An anisotropic, superconvergent nonconforming plate finite element

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1 Journal of Computational and Applied Mathematics 220 ( An anisotropic, superconvergent nonconforming plate finite element Shaochun Chen a,liyin a, Shipeng Mao b, a Department of Mathematics, Zhengzhou University, , China b Institute of Computational Mathematics, Academy of Mathematics and System Science, Chinese Academy of Sciences, P.O. Box 279, Beijing 00080, China Received 4 February 2007; received in revised form July 2007 Abstract The classical finite element convergence analysis relies on the following regularity condition: there exists a constant c independent of the element and the mesh such that h /ρ c, where h and ρ are diameters of and the biggest ball contained in, respectively. In this paper, we construct a new, nonconforming rectangular plate element by the double set parameter method. We prove the convergence of this element without the above regularity condition. The key in our proof is to obtain the O(h 2 consistency error. We also prove the superconvergence of this element for narrow rectangular meshes. Results of our numerical tests agree well with our analysis Elsevier B.V. All rights reserved. eywords: Regularity condition; Double set parameter; Nonconforming plate element; Anisotropic convergence; Superconvergence. Introduction Suppose Ω R n, T h is a family of subdivisions of Ω. {T h }={,..., N }, Ω= i T h i, i is the element. The convergence of the classical finite element requires that the mesh satisfies the following regularity [0] or nondegenerate [4] condition: there exists a constant c independent of T h and T h, such that h /ρ c, (. where h is the diameter of, ρ is the diameter of the biggest ball contained in. But recent research shows that (. is not a necessary condition for the convergence of the finite element methods, see for example [2,3,7,8,5,6,22]. Among them, Apel et al. have studied anisotropic Lagrange interpolations. Their fundamental results are of [2, Lemmas 3 and 4] or of [3, Lemmas 2.2 and 2.3], which give the conditions for which the following estimate holds: D α (u Iu m, C D α u l,, (.2 This work is supported by NSFC (04733 and Corresponding author. addresses: shchchen@zzu.edu.cn (S. Chen, yinli@zzu.edu.cn (L. Yin, maosp@lsec.cc.ac.cn (S. Mao /$ - see front matter 2007 Elsevier B.V. All rights reserved. doi:0.06/j.cam

2 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( α where α=(α,...,α n is a multi-index, D α = x α, α = n α i, 0 m l, and I is the finite element interpolation xαn n operator. We call the above property (.2 anisotropism of the element. In [7] we introduced a general theorem for anisotropic interpolation, which yields a new criterion to obtain (.2, improving Apel s result and which is easier to use. The research on anisotropic elements mainly concentrate on conforming elements. For nonconforming elements, the consistency error should also be estimated. For classical elements, the consistency error estimate still relies on (.. As far as we know, only few published articles discuss nonconforming elements on anisotropic meshes. The superconvergence of finite elements has been widely analyzed mathematically because of its practical importance in engineering computations, the reader is referred to [5,,2]. The term superconvergence also includes accelerated convergence achieved by means of various recovery (or post-processing techniques (refer to [4]. However, the studies mainly concentrated on second order problems and conforming C 0 elements. As for the superconvergence results of nonconforming finite elements, there seem to be few articles except Wilson s element for the second order problem (refer to [5,6,8]. Meanwhile, the meshes are required to be regular (sometimes quasi-uniform, namely, satisfy (. and an inverse assumption. In this paper, a new nonconforming rectangular plate element is presented, and its convergence is proved for anisotropic meshes. In addition, we prove that its anisotropic consistence error is O(h 2, which is the same order as that of Adini s nonconforming element on rectangular meshes (cf. [4], however, the shape space of our element is only exact for P 2 (. Its natural superconvergence and global superconvergence are derived without the regularity assumption (.. As far as the authors knowledge, this is the first time to find such properties for fourth order problems on anisotropic meshes. Some numerical examples supporting our theoretical analysis are given. 2. Basic theory Let ˆ be the reference element, shape function space Pˆ be a polynomial space on ˆ, the dimension of Pˆ be m, P ˆ be the dual space of P ˆ. {ˆp,..., ˆp m } Pˆ and { ˆN,..., ˆN m } P ˆ are a pair of dual basis for Pˆ and P ˆ, respectively, i.e., ˆN i ( ˆp j = δ ij, i, j m. Suppose the interpolation operator I ˆ : H k ( ˆ P,k ˆ is defined such that ˆN i ( Iˆ ˆv = ˆN i ( ˆv, i =,...,m, ˆv H k ( ˆ. (2. Obviously, and ˆ I ˆv = m ˆ I ˆv =ˆv, ˆN i ( ˆv ˆp i ˆv ˆ P. (2.3 Suppose α = (α,...,α n is a multi-index, then ˆD α Pˆ is also a polynomial space on ˆ. Let dim ˆD α P ˆ = r and {ˆq i } r be the set of basis functions of ˆD α P ˆ, suppose ˆD α ˆp i = r j= α ij ˆq j, i m, then ˆD α Iˆ ˆv can be expressed by where ˆD α ˆ I ˆv (2.2 = β j ( ˆv = m ˆN i ( ˆvD α ˆp i = (2.2 r β j ( ˆvˆq j, (2.4 j= m α ij ˆN i ( ˆv. (2.5

3 98 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( From (2.5 and (2. we have m m β j ( ˆv = α ij ˆN i ( ˆv = α ij ˆN i ( Iˆ ˆv = β j ( Iˆ ˆv. (2.6 Theorem 2. (Chen et al. [7], Chen et al. [8]. Let α be a multi-index, P l ( ˆ ˆD α Pˆ and I ˆ : W α +l+,p ( ˆ Pˆ be the above finite element interpolation operator defined by (2., (2.2 and satisfy: Iˆ L(W α +l+,p ( ˆ,W α +m,q ( ˆ the space of continuous linear mappings from (W α +l+,p ( ˆinto W α +m,q ( ˆ, and W l+,p ( ˆ W m,q ( ˆ. If β j ( ˆv of (2.5 can be expressed by where β j ( ˆv = F j ( ˆD α ˆv, j r, (2.7 F j (W l+,p ( ˆ, j r. (2.8 Then ˆD α ( ˆv ˆ I ˆv m,q, ˆ C( ˆ I, ˆ ˆD α ˆv l+,p, ˆ, ˆv W α +l+,p ( ˆ. (2.9 Remark 2.. In [2,3], Apel et al. show that one necessary condition for (2.9 to hold is ˆD α û = 0 ˆD α Iû ˆ = 0. (2.0 (2.0 is useful to prove that the finite element interpolation does not have anisotropic behavior ( Anisotropic rectangular plate element with double set parameters Let the reference element ˆ be a square in the (ξ, η plane, and let its nodes be â (,, â 2 (,, â 3 (,, â 4 (,, its middle points of 4 sides be â 5 (0,, â 6 (, 0, â 7 (0,, â 8 (, 0, and its 4 sides be ˆl =â â 2, ˆl 2 = â 2 â 3, ˆl 3 =â 3 â 4, ˆl 4 =â 4 â. The degrees of freedom are taken as D(ˆv = ( ˆv, ˆv 2, ˆv 3, ˆv 4, ˆv 5η, ˆv 6ξ, ˆv 7η, ˆv 8ξ, (3. where ˆv i =ˆv(â i, i 4 are function values at 4 nodes; ˆv iη = derivatives at the middle points of 4 sides. The shape function space is taken as ˆv η (â ˆv i, i = 5, 7, ˆv iξ = ξ (â i, i = 6, 8, are normal ˆ P = P 2 ( ˆ span{ξ 3, η 3 }=span{ˆp,..., ˆp 8 }, (3.2 where ˆp = 4 ( ξ( η, ˆp 2 = 4 (+ξ( η, ˆp 3 = 4 (+ξ(+η, ˆp 4 = 4 ( ξ(+η, ˆp 5 = ξ 2, ˆp 6 = η 2, ˆp 7 = ξ( ξ 2, ˆp 8 = η( η 2. ˆv P ˆ, suppose ˆv = 8 β i ˆp i. (3.3 Substituting (3.3 into (3.2 we get β i =ˆv i, i 4, β 5 = 4 ( ˆv 8ξ ˆv 6ξ, β 6 = 4 ( ˆv 5η ˆv 7η, β 7 = 8 ( ˆv +ˆv 2 +ˆv 3 ˆv 4 2 ˆv 8ξ 2 ˆv 6ξ, β 8 = 8 ( ˆv ˆv 2 +ˆv 3 +ˆv 4 2 ˆv 5η 2 ˆv 7η. (3.4

4 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( Using the Double Set Parameter Method [9], we take another set of nodal parameters: Q( ˆv = ( ˆv, ˆv ξ, ˆv η,..., ˆv 4, ˆv 4ξ, ˆv 4η, (3.5 i.e., the function values and the first order partial derivatives at 4 nodes, which are the real degrees of freedom. Approximating the degrees of freedom D(ˆv by the linear combinations of nodal parameters Q( ˆv as follows: { ˆvi =ˆv i, i 4, ˆv 5η = 2 ( ˆv η +ˆv 2η, ˆv 6ξ = 2 ( ˆv 2ξ +ˆv 3ξ, ˆv 7η = 2 ( ˆv 3η +ˆv 4η, ˆv 8ξ = 2 ( ˆv (3.6 ξ +ˆv 4ξ. For ˆv P ˆ = P 2 ( ˆ {ξ 3, η 3 }, these discretizations are exact. Substituting (3.6 into (3.4, we get the expressions of β i of (3.3 in Q( ˆv: β i =ˆv i, i 4, β 5 = 8 ( ˆv ξ ˆv 2ξ ˆv 3ξ +ˆv 4ξ, β 6 = 8 ( ˆv η +ˆv 2η ˆv 3η ˆv 4η, β 7 = 8 ( ˆv (3.7 +ˆv 2 +ˆv 3 ˆv 4 ˆv ξ ˆv 2ξ ˆv 3ξ ˆv 4ξ, β 8 = 8 ( ˆv ˆv 2 +ˆv 3 +ˆv 4 ˆv η ˆv 2η ˆv 3η ˆv 4η. Let T h be a rectangle subdivision of Ω, = Ω. Suppose a general rectangle element is on (x, y plane with center (x,y and side lengths 2h and 2h 2, respectively, its 4 nodes are: a (x h,y h 2, a 2 (x + h,y h 2, a 3 (x +h,y +h 2, a 4 (x h,y +h 2. The mapping from ˆ to denoted by x =F(ˆx is x = h ξ + x, y = h 2 η + y. (3.8 The shape function space on is defined as P ={v =ˆv F ;ˆv ˆ P is determined by (3.3, (3.7}. Let ˆ I be the finite element interpolation operators deduced by ˆ P, then ˆ I ˆv = 8 β i ˆp i, ˆv H 3 ( ˆ, (3.9 where the expressions of β,...,β 8 are (3.7. From (3.3 and (3.7, it is easy to see that ( Iˆ ˆv i =ˆv i, i 4, ( Iˆ ˆv 5η = 2 ( ˆv η +ˆv 2η, ( Iˆ ˆv 6ξ = 2 ( ˆv 2ξ +ˆv 3ξ, ( Iˆ ˆv 7η = 2 ( ˆv 3η +ˆv 4η, ( Iˆ ˆv 8ξ = 2 ( ˆv ξ +ˆv 4ξ. Define the interpolation on a general element as I v = ( Iˆ ˆv F, v H 3 (. The relations of the nodal parameters on ˆ and are (3.0 v i =ˆv i, v ix =ˆv iξ h, v iy =ˆv iη h 2, i 4. (3. From now on, the sign ĉ denotes a general constant which is independent of h /h 2,h/h, T h, here h = max Th h, h = max{h,h 2 }. Theorem 3.. The interpolation operator ˆ I is anisotropic for the fourth order problem, that is to say, α,, ˆD α ( ˆv ˆ I ˆv 0, ˆ ĉ ˆD α ˆv, ˆ, ˆv H 3 ( ˆ. (3.2

5 00 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( Proof. It is only needed to check the conditions of Theorem 2.. Let α = (2, 0, ˆD α ˆ I ˆv (3.9 = 8 β i 2 ˆp i ξ 2 = 2β 5 6ξβ 7, (3.3 { 2, 6ξ} are the bases of ˆD α P ˆ = Span{, ξ}, ( β 5 ( ˆv (3.7 = 2 ˆv 2 ˆv (ξ, dξ (ξ, dξ F ( ˆD α ˆv, ξ ξ [ ( β 7 ( ˆv (3.7 = ξ 2 ˆv 2 ˆv dξ (ξ, dξ (ξ, dξ ξ ξ ξ ( ξ 2 ] ˆv 2 ˆv + dξ (ξ, dξ (ξ, dξ F ( ˆD α ˆv. (3.4 ξ ξ ξ From the Hölder inequality and trace theorem [],wehave F i (ŵ ĉ ŵ 0, ˆ ĉ ŵ, ˆ, i =, 2. (3.5 For α = (0, 2 and α = (, the same properties as α = (2, 0 hold. Hence (3.2 follows from (2.8. Remark 3.. ( If using classical forms, i.e., (3.4 for β i, (2.9 does not hold, so this 8 degrees of freedom rectangular plate element (denoted by 8-2 element is not anisotropic and (3.2 does not hold. In fact, for α = (2.0, let ˆv = ξη 2, then ˆD α ˆv = 0, but β 5 = 0, β 7 = 2, ˆD α Iˆ ˆv = 2β 5 6ξβ 7 = 3ξ = 0. (2 For the initial 8 degrees of freedom rectangular plate element (denoted by 8- element, the degrees of freedom are also (3., but the shape function space P ˆ =P 2 ( ˆ {ξ 2 η, ξη 2 }. Under the regularity condition (., its convergence is correctly proven in [9]. It is easy to prove that this element does not satisfy (2.9. It should be pointed out that its double set parameter form (denoted by 8-2- element does not satisfy (2.9, either. 4. Anisotropic convergence of element for the plate bending problems Consider the plate bending problem [0]: Find u H0 2 (Ω such that a(u, v = (f, v, v H0 2 (Ω, (4. where Ω is a rectangle domain, f L 2 (Ω and a(u, v = A(u, v dx dy, (f, v = fvdx dy, Ω Ω A(u, v = u v + ( σ(2u xy v xy u xx v yy u yy v xx, H0 2 (Ω ={v H 2 (Ω, v = v = 0, on Ω}, n here n is the direction normal to the boundary Ω, σ is the Poisson ratio, 0 < σ < 2, u xy = 2 u x y, etc. The corresponding differential equation of (4. is 2 u = f in Ω, u = u n = 0 on Ω. (4.2

6 The finite element space V h is defined as S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( V h ={v h ; v h =ˆv F, ˆv is defined by (3.3, (3.7, T h, v(a = v x (a = v y (a = 0, for all node a on Ω}. The discrete problem of (4. is: find u h V h, such that a h (u h,v h = (f, v h, v h V h, (4.3 where a h (u h,v h = A(u h,v h dx dy. Set h = /2. 2 2, It is easy to prove that h is a norm of V h, so the discrete problem (4.2 has an unique solution by Lax Milgram Lemma [4]. Let u and u h be the solutions of (4. and (4.3, respectively, by Strang s Lemma [4,0], ( a h (u, w h (f, w h u u h h ĉ inf u v h h + sup. (4.4 v h V h w h V h w h h The finite element interpolation operator I h : H 3 (Ω H 2 0 (Ω V h is defined by I h I, T h. Let α = (α, α 2,,h α = hα hα 2 2. Obviously, ˆD α û = h α Dα u, then inf u v h h u I h u h v h V h = (3.2 ĉ β = h 2α (h h 2 ˆD α (û ˆ h 2β Dβ u 2 2, /2 Iû 2 0, ˆ /2. (4.5 Now we are in a position to estimate the consistence error. Let Π h be the piecewise bilinear interpolation operator on Ω, Π h = Π, Π v = ˆΠ ˆv F, ˆΠ is the bilinear interpolation operator on ˆ. It is easy to see that ˆΠ is anisotropic for α =, and from Theorem 2. we have v Π v 0, ĉh 2 v 2,, v Π v, ĉh v 2,, v H 2 (. (4.6 Obviously w h V h, Π h w h C0 0 (Ω, by Green s Formula [0], f(π h w h = f Π h w h = 2 uπ h w h = u Π h w h. (4.7 Ω Ω Ω The well-known result [2,8] gives a h (u, w h = u w h + E (u, w h + E 2 (u, w h, (4.8

7 02 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( where E (u, w h = E 2 (u, w h = [ u ( σu ss] w h n ds, ( σu w h sn s ds, where ( s = s,( n = n, are the tangential and normal derivatives along element boundaries, respectively. Combining (4.7 and (4.8 yields a h (u, w h f(w h = [ ] u (w h Π w h + f(w h Π w h (4.9 + E (u, w h + E 2 (u, w h. (4.0 From (4.6 we have u (w h Π w h Δu, w h Π w h, and f(w h Π w h f 0, w h Π w h 0, ĉh Δu, w h 2, (4. ĉh 2 f 0, w h 2,. (4.2 By the definition of the element, it is easy to see that l i,w h V h, w h n l i P (l i, w h n middle point of l i,so w h l i n is continuous between elements and is zero on Ω. Set P i w = w ds, l i l i U = U 3 = u ( σu xx and U 2 = U 4 = u ( σu yy,wehave E (u, w h = 4 l i U i ( wh n P i w h ds n 4 I i. is continuous at the From the construction of the element we know that 2 w h x y is a constant, hence by the skill in [3,7], we can get x +h ( y +h 2 U (x, y I + I 3 = w(x dy dx, x h y h 2 y where w(x = 4h h 2 x +h x h ( x t y +h 2 y h 2 2 w h (r, y dr dy r y h h 2 w h 2,, (4.3 from which we have I + I 3 2h ( Δu, + ( σ x (u xy 0, w h 2,. dt

8 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( Similarly, I 2 + I 4 2h 2 ( Δu, + ( σ y (u xy 0, w h 2,. Hence E (u, w h ĉ (h 2 Δu 2, + Dα u 2 2, α = In the same way, we get E 2 (u, w h c α = Dα u 2 2, /2 /2 w h h. (4.4 w h h. (4.5 Substituting (4., (4.2, (4.4 and (4.5 into (4.0, we have a h (u, w h f(w h sup w h V h w h h ĉ h 2 Δu 2, + /2 Dα u 2 2, w h h. (4.6 α = A combination of (4.5 and (4.6 yields the main result of this section. Theorem 4.. Using rectangular element to solve the plate bending problem, we have u u h h ĉ h 2 Δu 2, + /2 Dα u 2 2, + h4 f 2 0,, (4.7 α = where u and u h are the solutions of (4. and (4.2, respectively. Remark 4.. ( The classical method to estimate the consistence error [2] is directly based on (4.9, using coordinate transformation, interpolation theory and trace theorem, through ˆ ˆ, then (4.4 and (4.5 are obtained. During ˆ ˆ the factor h i (h h 2 /2,i =, 2 appears due to the Jacobian determinants. This makes the constant in (4.4 and (4.5 dependent on h /ρ. (2 Getting the consistence error O(h by the classical method [2] should suppose u H 4 (Ω, here we only suppose u H 3 (Ω, this argument came from [20]. 5. Anisotropic superconvergence of element In this section, we discuss the superconvergence behavior of element for the biharmonic equation with anisotropic meshes. We consider the following biharmonic equation 2 u = f in Ω, u = u n = 0 on Ω. Its weak form is taken as: find u H0 2 (Ω such that ã(u, v = (f, v, v H0 2 (Ω, (5.2 (5.

9 04 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( where ã(u, v = (u xx v xx + 2u xy v xy + u yy v yy dx dy. (5.3 Ω Let V h be the finite element space of element, then the discrete problem of (5.2 is: find u h V h such that ã h (u h,v h = (f, v h, v h V h, (5.4 where ã h (u h,v h = (u hxxv hxx + 2u hxy v hxy + u hyy v hyy dx dy. First we prove that the consistence error for (5.4 is O(h 2. Similar to (4.20, we can easily prove that [2]: for any w h V h, ã h (u, w h (f, w h = Ẽ (u, w h + Ẽ 2 (u, w h + Ẽ 3 (u, w h, (5.5 where Ẽ (u,w h = u ss w h n ds, Ẽ 2 (u, w h = u sn w h s ds, Ẽ 3 (u, w h = u n w h ds. Put U = u ss, then we have Ẽ (u, w h = 4 ( wh (U P i U l i n P w h i ds = 4 I i, n where x +h [ ( I + I wh 3 = (U P U x h ( wh y P 3 y P w h y w h (x, y h 2 y ] (x, y + h 2 +(U P 3 U dx = x +h w(xq(x dx, (5.6 2h x h here x +h ( x y +h 2 Q(x = 2 U (r, y dr dy dt x h t y h 2 r y 2 u ĉh h h 2 x 2. 2, Substituting the above estimate and (4.3 into (5.6 yields I + I 3 ĉh 2 2 u x 2 w h 2,. 2, Similarly, I 2 + I 4 ĉh 2 2 u 2 y 2 w h 2,. 2, Then Ẽ (u, w h ĉ /2 Dα u 2 2, w h h. (5.7 Following the same argument, we can prove Ẽ 2 (u, w h ĉ Dα u 2 2, /2 w h h. (5.8

10 Similarly, Ẽ 3 (u, w h = where L + L 3 = S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( x +h x h ĉh 2 Δu 2, w h 2,. l i u n (w h Π w h ds = 4 L i, ( u y (x, y + h 2 u y (x, y h 2 (w h Π w h (x, y + h 2 dx As for L 2 + L 4, we have the same results, thus Ẽ 3 (u, w h ĉ /2 h 4 Δu 2 2, w h h. (5.9 Then a collection of (5.7 (5.9 and (5.5 gives ã h (u, w h (f, w h ĉ h 4 Δu 2 2, + Dα u 2 2, /2 w h h. (5.0 Remark 5.. The well known nonconforming rectangular plate element with 2 node parameters is Adini s C 0 element [2], but its consistence error is O(h 2 only for uniform meshes, i.e., all the elements are equal and regular condition (. holds. Now we prove the following anisotropic superclose result. Theorem 5.. Suppose u H 4 (Ω, we have I h u u h h ĉ h 4 Δu 2 2, + Dα u 2 2, Proof. First we prove ã h (u I h u, v h ĉ Dα u 2 2, /2 /2. (5. v h h, v h V h. (5.2 By the expression of ã h (,, it is only need to prove the following three inequalities, (u I h u xx v hxx dx dy ĉ /2 Dα u 2 2, v h 2,, (5.3 (u I h u yy v hyy dx dy ĉ Dα u 2 2, /2 v h 2,, (5.4 (u I h u xy v hxy dx dy = 0. (5.5

11 06 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( Firstly, by the scaling argument, (u I h u xx v hxx dx dy = h 4 (4h h 2 Put ŵ =û ξξ, then from (3.3 and (3.4, we have û ξξ ( ˆ Iû ξξ =ŵ F (ŵ F 2 (ŵξ l(ŵ. It is easy to see that l(ŵ = 0, ŵ P ( ˆ. Thus by Bramble Hilbert Lemma [4,0], l(ŵ ĉ ˆv hξξ 0, ˆ ŵ 2, ˆ =ĉ ˆv hξξ 0, ˆ û ξξ 2, ˆ ĉh 4 (4h h 2 Dα u xx 2 0, (û ξξ ( Iû ˆ ξξ ˆv hξξ dξ dη. (5.6 ˆ /2 v h 2,. Substituting the above result into (5.6 implies (5.3. Similarly, (5.4 can be proved. Since v hxy is a constant on, then (u I hu xy v hxy dx dy = 0. Thus we can obtain (5.2. Finally, I h u u h 2 h = ã h(i h u u h,i h u u h = ã h (I h u u, I h u u h + ã h (u, I h u u h (f, I h u u h (5.2(5.0 ĉ h 4 Δu 2 2, + /2 Dα u 2 2, I h u u h h, which completes the proof of the theorem. Now we will discuss the natural superconvergence results about the second order derivatives of element. Theorem 5.2. Under the assumption in Theorem 5., we have the following anisotropic superconvergence results at the central points, /2 D α (u u h (x,y 2 h h 2 ĉ h 4 Δu 2 2, + Dα u 2 2, Proof. First, we focus on α = (2, 0, due to the triangle inequality, /2. (5.7 D (2,0 (u u h (x,y 2 2( D (2,0 (u I h u(x,y 2 + D (2,0 (I h u u h (x,y 2. (5.8 By the scaling technique, D (2,0 (u I h u(x,y (3.8 = h 2 ˆD (2,0 (û ˆ Iû(0, 0 (3.3(3.4 = h 2 l( ˆD (2,0 û, where l(ŵ =ŵ(0, 0 + 2F (ŵ.

12 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( From (3.4 it can be easily checked that for all ŵ P ( ˆ, l(ŵ = 0, then l(ŵ ĉ ŵ 2, ˆ, ŵ H 2 ( ˆ and D (2,0 (u I h u(x,y (3.8(3. ĉ(h h 2 /2 β =2 h β Dβ u 2,. (5.9 Thanks to ˆD (2,0 ( ˆ Iû û h P ( ˆ and the equivalent norms on the finite dimensional space, we have D (2,0 (I h u u h (x,y (3.8(3. ĉ(h h 2 /2 Iu u h 2,. (5.20 Substituting (5.9, (5.20 and (5. into (5.8, we obtain /2 D (2,0 (u u h (x,y 2 h h 2 ĉ h 4 Δu 2 2, + Dα u 2 2, Similarly, we can get the same results for α = (0.2 and α = (,. Hence (5.7 holds. /2. (5.2 Furthermore, we can obtain the global superconvergence results of element by virtue of a proper postprocessing technique. For simplicity, we assume that the mesh J h is obtained by dividing every element of the coarser mesh J 3h into 9 congruent elements, 2,..., 9, the vertices of, 2,..., 9 are denote by Z ij,i,j =, 2, 3, 4. We consider the conventional bicubic Lagrange interpolation operator Π 3 3h : H 2 ( Q 3 (x, y characterized by Π 3 3h u(z ij = u(z ij, i, j =, 2, 3, 4, where Q 3 (x, y is the space of all polynomials which are of degree 3 with respect to x and y, respectively. According to [2], wehave Π 3 3h u u 2,Ω ĉ /2 Dα u 2 2,. (5.22 Obviously, β i ĉ ˆD α ˆv 2, ˆ ĉ ˆD α ˆv 0, ˆ, v V h, where the inverse inequality [] is used. Then D α Π 3 3h v 0, = h α h h 2 ˆD α ˆΠ 3 v 0, ˆ r ĉh α h h 2 β i ĉ D α v 0,, v V h. Hence Π 3 3h v h = D α Π 3 3h v 2 0, /2 ĉ v h, v V h. (5.23 Then we can get the following superconvergence theorem following the standard technique, cf. [5,4].

13 08 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( Theorem 5.3. Under the same assumptions as in Theorem 5., we have u Π 3 3h u h h ĉ h 4 Δu 2 2, + Dα u 2 2, 6. Numerical experiments /2. (5.24 In order to examine the numerical performance of the above element for anisotropic rectangular meshes, we carry out numerical tests for the following two models: Model. The classical unit square plate bending problem with clamped boundaries under a uniform load. The Poisson ratio is chosen σ = 0.3 and f =. The analytic value of deflection at the center is , the analytic value of bending moment at the center is This experiment is used to investigate the convergence for the classical plate bending problem under anisotropic meshes. Model 2. A biharnomic differential equation with f chosen such that the exact solution of problem (4.2 is u(x, y = sin 2 (πxsin 2 (πy. This experiment is used to investigate the convergence and superconvergence for an ordinary biharmonic problem under anisotropic meshes. In order to obtain anisotropic meshes, the unit square Ω =[0, ] [0, ] is subdivided in the following fashion: each edge of Ω is divided into n segments with n + points ( cos( iπ n /2,i= 0,,..., n 2,( + sin( iπ n π 2 /2,i= n 2 +,...,n. The mesh obtained in this way for n = 6 is illustrated in Fig.. The aspect ratio of this mesh is demonstrated by Table. For Model, Fig. 2 gives the deflection error and the moment error at the central point (i.e., (u u h ( 2, 2, (M M h ( 2, 2, which shows the anisotropic convergence of element for model. Fig.. The anisotropic mesh for the case n = 6. Table The aspect ratio of mesh 2 n n max {h /ρ } max {h/h }

14 S. Chen et al. / Journal of Computational and Applied Mathematics 220 ( (u u h (0.5,0.5 (M M h (0.5, Errors Numerber of elements Fig. 2. The errors for model. 0 0 u u h h / u h 0 2 I h u u h h / u h S(u u h u Π 3 3h u hh / u h Errors Numerber of elements Fig. 3. The errors for model 2. For Model 2, we compute the errors u u h h / u h, I h u u h h / u h,s(u u h ( Dα (u u h (x,y 2 h h 2 /2 / u h and u Π 3 3h u h h / u h. The numerical results are listed in Fig. 3. These results agree well with the theoretical analysis. Acknowledgments The authors would like to express their sincere thanks to an anonymous referee for his many helpful suggestions, together with many corrections of the English and typesetting mistakes. References [] R.A. Adams, Sobolev Space, Academic Press, New York, 975. [2] T. Apel, M. Dobrowolski, Anisotropic interpolation with applications to the finite element method, Computing 47 (

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