Superconvergence of both two and three dimensional rectangular Morley elements for biharmonic equations ∗ Jun Hu†, Zhongci Shi∗, Xueqin Yang† † LMAM and School of Mathematical Sciences, Peking University, Beijing 100871,P.R.China ∗ LESC, Institute of Computational Mathematics and Scientific/Engineering Computing, Academy of Mathematics and System Science, Chinese Academy of Sciences, 5 Beijing 100190,P.R.China 1 0 email: [email protected]; [email protected]; 2 [email protected] n a J 1 1 Abstract In the present paper, superconvergence of second order, after an appropriate postprocessing, is ] achievedforboththetwoandthreedimensionalfirstorderrectangularMorleyelementsofbiharmonic A equations. The analysis is dependent on superconvergence of second order for the consistency error N and a corrected canonical interpolation operator, which help to establish supercloseness of second . order for the corrected canonical interpolation. Then the final superconvergence follows a standard h postprocessing. For first order nonconforming finite element methods of both two and three dimen- t a sional fourth order elliptic problems, it is thefirst time that full superconvergenceof second order is m obtained without an extra boundary condition imposed on exact solutions. It is also the first time [ that superconvergence is established for nonconforming finite element methods of three dimensional fourthorderellipticproblems. Numericalresultsarepresentedtodemonstratethetheoreticalresults. 1 v Keywords: Biharmonic equation; rectangular Morley element; superconvergence 4 2 4 1 Introduction 2 0 . Becauseofsignificantapplications inscientific andengineeringcomputing, superconvergenceanalysis 1 0 of finite element methods has become an active subject since 70’s last century. However, most of atten- 5 tionshavebeenpaidonconformingandmixedfinite elementmethods ofsecondorderproblems,we refer 1 interested readers to [3, 4, 12, 14] for more details. Since conforming finite element methods of fourth : v order problems are very complicated, most of popularly used elements in practice are nonconforming, i for instance, [10, 19, 21, 25, 26, 27, 28, 29]. However, for nonconforming finite elements, due to noncon- X formity of both trial and test functions, it becomes much more difficult to establish superconvergence r properties and related asymptotic error expansions. For second order elliptic problems, there are a few a superconvergence results on rectangular elements. In [5, 23], superconvergence of the gradient was ob- tained atthe centersof elements for the Wilsonelement, which relieson the observationthat the Wilson elementspacecanbe splitintoaconformingpartandanonconformingpart. Due tosuperconvergenceof consistency errors, superconvergence of the nonconforming rotated Q element [20] and its variants was 1 derived, see [8, 13, 18]. For the plate bending problem, there are only few superconvergence results for nonconforming finite elements. In [3], Chen first established the supercloseness of the corrected interpo- lation of the incomplete biquadratic element [29, 21] on uniform rectangular meshes. By using similar corrected interpolations as in [3], Mao et al. [17] first proved one and a half-order superconvergence for the Morley element [19] and the incomplete biquadratic nonconforming element on uniform rectangular ∗ThefirstauthorwassupportedbytheNSFCProjects11271035, 91430213 and11421101. 1 meshes. In a recent paper [6], Hu and Ma proposed a new method by using equivalence between the Morley elementand the firstorderHellan-Herrmann-Johnsonelementand obtainedone anda half-order superconvergence for the Morley element on uniform mesh. That half order superconvergence can be improved to one order superconvergence if the third order normal derivative of exact solutions vanishes on the boundary of the domain under consideration. Based on the equivalence to the Stokes equations and a superconvergence result of Ye [30] on the Crouzeix–Raivart element, Huang et al. [7] derived the superconvergence for the Morley element, which was postprocessed by projecting the finite element solution to another finite element space on a coarser mesh. See Lin and Lin [11] for superconvergence of the Ciarlet–Raviartscheme of the biharmonic equation. Note that allof those results are only for fourth order problems in two dimensions. Superconvergence of nonconforming finite element methods cannot be found for fourth order problems in three dimensions. The purpose of the present paper is to analyze superconvergence of both the two-dimensional and three-dimensional rectangular Morley elements from [25]. Since both of them are nonconforming, one difficultyistoboundtheconsistencyerror. Anotherdifficultyisfromthecanonicalinterpolationoperator which does not admit supercloseness. To overcome the first difficulty, we use some special orthogonal propertyofthecanonicalinterpolationoperatorsofboththebilinearandtrilinearelementswhenapplied to the functions in the rectangular Morley element spaces. The other crucial observation is that the error between the (piecewise) gradient of functions in the discrete spaces and its mean is equal on two oppositeedges(faces)ofanelement. Inparticular,this leadstosuperconvergenceofsecondorderforthe consistencyerror. Todealwiththe seconddifficulty,wefollowthe ideafrom[3]to useacorrectionofthe canonical interpolation. Together with the asymptotic expansion results from [9], this yields superclose- ness of second order for such a corrected interpolation. Finally, based on the above superconvergence results, we follow the postprocessing idea from [14] to obtain a global superconvergent approximate so- lution, which converges at the second order convergence rate. It should be stressed that for first order nonconforming finite element methods of both two and three dimensional fourth order elliptic problems, it is the first time that full superconvergence of second order is obtained without an extra boundary condition imposed on exact solutions. It is also the first time that superconvergence is established for nonconforming finite element methods of three dimensional fourth order elliptic problems. This paper is organized as follows. In the following section, we shall present the model problem and the rectangular Morley element. In section 3, we analyze the superconvergence property of the consistency error for the two-dimensional situation. In section 4, we make a correction of the canonical interpolation and obtain the superconvergence result after the postprocessing. In section 5, we establish the superconvergence result for the three-dimensional cubic Morley element. In the last section 6, we present some numerical results to demonstrate our theoretical results. 2 The model problem and the rectangular Morley element 2.1 The model problem Weconsiderthemodelfourthorderellipticproblem: Givenf ∈L2(Ω),Ω⊂R2 isaboundedLipschitz domain, ∆2u=f, inΩ, (2.1) u= ∂u =0, on∂Ω. (cid:26) ∂n The variational formula of problem (2.1) is to find u∈V :=H2(Ω), such that 0 a(u,v):=(∇2u,∇2v)L2(Ω) =(f,v)L2(Ω), for any v ∈V. (2.2) where ∇2u denotes the Hessian matrix of the function u. 2.2 The two-dimensional rectangular Morley element To consider the discretization of (2.2) by the rectangular Morley element method, let T be a regular h uniformrectangulartriangulationofthe domainΩ. GivenK ∈T , let (x ,x )be the centerofK,the h 1,c 2,c 2 e 3 ✻ ✉ R ✉ e 4 ✛ ✲ e 2 R R ✉ ✉ ❄ e 1 R Figure 1: degrees of freedom meshsize h and affine mapping: x −x x −x ξ = 1 1,c, ξ = 2 2,c , for any(x ,x )∈K. (2.3) 1 h 2 h 1 2 On element K, the shape function space of the rectangular Morley element from [24] reads P(K):=P (K)+span{x3,x3}, (2.4) 2 1 2 hereandthroughoutthispaper,P (K)denotesthespaceofpolynomialsofdegree≤l overK. Thenodal l parameters are: for any v ∈C1(K), 1 ∂v D(v):= v(a ), ds , i,j =1,2,3,4, (2.5) i |e | ∂n (cid:18) j Zej ej (cid:19) where a are vertices of K and e are edges with unit normal vectors n of K, |e | denote measure i j ej j of edges e , see Figure 1. Let the reference element K be a square on (ξ , ξ ) plane, its vertices be j 1 2 a (−1,−1), a (1,−1), a (1,1), a (−1,1), and its sides be e =a a , e =a a , e =a a , e =a a . 1 2 3 4 1 1 2 2 2 3 3 3 4 4 4 1 The nonconforming rectangular Morley element spacbe is then defined by b b b b b b b b b b b b b b b b V := {v∈L2(Ω): v| ∈P(K), ∀K ∈T , v iscontinuousatallinternalverticesand h K h ∂v vanishesatallboundary vertices, and dsiscontinuousoninternaledges ∂n Ze e eandvanishesonboundary edgeseof T }. h The discrete problem of (2.2) reads: Find u ∈V , such that h h ah(uh,vh):=(∇2huh,∇2hvh)L2(Ω) =(f,vh)L2(Ω), for anyvh ∈Vh. (2.6) where the operator ∇2 is the discrete counterpart of ∇2, which is defined element by element since the h discrete space V is nonconforming. Define a semi-norm over V by h h |u |2 :=a (u ,u ), for anyu ∈V . (2.7) h h h h h h h Let u and u be the solutions of (2.2) and (2.6), respectively, by the second Strang Lemma ([2],[24]), we h have |a (u,w )−(f,w )| h h h |u−u | ≤C inf |u−v | + sup , (2.8) h h h h (cid:18)vh∈Vh 06=wh∈Vh |wh|h (cid:19) where the first term is the approximation error and the second one is the consistency error. Herein and throughout this paper, C denotes a generic positive constant which is independent of the meshsize and may be different at different places. 3 3 Superconvergence of the rectangular Morley element in 2D 3.1 Superconvergence of the consistency error Let I be piecewise bilinear interpolation operator on Ω, I :V →B , h h h h I v(P)=v(P), for any vertexP of T , (3.1) h h where B = v ∈H1(Ω), v| ∈Q (K), ∀K ∈T , (3.2) h K 1 h and Ql(K) denotes the space of all(cid:8)polynomials which are of degree≤l w(cid:9)ith respect to each variable xi, over K. Let the interpolation operator IKb be the counterpart of Ih on the reference element K. The bilinear interpolation opertor I has the following error estimate: h b b |v−Ihv|Hl(K) ≤Ch2−l|v|H2(K), l =0,1, (3.3) for any v ∈ H2(K). It is straightforward to see that I is well defined for any w ∈ V . By Green’s h h h formula, (f,I w )=(∆2u,I w )=− ∇∆u·∇I w dx dx . (3.4) h h h h h h 1 2 ZΩ The integration by parts yields ∂2u∂w h a (u,w ) = − ∇∆u·∇w dx dx + ds h h h 1 2 ∂n2 ∂n KX∈ThZK KX∈ThZ∂K ∂2u ∂w h + ds, (3.5) ∂s∂n ∂s KX∈ThZ∂K where ∂ and ∂ are tangential and normal derivatives along element boundaries, respectively. A com- ∂s ∂n bination of (3.4) and (3.5) yields a (u,w )−(f,w )=a (u,w )−(f,I w )+(f,I w −w ) h h h h h h h h h h =− ∇∆u·∇(w −I w )dx dx − f(w −I w )dx dx h h h 1 2 h h h 1 2 KX∈ThZK KX∈ThZK (3.6) ∂2u∂w ∂2u ∂w h h + ds+ ds. ∂n2 ∂n ∂s∂n ∂s KX∈ThZ∂K KX∈ThZ∂K The Cauchy-Schwarzinequality and the interpolation error estimate (3.3) lead to f(wh−Ihwh)dx1dx2 ≤Ch2||f||L2(Ω)|wh|h, (3.7) (cid:12)(cid:12)KX∈ThZK (cid:12)(cid:12) (cid:12) (cid:12) which indicates a suprcon(cid:12)vergence rate O(h2). (cid:12) In the following three lemmas, we will analyze superconvergence for the three remaining terms of (3.6). Lemma 3.1. Suppose that u∈H2(Ω) H4(Ω) and w ∈V . Then, 0 h h T ∇∆u·∇(wh−Ihwh)dx1dx2 ≤Ch2|u|H4(Ω)|wh|h. (3.8) KX∈ThZK Proof. On the reference element K, consider the following functional Bb(φ,w )= φ∂(wh−IKbwh) dξ dξ , (3.9) 1 h ZKb ∂ξ1 1 2 b b b b b b 4 Table 1: calculation of interpolation w 1 ξ ξ ξ ξ ξ2 ξ2 ξ3 ξ3 h 1 2 1 2 1 2 1 2 IKbwh 1 ξ1 ξ2 ξ1ξ2 1 1 ξ1 ξ2 b b b A simple calculation leads to the interpolations, see Table 1. It can be checked that B1(φ,wh)≤C||φ||L2(Kb)|wh|H2(Kb), ( B1(φ,wh)=0, ∀φ∈P0(K), ∀wh ∈Vh. b b b b The Bramble-Hilbert lemma gives b b b b b B1(φ,wh)≤Cpb∈Pin0f(Kb)||φ+p||L2(Kb)|wh|H2(Kb) ≤C|φ|H1(Kb)|wh|H2(Kb). (3.10) A substitution of φ=b∂∆bu into (3.10), plubs abscaling abrgument, yieldb b ∂x1 ∂∆u∂(w −I w ) ∂x h∂x h h dx1dx2 ≤Ch2|u|H4(K)|wh|h for anyK ∈Th. (3.11) ZK 1 1 A similar argument proves ∂∆u∂(w −I w ) ∂x h∂x h h dx1dx2 ≤Ch2|u|H4(K)|wh|h, (3.12) ZK 2 2 which completes the proof. Lemma 3.2. Suppose that u∈H2(Ω) H4(Ω) and w ∈V . Then, 0 h h T∂∂n2u2∂∂wnh ds≤Ch2|u|H4(Ω)|wh|h. (3.13) KX∈ThZ∂K Proof. Given K ∈ T , let e ,i = 1,··· ,4 be its four edges. Define Π0 w = 1 wds and R0 w = h i ei |ei| ei ei w−Π0 w, for any w ∈L2(K), then we have ei R R0 wds=0. (3.14) ei Zei Since ei ∂∂wnhds is continuous on internal edges ei and vanishes on boundary edges of Th, thus R ∂2u∂w 4 ∂2u∂w h h ds = ds ∂n2 ∂n ∂n2 ∂n KX∈ThZ∂K KX∈ThXi=1Zei 4 ∂2u ∂w = R0 h ds ∂n2 ei ∂n KX∈ThXi=1Zei 4 = J . i KX∈ThXi=1 We first analyze the following terms ∂2u ∂w ∂2u ∂w J +J = R0 h ds+ R0 h ds 2 4 ∂n2 e2 ∂n ∂n2 e4 ∂n KX∈Th KX∈Th(cid:18)Ze2 Ze4 (cid:19) ∂2u ∂w ∂2u ∂w = R0 h dx − R0 h dx ∂x2 e2 ∂x 2 ∂x2 e4 ∂x 2 KX∈Th(cid:18)Ze2 1 1 Ze4 1 1 (cid:19) x2,c+h ∂2u ∂w ∂2u ∂w = R0 h − R0 h dx . ∂x2 e2 ∂x ∂x2 e4 ∂x 2 KX∈ThZx2,c−h (cid:18) 1(cid:12)(cid:12)e2 1(cid:12)(cid:12)e2 1(cid:12)(cid:12)e4 1(cid:12)(cid:12)e4(cid:19) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 5 For the rectangular Morley element, we have the following crucial property ∂w ∂w ∂w ∂w R0 h =R0 h , R0 h =R0 h . e2 ∂x e4 ∂x e1 ∂x e3 ∂x 1(cid:12)e2 1(cid:12)e4 2(cid:12)e1 2(cid:12)e3 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) This implies (cid:12) (cid:12) (cid:12) (cid:12) x2,c+h ∂2u ∂2u ∂w J +J = − R0 h dx KX∈Th 2 4 KX∈ThZx2,c−h (cid:16)∂x21(cid:12)e2 ∂x21(cid:12)e4(cid:17) e2 ∂x1(cid:12)(cid:12)e2 2 = x2,c+hR0 ∂(cid:12)(cid:12)2u − ∂(cid:12)(cid:12)2u R0 ∂(cid:12)(cid:12)wh dx . The error estimate of the interpolatKiXo∈nThoZpxe2r,ac−tohrs Πe02(cid:16)y∂ixel21d(cid:12)(cid:12)(cid:12)se2 ∂x21(cid:12)(cid:12)(cid:12)e4(cid:17) e2 ∂x1(cid:12)(cid:12)(cid:12)(cid:12)e2 2 e2 ∂2u ∂2u J +J ≤Ch ∇ − |w | KX∈Th 2 4 KX∈Th(cid:12)(cid:12) (cid:16)∂x21(cid:12)e2 ∂x21(cid:12)e4(cid:17)(cid:12)(cid:12)L2(K) h h (3.15) (cid:12)(cid:12) (cid:12) (cid:12) (cid:12)(cid:12) ≤Ch2|u|H4(cid:12)((cid:12)Ω)|wh|h.(cid:12) (cid:12) (cid:12)(cid:12) By the same argument, we can get J1+J3 ≤Ch2|u|H4(Ω)|wh|h. (3.16) KX∈Th Then, a combination of (3.15) and (3.16) completes the proof. Lemma 3.3. Suppose that u∈H2(Ω) H4(Ω) and w ∈V . Then, 0 h h T∂2u ∂w ∂s∂n ∂sh ds≤Ch2|u|H4(Ω)|wh|h. (3.17) KX∈ThZ∂K The proof of this lemma can follow the similar procedure of Lemma 3.2, and herein we omit it. According to above lemmas, we can obtain the following error estimate. Theorem 3.4. Suppose that u∈H4(Ω), f ∈L2(Ω) and for all w ∈V , then the consistency error can h h be estimated as ah(u,wh)−(f,wh)≤Ch2(||f||L2(Ω)+|u|H4(Ω))|wh|h. 3.2 Asymptotic expansion of the canonical interpolation Given K ∈ T , we define the canonical interpolation operator Π : H3(K) → P(K) by, for any h K v ∈H3(K), ∂Π v ∂v K Π v(P)=v(P)and ds= ds, (3.18) K ∂n ∂n Ze e Ze e for any vertex P of K and any edge e of K. The interpolation operator Π has the following error K estimates: |v−ΠKv|Hl(K) ≤Ch3−l|v|H3(K), l =0,1,2,3, (3.19) provided that v ∈H3(K). Then the global version Π of the interpolation operator Π is defined as h K Π | =Π for anyK ∈T . (3.20) h K K h We need the following asymptotic expansion result from [9]. 6 Lemma 3.5. Suppose that u∈H4(Ω), then for all v ∈V , we have h h h2 ∂3u ∂3v h a (u−Π u,v ) ≤ dx dx h h h 3 ∂x ∂x2 ∂x3 1 2 KX∈Th ZK 1 2 1 h2 ∂3u ∂3v h + dx dx 3 ∂x2∂x ∂x3 1 2 KX∈Th ZK 1 2 2 +Ch2|u|H4(Ω)|vh|h. (3.21) It is straightforwardfrom Lemma 3.5 to derive that by the inverse inequality ∂3u ∂3u a (u−Π u,v )≤Ch + |v | . (3.22) h h h ∂x1∂x22 L2(Ω) ∂x21∂x2 L2(Ω) h h (cid:16)(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:17) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) Basedon the above analysisand Theore(cid:12)m(cid:12) 3.4, Le(cid:12)m(cid:12) ma 3.5,(cid:12)(cid:12)we can g(cid:12)e(cid:12)tthe following errorestimate of |Π u−u | . h h h Theorem 3.6. Suppose that u∈H4(Ω), then we have ∂3u ∂3u |Π u−u | ≤Ch + . (3.23) h h h ∂x1∂x22 L2(Ω) ∂x21∂x2 L2(Ω) (cid:16)(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:17) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) Proof. It follows from Theorem 3.4 and (3.2(cid:12)2(cid:12)) that (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) |Π u−u |2 = a (Π u−u ,Π u−u ) h h h h h h h h = a (u−u ,Π u−u )+a (Π u−u,Π u−u ) h h h h h h h h = [a (u,Π u−u )−(f,Π u−u )]+a (Π u−u,Π u−u ) h h h h h h h h h ≤ Ch2(||f||L2(Ω)+|u|H4(Ω))|Πhu−uh|h ∂3u ∂3u +Ch + |Π u−u | , ∂x1∂x22 L2(Ω) ∂x21∂x2 L2(Ω) h h h (cid:16)(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:17) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) which completes the proof. (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) 4 Supercloseness of the correction interpolation In the view of Theorem 3.6, we cannot expect a higher order error estimate of |Π u −u | . To h h h overcome this difficulty, we follow the idea of [3] to make a correction of the interpolation Π u. First of h all, we define the correction term as follows 8 R v = a (v)ϕ , v ∈H4(K), (4.1) K j j j=5 X where a (v) read j a =−h ∂3v dx , j =5,7, j 6 e2 ∂x1∂x22 2 (4.2)  a =−hR ∂3v dx , j =6,8,  j 6 e1 ∂x21∂x2 1 R  and the basis functions ϕ read j ϕ = h(ξ +1)2(ξ −1), 5 4 1 1  ϕ = h(ξ +1)2(ξ −1), 6 4 2 2  ϕ7 = h4(ξ1+1)(ξ1−1)2, ϕ = h(ξ +1)(ξ −1)2.  8 4 2 2 7 ξ = x1−x1,c, ξ = x2−x2,c, where we defined in (2.3). 1 h 2 h Then the global version R is defined as h R | =R , for anyK ∈T . (4.3) h K K h Define the correction interpolation Π∗ v as follows, for all v ∈H4(K), K Π∗ v =Π v−R v, K ∈T , (4.4) K K K h Then the global version Π∗ is defined as h Π∗| =Π∗ , for anyK ∈T . (4.5) h K K h Regarding the correction term R , we have the following lemma. h Lemma 4.1. Suppose that u∈H4(Ω), then for all v ∈V , we have h h 1 ∂3v 1 ∂3v h h a (R u,v )= (a +a ) dx dx + (a +a ) dx dx . (4.6) h h h 2 5 7 ∂x3 1 2 2 6 8 ∂x3 1 2 KX∈ThZK 1 KX∈ThZK 2 Proof. Let ξ and ξ be defined as in (2.3). It follows from the definition of P(K) that 1 2 ∂2v ∂2v ∂3v h h h = +h ξ , i=1,2. (4.7) ∂x2 ∂x2 ∂x3 i i i i The definition of a (·,·) yields h ∂2R u∂2v ∂2R u ∂2v K h K h a (R u,v )= dx dx +2 dx dx h h h ∂x2 ∂x2 1 2 ∂x ∂x ∂x ∂x 1 2 KX∈Th(cid:18)ZK 1 1 ZK 1 2 1 2 (4.8) ∂2R u∂2v K h + dx dx . ∂x2 ∂x2 1 2 ZK 2 2 (cid:19) We are in the position to calculate three terms on the right-hand side of (4.8). It follows the definition of R that K ∂2R u∂2v h−1 ∂2v K h h dx dx = (6ξ +2)a +(6ξ −2)a dx dx ∂x2 ∂x2 1 2 4 1 5 1 7 ∂x2 1 2 ZK 1 1 ZK(cid:20) (cid:21) 1 h−1 ∂2v ∂3v h h = 6ξ (a +a )+2(a −a ) +h ξ dx dx 4 1 5 7 5 7 ∂x2 ∂x3 1 1 2 ZK(cid:20) (cid:21)(cid:20) 1 1 (cid:21) h−1 ∂2v h−1 ∂3v = 6ξ (a +a ) h dx dx + 6ξ2(a +a )h h dx dx 4 1 5 7 ∂x2 1 2 4 1 5 7 ∂x3 1 2 ZK 1 ZK 1 h−1 ∂2v h−1 ∂3v h h + 2(a −a ) dx dx + 2(a −a )h ξ dx dx . 4 5 7 ∂x2 1 2 4 5 7 ∂x3 1 1 2 ZK 1 ZK 1 Since coefficients like ∂2vh and ∂3vh are constants, we can get that by parity of function and symmetry ∂x21 ∂x31 of domains: 6ξ (a +a )dx dx =0, 2(a −a )hξ dx dx =0. 1 5 7 1 2 5 7 1 1 2 ZK ZK Because of a =a , hence, only one nonzero term is left, which reads 5 7 h−1 ∂3v 1 ∂3v 6ξ2(a +a )h h dx dx = (a +a ) h dx dx . (4.9) 4 1 5 7 ∂x3 1 2 2 5 7 ∂x3 1 2 ZK 1 ZK 1 This yields ∂2R u∂2v 1 ∂3v K h h dx dx = (a +a ) dx dx . (4.10) ∂x2 ∂x2 1 2 2 5 7 ∂x3 1 2 ZK 1 1 ZK 1 8 A similar argument proves ∂2R u∂2v 1 ∂3v K h h dx dx = (a +a ) dx dx . (4.11) ∂x2 ∂x2 1 2 2 6 8 ∂x3 1 2 ZK 2 2 ZK 2 Note that the basis functions ϕj(j =5,6,7,8)have no mixed terms, which leads to ∂∂x21R∂Kx2 =0. Thus ∂2R u ∂2v K h dx dx =0, (4.12) ∂x ∂x ∂x ∂x 1 2 ZK 1 2 1 2 which completes the proof. Basedonthe aboveanalysis,we canestablishsupercloseresultsofthe rectangularMorleyelementby the correction interpolation Π∗u. h Theorem 4.2. Let u∈H4(Ω), u ∈V , be the solutions of (2.2) and (2.6), respectively, then we have h h |Π∗hu−uh|h ≤Ch2 ||f||L2(Ω)+|u|H4(Ω) . (4.13) Remark 4.3. Comparing with the incomplete biq(cid:0)uadratic plate elemen(cid:1)t [17], herein, the theorem does not require ∂3u to be zero on the boundary. Besides, the correction interpolation Π∗u still belongs to the ∂n3 h space V . Because of theboundary condition ∂u =0, it can bededuced that ∂u =0 and ∂u =0, h ∂n ∂Ω ∂x1 ej ∂x2 ei ei,ej ∈∂Ω, i=1,3, j =2,4. Thus, ∂x∂13∂ux22 ej =(cid:12)(cid:12) 0 and ∂x∂213∂ux2 ei =0, ei,ej ∈∂Ω,(cid:12)(cid:12)i=1,3, j =2,(cid:12)(cid:12)4. Proof. On the reference element K, conside(cid:12)(cid:12)r the functional (cid:12)(cid:12) 1 ∂3u ∂3v 1 ∂3u ∂3v h h B2(u,vh) = 3ZKb ∂ξ1∂ξ22b∂ξ13 dξ1dξ2+ 3ZKb ∂ξ12∂ξ2 ∂ξ23 dξ1dξ2 1 b ∂b3u ∂3v b 1 b ∂3u ∂3v b b − dξ h dξ dξ − dξ h dξ dξ 6ZKb (cid:18)Zbe2 ∂ξ1∂ξ22 2(cid:19) ∂ξ13 1 2 6ZKb (cid:18)Zbe1 ∂ξ12∂ξ2 1(cid:19) ∂ξ23 1 2 b b b b It can be checked that B2(u,vh)≤c||u||H3(Kb)|vh|H2(Kb), ( B2(u,vh)=0, ∀u∈P3(K), ∀vh ∈Vh. b b b b Hence, the Bramble-Hilbert lemma givebsb b b b B2(u,vh)≤C|u|H4(Kb)|vh|H2(Kb). (4.14) A scaling argument leads to b b b b h2 ∂3u ∂3v h2 ∂3u ∂3v h h dx dx + dx dx 3 ∂x ∂x2 ∂x3 1 2 3 ∂x2∂x ∂x3 1 2 ZK 1 2 1 ZK 1 2 2 1 ∂3v 1 ∂3v h h + (a +a ) dx dx + (a +a ) dx dx 2 5 7 ∂x3 1 2 2 6 8 ∂x3 1 2 ZK 1 ZK 2 ≤Ch2|u|H4(K)|vh|h. An application of Lemma 3.5 and Lemma 4.1 yields ah(u−Π∗hu,vh)=ah(u−Πhu,vh)+ah(Rhu,vh)≤Ch2|u|H4(Ω)|vh|h. (4.15) Then, together with Theorem 3.4 and (4.15), this gives |Π∗u−u |2 = a (Π∗u−u ,Π∗u−u ) h h h h h h h h = a (u−u ,Π∗u−u )+a (Π∗u−u,Π∗u−u ) h h h h h h h h = [a (u,Π∗u−u )−(f,Π∗u−u )]+a (Π∗u−u,Π∗u−u ) h h h h h h h h h ≤ Ch2(||f||L2(Ω)+|u|H4(Ω))|Π∗hu−uh|h, which completes the proof. 9 Figure 2: macro element K e Based on the superclose property, we can obtain the superconvergence result of the two-dimensional rectangular Morley element by a proper postprocessing technique. In order to attain the global super- convergence,we follow the idea of [14] to construct the postprocessing operator Π3 as follows. 3h 9 We merge9adjacentelements intoamacroelement, K = K ,(seeFigure2),suchthat, inthe macro i i=1 element K, S e Π3 w ∈Q (K), ∀w ∈C(K). (4.16) 3h 3 e We denote Z ,i,j =1,2,3,4as the vertices of the 9 adjacent elements. Then, the operatorΠ3 satisfies ij e e 3h Π3 w(Z )=w(Z ), i,j =1,2,3,4. (4.17) 3h ij ij Besides, the postprocessing operator Π3 has the following properties 3h Π3 (Π∗u)=Π3 u, ∀u∈H4(Ω), 3h h 3h |Π3 v | ≤C|v | , ∀v ∈V , (4.18)  3h h h h h h h  |u−Π33hu|h ≤Ch2|u|H4(Ω), ∀u∈H4(Ω).  Then, we can get the following global superconvergentresult. Theorem 4.4. Let u∈H4(Ω), u ∈V , be the solutions of (2.2) and (2.6), respectively, then we have h h |u−Π33huh|h ≤Ch2 ||f||L2(Ω)+|u|H4(Ω) . (4.19) (cid:0) (cid:1) Proof. It follows the properties (4.18) and Theorem 4.2 that |u−Π3 u | ≤ |u−Π3 Π∗u | +|Π3 (Π∗u−u )| 3h h h 3h h h h 3h h h h ≤ |u−Π3 u| +C|Π∗u−u | 3h h h h h ≤ Ch2 ||f||L2(Ω)+|u|H4(Ω) , (4.20) (cid:0) (cid:1) which completes the proof. 5 Superconvergence of the cubic Morley element In this section, we analyze the superconvergence property of the three-dimensional Morley element on cubic meshes with Ω ⊂ R3. Let T be a regular uniform cubic triangulation of the domain Ω ⊂ R3. h 10