THE POLYNOMIAL-PRESERVING RECOVERY FOR HIGHER ORDER FINITE ELEMENT METHODS IN 2D AND 3D. A. Naga and Z. Zhang 1. (Communicated by Zhiming Chen)

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1 DISCRETE AND CONTINUOUS Website: DYNAMICAL SYSTEMS SERIES B Volume 5, Number 3, August 2005 pp THE POLYNOMIAL-PRESERVING RECOVERY FOR HIGHER ORDER FINITE ELEMENT METHODS IN 2D AND 3D A. Naga and Z. Zhang 1 College of Mathematics and Computer Science Hunan Normal University Changsha, Hunan , China Department of Mathematics Wayne State University Detroit, MI 48202, USA (Communicated by Zhiming Chen) 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 Babuška 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, Babuška 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 2000 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 DMS and DMS

2 770 A. NAGA AND Z. ZHANG 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 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 Notations. Let s [0, ], p [1, ], m, k, l Z + (the set of positive integers), and A R d with d = 2 or 3. The spaces Wp s (A) and H s (A) are the classical Sobolev spaces equipped with the norms s,p,a and s,a, respectively, and the seminorms s,p,a and s,a, respectively. The measure of A in R d will be denoted by A. For vectors in R m, will denote the Euclidean length. If m = d, we may use instead of.. 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 P m (A) and n m will denote the dimension of P m (A) where n m = 1 d (m+i). Let P d! m(a) P m (A) i=1 denote the subset {p : p P m (A), p is homogeneous, and deg(p) = m}. The space of real matrices of order k l is denoted by R k l. The identity of R k k is denoted by I k, while 1 k (0 k ) denotes the k 1 column vector whose entries are all ones (zeros). Let σ 1 (K) σ 2 (K)... σ max(k,l) (K) 0 denote the singular values of a matrix K R k l. Recall that σm(k) 2 = σ m (K T K) = σ m (KK T ) for all 1 m min(k, l) and σ m (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 E 1 and E 2 be two mathematical quantities and let C > 0 be some constant independent of any essential ingredients of E 1 and E 2. Instead of E 1 CE 2, E 1 CE 2, and E 2 /C E 1 CE 2, we will use the notations E 1 E 2, E 1 E 2, and E 1 = E2, respectively Preliminaries. Consider the boundary value problem (D u) + αu = f in Ω ν (D u) = g N on Γ N (1.1) u = g D on Γ D where Ω R d is a bounded domain with Lipschitz boundary Ω = Γ N Γ D, the boundary segments Γ N and Γ D are disjoint, ν is the unit outward normal vector to Ω, and D is a d d symmetric positive definite matrix. If Γ N = Ω and α = 0, the compatibility condition Ω f + g = 0 must be satisfied and the condition Ω

3 THE POLYNOMIAL-PRESERVING RECOVERY 771 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 where B(u, v) = L(v) v V 0 (1.2) V = {v H 1 (Ω) : v ΓD = g D }, V 0 = {v H 1 (Ω) : v ΓD = 0}, B(u, v) = [(D u) v + αuv], L(v) = fv + g N v. Ω Ω Γ N If Γ D is empty, we take V = H 1 (Ω). Under suitable assumptions [13, pp ], 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 T h be a conforming partition of Ω. The elements in T h are triangles when d = 2 and tetrahedrons when d = 3. To simplify notations, we assume the elements in T h to be closed, i.e., T = T T T h. For T T h, let h T denote the diameter of T and set h = max{h T : T T h }. Also, let ϱ T denote the diameter of the largest ball (circle when d=2 or sphere when d=3) inscribed in T. The regularity of T is measured with the aspect ratio h T ϱ T. The mesh T h is said to be regular if there exists a finite positive constant γ, independent of h, such that For r Z +, define the finite element space h T ϱ T γ T T h. (1.3) S h = {v C(Ω) : v T P r (T ) T T h }. Let N h denote the set of the mesh nodes. A mesh node z N h 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 T h. The basis for S h is the standard Lagrange basis {φ z : z N h } where φ z (ź) = δ zź z, ź N h. Using this basis, a function v S h takes the form v = z N h v(z)φ z. Let I h : C 0 (Ω) S h denote the Lagrange interpolation operator where I h w = w(z)φ z w C 0 (Ω). z N h The finite element approximation of u is the solution u h S h of (1.2) when v varies over S h V, i.e., B(u h, v) = L(v) v S h V. (1.4) 2. The PPR and the ZZ-PPR. Let G h : S h d i=1 S h denote the PPR operator. As in the SPR, the structure of G h u h, the PPR-recovered gradient of u h, relies on the fact that every function in S h is uniquely defined by its values at the mesh nodes. If the values {(G h u h )(z) : z N h } are well-defined, then G h u h z N h (G h u h )(z)φ z. Before exploring the definitions of G h u h at the mesh nodes, we need the following notations. If A is a union of mesh elements in T h and v S h, 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

4 772 A. NAGA AND Z. ZHANG 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., where L(z, 0) {z}. L(z, n) = {T : T T h, T L(z, n 1) } 2.1. Definition of the PPR. Let z N h be a mesh vertex and let denote a patch of mesh elements around z. Let p z P r+1 ( ) be the polynomial that best fits u h at the mesh nodes in in discrete least-squares sense, i.e., (u h p z )( z) 2 = min (u h p)( z) 2. p P r+1 ( ) z N h z N h For easy referencing, p z will be called the least-squares polynomial approximation, LSPA, of u h at z. Then, (G h u h )(z) p z (z). To complete the definition of the PPR, we need to define. Note that besides the dependence of its definition on the location of z, must have at least n r+1 nodes distributed around z in a way that leads to a unique p z. For an internal mesh vertex z, we first define an initial patch,0 such that N(,0 ) n r+1. This is achieved with the following definition:,0 L(z, 1) if N(L(z, 1)) n r+1 L(z, 2) if d = 3, r = 1, and M(L(z, 1)) = 4 L(z, 1) {T T h : 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 = n 2. The second part of the,0 definition takes care of this situation. Although N(,0 ) n r+1, this does not ensure the uniqueness of p z. If,0 does not lead to a unique p z and L(z, n),0 L(z, n + 1), set,0 to L(z, n + 1) and recompute p z. The patch is defined to be the smallest,0 that leads to a unique p z. Next, we define at a boundary mesh vertex z. Let n 0 be the smallest positive integer such that L(z, n 0 ) has at least one internal mesh vertex. Then, L(z, n 0 ) {K z : z L(z, n 0 ) and z is an internal vertex}. This definition ensures the uniqueness of p z 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 G h u h is completely defined. If r > 1, let z N h 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 T h and τ is a d 1 -simplex whose vertices are in N h and d 1 d. Let z 0,..., z d1 be the vertices of τ, then (G h u h )(z) 1 d d 1 i=0 p zi (z) where p zi is the LSPA of u h at z i. This completes the definition of G h u h for r > 1..

5 THE POLYNOMIAL-PRESERVING RECOVERY 773 (a) d=2, r=1 (b) d=2, r=2 Fig. 1. Examples for patches used in the PPR Extend Out Extend Out (a) d=2, r=1 (b) d=3, r=1 Fig. 2. Examples for extending L(z, 1) out, where z is an internal mesh vertex

6 774 A. NAGA AND Z. ZHANG 2.2. The computational aspects of the PPR. Consider a mesh vertex z whose patch is, let v S h, and let p z be the LSPA of v at z. We now outline the details of computing p z. Let z 1, z 2,..., z N(Kz) denote the mesh nodes in. Without loss of generality, let z = z 1 and set h z = max{ z i z 1 : 2 i N( )}. To avoid the computational instability resulting from small h z, the computations will be carried out on the patch ω z where ω z F z ( ) and F z : x ˆx = x z 1 h z. (2.5) The patch ω z will be called the reference patch associated with z. For 1 i N( ), let ẑ i = F (z i ) and set v i = v(z i ). Let ˆp z P r+1 (ω z ) be such that p z = ˆp z F z. For ˆx ω z, ˆp z (ˆx) can be written in the form ˆp z (ˆx) = p(ˆx) T c z (2.6) where c z = [ ] T [ c z,1 c z,2 c z,3 c z,nr+1, p(ˆx) = p1 (ˆx) p 2 (ˆx) p nr+1 (ˆx) ] T, and the set {p l : 1 l n r+1 } is the monomials-basis of P r+1 (R d ). In other words, if y = (y 1,..., y d ) R d and 1 l n r+1, then p l (y) = y α l = d j=1 yα l,j j, where α l = (α l,1,..., α l,d ) is a multi-index. For technical reasons (see the proof Lemma 3.6), an ordering of these monomials must satisfy the condition α l α m and α l,1 α m,1 1 l m n r+1. From the definition of p z, one can show that c z is the solution of the linear system where A z Setting B z = A T z A z, we have p(ẑ 1 ) T p(ẑ 2 ) T.. p(ẑ N(Kz )) T A T z A z c z = A T z v Kz (2.7) and v v 1 v 2. v N(Kz). (2.8) c z = B 1 z A T z v Kz. (2.9) Note that B z is a real symmetric positive-definite matrix of order n r+1 n r+1. The following lemma lists the basic facts about discrete least-squares. For the proof of this lemma, see [22, pp ]. Lemma 2.1. Let z T h be a mesh vertex whose patch is. Then, the following are equivalent. 1. p z is unique. 2. Rank(A z ) = n r The matrix B z is invertible. 4. There exists no q P r+1 ( ) such that q(z i ) = 0 for i = 1, 2,..., N( ). The equivalence of the first three claims in Lemma 2.1 is somewhat obvious, at least from the definitions of p z, A z, and B z. 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 by a polynomial of degree r + 1 leads to a unique polynomial p z if and only if the nodes in 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.

7 THE POLYNOMIAL-PRESERVING RECOVERY The ZZ-PPR error estimator. Following the standard notations, let η h denote the ZZ-PPR error estimator of e h u u h. As usual, η h is obtained from the error indicators {η T,h : T T h } where η T,h G h u h u h 2 and η h ηt,h 2. T T T h Note that the ZZ-PPR, like the ZZ-SPR, is obtained from the ZZ error estimator by using the PPR-recovered gradient. By definition, η T,h is a good approximation of e h 1,T if and only if G h u h is a good approximation of of u in T Properties of the PPR. The PPR has the following properties. 1. By definition, G h is linear. 2. G h satisfies the consistency condition, i.e., G h (I h p) = p p P r+1 (Ω). (2.10) This fact is proved in [26]. Consequently, G h is a polynomial-preserving operator and enjoys the approximation property u G h (I h u) L2 (Ω) h r+1 u r+2,ω u H r+2 (Ω). For more details about proving this property, see [13, pp ]. 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 P r+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], G h u h superconverges to u if the following conditions are satisfied. (a) The recovery operator G h is bounded in the following sense: where G h v L2 (T ) v 1,KT T T h and v S h (2.11) K T { : z is a vertex of T } is the patch corresponding to T. This condition is called the boundedness condition. (b) u h enjoys superconvergence in the following sense: (I h u u h ) L2(Ω) C(u)h r+ρ (2.12) for some ρ (0, 1], i.e., u h is a perturbation of (I h 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 G h u h superconverges to u as u G h u h = u G h (I h u) + G h (I h u u h ). 4. As in any recovery technique, if G h u h 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 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.

8 776 A. NAGA AND Z. ZHANG 3. The Boundedness of G h. Definition 3.1. Let T h be a partition of Ω. Then, T h 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 L(z, n) z Ω ( Ω). Also, T h 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 σ nr+1 (B z ) C z Ω ( Ω) for some constant C > 0 that depends only on r and γ. If T h 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 section we will implicitly assume that T h satisfies the finite layer condition in Ω. In practical meshes, this condition is easily satisfied. Indeed, our numerical experiments showed that 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, 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 meshes as we shall see in Section 4. Unfortunately, establishing that in 3D 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 T h be a regular partition of Ω and let z be a mesh vertex whose patch is. Then, there exists a constant C = C(γ) > 0 such that h T Ch z T. 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 T h 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 σ 1 (B z ) C where z is any mesh vertex. Moreover, T h satisfies the minimum singular value condition on Ω.

9 THE POLYNOMIAL-PRESERVING RECOVERY 777 Proof. Let z N h be a mesh vertex whose patch is and whose reference patch is ω z. From the definition of B z and the fact that ω z is contained in the unit ball, it is easy to verify that B z (i, j) N( ) + 1 for 1 i, j n r+1. Hence, σ 1 (B z ) σ 1 ( B z ) cn( ) for some constant c > 0 that depends only on r. Since T h is regular, Lemma 3.5 implies that N( ) N 0 < for some constant positive integer N 0 = N 0 (γ), and the first claim is true. To prove the second claim, let z Ω be a mesh vertex. By construction, has at least one internal mesh vertex y such that K y. Set ŷ = (ŷ 1,..., ŷ d ) = F z (y) and set ω = F z (K y ). Without loss of generality, assume that ŷ is on the positive ˆx 1 -axis; otherwise rotate ω z. After reordering the nodes in if necessary, [ ] Ãy A z = where the rows in Ãy correspond to the mesh nodes in ω y. Hence, Ā z B z = A T z A z = ÃT y Ãy + ĀT z Āz = B y + B z where B y = ÃT y Ãy and B z = ĀT z Āz. Since B z, B y, and B z are positive semi-definite, Since F z ( ) = h y h z F y ( ) + ŷ, σ nr+1 (B z ) σ nr+1 ( B y ). (3.13) ω y = h y h z ω y + ŷ. (3.14) This relation induces a linear transformation Q : P r+1 (ω y ) P r+1 ( ω y ). Using the monomial bases of P r+1 (ω y ) and P r+1 ( ω y ), Q = Q 2 Q 1 where Q 1, Q 2 R n r+1 n r+1 ( ( )). hy The matrix Q 1 is diag p and it represents the scaling part of (3.14), while h z Q 2 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 Q 2 are monomials in ŷ 1, i.e., σ nr+1 (Q 2 ) is a continuous function of ŷ 1. Since det(q 2 ) = 1 for all ŷ 1 [0, 1], σ nr+1 (Q 2 ) 1. Using the definitions of Q, B y, and Ãy, one can verify that Therefore, Ã T y = Q 2 Q 1 A T y B y = Q 2 Q 1 B y (Q 2 Q 1 ) T. b T By b σ nr+1 (B y ) Q T 1 Q T 2 b 2 σ nr+1 (B y )σ 2 n r+1 (Q 1 ) Q T 2 b 2 σ nr+1 (B y )σ 2 n r+1 (Q 1 )σ 2 n r+1 (Q 2 ) b 2 where b R nr+1. By virtue of (3.13), we get σ nr+1 (B z ) σ nr+1 ( B y ) σ nr+1 (B y )σ 2 n r+1 (Q 1 )σ 2 n r+1 (Q 2 ). Since T h satisfies the minimum singular value condition in Ω, then σ nr+1 (Q 1 ) = ( ) nr+1 hy, and h y h z by Lemma 3.5, the second claim is true. h z The following two lemmas are the key tools in establishing the boundedness of G h.

10 778 A. NAGA AND Z. ZHANG Lemma 3.7. Let T h be a regular partition of Ω that satisfies the minimum singular value condition in Ω. Let z be a mesh vertex whose patch is and let v S h. If v = 0, then there is a constant C > 0 that depends only on γ and r such that v 1,Kz = h (d/2 1) z v Kz. Proof. The proof uses three facts. Firstly, mesh regularity and Lemma 3.5 imply ϱ T = hz T. (3.15) Secondly, for any T T h, there exists an invertible affine mapping F T : ˆT T where ˆT is a reference element. Set ˆv T = (v T ) F T and let J T = DF T where DF T R d d is the Jacobian of F T. Since T h is regular, then { ht σ d (J T ) σ 1 (J T ) h T det(j T ) = h d (3.16) T for all T T h. Note that N(T ) = N( ˆT [ ) = n r, ˆv T = (v T ) T ˆφ, ˆφ = ˆφ1 ˆφ2 ˆφ ] T nr, and ˆφ i is the standard Lagrange-basis function associated with the ith node in ˆT for i = 1, 2,..., n r. Finally, the third fact says that det(j T ) 1 2 σd (J 1 T ) ˆv T 1, ˆT v 1,T det(j T ) 1 2 σ1 (J 1 T ) ˆv T 1, ˆT (3.17) for any T T h and for any v S h. For a comprehensive discussion about the inequalities in (3.16) and (3.17), the reader is referred to [13, pp ]. Let T 1, T 2,..., T M(Kz) denote the mesh elements in. Using (3.15)-(3.17), v 1,Tk It is easy to verify that = h (d/2 1) z ˆv Tk 1, ˆT for k = 1, 2,..., M( ). (3.18) ˆv Tk 2 1, ˆT = (v T k ) T Qv Tk for k = 1, 2,..., M( ) where Q R n r n r and Q(i, j) = ˆT ˆφ i ˆφ j for i, j = 1, 2,..., n r. Using (3.18), v 2 1, M(K z) = h (d 2) z v T T k Qv Tk. (3.19) For 1 k M( ), there exists a Boolean matrix E k R n r N( ) such that v Tk = E k v Kz, where { 1 if node i in Tk is node j in K E k (i, j) = z 0 otherwise for 1 i n r and 1 j N( ). Therefore, (3.19) takes the form where Q v 2 1, M( ) k=1 k=1 M( ) = h (d 2) z v T (Ek T QE k )v Kz E T k QE k k=1 = h (d 2) z v T QvKz (3.20) is a symmetric positive semi-definite matrix of order N( ) N( ). Also, Q has only one zero eigenvalue corresponding to 1N(Kz ) as v 1,Kz = 0 v Kz is identically constant. Since the eigenvectors of Q form an orthonormal basis of R N(), we can write v Kz = v (1) + v (2)

11 THE POLYNOMIAL-PRESERVING RECOVERY 779 where v (1) c1 N(Kz ) for some constant c and v (2) is a combination of the rest of the eigenvectors of Q. Consequently, v Kz = v (1) + v (2) where v (1) = c and the nodal values of v (2) are the entries of v (2). Note that c = v (1) / N( ), and v (1) = v (2) as v = 0. From the definitions, it is straightforward to verify that v (1) = v(1) N(Kz ) and v (2) v (2). Therefore, Thus, v Kz 2 = v (1) 2 + v (2) 2 (1 + N( )) v (2) 2. σ (N(Kz ) 1)( Q) v Kz 2 (2) σ (N(Kz) 1)( Q) v K 1 + N( ) z 2 ( v (2) ) T ( ) Q v (2) = v T QvKz σ 1 ( Q) v Kz 2. (3.21) Note that σ (N(Kz ) 1)( Q) and σ 1 ( Q) depend only on r and N( ). Since N( ) is bounded by Lemma 3.5, combining (3.19)-(3.21) concludes the proof. Remark 3.8. Note that v 1,Kz and v Kz in Lemma 3.7 define norms of v Kz. Since S h C( ) 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 h z and depends only on r and the geometry of. 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 T h 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 } and the coefficients in this combination are bounded uniformly by a constant independent of z, h, and v. Lemma Let T h be a partition of Ω and let T h satisfy the assumptions in Lemma 3.6. Let z be a mesh vertex whose patch is and consider v S h. If p z P r+1 ( ) is the LSPA of v at z, then p z 1,,Kz Ch d/2 z v 1,Kz. where C is a bounded constant that depends only on r and γ. Proof. At first assume that v = 0. Let ω z be the reference patch corresponding to z and let ˆp z P r+1 (ω z ) be such that p z = ˆp z F z. Let x be a point in and set ˆx = F z (x). From (2.6) and from the fact that ˆx ω z, Using (2.9), p z (x) = h 1 z ˆp z (ˆx) h 1 c z c z σ 1 (B 1 z )σ 1 (A T z ) v z = (σ nr+1 (B z )) 1 σ 1 (B z ) v z Hence, and by Lemma 3.6, c z v z z

12 780 A. NAGA AND Z. ZHANG and, consequently, p z (y) h 1 z v z. Therefore, and by virtue of Lemma 3.7, p z (y) h d/2 z v 1,Kz. Thus, the lemma conclusion is true if v = 0. If v 0, set ṽ = v 1 v and let p z be the LSPA of ṽ at z. Since ṽ = 0, p z 1,,Kz h d/2 z ṽ 1,Kz. This concludes the proof as p = p and v = ṽ. The following theorem is the main result of this section. Theorem Let T h be a partition of Ω that satisfies the assumptions in Lemma 3.6. Let T be a mesh element whose patch is K T and let v S h. Then, G h v L2 (T ) C v 1,KT. where C is a constant that depends only on r and γ. Proof. Let {z i : 1 i n r } be the mesh nodes in T such that z 1,..., z d+1 are the vertices of T. Since (G h v) T P r (T ), then G h v L (T ) max{ (G h v)(z i ) : 1 i n r }. By Lemma 3.10 and the definitions of G h v at mesh nodes, Hence, G h v L (T ) max{h d/2 z i v 1,Kzi : 1 i d + 1}. G h v L2(T ) G h v L (T ) T max{h d/2 T h d/2 z i v 1,Kzi : 1 i d + 1}. Since T i for all 1 i d + 1, h T h zi. Therefore, G h v L2 (T ) max{ v 1,Kzi : 1 i d + 1} d+1 and the proof is complete by virtue of Lemma 3.5. i=1 v 2 1,i 4. Verifying the Minimum Singular Value Condition in 2D Meshes. We have seen that G h is bounded in the sense of (2.11) given that T h 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 G h 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 T h 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 T h is said to satisfy the angle condition in Ω if L(z, 1) satisfies the angle condition at every mesh vertex in Ω.

13 THE POLYNOMIAL-PRESERVING RECOVERY 781 (a) (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 T h 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(r 2 ) and let z 0 R 2. If p(z 0 ) = 0, then p is 0 at any point on the line through z 0 and the origin. 2. Let p P m (R 2 ), and consider the distinct points {z i R 2 : i = 1, 2,..., n m }. 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(z i ) = 0 for i = 1, 2,..., n m, then p is identically zero. Proof. For the first claim, note that any point on the line through z 0 and the origin is a multiple of z 0. For the proof of the second claim, see [11, pp. 64]. Let z be an internal mesh node and let {z i = (x 1,i, x 2,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 z 1 = z. For (x 1, x 2 ) L(z, 1), define the polynomials l j (x 1, x 2 ) = (x 1 x 1,1 )(x 2,j+1 x 2,1 ) (x 2 x 2,1 )(x 1,j+1 x 1,1 ) (4.22) 1 j M(L(z, 1)), and p z,l = {l j : 1 j M(L(z, 1)), l j is not a multiple of l k k < j}. Note that l j = 0 is the equation of the line passing through the nodes z 1 and z j+1. Also, note that p z,l is the product of linear polynomials representing non-collinear triangles sides meeting at z 1 = z and that 2 deg(p z,l ) M(L(z, 1)). The proof of the following theorem is in [20].

14 782 A. NAGA AND Z. ZHANG 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 P 2 (L(z, 1)) such that p(z i ) = 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 p z,l if M(L(z, 1)) = 4. Before we continue, let us go over some properties of the matrix r r+1 r r r 2 r 1 (r 1) r+1 (r 1) r (r 1) 2 (r 1) 1 D r+1 2 r (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 D 1 and D 2 be the sub-matrices of D obtained by omitting the r + 1st and rth columns, respectively. Then, Rank(D 1 ) = Rank(D 2 ) = r. Proof. We start with D 1 and proceed by contradiction. If Rank(D 1 ) r, then r 1 there exists a non-zero polynomial q P r+1 (R) of the form q(ξ) = ξ 2 a i ξ i such that q(j) = 0 for 1 j r. This is impossible as q has only r 1 non-zero roots. Next, consider D 2. If Rank(D 2 ) r, then there exists a non-trivial polynomial r q P r+1 (R) of the form q(ξ) = ξ a i ξ i such that q(j) = 0 for 1 j r and i=0 a 1 = 0. If a r = 0, one can proceed as in the previous case. So, assume that a r 0. Without loss of generality, set a r = 1. Using the zeros of q, we get r r q(ξ) = ξ a i ξ i = ξ (ξ j). Comparing the two expressions of q leads to a contradiction as r 0 = a 1 = ( 1) r 1 1 (r!) j 0. i=0 j=1 j=1 The following corollary is a direct result of Proposition 4.6. Corollary 4.7. The reduced-row echelon form of D is 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 P r+1 (L(z, 1)) such that p(z i ) = 0 for 1 i N(L(z, 1)). If deg(p) = r + 1 and deg(p z,l ) r + 1, then there exits a polynomial q such that p = qp z,l ; otherwise p is identically 0. i=0

15 THE POLYNOMIAL-PRESERVING RECOVERY 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 n r 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 = z 1 = r+1 (0, 0). Then, p = p k where p k P k (L(z, 1)). We claim that p k(z i ) = 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 z 1 z i has r + 1 uniformly spaced nodes of the form jz i /r for j = 0, 1,..., r. Since p is zero at these nodes, we have r+1 ( ) k j p k (z i ) = 0 for j = 1, 2,..., r. r k=1 Setting p k = r k p k, we get r+1 j k p k (z i ) = 0 for j = 1, 2,..., r. k=1 This system of equations is actually a linear system of the form p r+1 (z i ) p r (z i ) D. = 0 r (4.24) p 2 (z i ) p 1 (z i ) where D is the Vandermonde matrix defined in (4.23). By virtue of Corollary 4.7, there exists a constant β = β(r) 0 such that p 2 (z i ) + β p 1 (z i ) = 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 (z i ) in the linear system in (4.24). This leads to a homogeneous linear system whose coefficients matrix is D 1. By Proposition 4.6, D 1 is non singular. Hence, p k (z i ) = 0 for k = 2, 3,..., r + 1 and i = 2, 3,..., M(L(z, 1)) + 1. By the first part of Proposition 4.4, p z,l is a factor of p k for k = 2,..., r + 1. Consequently, p k is identically zero for k < deg(p z,l ) and p k = q k p z,l for some polynomial q k when k deg(p z,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 P r+1 (L(z, 1)) such that p(z i ) = 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(p z,l ) 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 = l 1 q for some q P r(l(z, 1)) and l 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 z 1 and z 2 as depicted in Fig. 4(a). Therefore, L(z, 1) has a triangle T that intersects with the line through

16 784 A. NAGA AND Z. ZHANG z 1 and z 2 only at z 1. Hence, l 1 (x 1, x 2 ) 0 for all (x 1, x 2 ) T \{z 1 }. 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(p z,l ) = 2, Lemma 4.8 implies that p = qp z,l for some q P 1 (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 n r+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 Let z be an internal mesh vertex, and assume that L(z, 1) satisfy the angle condition and has no degenerate triangles. Then, = L(z, 1) for all r > 1, i.e., B z is invertible. Proof. This is obvious from Lemma 2.1 and Theorem 4.9. Theorem Let T h be a regular partition of Ω, and assume that the sum of any two adjacent angles in T h is at most π. Then, T h satisfies the minimum singular value condition in Ω for all r > 1. Proof. Consider an internal mesh vertex z N h. By assumption, L(z, 1) satisfies the angle condition and has no degenerate triangles. Hence, = L(z, 1) by Corollary The regularity of T h implies that any triangle angle is at least θ 0 (0, π) where θ 0 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 < θ 0 θ π 2θ 0 where θ is any angle in any triangle in T h. Let Θ z R 3M(Kz) denote the vector of all the triangles angles in. Note that Θ z [θ 0, π 2θ 0 ] 3M(Kz). Consider the set S z defined by S z = {Θ z [θ 0, π 2θ 0 ] 3M(Kz) : L(z, 1) satisfies the angle condition}. We claim that S z is compact. This is true if S z is a closed subset of [θ 0, π 2θ 0 ] 3M(Kz) as the latter is compact. For that, proceed by contradiction and assume that S z is not closed. Then, there exists a convergent sequence {Θ (m) z : m > 0} such that Θ (0) z = lim m Θ(m) z corresponds to L(z, 1) (0) that does not satisfy the angle condition. Let L(z, 1) (m) correspond to the angles in Θ (m) z for m 1. Since L(z, 1) (0) does not satisfy the angle condition, it has two adjacent angles θ (0) 1 and θ (0) 2 such that θ (0) 1 + θ (0) 2 > π. Hence, there exists m 0 > 0 such that θ (m 0) 1 + θ (m 0) 2 > π.

17 THE POLYNOMIAL-PRESERVING RECOVERY 785 This is a contradiction as L(z, 1) (m0) satisfies the angle condition by assumption. Let ω z be the reference patch corresponding to z. Since ω z has at least one mesh vertex z on the unit circle, ω z is completely defined by z and the angles in ω z. Without loss of generality, assume that z = (1, 0). Then, B z is a continuous function of all the angles in ω z. By the compactness of S z and the angle condition, σ nr+1 (B z ) achieves a minimum value σ 0 > 0. Note that σ 0 > 0 depends only on M( ) which is finite by the regularity of T h. Hence, σ 0 is independent of z, and the proof is complete. Corollary Under the conditions in Theorem 4.11, G h 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 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. z θ 3 θ 1 θ 4 θ 2 θ 3 z θ 1 θ 4 θ 2 (a) (b) Fig. 5. Regular refinement preserves the angle condition Remark 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 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 G h, 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 G h 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 x 1 -axis. We set a 1 = a cos θ and a 2 = a sin θ. Let T be a triangle in T and let K T be its corresponding patch as in Fig. 6(a). The nodes in K T are labelled as z 1, z 2,..., z 12. 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 G xi v denote the x i -component of G v. Note that G xi v at any node z is a weighted sum of the nodal values of v in. Since T is translation invariant, the set

18 786 A. NAGA AND Z. ZHANG z 11 z 12 z 8 z 9 z 10 z 4 z 5 z 6 z 7 T x2 z 1 z 2 z 3 (a) T and K T x1 θ a 1 (b) Dimensions a 1 6a a 1 6a a 1 6a a 1 6a a 1 6a 2 (c) The PPR weights a 1 6a 2 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 G xi v. We claim that G xi v L2(T ) C i xi v L2(K T ) for i = 1, 2 (5.25) where C 1, C 2 are constants independent of a and θ. Establishing this claim will imply that the mesh regularity is not necessary for the boundedness of G h. To prove this claim, we need the following proposition. Proposition 5.1. If xi v = 0 in K T, then G xi v = 0 in T.

19 THE POLYNOMIAL-PRESERVING RECOVERY 787 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 K T that are parallel to x 2 -axis. By a fiber parallel to x 2 -axis, or simply x 2 -fiber, we mean a line segment l K T such that l is parallel to x 2 -axis and l K T. We have one of two cases. Case 1: θ = π/2. In this case, K T has 4 of its x 2 -fibers composed of edges in K T. Denote these fibers by l 1, l 2, l 3, and l 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 K T is linear and { v (m) 1 if z lm (z) = 0 otherwise for all z K T. Using the weights in Fig. 6(c), one can show that (G x2 v (m) ) T = 0. Therefore, (G x2 v) T = 0. Case 2: θ (0, π/2). In this case, none of the x 2 -fibers of K T is a union of edges in K T. Hence, v KT = c 1 + c 2 x 1 for some real constants c 1 and c 2. Since the PPR satisfies the consistency condition, (G x2 v) T = x2 (c 1 + c 2 x 1 ) = 0. Let v k = v(z k ) for 1 k 12 and let v = [ v 1 v 2 v 12 ] T. After some lengthy algebraic manipulations, one can show that G xi v 2 L 2(T ) = (a 2) 3 2i v T A i v and xi v 2 L 864 2(K T ) = (a 2) 3 2i v T B i v. 2 The matrices A 1 and B 1 are constant while the entries of A 2 and B 2 are quadratic polynomials in a 1. Indeed, and A 2 = a 2 1A 1 + a 1  2 + P A 1 P T B 2 = a 2 1B 1 + a 1 ˆB2 + P A 1 P T. The constant matrices A 1, B 1, Â2, and ˆB 2 are listed in [19, pp ], while the matrix P is a permutation matrix obtained from I 12 by interchanging the rows 1 with 12, 2 with 10, 3 with 7, 4 with 11, and 5 with 9. Note that K T is invariant under the reflection on the line connecting z 6 and z 8. This explains why P is part of the formulas of A 2 and B 2. Also, note that A i is positive semi-definite and G xi v L2(T ) = 0 if and only if v Null(A i ). Similarly, xi v L2(K T ) = 0 if and only if v Null(B i ). Let E i : R 12 Null(B i ), let m i denote the dimension of Null(B i ), and set v i = v E i v. Note that v i is a linear combination of the eigenvectors of B i that correspond to the nonzero eigenvalues. Taking these vectors as columns of a matrix Ẽi R 12 (12 mi), we write v i = Ẽiṽ i. Hence, and by virtue of Proposition 5.1, and G xi v 2 L 2(T ) = (a 2) 3 2i 864 v T i A i v i = (a 2) 3 2i ṽ T i à i ṽ i 864 xi v 2 L 2 (K T ) = (a 2) 3 2i v T i B i v i = (a 2) 3 2i 2 2 where Ãi = ẼT i A iẽi and B i = ẼT i B iẽi. Hence, ṽ T i B i ṽ i G xi v 2 L 2(T ) xi v 2 L 2(K T ) = f i(ṽ i )

20 788 A. NAGA AND Z. ZHANG where f i : R 12 mi R is defined by f i (ṽ i ) = ṽt i à i ṽ i ṽ T. i B i ṽ i By definition, it is clear that B i is nonsingular. Moreover, the maximum value of f i is the maximum eigenvalue, λ max, of the generalized eigenvalue problem à i ṽ i = λ B i ṽ i. (5.26) It is clear that λ max is a function of a 1. Consequently, (5.25) is true if there exists a constant λ > 0 such that λ max λ and λ is independent of a 1. According to i and θ, we have the following three cases. Case 1: i = 1, θ [0, π/2]. Since A 1 and B 1 are constant, Ã1 and B 1 are constant too, i.e., the eigenvalues of (5.26) are constant (indeed λ max = 38). Hence, G x1 v L2(T ) C x1 v L2(K T ) where C is independent of a and θ. Case 2: i = 2, θ = π/2. In this case a 1 = 0 and hence A 2 and B 2 are constant. Proceeding as in Case 1, there is a constant C > 0, independent of a, such that G x2 v L2(T ) C x2 v L2(K T ). Case 3: i = 2, θ (0, π/2). Getting the eigenvectors of B 2 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 x2 v L2 (K T ) = 0 if and only if v KT = c 1 +c 2 x 1 for some constants c 1, c 2 R. Consequently, if v = 0 and x1 v = 0, (5.27) K T K T then x2 v L2(K T ) = 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 B 2 corresponding to nonzero eigenvalues. Hence, the members of a basis of this null space can serve as the columns of a matrix E that is similar to Ẽ2 (see [19, pp. 83] for one possible form of E). Without loss of generality, set Ẽ2 = E. After some algebraic manipulations, the characteristic equation of Ã2ṽ 2 = λ B 2 ṽ 2 is where 0 = λ 7 [g 1 (λ) + g 2 (λ)g 3 (a 1 )] g 1 (λ) = 6λ 3 281λ λ 3136, g 2 (λ) = 2λ 3 90λ λ 702, and a 2 g 3 (a 1 ) = 1(1 + a 1 ) 2 a a a a a a The nonzero roots of g 1 (λ) + g 2 (λ)g 3 (a 1 ) are perturbations of the roots of g 1 (λ) as g 3 L (R 1 ) is very small (indeed g 3 (a 1 ) < 0.15 a 1 > 0). The roots of g 1 (λ) are 7 2 and 65± and the perturbations are within 0.1. Hence, G x2 v L2(T ) C x2 v L2(K T ) where C is a finite constant that is independent of a and θ.

21 THE POLYNOMIAL-PRESERVING RECOVERY 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 e h. 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, N h is partitioned into N h,1 N h,2 where N h,1 = {z N h : dist i (z, Ω) H i for 1 i d} for some positive constants H 1,..., H d, and dist i ( ) denotes the distance along x i - axis. Accordingly, Ω is partitioned into Ω 1,h Ω 2,h where Ω 1,h = {T T h : T has all of its vertices in N h,1 }. Let A Ω be a union of a set of mesh elements in T h. Then, the ZZ-PPR error indicator in A is defined by η h,a = G h u h u h L2 (A). We use the effectivity index κ h,a to measure the accuracy of η h,a, where η h,a κ h,a =. (u u h ) L2 (A) To trace the accuracy of the ZZ-PPR in each of the mesh elements in A, we will use the mean, µ h,a, and the standard deviation, σ h,a, of the effectivity indices in these elements. If the ZZ-PPR is asymptotically exact in each of the elements in A, then lim µ h,a = 1 and lim σ h,a = 0. Recall that h 0 h 0 µ h,a 1 M(A) T A κ h,t and σ 2 h,a 1 M(A) (κ h,t µ h,a ) 2. T A In all of the following examples, the nodes in N h are partitioned using H 1 = H 2 = H 3 = 1 8, except for the last example where we use H 3 = 1 4. Example 2. In this example, we consider the model problem { u + 100u = 0 in Ω = (0, 1) 2 u = g D on Ω and u = cosh(10x 1) + cosh(10x 2 ). This function has a peak at the corner (1, 1), 2 cosh(10) and this peak may reduce the accuracy of u h near (1, 1). The finite element solution u h is computed using the quadratic element, and T h is obtained by the Delaunay triangulation. The initial mesh is shown in Fig. 7. In successive iterations, the new mesh is obtained from the previous one by regular refinement of all elements. Regular refinement of a single triangle is depicted in Fig. 8(a). Note that the initial mesh in Fig. 7 satisfies the angle condition in Ω. By Corollary 4.12, G h is bounded. The results in Fig. 9(a) shows that I h u u h 1,Ω has superconvergence. This explains the superconvergence in G h u h. It is clear that the PPR is better than the SPR and that the PPR performs well in both Ω 1,h and Ω 2,h. Finally, Fig. 9(b) shows that the ZZ-PPR is almost exact everywhere. 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.

22 790 A. NAGA AND Z. ZHANG 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 T h satisfies the condition (ρ 1, ρ 2 ) for some ρ 1, ρ 2 > 0 if T h admits a partition consisting of two sets that satisfy the following. In the first set, every pair of adjacent triangles form an O(h 1+ρ 1 ) parallelogram for some ρ 1 > 0 (An O(h δ ) parallelogram, δ > 0, is a quadrilateral in which the distance between the mid-points of its diagonals is O(h δ ).) In the second set, the triangles total area is O(h ρ2 ) for some ρ 2 > 0. The ideal value for ρ 1 or ρ 2 is. Note that ρ 1 = ρ 2 = if any edge in any triangle in T h is parallel to one of three fixed directions. This ideal case was covered in [10] where the authors completely determined the expansion of u u h at mesh vertices. The case ρ 1 = 2 and ρ 2 = is another interesting case that was studied in [17, 16], while the general case was treated in [25]. If a triangular mesh T h satisfies the condition (ρ 1, ρ 2 ) for some ρ 1, ρ 2 > 0 and if S h consists of piecewise linear polynomials, then I h u u h 1,Ω is O(h 1+ρ ) where ρ = 1 2 min(1, 2ρ 1, ρ 2 ); see [25] for the details. Example 3. Let us now consider a 3D example. The model problem is { u = 3π 2 sin(πx 1 ) sin(πx 2 ) sin(πx 3 ) in Ω = (0, 1) 3. u = 0 on Ω The solution of this problem is u(x 1, x 2, x 3 ) = sin(πx 1 ) sin(πx 2 ) sin(πx 3 ). We will solve this problem using linear and quadratic elements on the 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, I h u u h 1,Ω has superconvergence, which may explain the the good performance for the PPR and the ZZ-PPR. Unfortunately, the superconvergence in I h u u h 1,Ω has not been justified theoretically yet.

23 THE POLYNOMIAL-PRESERVING RECOVERY 791 (a) (b) Fig. 9. Results for Example 2

24 792 A. NAGA AND Z. ZHANG (a) Cross section (b) Inside the mesh Fig. 10. The initial Delaunay mesh used in Example 3 (a) (b) Fig. 11. Results for Example 3 - linear element case

25 THE POLYNOMIAL-PRESERVING RECOVERY 793 Example 4. Our final example is about 3D problems in which u has a singularity. The model problem is { u = 0 in Ω = {( 1, 1) 2 \ [0.5, 1) 2 } (0, 0.5). u = g D on Ω Using a cylindrical coordinate system at ( 1 2, 1 2, 0), the boundary condition g D is chosen such that u(r, θ, z) = zr 2 2θ π 3 sin, π/2 θ 2π. 3 Note that u has an edge singularity along the edge connecting the vertices ( 1 2, 1 2, 0) and ( 1 2, 1 2, 1 2 ). 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 u h 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. The performance of the PPR in uniform meshes. All of our previous examples focused on the 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 u G h u h L2 (Ω) indicate that the PPR enjoys superconvergence. This is expected because G h is bounded and I h u u h 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. G h u h 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. G h u h ultra-converges to u at all internal mesh nodes in the quadratic element cases. This result has no theoretical justification even in 2D.

26 794 A. NAGA AND Z. ZHANG (a) (b) Fig. 12. Results for Example 3 - quadratic element case

27 THE POLYNOMIAL-PRESERVING RECOVERY 795 Fig. 13. The initial Delaunay mesh used in Example 4 (a) (b) Fig. 14. Results for Example 4

28 796 A. NAGA AND Z. ZHANG (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 Element I u h G u h h L ( Ω ) 1, h u G u h h L Ω ( ) 2 1, h I u h G u h h L ( Ω ) 2, h u G u h h L Ω ( ) 2 2, h 2 3 Linear Quadratic Linear Quadratic Table 1. Convergence rates for the PPR-recovered gradient in uniform meshes REFERENCES [1] M. Ainsworth and J. T. Oden, A Poseriori Error Estimation in Finite Element Analysis, Wiley Interscience, New York, [2] A. B. Andreev and R. D. Lazarov, Superconvergence of the gradient for the triangular finite element, Numer. Methods Partial Differential Equations, 4 (1988),