, pp. 121-133]. Basically, the PPR can be viewed as a generator of finite difference formulas for first order partial derivatives. The generated formulas recover the exact derivatives of polynomials in Pr+1 (Ω). 3. The PPR-recovered gradient may enjoy superconvergence. Establishing this property is not straightforward. Beside the previous properties, and according to the general framework in [1, Ch. 4], Gh uh superconverges to ∇u if the following conditions are satisfied. (a) The recovery operator Gh is bounded in the following sense: k Gh vkL2 (T ) . |v|1,KT ∀ T ∈ Th and ∀ v ∈ Sh where KT ,

(2.11)

[ {Kz : z is a vertex of T }

is the patch corresponding to T . This condition is called the boundedness condition. (b) ∇uh enjoys superconvergence in the following sense: k∇(Ih u − uh )kL2 (Ω) ≤ C(u)hr+ρ

(2.12)

for some ρ ∈ (0, 1], i.e., ∇uh is a “perturbation” of ∇(Ih u). This condition is called the superconvergence condition. If the conditions in (2.11) and (2.12) are satisfied, it is straight forward to show that Gh uh superconverges to ∇u as ∇u − Gh uh = ∇u − Gh (Ih u) + Gh (Ih u − uh ). 4. As in any recovery technique, if Gh uh superconverges to ∇u, the ZZ-PPR is asymptotically exact. Generally speaking, both the boundedness condition and the superconvergence condition rely on the mesh geometry. The superconvergence phenomenon has been in the focus of research and there are many results in this direction especially in 2D. We will go briefly over some of these results in Section 6. The main concern in this paper is establishing the boundedness condition.

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3. The Boundedness of Gh . Definition 3.1. Let Th be a partition of Ω. Then, Th is said to satisfy the finite layer condition in Ω (on ∂Ω) if there exists a constant n ∈ Z+ , independent of h and z, such that Kz ⊆ L(z, n) ∀ z ∈ Ω (∂Ω). Also, Th is said to satisfy the minimum singular value condition in Ω (on ∂Ω) if it satisfies the finite layer condition in Ω (on ∂Ω) and if σnr+1 (Bz ) ≥ C ∀ z ∈ Ω (∂Ω) for some constant C > 0 that depends only on r and γ. If Th satisfies the finite layer (the minimum singular value) condition in Ω and on ∂Ω, then it is said to satisfy the finite layer (the minimum singular value) condition in Ω. Remark 3.2. In the rest of this chapter we will implicitly assume that Th satisfies the finite layer condition in Ω. In practical meshes, this condition is easily satisfied. Indeed, our numerical experiments showed that Kz ⊆ L(z, 2) for all z ∈ Ω and for d=2,3. Also, L(z, 1) corresponding to a boundary mesh vertex z always has at least one internal mesh vertex. Hence, Kz ⊆ L(z, 3) for all z ∈ ∂Ω. Remark 3.3. Obviously, the number of patterns of the PPR patches corresponding to internal vertices in a translation-invariant mesh is finite. For example, there is only one pattern for all the PPR patches corresponding to internal vertices in the uniform mesh with the regular pattern; see Fig. 15(a). Consequently, the translation-invariant meshes satisfy the minimum singular value condition in Ω. Remark 3.4. The minimum singular value condition is directly linked to the mesh geometry. Hence, it is better to find a geometric condition that implies the minimum singular value condition. This is possible in 2D regular meshes as we shall see in Section 4. Unfortunately, establishing that in 3D regular meshes is not clear yet. The main product of this section is Theorem 3.10 which needs some preparation. The next lemma says that the PPR patches in regular meshes are quasi-uniform. Lemma 3.5. Let Th be a regular partition of Ω and let z be a mesh vertex whose patch is Kz . Then, there exists a constant C = C(γ) > 0 such that hT ≥ Chz ∀ T ⊂ Kz . Moreover, the number of elements in a patch corresponding to a particular mesh vertex and the number of patches that contain a particular mesh element are bounded by constants depending only on γ. Proof. This is a direct corollary of Definition 3.1 and [1, Theorem 1.6]. The next lemma says that the minimum singular value condition is satisfied on ∂Ω if it is satisfied in Ω. This result grants us a license to focus our efforts in Ω. Lemma 3.6. Let Th be a regular partition of Ω that satisfies the minimum singular value condition in Ω. Then, there exists a constant C = C(r, γ) > 0 such that σ1 (Bz ) ≤ C where z is any mesh vertex. Moreover, Th satisfies the minimum singular value condition on ∂Ω.


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Proof. Let z ∈ Nh be a mesh vertex whose patch is Kz and whose reference patch is ωz . From the definition of Bz and the fact that ωz is contained in the unit ball, it is easy to verify that |Bz (i, j)| ≤ N (Kz ) + 1 for 1 ≤ i, j ≤ nr+1 . Hence, σ1 (Bz ) ≤ σ1 (|Bz |) ≤ cN (Kz ) for some constant c > 0 that depends only on r. Since Th is regular, Lemma 3.5 implies that N (Kz ) ≤ N0 < ∞ for some constant positive integer N0 = N0 (γ), and the first claim is true. To prove the second claim, let z ∈ ∂Ω be a mesh vertex. By construction, Kz has at least one internal mesh vertex y such that Ky ⊆ Kz . Set yˆ = (ˆ y1 , . . . , yˆd ) = Fz (y) and set ω ˜ = Fz (Ky ). Without loss of generality, assume that yˆ is on the positive x ˆ1 -axis; otherwise rotate ωz . After reordering the nodes in Kz if necessary, · ¸ A˜y Az = A¯z where the rows in A˜y correspond to the mesh nodes in ω ˜ y . Hence, ˜y + B ¯z Bz = ATz Az = A˜Ty A˜y + A¯Tz A¯z = B ˜y , and B ¯z are positive semi-definite, ¯z = A¯Tz A¯z . Since Bz , B ˜y = A˜Ty A˜y and B where B ˜y ). σnr+1 (Bz ) ≥ σnr+1 (B Since Fz (·) =

(3.13)

hy Fy (·) + yˆ, hz ω ˜y =

hy ωy + yˆ. hz

(3.14)

This relation induces a linear transformation Q : Pr+1 (ωy ) → Pr+1 (˜ ωy ). Using the monomial bases of Pr+1 (ω ) and P (˜ ω ), Q = Q Q where Q , Q ∈ Rnr+1 ×nr+1 . y r+1 y 2 1 1 2 µ µ ¶¶ hy The matrix Q1 is diag p and it represents the scaling part of (3.14), while hz Q2 is a lower triangular matrix whose diagonal entries are 1’s and it represents the translation part of (3.14). Obviously, the nonzero entries in Q2 are monomials in yˆ1 , i.e., σnr+1 (Q2 ) is a continuous function of yˆ1 . Since det(Q2 ) = 1 for all yˆ1 ∈ [0, 1], ˜y , and A˜y , one can verify that σnr+1 (Q2 ) & 1. Using the definitions of Q, B ˜y = Q2 Q1 By (Q2 Q1 )T . A˜Ty = Q2 Q1 ATy ⇒ B Therefore, ˜y b ≥ σn (By )kQT1 QT2 bk2 ≥ σn (By )σn2 (Q1 )kQT2 bk2 bT B r+1 r+1 r+1 ≥ σnr+1 (By )σn2r+1 (Q1 )σn2r+1 (Q2 )kbk2 where b ∈ Rnr+1 . By virtue of (3.13), we get ˜y ) ≥ σn (By )σ 2 (Q1 )σ 2 (Q2 ). σnr+1 (Bz ) ≥ σnr+1 (B nr+1 nr+1 r+1 Since Th satisfies the minimum singular value condition in Ω, then σnr+1 (Q1 ) = µ ¶nr+1 hy , and hy & hz by Lemma 3.5, the second claim is true. hz The following two lemmas are the key tools in establishing the boundedness of Gh . The following lemmas are the key tools in establishing the boundedness of Gh :

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Lemma 3.7. Let Th be a regular partition of Ω that satisfies the minimum singular value condition in Ω. Let z be a mesh vertex whose patch is Kz and let v ∈ Sh . If R v = 0, then there is a constant C > 0 that depends only on γ and r such that Kz |v|1,Kz ∼ kv Kz k. = h(d/2−1) z Proof. The proof uses three facts. Firstly, mesh regularity and Lemma 3.5 imply %T ∼ (3.15) = hz ∀ T ⊂ K z . Secondly, for any T ∈ Th , there exists an invertible affine mapping FT : Tˆ → T where Tˆ is a reference element. Set vˆT = (v|T ) ◦ FT and let JT = DFT where DFT ∈ Rd×d is the Jacobian of FT . Since Th is regular, then ½ hT . σd (JT ) ≤ σ1 (JT ) . hT (3.16) | det(JT )| ∼ = hdT h iT ˆ φ ˆ = φˆ1 φˆ2 · · · φˆnr , for all T ∈ Th . Note that N (T ) = N (Tˆ) = nr , vˆT = (v T )T φ, and φî is the standard Lagrange-basis function associated with the ith node in Tˆ for i = 1, 2, . . . , nr . Finally, the third fact says that 1

1

vT |1,Tˆ . |v|1,T . | det(JT )| 2 σ1 (JT−1 )|ˆ vT |1,Tˆ | det(JT )| 2 σd (JT−1 )|ˆ

(3.17)

for any T ∈ Th and for any v ∈ Sh . For a comprehensive discussion about the inequalities in (3.16) and (3.17), the reader is referred to [13, pp. 121-131]. Let T1 , T2 , . . . , TM (Kz ) denote the mesh elements in Kz . Using (3.15)-(3.17), |ˆ vTk |1,Tˆ for k = 1, 2, . . . , M (Kz ). |v|1,Tk ∼ = h(d/2−1) z

(3.18)

It is easy to verify that |ˆ vTk |21,Tˆ = (v Tk )T Qv Tk for k = 1, 2, . . . , M (Kz ) R where Q ∈ Rnr ×nr and Q(i, j) = Tˆ ∇φî · ∇φˆj for i, j = 1, 2, . . . , nr . Using (3.18), M (Kz )

|v|21,Kz

∼ = hz(d−2)

X

v TTk Qv Tk .

(3.19)

k=1

For 1 ≤ k ≤ M (Kz ), there exists a Boolean matrix Ek ∈ Rnr ×N (Kz ) such that v Tk = Ek v Kz , where ½ 1 if node i in Tk is node j in Kz Ek (i, j) = 0 otherwise for 1 ≤ i ≤ nr and 1 ≤ j ≤ N (Kz ). Therefore, (3.19) takes the form M (Kz )

|v|21,Kz

∼ = h(d−2) z

X

˜ K v TKz (EkT QEk )v Kz = h(d−2) v TKz Qv z z

(3.20)

k=1 M (Kz )

˜ , where Q

X

EkT QEk is a symmetric positive semi-definite matrix of order

k=1

˜ has only one zero eigenvalue corresponding to 1N (K ) as N (Kz ) × N (Kz ). Also, Q z |v|1,Kz = 0 ⇔ v|Kz is identically constant. ˜ form an orthonormal basis of RN (Kz ) , we can write Since the eigenvectors of Q (1)

(2)

v Kz = v Kz + v Kz

THE POLYNOMIAL-PRESERVING RECOVERY (1)

779

(2)

where v Kz , c1N (Kz ) for some constant c and v Kz is a combination of the rest of ˜ Consequently, v|K = v (1) + v (2) where v (1) = c and the the eigenvectors of Q. z Kz Kz Kz p (2) (2) (1) nodal values of vKz are the entries of v Kz . Note that |c| = kv Kz k/ N (Kz ), and ¯ ¯Z ¯ ¯Z ¯ ¯ (1) ¯¯ (2) ¯¯ ¯ ¯ vKz ¯ = ¯ vKz ¯ ¯ as

R Kz

Kz

Kz

v = 0. From the definitions, it is straightforward to verify that ¯Z ¯ ¯Z ¯ (1) ¯ ¯ kv Kz k|Kz | (1) ¯¯ (2) ¯¯ (2) ¯ ¯ vKz ¯ = p and ¯ vKz ¯ ≤ kv Kz k|Kz |. ¯ N (K ) Kz

Kz

z

Therefore, (1)

(2)

(2)

kv Kz k2 = kv Kz k2 + kv Kz k2 ≤ (1 + N (Kz ))kv Kz k2 . Thus, ˜ σ(N (Kz )−1) (Q) (2) 2 ˜ kv Kz k2 ≤ σ(N (Kz )−1) (Q)kv Kz k 1 + N (Kz ) ³ ´T ³ ´ (2) ˜ v (2) ≤ v Kz Q Kz

(3.21)

2 ˜ K ≤ σ1 (Q)kv ˜ = v TKz Qv Kz k . z

˜ and σ1 (Q) ˜ depend only on r and N (Kz ). Since N (Kz ) is Note that σ(N (Kz )−1) (Q) bounded by Lemma 3.5, combining (3.19)-(3.21) concludes the proof. Remark 3.8. Note that |v|1,Kz and kv Kz k in Lemma 3.7 define norms of v|Kz . Since Sh ∩ C(Kz ) has a finite dimension, the conclusion of that lemma is expected. The value of Lemma 3.7 stems from the fact that the equivalency constant does not depend on hz and depends only on r and the geometry of Kz . Remark 3.9. There is another way to establish the equivalence in Lemma 3.7 for linear elements [20]. The idea is quite simple. In linear elements, ∇v|T in any element T ∈ Th is constant. The equivalence in Lemma 3.7 is easily achieved if one can show that the PPR-recovered gradient at a mesh vertex z is equivalent to a linear combination of {∇v|T : T ⊂ Kz } and the coefficients in this combination are bounded uniformly by a constant independent of z, h, and v. Lemma 3.10. Let Th be a partition of Ω and let Th satisfy the assumptions in Lemma 3.6. Let z be a mesh vertex whose patch is Kz and consider v ∈ Sh . If pz ∈ Pr+1 (Kz ) is the LSPA of v at z, then |pz |1,∞,Kz ≤ Ch−d/2 |v|1,Kz . z where C is a bounded constant that depends only on r and γ. R Proof. At first assume that Kz v = 0. Let ωz be the reference patch corresponding to z and let pˆz ∈ Pr+1 (ωz ) be such that pz = pˆz ◦ Fz . Let x be a point in Kz and set x ˆ = Fz (x). From (2.6) and from the fact that x ˆ ∈ ωz , |∇pz (x)| = h−1 pz (ˆ x)| . h−1 z |∇ˆ z kcz k Using (2.9), kcz k ≤ σ1 (Bz−1 )σ1 (ATz )kv z k = (σnr+1 (Bz ))−1 Hence, and by Lemma 3.6, kcz k . kv z k

p σ1 (Bz )kv z k

780


and, consequently, |∇pz (y)| . h−1 z kv z k. Therefore, and by virtue of Lemma 3.7, |∇pz (y)| . h−d/2 |v|1,Kz . z R R R Thus, the lemma conclusion is true if Kz v = 0. If Kz v 6= 0, set v˜ = v − |K1z | Kz v R −d/2 and let p˜z be the LSPA of v˜ at z. Since Kz v˜ = 0, |˜ pz |1,∞,Kz . hz |˜ v |1,Kz . This concludes the proof as ∇p = ∇˜ p and ∇v = ∇˜ v. The following theorem is the main result of this section. Theorem 3.11. Let Th be a partition of Ω that satisfies the assumptions in Lemma 3.6. Let T be a mesh element whose patch is KT and let v ∈ Sh . Then, k Gh vkL2 (T ) ≤ C|v|1,KT . where C is a constant that depends only on r and γ. Proof. Let {zi : 1 ≤ i ≤ nr } be the mesh nodes in T such that z1 , . . . , zd+1 are the vertices of T . Since (Gh v)|T ∈ Pr (T ), then k Gh vkL∞ (T ) . max{|(Gh v)(zi )| : 1 ≤ i ≤ nr }. By Lemma 3.10 and the definitions of Gh v at mesh nodes, k Gh vkL∞ (T ) . max{h−d/2 |v|1,Kzi : 1 ≤ i ≤ d + 1}. zi Hence, k Gh vkL2 (T ) ≤ k Gh vkL∞ (T )

p |T |

d/2

. max{hT h−d/2 |v|1,Kzi : 1 ≤ i ≤ d + 1}. zi Since T ⊂ Kzi for all 1 ≤ i ≤ d + 1, hT ≤ hzi . Therefore, k Gh vkL2 (T ) . max{|v|1,Kzi

v ud+1 uX |v|2 : 1 ≤ i ≤ d + 1} . t i=1

1,Kzi

and the proof is complete by virtue of Lemma 3.5. 4. Verifying the Minimum Singular Value Condition in 2D Meshes. We have seen that Gh is bounded in the sense of (2.11) given that Th satisfies the minimum singular value condition in Ω. Obviously, this condition can not be used in mesh construction. It is better if it is replaced with a geometric condition. The key observation to achieve that is the equivalence between the first and the fourth items in Lemma 2.1 which links the uniqueness of LSPAs and the distribution of the nodes used in constructing these LSPAs. The authors [20] used this equivalence to establish the boundedness of Gh for the case r = 1. In fact, it was found that the PPR is bounded in meshes satisfying what is called the angle condition. Surprisingly, this condition leads to the same conclusion when r > 1. Definition 4.1. Let Th be a partition of Ω and let z be a mesh vertex. The layer L(z, 1) is said to satisfy the angle condition if the sum of any two adjacent angles in L(z, 1) is at most π. The partition Th is said to satisfy the angle condition in Ω if L(z, 1) satisfies the angle condition at every mesh vertex in Ω.


(a)

781

(b)

Fig. 3. L(z, 1) violates the Angle Condition

According to this definition, if z is an internal mesh vertex and L(z, 1) satisfies the angle condition, then M (L(z, 1)) ≥ 4. Moreover, if M (L(z, 1)) = 4, the triangles in L(z, 1) form a quadrilateral whose diagonals intersect at z; see Fig. 3. Remark 4.2. Nodes of the type shown in Fig. 3(a) rarely happen in practical meshes and, if they happen, they can be identified and removed. Also, a node of the type shown Fig. 3(b) may be fixed by moving it to the intersection of the quadrilateral diagonals. Remark 4.3. If z is a mesh vertex, one can show that L(z, 1) satisfies the angle condition if and only if every pair of adjacent triangles in L(z, 1) forms either a triangle or a convex quadrilateral. Indeed, a triangulation Th satisfies the angle condition in Ω if and only if any pair of adjacent triangles in Ω forms a triangle or a convex quadrilateral. Indeed, robust mesh generators target this type of meshes. Proposition 4.4. Let m ∈ Z+ . The following are true. 1. Let p ∈ P∗m (R2 ) and let z0 ∈ R2 . If p(z0 ) = 0, then p is 0 at any point on the line through z0 and the origin. 2. Let p ∈ Pm (R2 ), and consider the distinct points {zi ∈ R2 : i = 1, 2, . . . , nm }. Assume that these points are distributed on m + 1 non-overlapping parallel lines in such a way that the kth line has exactly k points for k = 1, 2, . . . , m+1. If p(zi ) = 0 for i = 1, 2, . . . , nm , then p is identically zero. Proof. For the first claim, note that any point on the line through z0 and the origin is a multiple of z0 . For the proof of the second claim, see [11, pp. 64]. Let z be an internal mesh node and let {zi = (x1,i , x2,i ) : i = 1, 2, . . . , N (L(z, 1))} be the mesh nodes in L(z, 1) such that the first M (L(z, 1)) + 1 nodes are the mesh vertices in L(z, 1) and z1 = z. For (x1 , x2 ) ∈ L(z, 1), define the polynomials `j (x1 , x2 ) = (x1 − x1,1 )(x2,j+1 − x2,1 ) − (x2 − x2,1 )(x1,j+1 − x1,1 )

(4.22)

∀ 1 ≤ j ≤ M (L(z, 1)), and Y pz,` = {`j : 1 ≤ j ≤ M (L(z, 1)), `j is not a multiple of `k ∀ k < j}. Note that `j = 0 is the equation of the line passing through the nodes z1 and zj+1 . Also, note that pz,` is the product of linear polynomials representing non-collinear triangles sides meeting at z1 = z and that 2 ≤ deg(pz,` ) ≤ M (L(z, 1)). The proof of the following theorem is in [20].

782


Theorem 4.5. Let z be an internal mesh vertex and assume that L(z, 1) satisfies the angle condition and has no degenerate triangles. Let p ∈ P2 (L(z, 1)) such that p(zi ) = 0 for i = 1, 2, . . . , M (L(z, 1)) + 1. Then, p is identically 0 when M (L(z, 1)) > 4 and is a multiple of pz,` if M (L(z, 1)) = 4. Before we continue, let us go over some  rr+1 rr  (r − 1)r+1 (r − 1)r   .. .. D, . .   2r+1 2r 1 1

properties of the matrix ··· ··· .. .

r2 (r − 1)2 .. . 2

··· ···

2 1

r1 (r − 1)1 .. . 1

2 1

      

(4.23)

where r ≥ 2. Note that D is a Vandermonde matrix of order r × (r + 1). This matrix will play a key rule in Lemma 4.8. Proposition 4.6. Let D1 and D2 be the sub-matrices of D obtained by omitting the r + 1st and rth columns, respectively. Then, Rank(D1 ) = Rank(D2 ) = r. Proof. We start with D1 and proceed by contradiction. If Rank(D1 ) 6= r, then r−1 X there exists a non-zero polynomial q ∈ Pr+1 (R) of the form q(ξ) = ξ 2 ai ξ i such i=0

that q(j) = 0 for 1 ≤ j ≤ r. This is impossible as q has only r − 1 non-zero roots. Next, consider D2 . If Rank(D2 ) 6= r, then there exists a non-trivial polynomial r X q ∈ Pr+1 (R) of the form q(ξ) = ξ ai ξ i such that q(j) = 0 for 1 ≤ j ≤ r and i=0

a1 = 0. If ar = 0, one can proceed as in the previous case. So, assume that ar 6= 0. Without loss of generality, set ar = 1. Using the zeros of q, we get r r X Y q(ξ) = ξ ai ξ i = ξ (ξ − j). i=0

j=1

Comparing the two expressions of q leads to a contradiction as r X 1 0 = a1 = (−1)r−1 (r!) 6= 0. j j=1 The following corollary is a direct result of Proposition 4.6. Corollary 4.7. The reduced-row echelon  1 ∗ ···  0 1 ···   .. .. . .  . . .   0 0 ··· 0 0 ···

form of D is  ∗ ∗ ∗ ∗ ∗ ∗   .. .. ..  . . . .   1 ∗ ∗  0 1 β

for some nonzero constant β that depends only on r. We are now ready for the next crucial lemma. Lemma 4.8. Let z be an internal mesh vertex and assume that L(z, 1) satisfies the angle condition and has no degenerate triangles. Let p ∈ Pr+1 (L(z, 1)) such that p(zi ) = 0 for 1 ≤ i ≤ N (L(z, 1)). If deg(p) = r + 1 and deg(pz,` ) ≤ r + 1, then there exits a polynomial q such that p = qpz,` ; otherwise p is identically 0.


783

Proof. By Theorem 4.5, our claim is true for r = 1. So, we need to consider r ≥ 2. Note that any triangle in L(z, 1) has nr nodes arranged as in the second part of Proposition 4.4. Hence, p is identically 0 if deg(p) < r + 1. So, let us assume that deg(p) = r + 1. Without loss of generality, assume that z = z1 = r+1 X (0, 0). Then, p = pk where pk ∈ P∗k (L(z, 1)). We claim that pk (zi ) = 0 for k=1

i = 2, 3, . . . , M (L(z, 1)) + 1 and for k = 1, 2, . . . , r + 1. To prove this claim, fix 2 ≤ i ≤ M (L(z, 1)) + 1. Using the definition of the Lagrange element of order r, the edge z1 zi has r + 1 uniformly spaced nodes of the form jzi /r for j = 0, 1, . . . , r. Since p is zero at these nodes, we have r+1 µ ¶k X j pk (zi ) = 0 for j = 1, 2, . . . , r. r k=1

k

Setting pk = r p¯k , we get r+1 X

j k p¯k (zi ) = 0 for j = 1, 2, . . . , r.

k=1

This system of equations is actually a linear system of the form   p¯r+1 (zi )  p¯r (zi )      .. D  = 0r .    p¯2 (zi )  p¯1 (zi )

(4.24)

where D is the Vandermonde matrix defined in (4.23). By virtue of Corollary 4.7, there exists a constant β = β(r) 6= 0 such that p¯2 (zi ) + β p¯1 (zi ) = 0. This is true for all 1 ≤ i ≤ M (L(z, 1)) + 1. Hence, the quadratic polynomial q¯ = p¯2 + β p¯1 passes through all the vertices in L(z, 1). By assumption, p¯2 is a homogeneous quadratic polynomial. Since L(z, 1) satisfies the angle condition, Theorem 4.5 implies that either q¯ is identically 0 when M (L(z, 1)) > 4 or q¯ is a homogeneous quadratic polynomial when M (L(z, 1)) = 4. Consequently, p¯1 is identically 0 for all M (L(z, 1)) ≥ 4. Hence, one can drop p¯1 (zi ) in the linear system in (4.24). This leads to a homogeneous linear system whose coefficients matrix is D1 . By Proposition 4.6, D1 is non singular. Hence, p¯k (zi ) = 0 for k = 2, 3, . . . , r + 1 and i = 2, 3, . . . , M (L(z, 1)) + 1. By the first part of Proposition 4.4, pz,l is a factor of pk for k = 2, . . . , r + 1. Consequently, pk is identically zero for k < deg(pz,l ) and pk = qk pz,l for some polynomial qk when k ≥ deg(pz,l ). Theorem 4.9. Let z be an internal mesh vertex, and assume that L(z, 1) satisfies the angle condition and has no degenerate triangles. Let p ∈ Pr+1 (L(z, 1)) such that p(zi ) = 0 for i = 1, 2, . . . , N (L(z, 1)). Then, p is identically 0 for r ≥ 2. Proof. Using Lemma 4.8, we may assume that deg(p) = r + 1 and deg(pz,` ) ≤ r + 1. Based on M (L(z, 1)), which is at least 4 by the angle condition, we have 3 cases: Case 1: M (L(z, 1)) > 4.M (L(z, 1)) > 4. By Lemma 4.8, p = `1 q for some q ∈ P∗r (L(z, 1)) and `1 is defined in (4.22). Since M (L(z, 1)) > 4, L(z, 1) must have at least 2 vertices on one side of the line through z1 and z2 as depicted in Fig. 4(a). Therefore, L(z, 1) has a triangle T that intersects with the line through

784


z1 and z2 only at z1 . Hence, `1 (x1 , x2 ) 6= 0 for all (x1 , x2 ) ∈ T \ {z1 }. Consequently, q(˜ z ) = 0 ∀ z˜ ∈ T . By the part 2 of Proposition 4.4, q, and hence p, is identically 0. Case 2: M (L(z, 1)) = 4 and r = 2. Since deg(pz,` ) = 2, Lemma 4.8 implies that p = qpz,` for some q ∈ P1 (L(z, 1)). Note that q must be 0 at the four mid-sides of the quadrilateral formed with the triangles in L(z, 1). Since these nodes are non-collinear (see Fig. 4(b)), q must be identically 0. Case 3: M (L(z, 1)) = 4 and r > 2. Using Fig. 4(c), one can show that L(z, 1) has nr+1 nodes distributed as in part 2 of Proposition 4.4, i.e., p must be 0.

(a) case 1

(b) case 2

(a) case 3

Fig. 4. Various cases in the proof of Theorem 4.9

Corollary 4.10. Let z be an internal mesh vertex, and assume that L(z, 1) satisfy the angle condition and has no degenerate triangles. Then, Kz = L(z, 1) for all r > 1, i.e., Bz is invertible. Proof. This is obvious from Lemma 2.1 and Theorem 4.9. Theorem 4.11. Let Th be a regular partition of Ω, and assume that the sum of any two adjacent angles in Th is at most π. Then, Th satisfies the minimum singular value condition in Ω for all r > 1. Proof. Consider an internal mesh vertex z ∈ Nh . By assumption, L(z, 1) satisfies the angle condition and has no degenerate triangles. Hence, Kz = L(z, 1) by Corollary 4.10. The regularity of Th implies that any triangle angle is at least θ0 ∈ (0, π) where θ0 is a constant that depends only on γ defined in (1.3). Since the sum of the angles in any triangle is π, we have 0 < θ0 ≤ θ ≤ π − 2θ0 where θ is any angle in any triangle in Th . Let Θz ∈ R3M (Kz ) denote the vector of all the triangles angles in Kz . Note that Θz ∈ [θ0 , π − 2θ0 ]3M (Kz ) . Consider the set Sz defined by Sz = {Θz ∈ [θ0 , π − 2θ0 ]3M (Kz ) : L(z, 1) satisfies the angle condition}. We claim that Sz is compact. This is true if Sz is a closed subset of [θ0 , π − 2θ0 ]3M (Kz ) as the latter is compact. For that, proceed by contradiction and assume (m) that Sz is not closed. Then, there exists a convergent sequence {Θz : m > 0} (0) (m) (0) such that Θz = lim Θz corresponds to L(z, 1) that does not satisfy the m→∞

(m)

angle condition. Let L(z, 1)(m) correspond to the angles in Θz for m ≥ 1. Since (0) L(z, 1)(0) does not satisfy the angle condition, it has two adjacent angles θ1 and (0) (0) (0) θ2 such that θ1 + θ2 > π. Hence, there exists m0 > 0 such that (m0 )

θ1

(m0 )

+ θ2

> π.


785

This is a contradiction as L(z, 1)(m0 ) satisfies the angle condition by assumption. Let ωz be the reference patch corresponding to z. Since ωz has at least one mesh vertex z˜ on the unit circle, ωz is completely defined by z˜ and the angles in ωz . Without loss of generality, assume that z˜ = (1, 0). Then, Bz is a continuous function of all the angles in ωz . By the compactness of Sz and the angle condition, σnr+1 (Bz ) achieves a minimum value σ0 > 0. Note that σ0 > 0 depends only on M (Kz ) which is finite by the regularity of Th . Hence, σ0 is independent of z, and the proof is complete. Corollary 4.12. Under the conditions in Theorem 4.11, Gh is bounded. Also, uniform regular mesh-refinement preserves the minimum singular value condition. Proof. The first claim is a direct consequence of theorems 3.11 and 4.11. The second claim follows directly from the fact that uniform regular mesh-refinement preserves the angle condition. The proof of this fact is elementary and is based on Definition 4.1 and Fig. 5. θ θ θ θ θ θ θ θ (a)

(b)

Fig. 5. Regular refinement preserves the angle condition

Remark 4.13. It is straight forward to generalize the arguments in this section to 3D meshes. However, one needs to establish a 3D version of Theorem 4.5. Remark 4.14. It is possible to relax the angle condition, but we avoided that in order to simplify our arguments. For example, the angle condition is not needed if L(z, 1) has three vertices lying on one straight line. Also, it is possible to allow the sum of two adjacent angles in L(z, 1) to exceed π, at most, at two of its vertices. 5. Boundedness of the PPR in Anisotropic Meshes. In establishing the boundedness of Gh , it was always assumed that the mesh is regular. This assumption may not be necessary as we shall see in the following example. Indeed, we believe that Gh is bounded in anisotropic meshes. Example 1. Consider the patch of triangles T shown in Fig. 6(a). Note that T is a union of translations of the parallelogram in Fig. 6(b). This parallelogram has a horizontal side of length 1 and its other side is of length a and makes an angle θ ∈ (0, π/2) with the positive x1 -axis. We set a1 = a cos θ and a2 = a sin θ. Let T be a triangle in T and let KT be its corresponding patch as in Fig. 6(a). The nodes in KT are labelled as z1 , z2 , . . . , z12 . Let v be a continuous function whose restriction to any triangle in T is linear, and let G v denote its PPR-recovered gradient. In the rest of this example, i is either 1 or 2. Let Gxi v denote the xi -component of G v. Note that Gxi v at any node z is a weighted sum of the nodal values of v in Kz . Since T is translation invariant, the set

786


(

(a)

2

and

(

1

θ

(b) Dimensions -1 6 2+a 1 6a2

"% # %% -1 $$$ %% 3 $$$ %%1+2a $$ 1 $$ %%& $ 6a '$

0 0

2

-1 6 -1+a 1 6a2 !

1 6 1-a 1 6a2

"% # %% 1 $$$ %% 6 $$$ %%%-2-a1 $$$ &% 6a '$$$ 2

"% # $$ 1 $ %%% %% 3 $$$$ %%-1-2a1 $$ %& $$ 6a2 '$

(c) The PPR weights Fig. 6. Geometry, patches, and the PPR weights for Example 1

of weights does not change from one node to another. The values of these weights are shown in Fig. 6(c). Note that the weight at any node has two components; the ith component is for Gxi v. We claim that k Gxi vkL2 (T ) ≤ Ci k∂xi vkL2 (KT ) for i = 1, 2

(5.25)

where C1 , C2 are constants independent of a and θ. Establishing this claim will imply that the mesh regularity is not necessary for the boundedness of Gh . To prove this claim, we need the following proposition.


787

Proposition 5.1. If ∂xi v = 0 in KT , then Gxi v = 0 in T . Proof. Without loss of generality, and to simplify the notations, we consider the case i = 2. Similarly, one can treat the case i = 1. If (∂x2 v)|KT = 0, then v is constant along the fibers of KT that are parallel to x2 -axis. By a fiber parallel to x2 -axis, or simply x2 -fiber, we mean a line segment ` ⊂ KT such that ` is parallel to x2 -axis and ∂` ⊂ ∂KT . We have one of two cases. Case 1: θ = π/2. In this case, KT has 4 of its x2 -fibers composed of edges in KT . Denote these fibers by `1 , `2 , `3 , and `4 . In this case, v|KT is a linear combination of the functions v (1) , v (2) , v (3) , and v (4) where v (m) , 1 ≤ m ≤ 4, is a continuous function whose restriction to any triangle in KT is linear and ½ 1 if z ∈ `m v (m) (z) = 0 otherwise for all z ∈ KT . Using the weights in Fig. 6(c), one can show that (Gx2 v (m) )|T = 0. Therefore, (Gx2 v)|T = 0. Case 2: θ ∈ (0, π/2). In this case, none of the x2 -fibers of KT is a union of edges in KT . Hence, v|KT = c1 + c2 x1 for some real constants c1 and c2 . Since the PPR satisfies the consistency condition, (Gx2 v)|T = ∂x2 (c1 + c2 x1 ) = 0. T

Let vk = v(zk ) for 1 ≤ k ≤ 12 and let v = [ v1 v2 · · · v12 ] . After some lengthy algebraic manipulations, one can show that (a2 )3−2i T (a2 )3−2i T v Ai v and k∂xi vk2L2 (KT ) = v Bi v. 864 2 The matrices A1 and B1 are constant while the entries of A2 and B2 are quadratic polynomials in a1 . Indeed, A2 = a21 A1 + a1 Aˆ2 + P A1 P T k Gxi vk2L2 (T ) =

and

ˆ 2 + P A1 P T . B2 = a21 B1 + a1 B ˆ2 are listed in [19, pp. 80-81], while the The constant matrices A1 , B1 , Aˆ2 , and B matrix P is a permutation matrix obtained from I12 by interchanging the rows 1 with 12, 2 with 10, 3 with 7, 4 with 11, and 5 with 9. Note that KT is invariant under the reflection on the line connecting z6 and z8 . This explains why P is part of the formulas of A2 and B2 . Also, note that Ai is positive semi-definite and k Gxi vkL2 (T ) = 0 if and only if v ∈ Null(Ai ). Similarly, k∂xi vkL2 (KT ) = 0 if and only if v ∈ Null(Bi ). Let Ei : R12 → Null(Bi ), let mi denote the dimension of Null(Bi ), and set v i = v − Ei v. Note that v i is a linear combination of the eigenvectors of Bi that correspond to the nonzero eigenvalues. Taking these vectors ˜i ∈ R12×(12−mi ) , we write v i = E ˜i v ˜ i . Hence, and by virtue as columns of a matrix E of Proposition 5.1, k Gxi vk2L2 (T ) =

(a2 )3−2i T ˜ (a2 )3−2i T ˜i ˜ i Ai v v i Ai v i = v 864 864

and

(a2 )3−2i T (a2 )3−2i T ˜ ˜i ˜ i Bi v v i Bi v i = v 2 2 ˜ T Ai E ˜i and B ˜i = E ˜ T Bi E ˜i . Hence, where A˜i = E i i k∂xi vk2L2 (KT ) =

k Gxi vk2L2 (T ) k∂xi vk2L2 (KT )

=

1 fi (˜ vi ) 432

788


where fi : R12−mi → R is defined by fi (˜ vi ) =

˜ Ti A˜i v˜i v . ˜i v ˜ Ti B ˜i v

˜i is nonsingular. Moreover, the maximum value of By definition, it is clear that B fi is the maximum eigenvalue, λmax , of the generalized eigenvalue problem ˜i v ˜ i = λB ˜i. A˜i v (5.26) It is clear that λmax is a function of a1 . Consequently, (5.25) is true if there exists ˜ > 0 such that λmax ≤ λ ˜ and λ ˜ is independent of a1 . According to i a constant λ and θ, we have the following three cases. ˜1 are constant Case 1: i = 1, θ ∈ [0, π/2]. Since A1 and B1 are constant, A˜1 and B too, i.e., the eigenvalues of (5.26) are constant (indeed λmax = 38). Hence, k Gx1 vkL2 (T ) ≤ Ck∂x1 vkL2 (KT ) where C is independent of a and θ. Case 2: i = 2, θ = π/2. In this case a1 = 0 and hence A2 and B2 are constant. Proceeding as in Case 1, there is a constant C > 0, independent of a, such that k Gx2 vkL2 (T ) ≤ Ck∂x2 vkL2 (KT ) . Case 3: i = 2, θ ∈ (0, π/2). Getting the eigenvectors of B2 is very difficult in this case and one has to pursue another direction. Since θ ∈ (0, π/2), the second case in the proof of Proposition 5.1 leads to k∂x2 vkL2 (KT ) = 0 if and only if v|KT = c1 +c2 x1 for some constants c1 , c2 ∈ R. Consequently, if Z Z v = 0 and ∂x1 v = 0, (5.27) KT

KT

then k∂x2 vkL2 (KT ) = 0 ⇔ v|KT = 0. Hence, the null space of the equations in (5.27), after writing them in terms of v, is the same as the space spanned by the eigenvectors of B2 corresponding to nonzero eigenvalues. Hence, the members of a basis of this null space can serve as the ˜2 (see [19, pp. 83] for one possible form columns of a matrix E that is similar to E ˜ of E). Without loss of generality, set E2 = E. After some algebraic manipulations, ˜2 v ˜ 2 = λB ˜ 2 is the characteristic equation of A˜2 v 0 = λ7 [g1 (λ) + g2 (λ)g3 (a1 )] where g1 (λ) = 6λ3 − 281λ2 + 1806λ − 3136, g2 (λ) = 2λ3 − 90λ2 + 486λ − 702, and

a21 (1 + a1 )2 . + + 6a41 + 7a31 + 6a21 + 3a1 + 1 The nonzero roots of g1 (λ) + g2 (λ)g3 (a1 ) are perturbations of the roots of g1 (λ) as kg3 kL∞ (R1 ) is very small (indeed |g3 (a1 )| < 0.15 ∀ a1 > 0). The roots of g1 (λ) are √ 65± 2881 7 and the perturbations are within 0.1. Hence, 2 and 3 g3 (a1 ) =

a61

3a51

k Gx2 vkL2 (T ) ≤ Ck∂x2 vkL2 (KT ) where C is a finite constant that is independent of a and θ.


789

6. Numerical Experiments. In this section we will go over some numerical experiments which will show that the PPR-recovered gradient enjoys superconvergence and that the ZZ-PPR accurately measures eh . The focus will be on quadratic elements in 2D, and linear and quadratic elements in 3D. Also, a comparison is held between the PPR and the SPR. This comparison is limited to 2D examples. The efficiency of many gradient recovery techniques deteriorates near ∂Ω. This suggests that we should separately study the performance of the PPR inside Ω and near ∂Ω. To do that, Nh is partitioned into Nh,1 ∪ Nh,2 where Nh,1 = {z ∈ Nh : disti (z, ∂Ω) ≥ Hi for 1 ≤ i ≤ d} for some positive constants H1 , . . . , Hd , and disti (·) denotes the distance along xi axis. Accordingly, Ω is partitioned into Ω1,h ∪ Ω2,h where [ Ω1,h = {T ∈ Th : T has all of its vertices in Nh,1 }. Let A ⊆ Ω be a union of a set of mesh elements in Th . Then, the ZZ-PPR error indicator in A is defined by ηh,A = k Gh uh − ∇uh kL2 (A) . We use the effectivity index κh,A to measure the accuracy of ηh,A , where ηh,A κh,A = . k∇(u − uh )kL2 (A) To trace the accuracy of the ZZ-PPR in each of the mesh elements in A, we will use the mean, µh,A , and the standard deviation, σh,A , of the effectivity indices in these elements. If the ZZ-PPR is asymptotically exact in each of the elements in A, then lim µh,A = 1 and lim σh,A = 0. Recall that h→0

µh,A

h→0

X X 1 1 2 , , κh,T and σh,A (κh,T − µh,A )2 . M (A) M (A) T ⊂A

T ⊂A

In all of the following examples, the nodes in Nh are partitioned using H1 = H2 = H3 = 18 , except for the last example where we use H3 = 14 . Example 2. In this example, we consider the model problem ½ −4u + 100u = 0 in Ω = (0, 1)2 u = gD on ∂Ω cosh(10x1 ) + cosh(10x2 ) . This function has a peak at the corner (1, 1), 2 cosh(10) and this peak may reduce the accuracy of uh near (1, 1). The finite element solution uh is computed using the quadratic element, and Th is obtained by Delaunay triangulation. The initial mesh is shown in Fig. 7. In successive iterations, the new mesh is obtained from the previous one by uniform regular refinement. Regular refinement of a single triangle is depicted in Fig. 8. Note that the initial mesh in Fig. 7 satisfies the angle condition in Ω. By Corollary 4.12, Gh is bounded. The results in Fig. 9(a) shows that |Ih u − uh |1,Ω has superconvergence. This explains the superconvergence in Gh uh . It is clear that the PPR is better than the SPR and that the PPR performs well in both Ω1,h and Ω2,h . Finally, Fig. 9(b) shows that the ZZ-PPR is almost exact everywhere. and u =

Remark 6.1. The slopes, or convergence rates, in the figures corresponding to the previous example, and the next ones, are computed using the last two points. In uniform meshes, the slope is rounded to the nearest whole number.

790


Regular refinement

(a)

Regular refinement

(b)

Fig. 7. The initial Delaunay mesh used in Example 2

Fig. 8. Regular refinement

Remark 6.2. A triangular mesh Th satisfies the condition (ρ1 , ρ2 ) for some ρ1 , ρ2 > 0 if Th admits a partition consisting of two sets that satisfy the following. In the first set, every pair of adjacent triangles form an O(h1+ρ1 ) parallelogram for some ρ1 > 0 (An O(hδ ) parallelogram, δ > 0, is a quadrilateral in which the distance between the mid-points of its diagonals is O(hδ ).) In the second set, the triangles’ total area is O(hρ2 ) for some ρ2 > 0. The ideal value for ρ1 or ρ2 is ∞. Note that ρ1 = ρ2 = ∞ if any edge in any triangle in Th is parallel to one of three fixed directions. This ideal case was covered in [10] where the authors completely determined the expansion of u − uh at mesh vertices. The case ρ1 = 2 and ρ2 = ∞ is another interesting case that was studied in [17, 16], while the general case was treated in [25]. If a triangular mesh Th satisfies the condition (ρ1 , ρ2 ) for some ρ1 , ρ2 > 0 and if Sh consists of piecewise linear polynomials, then |Ih u − uh |1,Ω is O(h1+ρ ) where ρ = 21 min(1, 2ρ1 , ρ2 ); see [25] for the details. Example 3. Let us now consider a 3D example. The model problem is ½ −4u = 3π 2 sin(πx1 ) sin(πx2 ) sin(πx3 ) in Ω = (0, 1)3 . u=0 on ∂Ω The solution of this problem is u(x1 , x2 , x3 ) = sin(πx1 ) sin(πx2 ) sin(πx3 ). We will solve this problem using linear and quadratic elements on Delaunay meshes, and the initial mesh is shown in Fig. 10. In successive iterations, the new mesh is obtained from the previous one by uniform regular refinement. Regular refinement of a single tetrahedron is depicted in Fig. 8(b); see e.g. [18] for a comprehensive discussion about this kind of refinement. The numerical results are shown in Fig. 11 and Fig. 12. In both the linear element and the quadratic element, |Ih u − uh |1,Ω has superconvergence, which may explain the the good performance for the PPR and the ZZ-PPR. Unfortunately, the superconvergence in |Ih u−uh |1,Ω has not been justified theoretically yet.


(a)

(b) Fig. 9. Results for Example 2

791

792


(a) Cross section

(b) Inside the mesh


(a)

(b)

Fig. 11. Results for Example 3 - linear element case


793

Example 4. Our final example is about 3D problems in which u has a singularity. The model problem is ½ −4u = 0 in Ω = {(−1, 1)2 \ [0.5, 1)2 } × (0, 0.5) . u = gD on ∂Ω Using a cylindrical coordinate system at ( 12 , 21 , 0), the boundary condition gD is chosen such that 1 2θ − π u(r, θ, z) = zr 3 sin , π/2 ≤ θ ≤ 2π. 3 Note that u has an edge singularity along the edge connecting the vertices ( 21 , 12 , 0) and ( 21 , 12 , 21 ). This singularity, beside affecting the smoothness of u, creates pollution error if the mesh near this edge is not sufficiently refined. The initial mesh used in computing uh is shown in Fig. 13. This mesh is finer near the edge singularity to reduce the effect of the pollution error. The numerical results are shown in Fig. 14 where we can see that the PPR continues to perform good and ZZ-PPR is almost exact every where. 6.1. The Performance of the PPR in Uniform Meshes. All of our previous examples focused on Delaunay partitions. Let us now study the performance of the PPR when we switch to uniform meshes. In 2D, we consider solving the problem in Example 2 using uniform meshes constructed by dividing Ω into m × m equal squares and each of these squares is divided into two triangles arranged as in Fig. 15(a). In successive iterations, we use m = 16, 32, 64. For the 3D case, we consider solving the problem in Example 3 using uniform meshes similar to the one depicted in Fig. 15. To construct one of such meshes, divide Ω into m × m × m equal cubes, and then divide each of these cubes into 6 tetrahedrons using Kuhn’s partition illustrated in Fig. 16. For our purposes, we use m = 8, 16, 32 in successive iterations. Before we continue, note that the uniform meshes used in the 2D and the 3D cases are translation invariant, and hence the PPR is bounded in both cases. Various convergence rates of the PPR-recovered gradient are listed in Table 1. From this table, we may observe the following: 1. The rates of k∇u − Gh uh kL2 (Ω) indicate that the PPR enjoys superconvergence. This is expected because Gh is bounded and |Ih u − uh |1,Ω has superconvergence in all of the four cases (see [21] for the linear 2D case, [2] for the 2D quadratic case, [14] for the 3D linear case, and [12] for the 3D quadratic case). 2. Gh uh superconverges to ∇u at all mesh nodes in the linear element cases. A proof of this result in 2D is given in [26], and the argument in this proof can be extended to cover the 3D linear case. 3. Gh uh ultra-converges to ∇u at all internal mesh nodes in the quadratic element cases. This result has no theoretical justification even in 2D.

794


(a)

(b) Fig. 12. Results for Example 3 - quadratic element case



(a)

(b)

Fig. 14. Results for Example 4

795

796


(a) Cross section

(b) Inside the mesh

Fig. 15. A uniform partition of Ω = (0, 1)3

Partition the cube into two prisms

Partition each prism into 3 tetrahedrons

Fig. 16. Kuhn’s partition of the unit cube

Example

2

3

Element

∇ −

∞

Ω

∇ −

Ω

∇ −

Ω

∞

∇ − &

" $

Linear

2

2

2

2

Quadratic

4

3

3

3

Linear

2

2

2

2

Quadratic

4

3

3

3

& $ %

! #

Ω

Table 1. Convergence rates for the PPR-recovered gradient in uniform meshes

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797

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Received October 2004; revised February 2005. E-mail address: [email protected] E-mail address: [email protected]