Polynomial Preserving Recovery for Quadratic Elements on Anisotropic Meshes Can Huang, 1 Zhimin Zhang 1, 1 Department of Mathematics, Wayne State University, Detroit, Michigan 480 College of Mathematics and Computational Science, Sun-Yat-Sen University, Guangzhou, Guangdong 51075, China Received 1 November 010; accepted 15 November 010 Published online in Wiley Online Library (wileyonlinelibrary.com. DOI 10.100/num.0669 Polynomial preserving gradient recovery technique under anisotropic meshes is further studied for quadratic elements. The analysis is performed for highly anisotropic meshes where the aspect ratios of element sides are unbounded. When the mesh is adapted to the solution that has significant changes in one direction but very little, if any, in another direction, the recovered gradient can be superconvergent. The results further explain why recovery type error estimator is robust even under nonstandard and highly distorted meshes. 011 Wiley Periodicals, Inc. Numer Methods Partial Differential Eq 000: 000 000, 011 Keywords: anisotropic mesh; finite element; recovery; superconvergence I. INTRODUCTION Finite element recovery techniques are postprocessing methods that reconstruct numerical approximations from finite element solutions to achieve better results. Recent years witness a revitalization of this field, especially those years when Zieniewicz-Zhu introduced their estimator based on superconvergence patch recovery (SPR, where they applied least squares fitting over a set of elements surrounding the vertex to smooth the stress (gradient computed from finite element method [1, ]. Later, another gradient recovery method, called polynomial preserving recovery (PPR was proposed [3, 4]. Both theoretical analysis and numerical tests reveal that PPR has better or at least the same properties as SPR [5]. In this article, we study PPR for quadratic elements under anisotropic meshes. This is the further extension of a recent study for linear element [6]. Nevertheless, this extension is by no means straightforward. A newly developed tool by Huang Xu [7] has to be adopted in order to carry on the needed analysis. Correspondence to: Can Huang, Department of Mathematics, Wayne State University, Detroit, MI 480, USA (e-mail: canhuang007@gmail.com Contract grant sponsor: US National Science Foundation; contract grant number: DMS-061908 (to Z.Z. Contract grant sponsor: Guangdong Provincial Government of China ( Computational Science Innovative Research Team program (to Z.Z. 011 Wiley Periodicals, Inc.
HUANG AND ZHANG In this article, we consider superconvergence for quadratic elements. We now that the optimal convergence rate of gradient approximation by quadratic element is O(h. Our theoretical results shows the convergence rate can reach O(h 3 for mildly structured meshes as well as anisotropic meshes with high aspect ratio in -D. As for references regarding a posteriori error estimates and superconvergence related to this article, the reader is referred to [1,, 8 19]. II. PRELIMINARY KNOWLEDGE A. Model Problem On a bounded polytopic domain R n, we consider the boundary value problem: To find u H 1 ( that satisfies some well-posed boundary conditions and for linear continuous functional f over H 1 B(u, v = f(v, v H 1 (, with bilinear form defined by B(u, v = = n i,j=1 a ij (x u v dx + x i x j n i=1 u A(x vdx + (u, b v + (cu, v. b i (xu v dx + cuvdx x i We assume the usual strong elliptic condition on A = A(x and sufficient regularity on all input data, which will be specified later when needed. Let T h be a simplicial partition of. The quadratic finite element space is then defined by S h ={v : v H 1 (, v P, T h }, where P is the set of polynomials of degree no more than. Then the finite element approximation u h S h satisfies B(u h, v = f(v, v S h. B. Simplification Let A 0 be a piecewise constant function such that each element T h A 0 = 1 A(xdx. Now we define a(u, v = u A(x vdx, a (u, v = u A 0 vdx, e h = u h h, Assume that a ij C 0,α (, it is easy to show that a(e h, v a (e h, v Ch α e h 1, v 1,, α>0.
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 3 Therefore, we shift our analysis to a (e h, v. Since A(x is symmetric positive definite, so is A 0. Then there exists an orthogonal matrix Q such that A 0 = Q T D Q with D = diag(d1,..., d n. By changing of variable x = Q z, wehave a (e h, v = z e h D z v detq z dz. z Note that z = Q z x, detq z =±1, and z is obtained by rotating. Therefore, without loss of generality, we may concentrate on the second-order form and estimate the bilinear form e h D vdx = B(e h, v = Th n i=1 d i e h v dx, d i > 0. x i x i e h D vdx + e h (b v + cvdx. Now, let s focus on domain variation. Following [6], we assume that T h can be separated into two parts T h = T 0,h T 1,h, = i,h, = 0,h 1,h, T i,h such that we have the following ɛ-σ condition: 1. Any two triangles that share a common edge in T 0,h form a convex quadrilateral which is an ɛ-perturbation from a parallelogram.. 1,h has a small measure: 1,h =O(h σ (σ > 0. Moreover, in the y-coordinate system, the image of 0,h under the triangulation T 0,h is formed by triangles, where each pair of adjacent elements forms a parallelogram. Under this y-coordinate system, we define a broen norm u, y = T 0,h u,y. By a similar argument as in [6], we have the following theorem. Theorem.1. Let u H 4 ( W 3 ( and u I S h be the solution of the model problem and its quadratic finite element interpolant, respectively. Then e I D vdx (h5/ ( u 4, 0,h + ɛ + h ɛ( u 3, 0,h + ɛ v 1,. (.1 T 0,h Proof. Let e I = u u I, then, we have e I D vdx = I 0 (r + εi 1 (r + ε I (r. (.
4 HUANG AND ZHANG Here, I 0 ( = I 1 ( = I ( = n i=1 d i n i=1 n i=1 d i y y d i y e I v dy = y e I D y vdy; y i y i y ( e I v div x η + e n I v η l + v y i y i y i y l=1 l x i y i ( n e I η n v η l e n I y x i l=1 y l x i y i =1 l=1 n e I η dy; y x i n e I η dy; y x i =1 v y l η l x i div x η v y i =1 ɛ is involved in the conformal transformation of domain variation y = x + ɛη(x, where η is a piecewise linear function such that x η is bounded uniformly in ɛ and h. Then, by the standard finite element superconvergence analysis, we are able to derive I 0 ( T 0,h h5/ u 4, y v 1, y h 5/ ( u 4, 0,h + ɛ v 1,, (.3 due to cancellations between parallel sides of adjacent triangles. For the term I 1 (, wehave I 1 ( T 0,h h u 3, y v 1, y h ( u 3, 0,h + ɛ v 1,. (.4 Combining (.3 (.4, we obtain e I D vdx (h5/ ( u 4, 0,h + ɛ + h ɛ( u 3, 0,h + ɛ v 1,. (.5 T 0,h Polynomial Preserving Gradient Recovery. Following [3, 4], we introduce a gradient recovery operator G h : S h S h S h, which has the two properties below: 1. Polynomial preserving: G h (p I = p, p P 3. Consequently, it owns the approximation property u G h u I h 3 u 4,, u H 4 (.. Boundedness: When there are no two adjacent angles on an element patch adding up to exceed π, wehave G h v v 1,, v S h.
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 5 Theorem.. Let u H 4 ( W 3 ( and u h S h be the solution of the model problem and its quadratic finite element approximation respectively. Assume (a the ɛ-σ mesh condition, (b the maximum angle condition, and (c the discrete inf-sup condition. Then the polynomial preserving gradient recovery operator G h leads to superconvergence in the sense that: u G h u h h 3 u 4, + h 5/ ( u 4, 0,h + ɛ + h ɛ( u 3, 0,h + ɛ + h +σ/ u 3,,. Proof. By the triangle inequality and the polynomial preserving property, u G h u h u G h u I + G h (u I u h h 3 u 4, + u h u I 1,. (.6 The analysis for the second term on the right-hand side under conditions (a and (b is proceeded as the following: e I D vdx h u 3,, v dx h +σ/ u 3,, v 0,, (.7 T 1,h T 1,h and (by the standard approximation theory e I (b v + cvdx h3 u 3, v 1,. (.8 Based on (.5, (.7, and (.8, we derive B(e I, v (h 5/ ( u 4, 0,h + ɛ + h ɛ( u 3, + ɛ + h +σ/ u 3,, v 1,. (.9 Using the discrete inf-sup condition and the strong elliptic assumption, we then have B(u h u I, v B(e I, v u I u h 1, sup = sup v S h v 1, v S h v 1, h 5/ ( u 4, 0,h + ɛ + h ɛ( u 3, + ɛ + h +σ/ u 3,,. (.10 The conclusion follows by substituting (.10 into (.6. III. ERROR ESTIMATES UNDER ANISOTROPIC MESHES We analyze errors under triangulation of the regular and Union-Jac patterns. We consider the worst case for the regular pattern where the maximum angle condition is violated at the extremal case θ 0, where a patch is compressed at a fixed ratio, i.e., θ = θ, θ 3 = 3 θ, and, 3 > 0. To simplify the matter, we analyze the case b = 0 and c = 0. We need the following integral identity [7] for v Q P (, (u u I v Q = 3 3 (a 3 s ( +b s ( u =1 s=0 l n s t + O(h 4 j u 4,p, v Q j,q,, j = 0, 1.
6 HUANG AND ZHANG where is modulo 3, angle θ is opposite of side l (with length l, t (n is the counter-clocwise unit tangential (outer-normal vector on side l. Denote We can express M = sin θ sin θ 1 + sin θ + sin θ 3. and a 0 ( = M ( l 1 l 180 cos θ +1 + l 1l l 3 10 cos θ 1 cos θ +1 l l +1 180 cos θ 1 l +1 l 1 cos θ 1 cos θ +1, 180 ( a 1 ( = M l 1 l 90 sin θ +1 cos θ +1 l 1l l 3 10 cos θ 1 sin θ +1 a ( = M cos θ 1 sin θ 1 cos θ +1 + l 1l l 3 10 cos θ +1 sin θ 1 180 sin θ 1 + l +1 l 1 cos θ 1 sin θ +1, 180 l +1 l 1 90 l l +1 ( l 1 l 180 sin θ +1 l 1l l 3 10 sin θ 1 sin θ +1 + l +1 l 1 90 cos θ 1 sin θ 1 sin θ +1 l +1 l 1 180 a 3 ( = M l+1 l 1 sin θ 1 sin θ +1, 180 sin θ 1 cos θ +1, ( b 0 ( = 1 ( l 4 l +1 l 1 + l4 1 cos3 θ +1 l4 +1 cos3 θ 1, 1440 sin θ +1 sin θ 1 b 1 ( = 1 1440 b ( = 1 1440 b 3 ( = 1 1440 ( 3l 4 1 cos θ +1 3l 4 +1 cos θ 1, ( 3l 4 1 cos θ +1 sin θ +1 3l 4 +1 cos θ 1 sin θ 1, ( l 4 +1 sin θ 1 l 4 1 sin θ +1. In Fig. 1, we let θ = θ, θ 3 = 3 θ. It is straightforward to verify t t = 3 u 3 1 x, 3 = cos 3 t 3 3 θ 3 u x + (3 3 cos 3 θ sin 3 θ 3 u x y (3 cos 3 θ sin 3 θ 3 u x y + sin3 3 θ 3 u y, 3
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 7 FIG. 1. Anisotropic mesh. t t 3 = (cos3 θ 3 u 3 x 3 3(sin θ(cos θ 3 u x y 3(sin θ(cos θ 3 u x y (sin3 θ 3 u y, 3 n n 1 t t = 3 u 1 x y, n n t t = (cos 3 θ(sin 3 θ 3 u x 3 +[(cos3 3 θ ( cos 3 θ(sin 3 θ] x y [(cos 3 θ(sin 3 θ (sin 3 3 θ] x y + (sin 3 θ(cos 3 θ 3 u y, 3 n n 3 t t = (cos θ(sin θ 3 u 3 x 3 +[(cos3 θ (sin θ(cos θ] x y +[(cos θ(sin θ (sin 3 θ] x y + (sin θ(cos θ 3 u y, 3 Note that sin θ 1 = sin(π θ θ 3 = sin( + 3 θ. Let l 1 = h, l h, l 3 h. Therefore, as θ 0, for = 1,, 3, we have 1 :a 0 h3 θ, :a 1 h3 θ 1, 3 :a h3, 4 :a 3 h3 θ, 5 :b 0 h4 θ 1, 6 :b 1 h4 7 :b h4 θ, 8 :b 3 h4 θ. (3.1 To balance a 0, a1 and b0, we need following conditions: x 3 = C(u, θ, 0, x 3 = C(u, θ, 0, x y = C(u, θ. (3. 0, Here and hereafter, C(u, (C(u, means a generic constant that depends on u and (, but independent of h and θ. Lemma 3.1. If a 0, a1, b0, = 1,, 3, are balanced, then the coefficients as ( and bs ( given in above have the following properties: (1 a s ( = O(h3, b s ( = O(h4 ;
8 HUANG AND ZHANG ( Assume that and are two adjacent equivalent triangles. Then a s ( = as (, b s ( = bs ( ; (3 If and form a pair of O(h approximate equivalent triangles (namely the lengths of any two opposite edges differ only by O(h, then a s ( as ( = O(h 4, b s ( bs ( = O(h 5. Also, by direct computation, one easily get t 1 = v Q x, t = v Q x cos θ 3 v Q x y cos θ 3 sin θ 3 + v Q y sin θ 3, t 3 = v Q x cos θ + v Q x y cos θ sin θ + v Q y sin θ. Noting that the scale in the x-direction is h while in the y-direction, the scale is hθ, one gets v Q y sin θ i = O ( vq, v Q x y sin θ i = O ( Therefore, under the regular partition of anisotropic mesh, see Fig. 1, x. 1, = 1,, 3. (3.3 h Similarly, v Q 1, = 1,, 3. (3.4 t h Thus, L ( h 1 v Q t L (, = 1,, 3. (3.5 Hence, by Lemma 3.1, Hölder s inequality and (3.3 (3.5, we have (a 3 01 ( a 0 1 ( u h 3 θ C(u, θ v t Q 1, h 3 C(u, v Q 1,. 1 t 3 1 Similarly, we obtain the estimate for other cases. Hence, for = 1,, 3 and s = 0, 1,, we have (a 3 s ( a s ( u n s t h 3 C(u, v Q 1,.
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 9 However, we can not obtain a similar estimate for boundary terms under such conditions as above because v Q t h 1/ θ 1/ v Q L (γ t. (3.6 L ( To get rid of θ 1/, condition (3. is refined to x 3 = C(u, θ, 0, x y = C(u, θ, 0, x 3 = C(u, θ 3/, 0, x y = C(u, θ 1/. (3.7 0, Thus, assume condition (3.7, and (3.3 (3.5, we obtain: ( 3 b 0 1 ( b0 1 ( u h 5 θ 1 C(u, l t 1 θ 3/ (By (3.1 1 t1 l 1 t 3 1 h 9/ C(u, l 1 t 1 L ( L (l 1 (By (3.6 h 7/ C(u, l 1 v Q 1, (By (3.5 h 7/ C(u, l 1 v Q 1,. By following the same method, we can obtain the estimate for the rest of cases. For the sae of demonstration, we provide the proof of one more case here. ( 3 b 1 ( b1 ( u h 5 C(u, l l n t t θ 1/ t (By (3.1 L (l 1 h 9/ C(u, l (By (3.6 t Hence, for any = 1,, 3 and s = 0, 1,, we have ( 3 b s ( bs ( u l h 7/ C(u, l v Q 1, h 7/ C(u, l v Q 1,. n s t L ( h 7/ C(u, γ v Q 1,, (By(3.5 where γ is the common edge of and. In addition, due to cancellations between parallel sides of adjacent triangles, we have the following identity [7]: ( (u u I, v h = e= s=0 3 (a se ( a s e ( + [ b s e ( bs e ( ] + O(h 4 s u e n s t t 4,p, v h s,q,. Following the above identity, we derive the following theorem.
10 HUANG AND ZHANG FIG.. Union-Jac Pattern. Theorem 3.1. Let u H 4 ( a and let a contain anisotropic uniform triangles of type in Fig. 1 with θ (0, π/4] (θ (π/4, π/ can be treated similarly by shifting the focus directions. Assume that x 3 = C(u, θ, 0, x y = C(u, θ, 0, x 3 = C(u, θ 3/, 0, x y = C(u, θ 1/. 0, Then we have e I D v h h 3 v h 1, a, a v h S 0 h ( a. Remar 1. From Theorem 3.1, we see that if u has very little activity in the x-direction, the degenerating limit θ = 0 in a sample triangle can be allowed, which is equivalent to allow the maximum angle in be as close as possible to π. The rate of convergence is maintained at O(h 3. Remar. In this case, two adjacent triangles form a parallelogram or a quasi-parallelogram, so cancellation instead of doubling between parallel sides occurs. Remar 3. In the result, the constant depends on u and a, but it is independent of h and θ. Next, let us consider the model problem under Union-Jac Patch, see Fig.. In this case, we can not analyze the problem in a similar fashion as we did in the previous case since some two adjacent triangles, say 1 and 8 does not form a quasi-parallelogram. The integral on the side shared by these two triangles doubles instead of canceling. However, if we group triangles 1 and 5, and 6, 3 and 7, 4, and 8, cancelations also happens under certain conditions. Here we only focus on 1 and 5 and we allow them to form a pair of O(h equivalent triangles, but we restrict θ = π. For other pairs, we can analyze them in exactly the same way (Fig. 3. From the graph above, we now that θ 1, θ, and θ 3 of these two triangles are almost the same. Moreover, both of these pairs are in count-clocwise order. Hence, a s ( 1 a s ( 5 = O(h 4, b s ( 1 b s ( 5 = O(h 5, = 1,, 3; s = 0, 1,, 3. Also, both the unit tangent and out-normal vectors on the corresponding sides are opposite.
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 11 FIG. 3. A group: 1 and 5. We also assume that the patch is compressed at a fixed ratio as in the case of regular mesh, i.e., θ 3 = θ. Hence, l 1 = h, l = O(h, l 3 = O(hθ,asθ 0. By simple computation, we obtain 1 :a1 0 h3, a 0 = 0, a0 3 h3 θ 3, :a1 1 h3 θ, a 1 = 0, a1 3 h3 θ, 3 :a1 h3 θ, a = 0, a 3 h3 θ, 4 :a1 3 h3 θ 3, a 3 = 0, a3 3 h3, 5 :b1 0 h4 θ 1, b 0 h4 θ 1, b3 0 h4 θ 3, 6 :b1 1 h4, b 1 h4, b3 1 h4 θ, 7 :b1 h4 θ, b h4 θ, b3 h4 θ, 8 :b1 3 h4 θ, b 3 h4 θ, b3 3 h4. (3.8 On the other hand, t t = 3 u 3 1 x, 3 u n n 3 = 3 3 x, 3 = cos 3 θ 3 u t 3 x + (3 3 cos θ sin θ 3 u x y (3 cos θ sin θ 3 u x y + sin3 θ 3 u y, 3 n n t = cos θ sin θ 3 u x + 3 (cos3 θ sin θ cos θ 3 u x y + (sin 3 θ cos θ sin θ 3 u x y + sin3 θ cos θ 3 u y, 3 It follows that if x 3 = C(u, θ, (3.9
1 HUANG AND ZHANG then b1 0 and b0 are balanced. From Appendix, we see that under the Union-Jac mesh, v t h 1 θ 1 v 1, h 1 θ 1 v 1,, = 1,, 3; L ( v h 3/ θ 1 v 1, h 3/ θ 1 v 1,, = 1,, 3. (3.10 L (γ Thus, we require condition (3.9 to be refined to x 3 = C(u, θ, 0, x 3 = C(u, θ, 0, x y = C(u, θ. (3.11 0, Hence, if u W 4,p ( and condition (3.11 holds, then we obtain ( { a s ( 1 + a s ( 5 1 5 n s t = a s t t t ( v Q 1 1 n s t t t t + + ( a s ( 5 a s ( 1 5 n s t t = I 1 + I. 5 t } n s t t t t Since n and t of corresponding sides of 1 and 5 are opposite, we conclude that the sign of v Q 1 n s t t t t and 5 n s t t are opposite. Therefore, for = 1 and s = 0, we obtain I 1 h a 0 1 ( C(u, 1 5 θ (By (3.8 and (3.11 t 0, 1 5 h 4 C(u, 1 5 θ t 0, 1 5 t h 3 C(u, 1 5 v 1,1 5. (By (3.10 I h 4 C(u, 1 5 θ (By (3.8 and (3.11 0, 1 5 h 4 C(u, 1 5 θ 0, 1 5 h 3 C(u, 1 5 v 1,1 5. (By (3.10 By following the same argument, we can obtain the same estimate for all other s and s s. Hence, I i h 3 C(u, 1 5 v 1,1 5, = 1,, 3, s = 0, 1,, 3, i = 1,. t
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 13 Summing up I 1 and I, we obtain, ( a s ( 1 + a s ( 5 1 5 n s t h 3 C(u, 1 5 v 1,1 5. Now, let s consider terms involving integration on the boundary. ( b s ( 1 + b s ( 5 l 1 l 5 nn n s t t t t { = b s ( 1 l 1 nn n s t t v Q t t + + ( b s ( 5 b s ( 1 l 5 n n s tt = L 1 + L. l 5 t n n s tt t t t } t t By the same analysis as I 1, we now that the sign of l 1 n s t and l 5 n s t are opposite. Hence, for = 1 and s = 0, L 1 h b 0 1 ( C(u, l1 l 5 θ, (By (3.8 and (3.11 t1 0, l 1 l 5 h 7/ C(u, l 1 l 5, (By (3.10 t1 0, h 7/ C(u, l 1 l 5 v Q 1,. L h 5 θ 1 C(u, l 1 l 5 θ, (By (3.8 and (3.11 t1 0, l 1 l 5 h 7/ v Q C(u, l 1 l 5, (By (3.10 0, t 1 h 7/ C(u, l 1 l 5 v Q 1,. Thus, ( b 0 1 ( 1 + b 0 1 ( 5 l 1 l 5 t 3 t 1 h 7/ C(u, l 1 l 5 v Q 1,. By the same argument as above, we can obtain ( b s ( 1 + b s ( 5 l 1 l 5 n s 3 t h 7/ C(u, l 1 l 5 v Q 1,1 5 = 1,, 3, s = 0, 1,, 3.
14 HUANG AND ZHANG Again, we provide the proof of one more case for demonstration. For =, s = 1, L 1 h b 1 ( C(u, l 1 l 5 θ, (By (3.8 and (3.11 t 0, l 1 l 5 h 7/ C(u, l 1 l 5, (By (3.10 t 0, h 7/ C(u, l 1 l 5 v Q 1,. L h 5 C(u, l 1 l 5 θ, (By (3.8 and (3.11 0, l 1 l 5 h 7/ C(u, l 1 l 5 t t h 7/ C(u, l 1 l 5 v Q 1,. 0,, (By (3.10 Our result can be concluded as the following theorem: Theorem 3.. Let u H 5 ( a and let contain anisotropic uniform triangles of type in Figure with θ (0, π/4] (θ (π/4, π/ can be treated similarly by shifting the focus directions. Assume that x 3 = C(u, θ, 0, x 3 = C(u, θ, 0, x y = C(u, θ. 0, Then we have e I D v h h 3 v h 1, a, a v h S 0 h ( a. By the similar argument as in Theorem., we have following theorems. Theorem 3.3. Let u H 4 ( W 3 ( and u h S h be the solution of the model problem and its quadratic finite element approximation, respectively. Assume (a ɛ-σ mesh condition, (b a 0,h contains anisotropic uniform triangles of type in Fig. 1 with θ (0, π/4] and elements in \ a satisfy the maximum angle condition, and (c the discrete inf-sup condition. Furthermore, assume that the condition in Theorem 3.1 is satisfied. Then the polynomial preserving gradient recovery operator G h has the following error bound: u G h u h h 3 C 1 (u, + h 5/ (C (u, 0,h + ɛ + h ɛ(c 3 (u, ɛ + ɛ + h +σ/ C 4 (u,. Theorem 3.4. Let u H 5 ( W 3 ( and u h S h be the solution of the model problem and its quadratic finite element approximation, respectively. Assume (a ɛ-σ mesh condition, (b a 0,h contains anisotropic uniform triangles of type in Fig. with θ (0, π/4] and elements in \ a satisfy the maximum angle condition, and (c the discrete inf-sup condition. Furthermore, assume that the condition in Theorem 3. is satisfied. Then the polynomial preserving gradient recovery operator G h has the following error bound: u G h u h h 3 C 1 (u, + h 5/ (C (u, 0,h + ɛ + h ɛ(c 3 (u, ɛ + ɛ + h +σ/ C 4 (u,.
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 15 TABLE I. L errors for both meshes on [0,1]. Partition Regular Union-Jac 0.60e + 000 3.408e + 000 4 40 3.800e 001 4.4553e 001 8 80 5.0715e 00 4.80e 00 16 160 5.6078e 003 7.97e 003 3 30 6.3853e 004 1.4361e 003 FIG. 4. L error for regular meshes and L error for Union-Jac meshes. IV. NUMERICAL EXPERIMENT We compute the Neumann boundary equation below on the unit square domain: u + u = f, u µ = g. (4.1 Here, we choose proper functions f and g such that u = sin(πx sin(10πy. The L Error on the whole domain under regular and Union-Jac patterns are shown in Table I: Corresponding graph is in Fig. 4. APPENDIX: INVERSE INEQUALITY FOR UNION-JACK ANISOTROPIC MESHES Let us focus on one in Fig. A1 since we can obtain the results for regular mesh in a similar way. We observe that h = θh as θ 0. Theorem. Let γ be the boundary of, and v S h, then v 1 Hθ v 1, L (
16 HUANG AND ZHANG FIG. A1. One element of Union-Jac pattern. and v L (γ 1 H 3/ θ v 1, Proof. Without loss of generality, we assume that vertexes of are A(0, 0, B(H,0, and C(0, h. Then v is of the form ( x ( x ( y ( y x v = a + b + c + d H H h h H + e y h + f, where a, b, c, d, e, f are constants. The results are obvious if v is a constant or a linear function, then the results follow from the classic inverse inequality. From now on, we assume that the coefficient of y term is not 0, i.e., c = 0. v x = ax H + by Hh + d H, v xx = a H, v xy = b hh, v t 1 v y = cy h v yy = c h, + bx Hh + e h, = v x, v = v t3 y, v = cos θ v t x cos θ sin θ v x y + sin θ v y. Therefore, v t1 v t1 v t L ( L (l 1 L ( H hh = h1/ H 3/, H H 1/ = H 3/, v x + θ v L ( x y + v L ( y θ H hh + θ L ( 1 hh hh + 1 h θ hh h1/ H 3/,
POLYNOMIAL PRESERVING RECOVERY FOR QUADRATIC ELEMENTS 17 v t v t3 v t 3 L (l L ( L (ł 3 v x + θ v L (l x y + v L (l y θ H H + θ 1 H, 3/ = v y = v y Now, let s compute v 1,. [ (ax v 1, = H + by hh + d H [ (ax = 1 H h h H x 0 0 H h 0 0 H [ (ax H = 4a H 3 h + b h 3 H H 4 3 h H 3 + 4c h 4 h H + H h h 3 H 3 + b L ( L (ł 3 1 hh H + 1 h θ H h hh H 1/ h 3/, h h 1 h 3/. ( cy + + bx h hh + e ] dxdy h L (l + by hh + d ( cy + + bx H h hh + e ] dxdy h + by hh + d ( cy + + bx H h hh + e ] dxdy h + d 4ab H h hh + + 4ad H h + bd h H H hh 3 4 H 3 hh H 3 h + e h H 3 H ( h H 0asθ 0 4bc h H hh + h 3 H 4 + 4ec h 3 h H + be H h h H = H h Hence, v L ( h H. we obtain v t v L ( L γ 1 h v 1, 1 Hθ v 1,, = 1,, 3. 1 h H v 1, 1 H 3/ θ v 1,, = 1,, 3. References 1. O. C. Zieniewicz and J. Z. Zhu, The superconvergence patch recovery and a posteriori error estimates, Part 1: the recovery technique, Int J Numer Methods Eng 33 (199, 1331 1364.
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