THE PATCH RECOVERY FOR FINITE ELEMENT APPROXIMATION OF ELASTICITY PROBLEMS UNDER QUADRILATERAL MESHES. Zhong-Ci Shi and Xuejun Xu.

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1 DISCRETE AND CONTINUOUS Website: DYNAMICAL SYSTEMS SERIES B Volume 9, Number, January 2008 pp THE PATCH RECOVERY FOR FINITE ELEMENT APPROXIMATION OF ELASTICITY PROBLEMS UNDER QUADRILATERAL MESHES Zhong-Ci Shi and Xuejun Xu LSEC, Institute of Computational Mathematics Chinese Academy of Sciences P.O. Box 279, Beijing 00080, China Zhimin Zhang College of Mathematics and Computer Science Hunan Normal University, Changsha, Hunan 4008, China Department of of Mathematics Wayne State University, Detroit, MI 48202, USA (Communicated by Yanping Lin) Abstract. In this paper, some patch recovery methods are proposed and analyzed for finite element approximation of elasticity problems using quadrilateral meshes. Under a mild mesh condition, superconvergence results are established for the recovered stress tensors. Consequently, a posteriori error estimators based on the recovered stress tensors are asymptotically exact.. Introduction. In recent years, the adaptive method based on a posteriori error control has become one of the most popular numerical methods for solving PDEs. In this paper, we will focus our attention on a posteriori error estimators based on gradient recovery techniques. The most well-known gradient recovery method is the ZZ patch recovery method developed by Zienkiewicz-Zhu [23]. Recently, Zhang- Naga proposed a new recovery technique [7], which is different from the ZZ patch recovery. The main advantages of the new recovery technique are as follows: It is polynomial preserving under arbitrary meshes, a property not shared by the ZZ; it is superconvergent under minor mesh restrictions. In this paper, we will use the recovery techniques developed by ZZ and Zhang-Naga to construct a posteriori error estimators for linear elasticity problems. While residual type estimators have been well researched even for elasticity problems, (cf. [7],[8]), we have not seen theoretical analysis on recovery type error estimates for these problems. Nevertheless, recovery type estimators are widely used 2000 Mathematics Subject Classification. Primary: 65N30. 65N5; Secondary: 65N2, 65D0, 4A0. ey words and phrases. Finite element, Superconvergence, Gradient recovery, Linear elasticity. The first two authors are supported in part by the special funds for major state basic research projects (973) under grant 2005CB3270 and the National Natural Sciences Foundation of China under grant The third author is supported in part by the US National Science Foundation under grants DMS and DMS , and by the National Natural Sciences Foundation of China under grant

2 64 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG in engineering application. It is confirmed numerically that these type error estimators are very effective for computation of elasticity problems [2, 23, 24]. In this paper, we will present some recovered stress tensors and show that the relevant a posteriori error estimators are asymptotically exact. The theoretical results given in this paper are numerically supported by ZZ in [2, 23, 24]. On the other hand, the analysis here generalizes the results in [20] for the second order elliptic problem. Analysis of other recovery techniques including the ZZ patch recovery for the second order elliptic equation with triangular meshes can be found in [5]. This paper is organized as follows: Section 2 describes the model problem under consideration. Section 3 introduces the finite element method followed by the derivation of the superconvergence error estimate in Section 4. Section 5 is devoted to construct the recovered stress tensors, and derive their corresponding superconvergence errors. Finally, in the last section, we will give the reliable a posteriori error estimators based on the recovered stress tensors. 2. Elasticity problems. Let Ω R 2 be a bounded convex polygonal domain with vertices S i, edges Γ ij connecting S i and S j. Here indices on Γ ij, S i are understood as integers modulo n. In this paper, we consider quadrilateral bilinear finite element approximation of pure displacement and pure traction elasticity problems. For any v = (v, v 2 ) V ˆ=(H (Ω)) 2, we set ε ij (v) = 2 ( v i x j + v j x i ), i, j 2, σ ij (v) = λ(divv)δ ij + 2µε ij (v), i, j 2, where the constants λ 0 and µ > 0 are the Lame coefficients, and δ ij = 0(i j), δ ii =. Moreover, we set for any v V, the expressions ( v 2 m + v 2 2 m) 2 and ( v 2 m + v 2 2 m) 2 for (m = 0, ) will still be denoted by v m and v m, the expressions m and m represent respectively a norm and a seminorm over the Sobolev space H m (Ω) (cf. [9] for details). We define the bilinear form a(, ) on V V by a(u, v) = = λ 2 σ ij (u)ε ij (v)dx Ω i,j= Ω divudivvdx + 2µ 2 ε ij (u)ε ij (v)dx Ω i,j= := λ(divu, divv) + 2µ(ε(u), ε(v)), and the linear form 2 (f, v) = f i v i dx Ω i= f i L 2 (Ω). For any k, define the Sobolev space Ĥ k (Ω) = {v H k (Ω) 2 : vdx = 0, Ω Ω rotvdx = 0},

3 PATCH RECOVERY OF ELASTICITY PROBLEMS 65 where rotv = v 2 v. Then the weak formulation of the pure traction elasticity x x 2 problem is to find unknown displacement u Ĥ (Ω) such that a(u, v) = (f, v) + g, v v Ĥ (Ω), (2.) where the inner product, defines the boundary integration n g, v = g i vds. Γ i i= For the above pure traction problem to be uniquely solvable, the following compatibility condition must be satisfied (f, v) + g, v = 0 v RM, where the space of infinitesimal rigid motions RM is defined as follows: RM = {v : v T = (a + bx 2, c bx ), a, b, c R}. For the pure displacement problem, the boundary value problem can be written as: { µ u + (λ + µ)grad(divu) = f in Ω (2.2) u = 0 on Ω. It has the following weak formulation: Find u H 0(Ω) such that µ(gradu, gradv) + (µ + λ)(divu, divv) = (f, v) v H 0(Ω). (2.3) By the Poincare inequality, we know that there exists a unique solution to the above pure displacement problem. 3. Finite element discretization. Let T h be a partition of the domain Ω by convex quadrilaterals with the mesh size h := max Th h, h the longest edge of. Let ˆ = [, ] [, ] be the reference square with vertices Ẑi, and let be a convex quadrilateral with vertices Zi (x i, y i ), i =, 2, 3, 4. It is known that there exists a unique bilinear mapping F such that F ( ˆ) =, F (Ẑi) = Zi given by 4 4 x = x i N i, x 2 = yi N i, where i= i= N = 4 ( ξ)( η), N 2 = ( + ξ)( η), 4 N 3 = 4 ( + ξ)( + η), N 4 = ( ξ)( + η). 4 By a simple manipulation, we can explicitly express the mapping F as: x = a 0 + a ξ + a 2 η + a 3 ξη, x 2 = b 0 + b ξ + b 2 η + b 3 ξη, where by suppressing the index, a 0 = (x + x 2 + x 3 + x 4 )/4, b 0 = (y + y 2 + y 3 + y 4 )/4; a = ( x + x 2 + x 3 x 4 )/4, b = ( y + y 2 + y 3 y 4 )/4; a 2 = ( x x 2 + x 3 + x 4 )/4, b 2 = ( y y 2 + y 3 + y 4 )/4; a 3 = (x x 2 + x 3 x 4 )/4, b 3 = (y y 2 + y 3 y 4 )/4.

4 66 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG The Jacobi matrix of the mapping F is ] [ ] a + a (DF )(ξ, η) = = 3 η b + b 3 η a 2 + a 3 ξ b 2 + b 3 ξ [ x x x 2 x 2 The determinant of the Jacobi matrix is where x x 2 J = J (ξ, η) = J 0 + J ξ + J 2 η, (3.) 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 [ ] = (DF ) = J x x 2 It can be shown that J 0 [ b2 + b 3 ξ b b 3 η ] a 2 a 3 ξ a + a 3 η := J X. (3.2) := J (0, 0) = 4, J (ξ, η) > 0. (3.3) For any function v(x, x 2 ) defined on, we associate ˆv(ξ, η) by It is straightforward to verify that ˆv(ξ, η) = v(x (ξ, η), x 2 (ξ, η)), or ˆv = v F. v = ˆv + ˆv x x x here B := [b 2 + b 3 ξ, b b 3 η]. Similarly, = J [b 2 + b 3 ξ, b b 3 η][ ˆv, ˆv ]T = J B ˆ ˆv, v = ˆv + ˆv x 2 x 2 x 2 = J [ a 2 a 3 ξ, a + a 3 η][ ˆv, ˆv ]T = J A ˆ ˆv, here A = [ a 2 a 3 ξ, a + a 3 η]. We denote the midpoints of Z 2 Z 4 and Z Z 3 as O and O 2, respectively and let P i is the midpoint of edge Z i Z i+, i =, 2, 3, 4, (Figure ). We have P 2 P 4 = 2 (x 2 + x 3 x 4 x, y 2 + y 3 y 4 y ) = 2(a, b ), P 3 P = 2 (x 3 + x 4 x x 2, y 3 + y 4 y y 2 ) = 2(a 2, b 2 ), O O 2 = 2 (x + x 3 x 2 x 4, y + y 3 y 2 y 4 ) = 2(a 3, b 3 ). Then P 2 P 4 = 2(a 2 + b 2 ) 2, P P 3 = 2(a b 2 2) 2, O O 2 = 2(a b 2 3) 2,

5 PATCH RECOVERY OF ELASTICITY PROBLEMS 67 Z 3 P 3 Z 4 P 4 O 2 O P 2 Z Z 2 P Figure. Quadrilateral Moreover,it is easy to prove that J 0 = 4 P 2P 4 P P 3 = 4 P 2P 4 P P 3 sinα, (3.4) J = 4 P 2P 4 O O 2 = 4 P 2P 4 O O 2 sinβ, (3.5) J 2 = 4 P P 3 O O 2 = 4 P P 3 O O 2 sinγ, (3.6) where α is the acute angle between two lines P P 3 and P 2 P 4, β is the acute angle between two lines P 2 P 4 and O O 2, and γ is the acute angle between two lines P P 3 and O O 2. Definition 3.. A convex quadrilateral is said to satisfy the diagonal condition if d = O O 2 = O(h +α ), α > 0. Note that is a parallelogram if and only if d = 0. Remark. Using d to characterize asymptotic quality of quadrilaterals in analysis can be traced back at least 20 years, cf. [4], also see [] for more discussions on mesh conditions. Note that all quadrilaterals produced by a bi-section scheme of mesh subdivisions have the property α =. Definition 3.2. A partition T h is said to satisfy condition α if there exist α > 0 such that ) any T h satisfies the diagonal condition; and 2) any two, 2 in T h that share a common edge satisfy a neighboring condition: For j =, 2 a j = a 2 j ( + O(h α + h α 2 )), b j = b 2 j ( + O(h α + h α 2 )). (3.7) Remark 2. Let us comment on the geometric meaning of condition α. Observe that a j = a 2 j, b j = b 2 j, j =, 2 implies P 2 P 4 = 2(a, b ) = P 2 2 P 2 4, P 3 P = 2(a 2, b 2 ) = P 2 3 P 2.

6 68 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG Therefore, the neighboring condition characterizes how similar the neighbor quadrilaterals are. Together with the diagonal condition, quadrilaterals under condition α cannot deviate from parallelograms too much asymptotically. It is well known from the literature that parallelogram mesh means superconvergence. Remark 3. We may use another parameter, say β, in Definition 3.2 for the neighboring condition. Then we need to carry both parameters all the way in later statements and proofs, and have γ = min(α, β) for the convergence rate. In order to simplify the matter, we understand from the very beginning that the α is the smaller power in the diagonal condition and the neighboring condition. To obtain optimal order error estimate in the H -norm for the bilinear isoparametric interpolation on a convex quadrilateral, namely, the estimate v v I 0 + h v v I, Ch 2 u 2,, v H (), (3.8) we need the following degeneration condition, which was introduced by Acosta and Duran []. Definition 3.3. 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 diagonal, 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). By the definition of J, and the geometric relations of (2.3),(3.)-(3.3), we can easily check that J = + O(h α ), J J = O(h α ), J 2 J = O(h α ), (3.9) Define We denote J = + O(h α ), J = O(h α ), J2 A 0 = [ a 2, a ], A = [ a 3 ξ, a 3 η] B 0 = [b 2, b ], B = [b 3 ξ, b 3 η]. = O(h α ), (3.0) A = A 0 + A, B = B 0 + B. (3.) Lemma 3.4. Let a convex quadrilateral satisfy the diagonal condition. Then (a) B 0 X 2 + O(h α ), B X 2 O(h α ), (b) A 0 X 2 + O(h α ), A X 2 O(h α ). Proof. We only prove (a). The proof of (b) is similar. It is straightforward to verify that [ ] [ ] B 0 X a b = [b 2, b ] ( a3 η b + 3 η ) J a 2 b 2 a 3 ξ b 3 ξ = J 0 [, 0] + [b 2 a 3 b a 3 ξ, b 2 b 3 η b b 3 ξ], J J B X = [b 3 ξ, b 3 η] J ( [ a b a 2 b 2 ] + = J [a b 3 ξ a 2 b 3 η, b 3 b ξ b 2 b 3 η]. [ a3 η b 3 η a 3 ξ b 3 ξ ] )

7 PATCH RECOVERY OF ELASTICITY PROBLEMS 69 Then B 0 X 2 J 0 J + C h2+α J + O(h α ), (3.2) B X 2 J [a b 3 ξ a 2 b 3 η, b 3 b ξ b 2 b 3 η] 2 Ch α. Next we define the finite element space S h on T h as follows: S h = {v H (Ω) : ˆv = v F Q ( ˆ), T h }, V h = S h S h. Here Q ( ˆ) is the bilinear function space on the reference element. 4. Superconvergence analysis. Define v x and v x := J 0 B 0 ˆ ˆv, v x 2 as follows: v x 2 := J 0 A 0 ˆ ˆv, (4.) where B 0 = [b 2, b ], A 0 = [ a 2, a ]. Based on the definition, we can define the modified divergence and strain tensor by and for any v = (v, v 2 ). divv = v x + v 2 x 2, ε(v) = [ε ij (v)] = [ 2 ( v i x j + v j x i )], i, j =, 2, Next, we define some new bilinear forms based on as follows: and (ε(w), ε(w)) := J 0 (div(w), div(v)) := J 0 ˆ ˆ v x and v x 2 ε(w) : ε(v)dξdη, div(w) div(v)dξdη. on element Theorem 4.. Assume that satisfies the diagonal condition. Then for any v = (v, v 2 ), w = (w, w 2 ) V h, there exists a constant C independent of mesh sizes h such that (a) (ε(w), ε(v)) ( ε(w), ε(v)) Ch α w, v,, (b) (div(w), div(v)) ( div(w), div(v)) Ch α w, v,. Proof. First we prove (a). The definition of ε, and ε gives = (ε(w), ε(v)) ( ε(w), ε(v)) 2 [(ε ij (w), ε ij (v) ( ε ij (w), ε ij (v) ]. i,j=

8 70 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG We estimate these terms one by one. For the first term, we have (ε (w), ε (v)) ( ε (w), ε (v)) = ( w, v w ) J v 0 dξdη x x ˆ x x = ( B ˆ J ˆ ŵ ) T ( B J ˆ ˆv )J dξdη ( ˆ B 0 ˆ ŵ ) T ( B 0 ˆ ˆv )dξdη = (B ˆ J ˆ ŵ ) T (B ˆ ˆv )dξdη ˆ (B 0 ˆ ŵ ) T (B 0 ˆ ˆv )dξdη = [ (B ˆ J ˆ ŵ ) T (B ˆ ˆv )dξdη ˆ (B ˆ ŵ ) T (B ˆ ˆv )dξdη] +[ (B ˆ ŵ ) T (B ˆ ˆv )dξdη (B 0 ˆ ŵ ) T (B 0 ˆ ˆv )dξdη] ˆ := E + E 2 ˆ For the term E, we can estimate it as follows: w v J E = dx dx 2 x x = ( J ) w v dx dx 2 x x = ( J ξ + J 2 Ch α w v. η) w x v x dx dx 2 w x v x dx dx 2 In the last inequality, we have used (3.0). Next, we estimate the term E 2. Because B = B 0 + B, we have E 2 = [ (B ˆ ŵ ) T (B 0 ˆ ˆv )dξdη + (B 0 ˆ ŵ ) T (B ˆ ˆv )dξdη ˆ ˆ + (B ˆ ŵ ) T (B ˆ ˆv )dξdη]. ˆ We only estimate the first term in the above formulation. The estimation of the other two terms is similar. Using Lemma 3.4 and (3.7) yields (B ˆ ŵ ) T (B 0 ˆ ˆv )dξdη ˆ = = = = J ˆ J ˆ J ˆ ( ˆ ŵ ) T (B T B 0 )( ˆ ˆv )dξdη ˆ Ch α w, v,. ( J X ˆ ŵ ) T (X ) T (B T B 0 )X ( J X ˆ ˆv )dx dx 2 ( J X ˆ ŵ ) T (B X ) T )(B 0 X )( J X ˆ ˆv )dx dx 2 ( w ) T (B X ) T )(B 0 X )( v )dx dx 2

9 PATCH RECOVERY OF ELASTICITY PROBLEMS 7 Hence, we can obtain (ε (w), ε (v)) ( ε (w), ε (v)) Ch α w, v,. (4.2) Similarly, we have (ε 22 (w), ε 22 (v)) ( ε 22 (w), ε 22 (v)) Ch α w 2, v 2,. (4.3) For the term, we can derive that (ε 2 (w), ε 2 (v)) ( ε 2 (w), ε 2 (v)), (ε 2 (w), ε 2 (v)) ( ε 2 (w), ε 2 (v)) = ( w 2 + w )( v 2 + v )dx dx 2 4 x v 2 x v 2 ( w 2 + w )( v 2 + v )dξdη 4 ˆ x v 2 x v 2 = 4 [ w 2 v w dx dx 2 J 2 v 0 dξη] x x 2 ˆ x x [ w v 2 w dx dx 2 J v2 0 dξη] x 2 x ˆ x 2 x + 4 [ w 2 v w dx dx 2 J 2 v 0 dξη] x x 2 ˆ x x [ w 2 v 2 w dx dx 2 J 2 v2 0 dξη] x x ˆ x x := H + H 2 + H 3 + H 4. Using the same techniques as the estimation for (4.2), we can estimate H i, (i =,..., 4) one by one, and finally we can get 4 H i Ch α w, v,. (4.4) i= Then combining above two equalities yields Similarly, we have (ε 2 (w), ε 2 (v)) ( ε 2 (w), ε 2 (v)) Ch α w, v,. (4.5) (ε 2 (w), ε 2 (v)) ( ε 2 (w), ε 2 (v)) Ch α w, v,, (4.6) which, together with (4.2),(4.3),(4.5), gives (a). The proof of (b) is similar to the proof of (a). We then define a modified bilinear form over the finite element space V h follows: a h (u, v) = 2µ ( ε(u), ε(v)) + λ ( divu, divv), u, v V h. as Theorem 4.2. Assume the partition T h satisfy the diagonal condition and RDP(N,Ψ), and let u I be the finite element interpolation of the function u =

10 72 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG (u, u 2 ) H 3 (Ω) H 0(Ω) under quadrilateral meshes. Then for any v = (v, v 2 ) V h = S h S h, a h (u u I, v) C(h +α u 2,Ω + h 2 u 3,Ω ) v,ω. (4.7) Proof. By the definition, we have a h (u u I, v) = 2µ ( ε(u u I ), ε(v)) + λ ( div(u u I ), divv), (4.8) For the first term in the right hand side of the above equality, we have ( ε(u u I ), ε(v)) = By simple calculation, we get 2 ( ε ij (u u I ), ε ij (v)). i,j= ( ε (u u I ), ε (v)) (u = u I ) v dx dx 2 x x = ˆ (û û I ).( B0 T B 0 ). ˆ v dξdη ˆ = b2 2 J 0 ˆ [ b 2b J 0 J 0 (û û I ) ˆ v dξdη + b2 J 0 (û û I ) v dξdη + b 2b ˆ (û û I ) ˆ v dξdη (û û I ) v dξdη]. We estimate the terms on the right hand side of the above equality one by one. For the first term, if û P 2 ( ˆ), it is easy to see that ˆ (û û I ) v dξdη = 0. So by the Bramble-Hilbert Lemma and the fact b2 2 b2 2 J 0 ˆ (û û I ) v dξdη = O(), we have C D 3 û 0, ˆ ˆv, ˆ C(h +α u 2, + h 2 u 3, ) v,. Similarly, b2 J 0 ˆ (û û I ) v dξdη C(h+α u 2, + h 2 u 3, ) v,. Next we estimate the last term. For any v S h, we can express ˆv = ˆv (0, 0) + η 2ˆv, ˆv = ˆv (0, 0) + ξ 2ˆv,

11 PATCH RECOVERY OF ELASTICITY PROBLEMS 73 Then we can write b 2 b [ = b 2b J 0 ˆ (û û I ) [ v (0, 0) + b 2b [ 2 v ˆ ˆ v dξdη + ˆ (û û I ) (û û I ) v dξdη + v (0, 0) ξ (û û I ) dξdη + 2 v Since for any û P 2 ( ˆ), it is easy to check that 2 v (û û I ) 2 v dξdη = 0, ˆ Applying Bramble-Hilbert Lemma gives b 2b [ v (û û I (0, 0) ) ˆ ˆ ˆ dξdη] ˆ dξdη + v (0, 0) ˆ (û û I ) dξdη] η (û û I ) dξdη]. (û û I ) dξdη = 0. C D 3 û 0, ˆ ˆv, ˆ C(h +α u 2, + h 2 u 3, ) v,. (û û I ) dξdη] For the term 2 v ξ (û û I ) dξdη = 2 v (ξ 2 ) (û û I ) dξdη ˆ 2 ˆ = 2 v (ξ 2 ) 2 (û û I ) 2 ˆ 2 dξdη = 2 v (ξ 2 ) 2 û 2 ˆ 2 dξdη = 2 + Similarly, (ξ 2 )( 2 û 2 (ξ 2 ) 3 û ˆ 2 2 v = 2 + ˆ ˆv )(ξ, )dξ 2 ˆv dξdη. η (û û I ) dξdη (η 2 )( 2 û ˆv 2 η )(, η)dξ 2 (ξ 2 ) 3 û ˆv ˆ 2 dξdη. Then combining above two equalities yields b 2 b [ 2 v J 0 = b 2b 4 { J 0 i= +[ b 2b J 0 ˆ l i 2 (ξ 2 ) 3 û ˆ 2 4 i= (ξ 2 )( 2 û ) ˆv 2 (ξ, )dξ ξ (û û I ) dξdη + 2 v l i (t(s) 2 ) 2 u s 2 v s ds ˆv dξdη + l i (t(s) 2 ) 2 u v l i s 2 s ds +C(h +α u 2, + h 2 u 3, ) v,, (η 2 )( 2 û ) ˆv 2 (, η)dξ ˆ η (û û I ) dξdη] (ξ 2 ) 3 û ˆv ˆ 2 dξdη]}

12 74 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG here l i are four edges of. Using the neighboring condition, for any two elements, 2, that shares a common edge, and the trace theorem, we have ( b b 2 J b 2 b J ) l 2 (t(s) 2 ) 2 u v 2 0 l i s 2 s ds Ch α l 2 (h u 2, v, + h u 3, v, ) C(h +α u 2, + h 2 u 3, ) v,. Adding all elements T together gives ( ε (u u I ), ε (v)) Similarly, we have For the term C(h +α u 2 + h 2 u 3 ) v. (4.9) ( ε 22 (u u I ), ε 22 (v)) C(h +α u 2 + h 2 u 3 ) v. (4.0) ( ε 2 (u u I ), ε 2 (v)) = ( (u 2 u I 2) dx dx 2 + (u u I ) )( v 2 + v )dx dx 2 4 x x 2 x x 2 = [ ( (u 2 u I 2) v2 dx dx 2 + (u 2 u I 2) v dx dx 2 4 x x x x 2 + ( (u u I ) v2 dx dx 2 + (u u I ) v dx dx 2 ] x 2 x x x 2 := I + I 2 + I 3 + I 4. Using same arguments as in the proof of (4.9), we can estimate I i one by one, and finally we can get ( ε 2 (u u I ), ε 2 (v)) Similarly, we have C(h +α u 2 + h 2 u 3 ) v. (4.) ( ε 2 (u u I ), ε 2 (v)) C(h +α u 2 + h 2 u 3 ) v. (4.2) Using the same technique, we can prove that ( div(u u I ), divv) C(h +α u 2 + h 2 u 3 ) v. Conclusion follows by combining (4.8)-(4.2).

13 PATCH RECOVERY OF ELASTICITY PROBLEMS 75 From the proof of the above theorem, we know that the following result is also valid. Proposition 4.. Assume the partition T h satisfy the diagonal condition and RDP(N,Ψ), and let u I be the finite element interpolation of the function u H 3 (Ω), under the quadrilateral meshes. Then for any Ω 0 Ω, v V h = S h S h, we have a h,ω0 (u u I, v) C(h +α u 2,Ω0 + h 2 u 3,Ω0 ) v,ω. (4.3) Theorem 4.3. Assume that T h satisfies diagonal condition, and RDP (N, Ψ). Let u H 3 (Ω) H 0(Ω) be the solution of (2.), let u h and u I be the finite element approximation and the finite element interpolation of u, respectively. Then () u I u h C(h +α u 2,Ω + h 2 u 3,Ω ) (2) σ(u I ) σ(u h ) 0 C(h +α u 2,Ω + h 2 u 3,Ω ). Proof. By Theorem 4., and Theorem 4.2, we can derive a(u u I, v) a(u u I, v) a h (u u I, v) + a h (u u I, v) Ch α u u I, v + a h (u u I, v) Using orn equality gives C(h +α u 2,Ω + h 2 u 3,Ω ) v. u I u h C u I u h a(u I u h, v) a(u I u, v) C sup = sup v V h v v V h v C(h +α u 2,Ω + h 2 u 3,Ω ), where = a(, ) 2. By the definition of the stress tensor, we have σ(u I ) σ(u h ) 0 2µ ε(u I ) ε(u h ) 0 + λ divu I divu h 0 We complete the proof. C u I u h C(h +α u 2,Ω + h 2 u 3,Ω ). 5. Stress tensor recovery. In this section, we will introduce some patch recovery methods, which were proposed in [23],[7] for the second order elliptic problems. First we define a gradient recovery operator G h : S h S h S h, where S h is the bilinear element space under quadrilateral meshes, as follows: Given the finite element function v h, we first define G h v h at all vertices, and then obtain G h v h on the whole domain by interpolation using the original nodal shape functions of S h. Given an interior vertex z i, we select an element patch ω i, where ω i = Th,z. We denote all nodes on ω i (including z i ) as z ij, j =, 2,..., n( 6). In the following, we will introduce several fitting methods to recover the gradient of a function v h S h at z i by values of z ij. To demonstrate the idea, we write the recovery procedure explicitly on the element patch shown by Fig. 2. For general quadrilateral meshes, a computer algorithm will be needed. We use local coordinates (x, x 2 ) with z i as the origin.

14 76 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG ) Fitting p x (x, x 2 ) = p T a = (, x, x 2 )(a, a 2, a 3 ) T with respect to the four derivative values at the center of each element on the patch results in Q T Q a = Q T σ where σ T = (σ,, σ 4 ) (here we use σ to represent x v h since the case for x2 v h is the same) and Q T = /2 /2 /2 /2. /2 /2 /2 /2 It is easy to calculate Therefore, p x (x, x 2 ) = 4 We then have Similarly, Q T Q = σ j + 2 ( σ + σ 2 + σ 3 σ 4 )x + 2 ( σ σ 2 + σ 3 + σ 4 )x 2. j= p x (0, 0) = 8h [2(v v 5 ) + v 2 v 4 + v 8 v 6 ]. p x2 (0, 0) = 8h [2(v 3 v 7 ) + v 2 v 8 + v 4 v 6 ], where v i, (i =,..., 8) denote the values of the function v h at the vertices of ω z (cf. Figure 2 for details). Now define 2) Fitting G h v h (z i ) = (G hv h (z i ), G 2 hv h (z i )) T = (p x (0, 0), px 2 (0, 0))T. q x (x, x 2 ) = p T a = (, x, x 2, x x 2 )(a, a 2, a 3, a 4 ) T with respect to the four gradient values at the center of each element on the patch is the same as the interpolation. Therefore, q x (x, x 2 ) = 4 σ j ( σ + σ 2 + σ 3 σ 4 )x + 2 ( σ σ 2 + σ 3 + σ 4 )x 2 j= Note that q x (0, 0) = px (0, 0). Then we define p x 2 +(σ σ 2 + σ 3 σ 4 )x x 2 = p (x, x 2 ) + (σ σ 2 + σ 3 σ 4 )x x 2. (0, 0), the same as in ). Similarly, we have qx (0, 0) = G h v h (z i ) = (G hv h (z i ), G 2 hv h (z i )) T = (q x (0, 0), qx2 3) Fitting (0, 0))T = (p x p 2 (x, x 2 ) = (, x, x 2, x 2, x x 2, x 2 2)(a,, a 6 ) T with respect to the nine nodal values on the patch. Now e = (,,,,,,,, ) T, x = (0,,, 0,,,, 0, ) T, y = (0, 0,,,, 0,,, ) T, A = ( e, x, y, x 2, x y, y 2 ), 2 (0, 0), px2 (0, 0))T.

15 PATCH RECOVERY OF ELASTICITY PROBLEMS 77 and Now define (Q T Q) Q T = diag( 9, 6, 6, 6, 4, 6 ) ) Fitting p 2 (0, 0) = x 6h (v v 5 + v 2 v 4 + v 8 v 6 ), p 2 (0, 0) = x 2 6h (v 2 v 8 + v 3 v 7 + v 4 v 6 ). G h v h (z i ) = (G hv h (z i ), G 2 hv h (z i )) T = p 2 (0, 0; z i ). q 2 (x, x 2 ) = (, x, x 2, x 2, x x 2, x 2 2, x 2 x 2, x x 2 2, x 2 x 2 2)(a,, a 9 ) T with respect to the nine nodal values on the patch is the same as interpolation. This results in q 2 (0, 0) = x 2h (v v 5 ), (5.) and q 2 (0, 0) = x 2 2h (v 3 v 7 ), (5.2) which is different from all above. Fitting q 2 (x, x 2 ) = (, x, x 2, x 2, x x 2, x 2 2, x 2 x 2, x x 2 2)(a,, a 8 ) T with respect to the nine nodal values on the patch also produces (5.),(5.2). Next define G h v h (z i ) = (G hv h (z i ), G 2 hv h (z i )) T = q 2 (0, 0; z i ). Remark 4. Fitting strategies ) & 2) are the well known ZZ patch recovery, and strategies 3) & 4) were proposed by Zhang and Naga in [7]. Then for any u h = (u h, u2 h ), we define the Stress Tensor Recovery as follows where and σ h(u h ) = 2µε (u h ) + λdiv (u h ), (5.3) ε (u h ) = 2 i,j= 2 (Gj h ui h + G i hu j h ), div (u h ) = G hu h + G 2 hu 2 h, where G h, G2 h are defined by the above four fitting methods. When we consider the pure traction problem, we do not need to do stress tensor recovery on the boundary. However, if we consider the pure displacement boundary value problem, the recovered strain tensor on a boundary node z can be determined from an element patch ω i such that z ω i in the following way: As an example, we

16 78 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG z Figure 2. Element Patch ω z consider the fitting strategy 3). Let the relative coordinates of z with respect to z i is (h, h), then G h v h (z i ) = (G hv h (z i ), G 2 hv h (z i )) T = p 2 (h, h; z i ). Putting the gradient recovery to the (5.3) yields the stress tensor recovery. If z i is covered by more than one element patches, then some averaging may be applied (cf. [7] for details). Theorem 5.. Let T h satisfy the diagonal condition. Then there exists a constant C independent of h such that σ (v h ) 0 C v h v h V h. Proof. We first consider an element patch that contains four uniform square elements (Figure 2). We discuss only the fitting strategy 3), the proof for the another three strategies is similar. Following the recovery procedure of the above, it is straightforward to derive G hv(z) = h ˆpv 2(0, 0) = 6h (v v 5 + v 2 v 4 + v 8 v 6 ) = c h jv j, (5.4) where the weighted c j are the second row of (QT 0 Q 0 ) Q T 0 for the x -derivative and similarly Ghv(z) 2 = h ˆpv 2(0, 0) = 6h (v 2 v 8 + v 3 v 7 + v 4 v 6 ) = c 2 h jv j, (5.5) where the weighted c 2 j are the third row of (QT 0 Q 0 ) Q T 0 for the x 2 -derivative. By inverse inequality, Similarly v v v v(z) v(z) v 5 2 Ch 2 ( v 2,, 2 + v 2,, ) C( v 2, 2 + v 2, ) = C v 2, 2. v 2 v 4 2 C v 2, 3 4, v 8 v 6 2 C v 2, 2, j j

17 PATCH RECOVERY OF ELASTICITY PROBLEMS 79 Then we have G hv(z) Ch ( v v 5 + v 2 v 4 + v 8 v 6 ) Ch ( v v v 2 v v 8 v 6 2 ) 2 Ch v,ωz. We can also get G 2 hv(z) Ch v,ωz. By linear mapping, these results are also valid for four uniform parallelogram. We observe that the diagonal condition together with the neighboring condition imply that for any given node z i, there are four elements attached to it when h is sufficiently small. In addition, we can decompose the least square fitting matrix Q associated with those four quadrilateral elements as follows: Q = Q 0 + h α Q where Q 0 is the least-square fitting matrix associated with those four parallelograms. After some simple calculations, we can show that where I is identity matrix. Then G hv(z) = h (Q T Q) Q T = (Q T 0 Q 0 ) Q T 0 + O(h α )I, ( c j + O(h α ))v j. j Notice that c is constant vector, so for sufficiently small h, we always have G hv(z) ( c h j)v j Ch v,ωz. (5.6) Then for any T h, define j ω = T h,. Let z i, i =, 2, 3, 4 be the four vertices of the element, by scaling argument, it is straightforward to show that 4 G hv 2 0, Ch 2 [G hv(z i )] 2. Combining above two equalities and summing all T h together yields i= G hv 2 0 = G hv 2 0, 4 h 2 i=[g hv(z i )] 2 Ch 2 h 2 v 2,ω, which means G hv 0 C v. Similarly, G 2 hv 0 C v. By the definition of the recovery stress tensor and using the above two equalities, we see that σ (v h ) 0 C v h, v h V h. We complete the proof.

18 80 ZHONG-CI SHI, XUEJUN XU AND ZHIMIN ZHANG Remark 5. Another important feature of the recovery operator by the fitting strategies 3) & 4) is the polynomial preserving property, which have already been observed in [7],[9]. More precisely, let T, and u be a quadratic polynomial on ω. Assume that and all elements adjacent to are convex. Then σ (u) = σ(u), (5.7) where u = (u, u 2 ). Actually, the fitted polynomial p 2 = u i when u i (i =, 2) is a quadratic polynomial [7]. As a consequence, p 2 = u i, i =, 2. Theorem 5.2. Let T h satisfy the diagonal condition and RDP (N, Ψ). Let u H 3 (Ω) H 0(Ω) be the solution of (2.), and let u h S h be the finite element approximation of u. Then the recovered Stress Tensor is superconvergent in the sense σ(u) σ (u h ) 0 C(h +α u 2 + h 2 u 3 ), where C is constant independent of u and h. Proof. We first consider the fitting strategies 3) & 4). We can decompose the error into σ(u) σ (u h ) = σ(u) σ (u) + σ (u I u h ), (5.8) here we have used the fact σ (u) = σ (u I ). By the polynomial preserving property for the fitting strategies 3) & 4), and the Bramble-Hilbert Lemma, we can get σ(u) σ (u) 0 Ch 2 u 3. (5.9) For the fitting strategies ) & 2), we continue to split the term σ(u) σ (u) as σ(u) σ (u) = σ(u) [σ(u)] I + [σ(u)] I σ (u), where [σ(u)] I is the bilinear interpolation. By the interpolation estimate, we have σ(u) [σ(u)] I 0 Ch 2 u 3. Moreover, using the similar techniques developed by [5], [9], we can prove that [σ(u)] I σ (u) 0 Ch 2 u 3. So for the strategies ) & 2), we also have σ(u) σ (u) 0 Ch 2 u 3. (5.0) On the other hand, by Theorems 4.2 and 5., we can derive σ (u I u h ) 0 u I u I C(h +α u 2 + h 2 u 3 ), (5.) which, combining (5.8) (5.0), yields Theorem 5.2. For the pure traction problem, we can consider so-called interior estimation, which was first investigated by Nitsche and Schatz [2]. Using this technique, the global regularity u H 3 (Ω), which may not hold in general, is not required. In the following, we will provide a local result based on the interior estimate. We consider Ω 0 Ω Ω where Ω 0 and Ω are compact polygonal subdomains that can be decomposed into quadrilaterals. Here compact subdomains means that dis(ω 0, Ω ) and dis(ω, Ω) are of order O(). Combining the results developed in Section 4 and the standard interior estimate technique in [2], we immediately obtain the following result.

19 PATCH RECOVERY OF ELASTICITY PROBLEMS 8 Theorem 5.3. Let Ω R 2 be a polygonal domain and Ω 0 Ω Ω. Assume that T h satisfy the diagonal condition and RDP (N, Ψ) on Ω. Let u H 3 (Ω) H 0(Ω) be the solution of (2.), and let u h S h be the finite element approximation of u. Then σ(u) σ (u h ) 0,Ω0 C(h +α u 2,Ω + h 2 u 3,Ω0 ), where C is constant independent of u and h. 6. A posterior error estimate. Let e h = u u h, our purpose is to estimate the error σ(e h ) 0,Ω0 by a computable quantity η h. We can define the error estimator by the recovered stress tensor We assume that η h = σ (u h ) σ(u h ) 0,Ω0. σ(e h ) 0,Ω0 Ch. (6.) Theorem 6.. Assume that the same hypotheses as in Theorem 5.3. Let (6.) be satisfied. Then η h σ(e h ) 0,Ω0 = + O(h ρ ), ρ = min(, α). (6.2) Proof. By the triangle inequality, we have η h σ(u) σ (u h ) 0,Ω0 σ(e h ) 0,Ω0 η h + σ(u) σ (u h ) 0,Ω0. Divided the above inequality by σ(e h ) 0,Ω0 yields (6.2). Remark 6. In this paper, we just consider the compressible elasticity. For the incompressible elasticity, i.e., Poisson ratio ν 2, or equivalently λ, we can use the so-called enhanced strain finite element method [3] to cope with it. The advantage of this method is that it provides a stable locking-free element without the need of implementing the filter. Another important feature of this method is that we only need to solve a symmetric and definite algebraic system. The saddle point algebraic system will be avoided. Along this line, some previous results on non-conforming finite element methods for elasticity [6, 0, 8] would be useful. Remark 7. Generalization of our results to higher order quadrilateral elements is feasible. Mesh conditions would be more important since for non-tensor product space, degeneration of approximation order may happen [3]. Acknowledgements. We would like to thank the referees very much for their valuable comments and suggestions. REFERENCES [] G. Acosta and R. G. Duran, Error estimates for Q isoparameter element satisfying a weak angle condition, SIAM J. Numer. Anal., 38 (200), [2] M. Ainsworth and J. T. Oden, Aa posteriori Error Estimation in Finite Element Analysis, Wiley Interscience, New York, [3] D. N. Arnold, D. Boffi and R. S. Falk, Approximation by quadrilateral finite elements, Math. Comp., 7 (2002), [4] I. Babuška and T. Strouboulis, The Finite Element Method and its Reliability, Oxford University Press, London, 200. [5] S. C. Brenner and L. R. Scott, The Mathematical Theory of Finite Element Methods, Springer-Verlag, New York, 994. [6] S. C. Brenner and L.-Y. Sung, Linear finite element methods for planar linear elasticity, Math. Comp., 59 (992),

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