POLYNOMIAL PRESERVING GRADIENT RECOVERY AND A POSTERIORI ESTIMATE FOR BILINEAR ELEMENT ON IRREGULAR QUADRILATERALS

INTERNATIONAL JOURNAL OF NUMERICAL ANALYSIS AND MODELING Volume, Number, Pages 24 c 2004 Institute for Scientific Computing and Information POLYNOMIAL PRESERVING GRADIENT RECOVERY AND A POSTERIORI ESTIMATE FOR BILINEAR ELEMENT ON IRREGULAR QUADRILATERALS ZHIMIN ZHANG Abstract. A polynomial preserving gradient recovery method is proposed and analyzed for bilinear element under quadrilateral meshes. It has been proven that the recovered gradient converges at a rate O(h +ρ ) for ρ = min(α, ), when the mesh is distorted O(h +α ) (α > 0) from a regular one. Consequently, the a posteriori error estimator based on the recovered gradient is asymptotically exact. ey Words. Finite element method, quadrilateral mesh, gradient recovery, superconvergence, a posteriori error estimate.. Introduction A posteriori error estimation is an active research area and many methods have been developed. Roughly speaking, there are residual type error estimators and recovery type estimators. For the literature, readers are referred to recent books by Ainsworth-Oden [2] and by Babuška-Strouboulis [4], a conference proceeding [6], a survey article by Bank [5], an earlier book by Verfürth [23], and references therein. While residual type estimators have been analyzed extensively, there is only limited theoretical research on recovery type error estimators (see, e.g., [2, Chapter 4], [6, 7, 9, 0, 5, 22, 28, 29]). Yet, recovery type error estimators are widely used in engineering applications and their practical effectiveness has been recognized by more and more researchers. Currently, ZZ patch recovery is used in commercial codes, such as ANSYS, MCS/NASTRAN- Marc, Pro/MECHANICA (a product of Parametric Technology), and I- DEAS (a product of SDRC, part of EDS), for the purpose of smoothing and adaptive re-meshing. It is also used in NASA s COMET-AR (COmputational MEchanics Testbed With Adaptive Refinement). In a computer based investigation [4] by Babuška et al., it was found that among all error estimators tested (including the equilibrated residual error estimator, the ZZ patch recovery error estimator, and many others), the ZZ patch recovery error estimator based on the discrete least-squares fitting is the most robust. 2000 Mathematics Subject Classification. 65N30, 65N5, 4A0, 4A25, 4A27, 4A63. This research was partially supported by the National Science Foundation grants DMS-007430 and DMS-03807.

2 ZHIMIN ZHANG It is worth pointing out that the recovery type error estimator was originally based on finite element superconvergence theory, in hopes that a recovered gradient was superconvergent and hence could be used as a substitute of the exact gradient to measure the error. The reader is referred to [4,, 6, 8, 25, 34] for literature regarding superconvergence theory. In order to prove superconvergence, it is necessary to impose some strong restrictions on mesh, which are usually not satisfied in practice. Nevertheless, it is found that in many practical situations, recovery type error estimators perform astonishingly well under meshes produced by the Delaunay triangulation. Mathematically, this fact has not yet been rigorously justified. In a recent work, Bank-Xu [6, 7] introduced a recovery type error estimator based on global L 2 -projection with smoothing iteration of the multigrid method, and they established asymptotic exactness in the H -norm for linear element under shape regular triangulation. However, the recovery operator is a global one. On the other hand, Wang proposed a semi-local recovery [27] and proved its superconvergence under the quasi-uniform mesh assumption. The main feature of his method is to apply L 2 projection on a coarser mesh with size τ = Ch α with α (0, ). Consequently, there is no upper bound for the number of elements in an element patch when mesh size h 0. As for element-wise recovery operators, Schatz-Wahlbin et al. [5, 22] established a general framework which requests, for linear element, given a fixed 0 < ɛ <, that m = C ( (H h ) 2 ( h h ɛ + H ) ) ɛ ln H <. h Here h is the size of element τ, H 2h is the size of the patch ω τ (surrounding τ), where the recovery takes place, and C is an unknown constant which comes from the analysis. Let H = Lh. In order for m <, we need C(L 2 h ɛ + L ɛ ln L) <. Depending on C, this essentially asks for sufficiently large L and sufficiently small h, which implies many elements may be needed for the recovery operator. Nevertheless, in practice, many recovery operators work well with an H/h that is not large (usually 2). Therefore a theoretical justification for recovery that involves only a few elements surrounding a node is necessary. In other word, it is desired to study the case when H = 2h. The situation is further complicated by quadrilateral meshes where mappings between the reference element and physical elements are not affine. We encounter some delicate theoretical issue in analysis. See [, 3, 8, 3, 4, 9, 2, 30, 3, 37] for more details. In this article, we propose and analyze a gradient recovery method which is different from the ZZ recovery [35]. We show that the a posteriori estimate based on this new recovery operator is asymptotically exact under distortion O(h +α ) when α > 0 for quadrilateral meshes. Here α = represents the uniform mesh and α = 0 represents completely unstructured mesh. For counterpart results under triangular meshes, the reader is referred to [32, 33].

POLYNOMIAL PRESERVING GRADIENT RECOVERY 3 Due to non-affine mapping feature of the quadrilateral mesh, the analysis is much harder than the triangular mesh. The main feature of this new recovery operator is: () It is completely local just like the ZZ patch recovery; (2) It is polynomial preserving under practical meshes, a property not shared by the ZZ; (3) It is superconvergent under minorly restricted mesh conditions; (4) It results in an asymptotically exact error estimator when the mesh is not overly distorted. The error bound is in the form of (.) η h + O(h +ρ ) (u u h ) η h + O(h +ρ ), rather than C η h + higher order term (u u h ) Cη h + higher order term in most error bounds in the literature. Here C is an unknown constant, which may be very large and hence makes the error bound not very meaningful in practice. We comment that h α can be reduced to o() and still maintain the asymptotic exactness of the error estimator. If we give up the asymptotic exactness requirement and only ask for a reasonable error estimator, we may further reduce the condition to a sufficiently small constant γ > 0. The main results of this paper include a super-close property (Theorem 3.3), a global superconvergent recovery result (Theorem 4.2), and a local superconvergent recovery result (Theorem 4.3). The error bound (.) is a consequence of Theorems 4.2 or 4.3. All these results need a mesh assumption, Condition (α), which is introduced in Section 2. Basically, we allow quadrilaterals to be asymptotically distorted by O(h +α ) (α > 0) from parallelograms. Note that α = 0 represents arbitrary meshes. Therefore, the mesh considered here is next to arbitrary (with a little structure). Indeed, when a very practical mesh refinement strategy, bisection (link edge-center of each opposite side of a quadrilateral) is applied, we have α = (see Lemma 2.). 2. Geometry of the Quadrilateral Let = [, ] [, ] be the reference element with vertices Ẑi, and let be a convex quadrilateral with vertices Zi (x i, y i ), i =, 2, 3, 4. There exists a unique bilinear mapping F such that F ( ) =, F (Ẑi) = Zi given by 4 4 x = x i N i, y = yi N i, where i= N = ( ξ)( η), 4 N i= 2 = ( + ξ)( η), 4 N 3 = ( + ξ)( + η), 4 N 4 = ( ξ)( + η). 4

4 ZHIMIN ZHANG We can also express x = a 0 + a ξ + a 2 η + a 3 ξη, where by suppressing the index, y = b 0 + b ξ + b 2 η + b 3 ξη; 4a 0 = x + x 2 + x 3 + x 4, 4b 0 = y + y 2 + y 3 + y 4 ; 4a = x + x 2 + x 3 x 4, 4b = y + y 2 + y 3 y 4 ; 4a 2 = x x 2 + x 3 + x 4, 4b 2 = y y 2 + y 3 + y 4 ; 4a 3 = x x 2 + x 3 x 4, 4b 3 = y y 2 + y 3 y 4. To any function v(x, y) defined on, we associate ˆv(ξ, η) by ˆv(ξ, η) = v(x(ξ, η), y(ξ, η)), or ˆv = v F. The Jacobi matrix of the mapping F is ( ) ( ) xξ y (DF )(ξ, η) = ξ a + a = 3 η b + b 3 η. x η y η a 2 + a 3 ξ b 2 + b 3 ξ Let v = ( x v, y v) T, it is straight forward to verify that (2.) ˆ ˆv = ( ξˆv, ηˆv) T = DF v, (2.2) r ξ r ˆv = [(a + a 3 η) x + (b + b 3 η) y ] r v, (2.3) r+ ξ r η ˆv = r[(a + a 3 η) x + (b + b 3 η) y ] r (a 3, b 3 ) v +[(a + a 3 η) x + (b + b 3 η) y ] r [(a 2 + a 3 ξ) x + (b 2 + b 3 ξ) y ]v, and η r r ˆv and r+ ξη ˆv can be expressed in a similar way. The determinant of r the Jacobi matrix is where J = J (ξ, η) = J 0 + J ξ + J 2 η, J 0 = a b 2 b a 2, J = a b 3 b a 3, J 2 = b 2 a 3 a 2 b 3. The inverse of the Jacobi matrix is ( ) ξx η x = (DF ξ y η ) = y J ( b2 + b 3 ξ b b 3 η a 2 a 3 ξ a + a 3 η Note that a 3 = b 3 = 0 when is a parallelogram in which case F is an affine mapping, and further a 3 = b 3 = a 2 = b = 0 when is a rectangle. Starting from Z, we express the four edges (with the midpoint P i ) as four vectors v i, i =, 2, 3, 4, pointing counter-clock-wisely (Figure ). We denote the midpoints of Z 2 Z 4 and Z Z 3 as O and O 2, respectively. For analysis purpose, it is convenient to identify 2-D vectors as 3-D vectors by adding the third component 0. We can verify that P 4 P 2 = 2 (x 2 + x 3 x 4 x, y 2 + y 3 y 4 y, 0) = 2(a, b, 0), P P 3 = 2 (x 3 + x 4 x x 2, y 3 + y 4 y y 2, 0) = 2(a 2, b 2, 0), O O 2 = 2 (x + x 3 x 2 x 4, y + y 3 y 2 y 4, 0) = 2(a 3, b 3, 0). ).

POLYNOMIAL PRESERVING GRADIENT RECOVERY 5 5 4 Z 3 P 3 3 Z 4 v3 v 2 2 o 2 o θ 2 0 O 2 P 2 P 4 O O 2 θ 3 v 4 4 Z P v Z 2 5 6 4 2 0 2 4 6 Figure. Geometry of a quadrilateral Then (2.4) 2 a 2 + b2 = P 4P 2, 2 a 2 2 + b2 2 = P P 3, 2 a 2 3 + b2 3 = O O 2. (2.5) (2.6) (2.7) 4(a a 2 + b b 2 ) = P 4 P 2 P P 3 = P 4 P 2 P P 3 cos α, 4(a a 3 + b b 3 ) = P 4 P 2 O O 2 = P 4 P 2 O O 2 cos β, 4(a 2 a 3 + b 2 b 3 ) = O O 2 P P 3 = O O 2 P P 3 cos γ, where the meaning of angles α, β, and γ is obvious from the context. J0 i j k (2.8) k = a b 0 a 2 b 2 0 = 4 P 4P 2 P P 3 = 4 P 4P 2 P P 3 sin α, J i j k (2.9) k = a b 0 a 3 b 3 0 = 4 P 4P 2 O O 2 = 4 P 4P 2 O O 2 sin β, J2 i j k (2.0) k = a 2 b 2 0 a 3 b 3 0 = 4 P P 3 O O 2 = 4 P P 3 O O 2 sin γ. We could also express J = 2 (x 4 x 3 )(y 2 y ) (x 2 x )(y 4 y 3 ) = 2 v 3 v, J 2 = 2 v 4 v 2. Let h be the longest edge length of, we introduce the following condition: Definition. A convex quadrilateral is said to satisfy the diagonal condition if (2.) d = O O 2 = O(h +α ), α 0.

6 ZHIMIN ZHANG Note that is a parallelogram if and only if d = 0. Therefore, the distance between the two diagonal mid-points O and O 2 is a convenient measure for the deviation of a quadrilateral from a parallelogram. The two extremal cases α and α 0 represent parallelogram and completely unstructured quadrilateral, respectively. Anything in between will pose some restriction, especially α = is the well-known 2-strongly regular partition, see, e.g., [3, 37]. The diagonal condition was previously used by Chen [2] for triangular meshes, where two adjacent triangles form a quadrilateral that satisfies the condition. The following lemma states a known fact regarding the 2-strongly regular partition (α = ). Although this fact is widely used, we have not seen a formal proof of it in the literature. An elementary proof is therefore provided in the Appendix. Lemma 2.. Let o o 2 be the distance between two diagonal mid-points of any of four refined quadrilaterals through the bi-section of. Then o o 2 = 4 O O 2. Recall that the bi-section reduces the length of longest edge by half, which is h /2. Therefore, the diagonal condition (2.) is satisfied with α =. To measure this deviation, Rannarchar and Turek [2] used the quantity σ = max( π θ, π θ 2 ), where θ and θ 2 are the angles between the outward normals of two opposite sides of. Definition 2. A convex quadrilateral is said to satisfy the angle condition if (2.2) σ = O(h α ), α 0. Lemma 2.2. The diagonal condition (2.) and the angle condition (2.2) are equivalent in the sense d = O(h +α ) σ = O(h α ), α 0. A special case of this lemma has been proved in [9, Theorem 4.3] under some complicated mesh restrictions. Here we provide a direct and much simpler proof in the Appendix without any mesh assumption. Definition 3. A partition T h is said to satisfy Condition (α) if there exist α > 0 such that i) Any T h satisfies the diagonal condition (2.). ii) Any two, 2 in T h that share a common edge satisfy a neighboring condition: For j =, 2, (2.3) a j = a 2 j ( + O(h α + h α 2 )), b j = b 2 j ( + O(h α + h α 2 )). To assure optimal order error estimates in the H -norm for the bilinear isoparametric interpolation on a convex quadrilateral, namely, the

POLYNOMIAL PRESERVING GRADIENT RECOVERY 7 estimate (2.4) u u I 0, + h u u I, Ch 2 u 2,, we need a degeneration condition, which was introduced by Acosta and Durán []. Definition 4. A convex quadrilateral is said to satisfy the Regular decomposition property with constants N R and 0 < Ψ < π, or shortly RDP (N, Ψ), if we can divide into two triangles along one of its diagonals, which will always be called d, in such a way that d / d 2 N and both triangles satisfy the maximum angle condition with parameter Ψ (i.e., all angles are bounded by Ψ). Remark. This is a weaker condition than many other similar degenerate conditions, cf. e.g., [3, 4, 3, 37]. It was proved in [] that RDP (N, Ψ) is a sufficient condition for (2.4) to be hold, and the authors conjectured that it is also a necessary condition. Recently, Ming-Shi confirmed this conjecture by a simple counter-example [9]. We denote X = X(ξ, η) = X 0 + X where ( ) b2 b X 0 =, X a 2 a = X (ξ, η) = ( b3 a 3 ) (ξ, η). Lemma 2.3. Let a convex quadrilateral satisfy the diagonal condition. Then X 0 X 2 = + O(h α ), X X 2 = I X 0 X 2 = O(h α ). Proof: It is straightforward to verify that ( ) [( ) X 0 X b2 b = a b + a 2 a J a 2 b 2 = J 0 I + ( ) ( b2 b η J J a 2 a ξ where I is a 2-by-2 identity matrix; and X X = I X 0 X = ( J J ξ + J 2 J η)i J ( ] η (a ξ) 3, b 3 ) ) (a 3, b 3 ) ( ) ( b2 b η a 2 a ξ ) (a 3, b 3 ). By the definition of J and geometric relations of (2.4), (2.8) (2.0), we see that J0 = + O(h α J J ), = O(h α J2 J ), = O(h α J ), by the diagonal condition (2.). The desired conclusion follows. 3. Superconvergence Analysis We consider the variational problem: Find u H (Ω) such that (3.) a(u, v) = ( u, A v) + (b u, v) + (cu, v) = (f, v), v H (Ω),

8 ZHIMIN ZHANG where A is a 2-by-2 symmetric positive definite matrix and Ω is a polygonal domain which allows a quadrilateral partition T h with h = max h. We T h assume that all functions are sufficiently smooth, in particular, (3.2) A A 0 0,, = O(h α ), b b 0 0,, = O(h α ), where A 0 and b 0 are piece-wisely constant functions that on each T h, A 0 = A(x, y)dxdy, b 0 = b(x, y)dxdy. We also assume that a(, ) satisfies the inf-sup condition to insure that (3.) has a unique solution. Using v = J X ˆ ˆv, we write ( w, A v) = ( w) T A vdxdy = (X J ˆ ŵ) T Â(X ˆ ˆv)dξdη, (b w, v) = vb wdxdy = ˆvˆb X ˆ ŵdξdη; and define (3.3) ( w, A v) = = J 0 (X 0 ˆ ŵ) T A 0 (X 0 ˆ ˆv)dξdη ( ˆ ŵ) T B ˆ ˆvdξdη, (3.4) (b w, v) = b 0 X 0 where ˆv ˆ ŵdξdη, B = J0 (X0 ) T A 0 X0. We introduce the following lemma, which can be verified by straightforward calculation. Lemma 3.. Under the condition (2.3) and (3.2), we have J 0 = J 2 0 ( + O(h α + h α 2 )), B B 2 = O(h α + h α 2 ). Theorem 3.. Let the assumption (3.2) be satisfied, and let satisfy the diagonal condition. Then there exists a constant C independent of u and, such that (3.5) ( w, A v) ( w, A v) Ch α w 0, v 0,, (3.6) (b w, v) (b w, v) Ch α w 0, v 0,.

(3.7) Proof: We decompose By (3.2) POLYNOMIAL PRESERVING GRADIENT RECOVERY 9 ( w, A v) ( w, A v) = ( w, (A A 0 ) v) + [(X J ˆ ŵ) T A 0 (X ˆ ˆv) (X 0 ˆ ŵ) T A 0 (X 0 ˆ ˆv)]dξdη + ( J J0 )(X 0 ˆ ŵ) T A 0 (X 0 ˆ ˆv)dξdη. (3.8) ( w, (A A 0 ) v) Ch α w 0, v 0,. Using X = X 0 + X, we express [(X J ˆ ŵ) T A 0 (X ˆ ˆv) (X 0 ˆ ŵ) T A 0 (X 0 ˆ ˆv)]dξdη = [(X 0 ˆ ŵ) T A 0 (X ˆ ˆv) + (X ˆ ŵ) T A 0 (X 0 ˆ ˆv) J + (X ˆ ŵ) T A 0 (X ˆ ˆv)]dξdη. The first term can be estimated as (X 0 ˆ ŵ) T A 0 (X ˆ ˆv)dξdη J = ( X J ˆ ŵ) T X T X0 T A 0 X X ( X J ˆ ˆv)J dξdη = ( w) T (X 0 X ) T A 0 X X vdxdy Ch α w 0, v 0,. Note that (X 0 X ) T A 0 X X = O(h α ) by Lemma 2.3. The other two terms can be estimated similarly. Then we derive (3.9) [(X J ˆ ŵ) T A 0 (X ˆ ˆv) (X 0 ˆ ŵ) T A 0 (X 0 ˆ ˆv)]dξdη Ch α w 0, v 0,. Next (3.0) = = ( J J0 )(X 0 ˆ ŵ) T A 0 (X 0 ˆ ˆv)dξdη ( J J0 )( X J ˆ ŵ) T X T X0 T A 0 X 0 X ( X J ˆ ˆv)J dξdη ( J J0 ξ(x, y) J 2 J0 Ch α w 0, v 0,. Note that by Lemma 2.3, J J0 η(x, y))( w) T (X 0 X ) T A 0 X 0 X vdxdy ξ(x, y) + J 2 J0 η(x, y) = O(h α ), (X 0 X ) T A 0 X 0 X = + O(h α ). We then obtain (3.5) by applying (3.8)-(3.0) to the right hand side of (3.7).

0 ZHIMIN ZHANG Now we write the convection term as following: (3.) = (b w, v) (b w, v) ˆvˆb (X X 0 ) ˆ ŵdξdη + ˆv(ˆb b 0 ) X 0 ˆ ŵdξdη. We estimate the two terms separately. (3.2) ˆvˆb (X X 0 ) ˆ ŵdξdη = ˆvˆb (I X 0 X )X ˆ ŵdξdη = vb (I X 0 X ) wdxdy by Lemma 2.3. (3.3) Ch α w 0, v 0,, = = ˆv(ˆb b 0 ) X 0 ˆ ŵdξdη ˆv(ˆb b 0 ) X 0 X (X ˆ ŵ)dξdη v(b b 0 ) X 0 X wdxdy Ch α w 0, v 0,, by Lemma 2.3 and (3.2). Applying (3.2) and (3.3) to (3.), we obtain (3.6). We then define two modified bilinear forms where a h (u, v) = a h (u, v), b h (w, v) = b h (u, v) (3.4) a h (u, v) = ( u, A v) + (b u, v) + (cu, v), (3.5) b h (u, v) = ( u, A v) + (b w, v) + (cu, v). Given a quadrilateral partition T h on a polygonal domain Ω, we define the bilinear finite element space S h = {v H (Ω) : ˆv = v F Q ( ), T h }. Theorem 3.2. Let T h satisfy the condition (α) and RDP (N, Ψ), and let u I S h be the bilinear interpolation of u H 3 (Ω) H0 (Ω). Then there exists a constant C independent of h and u, such that for any v S h, a h (u u I, v) + b h (u u I, v) C(h +α u 2,Ω + h 2 u 3,Ω ) v,ω. Proof: For convenience, we set w = u u I. By (3.4) and (3.5), we can express a h (w, v) b h (w, v) = T h [(b w, v) (b w, v) ].

Recall (3.6), and we have POLYNOMIAL PRESERVING GRADIENT RECOVERY (3.6) a h (w, v) b h (w, v) C T h h α w 0, v 0, Ch +α u 2,Ω v 0,Ω, since by the RDP (N, Ψ) assumption, (2.4) is valid. Therefore, we only need to estimate b h (w, v). Again, by the RDP (N, Ψ) assumption, we have (3.7) (cw, v) Ch 2 u 2,Ω v 0,Ω. Hence, our task is narrowed down to estimate ( u, A v), and (b w, v) for T h. By the definition (3.3) and (3.4), we see that all coefficients are constants now and we only need to estimate following terms ξ ŵ ξˆv, ξ ŵ ηˆv, η ŵ ξˆv, η ŵ ηˆv, ˆv ξ ŵ, ˆv η ŵ. a) Let û P 2 ( ). There are only two terms ξ 2, η 2 not in the reference space of the bilinear interpolation, therefore, ξ ŵ ξˆv = 0, v S h. By the Bramble-Hilbert Lemma, (3.8) ξ ŵ ξˆv C D 3 û L 2 ( ) ξˆv L 2 ( ) C(h +α u 2, + h 2 u 3, ) v,. We have used (2.2) and (2.3) in the last step. Similarly, (3.9) η ŵ ηˆv C(h +α u 2, + h 2 u 3, ) v,. Next we discuss the cross terms. For any v S h, we can express ξˆv = ξˆv(0, 0) + η 2 ξηˆv, ηˆv = ηˆv(0, 0) + ξ 2 ξηˆv. Note that 2 ξηˆv is a constant. We write ( ξ ŵ ηˆv ± η ŵ ξˆv) = ηˆv(0, 0) ξ ŵ ± ξˆv(0, 0) η ŵ + ξηˆv( 2 Since for û = ξ 2, or û = η 2, ξ ŵ = 0, η ŵ = 0. Therefore, by the Bramble-Hilbert Lemma, (3.20) ηˆv(0, 0) ξ ŵ ± ξˆv(0, 0) η ŵ ξ ξ ŵ ± η η ŵ). C D 3 û L 2 ( ) ˆ ˆv L 2 ( ) C(h+α u 2, + h 2 u 3, ) v,.

2 ZHIMIN ZHANG Next we consider, ξ ξ ŵ = 2 (ξ 2 ) ξ ŵ = (ξ 2 ) ξ 2 û, 2 2 ξηˆv 2 ξ ξ ŵ = (ξ 2 ) ξ 2 û 2 ξηˆv 2 2 Similarly, = 2 2 2 ξηˆv (ξ 2 )( 2 ξ 2 û ξˆv)(ξ, )dξ (ξ 2 )( 2 ξ 2 û ξˆv)(ξ, )dξ + 2 η η ŵ = 2 (η 2 )( 2 η 2 û ηˆv)(, η)dη (ξ 2 ) ξ 3 2ηû ξˆv. 2 (η 2 )( η 2 û ηˆv)(, η)dη + 2 2 Therefore, we have ξηˆv( 2 ξ ξ ŵ ± η η ŵ) = (3.2) (t 2 ) 2 2 t û tˆvdt + [(ξ 2 ) ξ 3 2 2ηû ξˆv ± (η 2 ) ξη 3 û ηˆv], 2 (η 2 ) ξη 3 û ηˆv. 2 where indicates a sign influence whenever it applies. For the second term on the right hand side of (3.2), we have, from (2.3), (3.22) [(ξ 2 ) ξ 3 2 2ηû ξˆv ± (η 2 ) ξη 3 û ηˆv] 2 C(h +α u 2, + h 2 u 3, ) v,. In light of (3.8) (3.22), we can express ( ˆ ŵ) T B b ˆ ˆvdξdη 4 = 2 (3.23) l j 2 (t(s) 2 l 2 ) s 2 u s vds j + (O(h +α ) u 2, + O(h 2 ) u 3, ) v,, where l j are four sides of. By the neighboring condition (2.3), any two adjacent elements, 2 that share a common edge satisfy (see Lemma 3.) j= B B 2 = O(h α ), Therefore, we have, by the trace theory, b 2 b 2 2 (3.24) l 2 (t(s) 2 ) s 2 u s vds 2 l Ch α l 2 (h D 2 udv + D 3 udv + D 2 ud 2 v ) C(h +α u 2, + h 2 u 3, ) v,.

POLYNOMIAL PRESERVING GRADIENT RECOVERY 3 In the last step, we have used the inverse inequality. Adding up (3.23) with the edge integral estimated by (3.24), we obtain, under the homogeneous Dirichlet boundary condition, (3.25) T h ( w, A v) C(h +α u 2,Ω + h 2 u 3,Ω ) v,ω. b) we now consider ˆv ξ ŵ where we can express ˆv = ˆv(0, 0) + ξˆv(0, 0)ξ + ηˆv(0, 0)η + 2 ξηˆvξη. Since for any û P 2 ( ), we have ξ ŵ(ˆv(0, 0) + ηˆv(0, 0)η) = 0, by the same argument as in b), we have (3.26) ξ ŵ(ˆv(0, 0) + ηˆv(0, 0)η) C D 3 û 0, ˆv, C(h α u 2, + h u 3, ) v,. Next, by identities ξ ŵξ = ξ 2 û(ξ 2 ), 2 2 ξ ŵξη = ξ ŵ(ξ 2 ) (η 2 ) = ξ 3 4 4 2ηû(ξ2 )(η 2 ), we have (3.27) ξ ŵ( ξˆv(0, 0)ξ + ξηˆvξη) ξ 2 ŵ 2 0, ξˆv 0, + ξ 3 2ηŵ 0, 2 ξηˆv 0, C(h α u 2, + h u 3, ) v,. Again, we have used (2.2), (2.3), and the inverse inequality in the last step. Combining (3.26) and (3.27), we obtain (3.28) ˆv ξ ŵ C(h α u 2, + h u 3, ) v,. Similarly, we have (3.29) Note that X 0 (3.30) ˆv η ŵ C(h α u 2, + h u 3, ) v,. = O(h ), therefore, (b 0 w, v) = b 0 X 0 ˆv ŵ Ch (h α u 2, + h u 3, ) v,. Adding up all T h and using the Cauchy inequality, we obtain (3.3) T h (b 0 w, v) C(h +α u 2,Ω + h 2 u 3,Ω ) v,ω.

4 ZHIMIN ZHANG Combining (3.7), (3.25), and (3.3), we establish the assertion for b h (w, v). Theorem 3.3. Assume that T h satisfies the condition (α) and RDP (N, Ψ). Let u H 3 (Ω) H 0 (Ω) solves (3.), let u h, u I S h be the finite element approximation and the bilinear interpolation of u, respectively, and let a(, ) satisfy the discrete inf-sup condition on S h. Then there exists a constant C independent of h and u, such that (3.32) a(u u I, v) C(h +α u 2,Ω + h 2 u 3,Ω ) v,ω, (3.33) u h u I,Ω C(h +α u 2,Ω + h 2 u 3,Ω ). Proof: Let w = u u I, and by Theorem 3., a(w, v) a h (w, v) = ( w, A v) ( w, A v) Ch α w 0, v 0,. Adding all T h and using (2.4) with the Cauchy-Schwarz inequality, we have a(w, v) a h (w, v) Ch +α u 2,Ω v,ω. Recall Theorem 3.2, and we obtain a(w, v) a(w, v) a h (w, v) + a h (w, v) C(h +α u 2,Ω + h 2 u 3,Ω ) v,ω, which establishes (3.32). We then complete the proof by the inf-sup condition in a(u h u I, v) c u h u I,Ω sup v S h v,ω a(u u I, v) = sup C(h +α u 2,Ω + h 2 u 3,Ω ). v S h v,ω 4. Gradient Recovery In this section, we introduce and analyze a polynomial preserving recovery method (PPR). We define a gradient recovery operator G h : S h S h S h, on bilinear finite element space under a quadrilateral partition T h in a following way: Given a finite element solution u h, we first define G h u h at all nodes (vertices), and then obtain G h u h on the whole domain by interpolation using the original nodal shape functions of S h. Given an interior node (vertex) z i, we select an element patch ω i, where ω i =. T h,z i We then denote all nodes on ω i (including z i ) as z ij, j =, 2,..., n( 6), and fit a quadratic polynomial, in the least-squares sense, to the finite element solution u h at those nodes. Using local coordinates (x, y) with z i as the origin, the fitting polynomial is p 2 (x, y; z i ) = P T a = ˆP T âa,

with POLYNOMIAL PRESERVING GRADIENT RECOVERY 5 P T = (, x, y, x 2, xy, y 2 ), ˆP T = (, ξ, η, ξ 2, ξη, η 2 ); a T = (a, a 2, a 3, a 4, a 5, a 6 ), âa T = (a, ha 2, ha 3, h 2 a 4, h 2 a 5, h 2 a 6 ), where the scaling parameter h = h i is the length of the longest element edge in the patch ω i. The coefficient vector âa is determined by the linear system (4.) Q T Qâa = Q T b h, where b T h = (u h(z i ), u h (z i2 ),, u h (z in )) and ξ η ξ 2 ξ η η 2 ξ 2 η 2 ξ2 2 ξ 2 η 2 η 2 2 Q =....... ξ n η n ξn 2 ξ n η n ηn 2 The condition for (4.) to have a unique solution is: Q has a full rank, which is always satisfied in practical situations. In fact, Q has a full rank if and only if z ij s are not all lying on a conic curve. In practice, this is not a restriction at all: Any interior node z i is a common vertex of at least three quadrilaterals. This makes n 7 > 6. As long as these n nodes are reasonably distributed, it is impossible to fit all of them on a conic curve. A simple example is that three nodes are on the same line and others are not form a line. Conjecture: A sufficient condition for Q to have a full rank is all quadrilaterals are convex. In the rest of this article, we will always assume the rank condition: Q has a full rank. Now we define (4.2) G h u h (z i ) = p 2 (0, 0; z i ). When Neumann boundary condition is post, there is no need to do gradient recovery on the boundary. However, if the Dirichlet boundary condition is post, the recovered gradient on a boundary node z can be determined from an element patch ω i such that z ω i in the following way: Let the relative coordinates of z with respect to z i is, say (h, h), then G h u h (z) = p 2 (h, h; z i ). If z is covered by more than one element patches, then some averaging may be applied. Remark 4.. In an earlier work [26], Wiberg-Li least-squares fitted solution values to improve and to estimate the L 2 -norm errors of the finite element approximation. Now, we demonstrate PPR on an element patch that contains four uniform square elements (Figure 2). Fitting ˆp 2 (ξ, η) = (, ξ, η, ξ 2, ξη, η 2 )(â,, â 6 ) T with respect to the nine nodal values on the patch. Now e = (,,,,,,,, ) T, ξ = (0,,, 0,,,, 0, ) T, η = (0, 0,,,, 0,,, ) T, Q = ( e, ξ, η, ξ 2, ξη, η 2 ),

6 ZHIMIN ZHANG 4 3 2 5 z 6 7 8 Figure 2 (4.3) (Q T Q) Q T = diag( 9, 6, 6, 6, 4, 6 ) 5 2 2 2 2 0 0 0 0 0 0 2 2 2. 0 0 0 0 0 2 2 2 G h u(z) = h ˆ ˆp 2 (0, 0) = 6h ( u u 5 + u 2 u 4 + u 8 u 6 u 2 u 8 + u 3 u 7 + u 4 u 6 ) = h c j u j. Note that the desired weights c j are the second row of (Q T Q) Q T for the x-derivative, and the third row of (Q T Q) Q T for the y-derivative, respectively. Moreover, c j = 0 and G h u(z) provides a second-order finite i difference scheme at z. Given v S h, it is straightforward to verify that v x (h 2, 2h 3 ) = 3h (v v 0 ) + 2 3h (v 2 v 3 ), v x ( h 2, 2h 3 ) = 3h (v 0 v 5 ) + 2 3h (v 3 v 4 ), v x ( h 2, 2h 3 ) = 3h (v 0 v 5 ) + 2 3h (v 7 v 6 ), j

Therefore, POLYNOMIAL PRESERVING GRADIENT RECOVERY 7 v x (h 2, 2h 3 ) = 3h (v v 0 ) + 2 3h (v 8 v 7 ). G x h v(z) = 4 [ v x (h 2, 2h 3 ) + v x ( h 2, 2h 3 ) + v x ( h 2, 2h 3 ) + v x (h 2, 2h 3 )]. The recovered y-derivative can be obtained similarly. Hence, in this special case, G h v(z) v,,ωz. By linear mapping, this is also valid for four uniform parallelograms in that (4.4) G h v(z) C v,,ωz, v S h. with C independent of h and v. Theorem 4. Let T h satisfy Condition (α). Then the recovery operator G h is a bounded linear operator on bilinear element space such that G h v 0,p,Ω C v,p,ω, v S h, p, where C is a constant independent of v and h. Proof: We observe that the diagonal condition together with the neighboring condition imply that for any given node z, there are four elements attached to it when h is sufficiently small. In addition, these four elements deviate from four parallelograms that attached to the same node in the following sense, Q = Q 0 + h α Q, where Q and Q 0 are least-square fitting matrices associated with those four quadrilateral elements and four parallelograms, respectively. We want to express (Q T Q) Q T in terms of (Q T 0 Q 0) Q T 0. Towards this end, we have where Therefore, Q T Q = Q T 0 Q 0 (I + h α E ), E = (Q T 0 Q 0 ) (Q T Q 0 + Q T 0 Q + h α Q T Q ). (Q T Q) Q T = (I +h α E ) (Q T 0 Q 0 ) (Q T 0 +h α Q T ) = (Q T 0 Q 0 ) Q T 0 +h α E 2, where We see that E 2 = (Q T 0 Q 0 ) Q T (h α E ) j E (Q T 0 Q 0 ) Q T. j=0 (4.5) (Q T Q) Q T = (Q T 0 Q 0 ) Q T 0 + O(h α ). Therefore, the fact that Q T 0 Q 0 is invertible guarantees that Q T Q is invertible for sufficiently small h. Moreover, by (4.5), we have G h v(z) = ( c j + O(h α ))v j h j where G h is the recovery operator under the quadrilateral mesh that satisfies the diagonal condition and the neighboring condition, and c j s are weights

8 ZHIMIN ZHANG for the related parallelogram mesh so that, by (4.4), c j is a bounded h operator on S h such that c j v j C v,,ωz. h j Therefore, in the quadrilateral case, (4.4) is also valid, provided Condition (α) is satisfied and h is sufficiently small. If (4.4) is valid for each node of, then we have, (4.6) G h v 0,, C v,,ω, v S h, where ω is defined as ω = T h, Note that (4.6) is true for all T h including boundary elements, since by our construction the boundary recovery is simply some averaging of nearby patches. Therefore, (4.7) G h v 0,,Ω C v,,ω, v S h. This establishes the assertion for p =. As for p <, we notice that all norms are equivalent for finite dimensional spaces, and with a scaling argument, G h v p C h 2 G h v p 0,, T h T h C 2 h 2 T h v p,, C 3 h 2 T h h 2. v p C T h v p,p, Here, all constants C j s and C are independent of p, v, and h. The conclusion then follows. Another important feature of the new recovery operator is the following polynomial preserving property: Lemma 4.. Let T h and u be a quadratic polynomial on ω. Assume that Q has a full rank. Then G h u = u on. Proof: The rank condition guarantees that the least-squares fitting has a unique solution. On each of four element patches, the recovery procedure results in a quadratic polynomial p 2 that least-squares fits u, a quadratic polynomial. Therefore, p 2 = u, and consequently, G h u = p 2 = u, a linear function, at each of the four vertices of. Therefore, G h u = u on. Remark 4.2. Note that we do not make any mesh assumptions in Lemma 4. except the rank condition, which is always satisfied in practice. Basically, j

POLYNOMIAL PRESERVING GRADIENT RECOVERY 9 as long as the least-squares fitting procedure can be carried out, the polynomial preserving property is satisfied. As a comparison, the ZZ recovery operator does not have this polynomial preserving property under general meshes, see [33] for more details. Theorem 4.2. Let T h satisfy the condition (α) and RDP (N, Ψ). Let u h S h be the finite element approximation of u H 3 (Ω) H0 (Ω), the solution of (3.), and let a(, ) satisfy the discrete inf-sup condition on S h. Then the recovered gradient is superconvergent in the sense u G h u h 0,Ω C(h +α u 2,Ω + h 2 u 3,Ω ), where C is a constant independent of u and h. Proof: We decompose the error into (4.8) u G h u h = u G h u + G h (u I u h ). Note that G h u = G h u I since u I = u at all vertices and the recovery operator G h is completely determined by nodal values of u. By the polynomial preserving property and the Bramble-Hilbert lemma, (4.9) u G h u 0,p,Ω Ch 2 u 3,p,Ω, p. By Theorem 4., G h is a bounded operator for all interior patches. Therefore, (4.0) G h (u I u h ) 2 0,Ω = T h G h (u I u h ) 2 0, C 2 T h u I u h 2, C 2 (h +α u 2,Ω + h 2 u 3,Ω ) 2 by Theorem 3.3. The conclusion then follows by applying (4.9) with p = 2 and (4.0) to (4.8). Theorem 4.2 assumes a global regularity u H 3 (Ω), which may not hold in general. However, higher regularity requirement is usually satisfied in an interior sub-domain. In the rest of this section, we shall prove a local result based on interior estimates. In order to concentrate on superconvergence analysis, treatments of curved boundaries and corner singularities will not be discussed here. We merely assume that they have been taken care of in the following sense, (4.) u u h,ω C(f, a, Ω)h +ρ, ρ = min(, α). The negative norm term is the only one in our analysis that takes into account what happens outside of a local region Ω. We shall show that under assumption (4.), superconvergent recovery will occur in an interior sub-domain. Toward this end, we consider Ω 0 Ω Ω where Ω 0 and Ω are compact polygonal sub-domains that can be decomposed into quadrilaterals. By compact sub-domains we mean that dist(ω 0, Ω ) and dist(ω, Ω) are of order O(). Outside Ω, we may have quadrilateral or triangular subdivisions. We may also have refined meshes near the corner singularities and curved elements on the boundary regions. We assume that all these together will result in (4.).

20 ZHIMIN ZHANG We define a cut-off function ω C 0 (Ω) such that ω = on Ω 0 and ω = 0 in Ω \ Ω. We decompose u into u = ũ + û, ũ = uω. Let ũ I be the bilinear interpolation of ũ and let ũ h S h (Ω ) = S h H 0 (Ω ) be the finite element approximation of ũ on Ω. There holds a(ũ ũ h, v) Ω = 0, v S h (Ω ). The index Ω indicates that the integrations in the bilinear form are performed on the subdomain. Further, we let û I = u I ũ I, û h = u h ũ h. Note that we have ũ I = u I on Ω 0 since ω = on Ω 0. However, ũ h u h on Ω 0 in general. Apply Theorem 3.3 on Ω, and we immediately obtain (4.2) ũ h ũ I,Ω Ch +ρ ũ 3,Ω. However, for any k 3, (4.3) ũ k,ω = uω k,ω Therefore, from (4.2), k D j ud k j ω L2 (Ω ) C(k, ω) u k,ω. j=0 (4.4) ũ h ũ I,Ω0 ũ h ũ I,Ω Ch +ρ u 3,Ω. Next, we consider û h û I. Since û = û I = 0 on Ω 0, there holds (4.5) û h û I,Ω0 = û h,ω0 = û h û,ω0. Note that for all v S h (Ω ), As a result, a(û h û, v) Ω = a(u h u, v) Ω a(ũ h ũ, v) Ω = 0. (4.6) û û h,ω0 C(h 2 û 3,Ω + û û h,ω ), by Nitsche and Schatz [20, Theorem 5.] (All the conditions of this theorem can be verified in the current situation, see Remark 4.3 below). With the same argument as in (4.3), we have (4.7) û 3,Ω = u ũ 3,Ω u 3,Ω + ũ 3,Ω C u 3,Ω. Observe that ũ ũ h,ω ũ ũ h 0,Ω Ch 2 u 2,Ω, therefore, by assumption (4.), (4.8) û û h,ω u u h,ω + ũ ũ h,ω Ch +ρ (C(f, a, Ω) + u 2,Ω ). Substituting (4.7) and (4.8) into (4.6), we derive (4.9) û û h,ω0 Ch +ρ ( u 3,Ω + C(f, a, Ω)).

POLYNOMIAL PRESERVING GRADIENT RECOVERY 2 Combining (4.9) with (4.4) and (4.5), we obtain (4.20) u h u I,Ω0 ũ h ũ I,Ω0 + û h û I,Ω0 Ch +ρ ( u 3,Ω + C(f, a, Ω)). Now, following the same argument as in Theorem 4.2, we immediately obtain the following result on a general polygonal domain. Theorem 4.3. Let Ω R 2 be a polygonal domain and Ω 0 Ω Ω. Assume that T h satisfy the condition (α) and RDP (N, Ψ) on Ω. Let u h S h be the finite element approximation of u H 3 (Ω ) H0 (Ω) that solves (3.) with a(, ) satisfying the discrete inf-sup condition on S h. Furthermore, let (4.) be satisfied. Then there exists a constant C independent of h such that G h u h u,ω0 Ch +ρ ( u 3,Ω + C(f, a, Ω)), ρ = min(, α). Remark 4.3. In the proof of Theorem 4.3, we used a result of Nitsche and Schatz [20, Theorem 5.], which requires following conditions from the underlining finite element space: R. Coercive and continuity of the bilinear form; A.. Approximation in an optimal sense; A.2. Superapproximation property; A.3. Inverse properties. In our situation, R is assured by the inf-sup condition and A.. is actually (2.4). Here we sketch a proof for A.2. under a quadrilateral mesh. The verification for A.3. is similar. We wand to show that for Ω 0 G Ω, v h S h and ω C 0 (Ω 0), there exists an η S h (G) such that (4.2) ωv h η,g Ch v h,g. By direct calculation over a quadrilateral element, we have ωv h η 2, = X J ˆ ( ωv h ˆη) 2 dξdη ) C ( 2 ξ 2 ( ωv h) 2 2 + L 2 ( ) η 2 ( ωv h) 2 L 2 ( ) Ch 2 v h 2 H (). Indeed, since ξ 2 ˆv h = 0 for a bilinear function on, we have 2 ξ 2 ( ωv h) = 2 ˆω ξ 2 v h + 2 ˆω ˆv h ξ ξ = v h [(a + a 3 η) x + (b + b 3 η) y ] 2 ω + 2DF ω DF v h. We then have the needed power for h. Another term ωv h η L 2 () can be similarly estimated. Finally, we simply add up G and taking the square root to obtain (4.2). 5. A Posteriori Error Estimates

22 ZHIMIN ZHANG Let e h = u u h, the task here is to estimate the error e h 0,Ω0 by a computable quantity η h. According to Zienkiewicz-Zhu [36], η h is the error estimator defined by the recovered gradient, We need the following assumption: η h = G h u u u h 0,Ω0. (5.) e h 0,Ω0 Ch. Theorem 5.. Assume the same hypotheses as in Theorem 4.3. Let (5.) be satisfied. Then η h e h 0,Ω0 = + O(h ρ ), ρ = min(, α). Proof: By the triangle inequality, η h u G h u h 0,Ω0 e h 0,Ω0 η h + u G h u h 0,Ω0. Dividing the above by e h 0,Ω0, the conclusion follows from Theorem 4.3 and (5.). Remark 5.. Theorem 5. indicates that the error estimator based on our polynomial preserving recovery is asymptotically exact on an interior region Ω 0. This result is valid for fairly general quadrilateral meshes. Remark 5.2. If we use o() to substitute O(h α ), the conclusion of Theorem 5. would be + o() and the error estimate would still be asymptotically correct. Furthermore, we may use a more practical term: a sufficiently small constant γ > 0, instead of o(). We would lose the asymptotic exactness, nevertheless, the effectivity index would still be in a reasonable range around, as observed in practice. Appendix Proof of Lemma 2.. Let the longest edge length of be h, then the longest edge length after one bisection refinement is h /2. We shall show that the distance between the two diagonal mid-points of any one of the four refined quadrilaterals is d /4, a quadratic reduction. ) The coordinates O show that P P 3, P 4 P 2 and O O 2 bisect each other at O. 2) O P 2 = Z 3 Z 4 /2 = Z 3 P 3 since O P 2 connects two edge centers in Z 2 Z 3 Z 4. 3) Z 3 Q = Q O since two triangles Q O P 2 and Q Z 3 P 3 are congruent. 4) Q Q 2 = OO /2 = d /4 since Q Q 2 connects two edge centers in Z 3 OO. Proof of Lemma 2.2. From v v 3 sin(π θ ) = v v 3 = 2 J = 8 P 4P 2 O O 2 = 8 P 4P 2 d sin β,

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