Journal of Computatonal Mathematcs Vol.29, No.2, 2011, 215 226. http://www.global-sc.org/jcm do:10.4208/jcm.1009-m3246 A NOTE ON THE NONCONFORMING FINITE ELEMENTS FOR ELLIPTIC PROBLEMS * Boran Gao School of Mathematcal Scences, Pekng Unversty, Bejng 100871, Chna Emal: ggggbr@163.com Shuo Zhang ICMSEC, AMSS, Chnese Academy of Scence, Bejng 100190, PR Chna Emal: szhang@lsec.cc.ac.cn Mng Wang LMAM, School of Mathematcal Scences, Pekng Unversty, Bejng 100871, Chna Emal: mwang@math.pku.edu.cn Abstract In ths paper, a class of rectangular fnte elements for 2m-th-oder ellptc boundary value problems n n-dmenson (m, n 1) s proposed n a canoncal fashon, whch ncludes the (2m 1)-th Hermte nterpolaton element (n = 1), the n-lnear fnte element (m = 1) and the Adn element (m = 2). A nonconformng trangular fnte element for the plate bendng problem, wth convergent order O(h 2 ), s also proposed. Mathematcs subject classfcaton: 65N30. Key words: Nonconformng fnte element, Ellptc boundary value problem, Plate bendng problem. 1. Introducton When the conformng fnte element s used for numercally dscretzng the ellptc problem, the convergence of the numercal soluton to the exact soluton depends on the approxmaton of the fnte element space only. But the strong contnuty requrement makes t dffcult to construct such a conformng fnte element. The dea of nonconformng fnte element les n that such dffculty can be overcome by loosng the request on the contnuty. However, the loss of contnuty wll brng n the so-called consstent error, and some fundamental contnuty of the fnte element space s stll necessary for well-posedness and convergence. Ths s the reason that most of the fnte elements, conformng or nonconformng, were constructed case by case, dependng on the order of the problem and sometmes the dmensons (cf. [1 3,5,7, 8, 12, 14, 15, 17]). A unfed approach of constructng fnte elements for general problems s stll of theoretcal and practcal nterest. Recently, a class of fnte elements was dscussed n a canoncal fashon n [16], for all n-dmensonal 2m-th-order ellptc problem wth n m 1. The well-known nonconformng lnear element for the second-order problem and the Morley element for fourth-order problem are examples of ths class. The class of fnte elements s establshed on smplces, and makes use of the pecewse polynomals of the lowest degree. The nodal parameters are the natural ones to guarantee the fundamental contnuty, and the consstency error can be controlled smultaneously. * Receved October 23, 2009 / Revsed verson receved May 17, 2010 / Accepted July 20, 2010 / Publshed onlne November 20, 2010 /
216 B.R. GAO, S. ZHANG AND M. WANG In ths paper, we wll dscuss the choce of nodal parameters that can be used to construct nonconformng fnte elements, wth admssble consstency error. We wll frst propose a class of rectangular fnte elements for n-dmensonal 2m-th-oder problems (m, n 1) n a canoncal fashon. The degrees of freedom are the values of functon and all dervatves up to (m 1)- th-oder at all vertces of n-rectangle. The basc fundamental contnuty s guaranteed and an O(h) convergence rate s shown. The (2m 1)-th Hermte nterpolaton element (n = 1), the n-lnear fnte element (m = 1) and the Adn element (m = 2) all belong to ths class. As almost all of the nonconformng fnte elements are convergent n energy norm wth order O(h), and the consstency error s the man lmt, we wll dscuss the possblty of mprovng the convergence rate by strengthenng the contnuty of the fnte element space. We choose the plate bendng problem as an example. There have been successful attempts va other approaches, lke conformng fnte element, quas-conformng fnte elements (cf. [4, 6, 11]) and the double set parameter element (cf. [9]). But most nonconformng element for the plate bendng problem, such as the Morley element [8], two Veubake elements [12], the NZT element [14], the rectangle Morley element (cf. [15]) and the Adn element (cf. [1]), are convergent wth order O(h). In ths work, a new nonconformng plate element wll be gven, wth a convergence rate of O(h 2 ) n energy norm. Fnally, based on the new plate element, a new Zenkewcz-type element wll be deduced and reported for comparson. The new Zenkewcz-type element s convergent for the plate bendng problem wth order O(h). Its consstent error s of order O(h 2 ) whch s better than the two dmensonal Zenkewcz-type element proposed n [14]. In fact, the phenomenon that the consstency error can perform better than the approxmaton error has seldom been reported n lteratures. The paper s organzed as follows. The rest of ths secton gves some basc notatons. Secton 2 gves the descrpton of the class of rectangular fnte elements. Secton 3 gves the descrpton of the new plate elements. Secton 4 shows ther convergence. Secton 5 gves some numercal results for the new plate element. Let n be a postve nteger. Gven a nonnegatve nteger k and a bounded doman G R n wth boundary G, let H k (G), H k 0(G), k,g and k,g denote the usual Sobolev spaces, norm and sem-norm respectvely. Let (, ) denote the nner product of L 2 (Ω). We wll use α,β,γ to denote n dmensonal mult-ndexes. Defne α = x α1 α, α = α 1 + +α n. 1 xαn n A fnte element can be represented by a trple (T,P T,D T ) wth T the geometrc shape, P T the shape functon space and D T the vector of degrees of freedom, provded that D T s P T -unsolvent (see [5]). Let Ω be a bounded polyhedron doman of R n. For mesh sze h wth h 0, let Th be a partton of Ω correspondng to a fnte element (T,P T,D T ), and let V h,v h0 be the fnte element spaces correspondng to the element and Th. Throughout ths paper, we assume that {Th} s shape regular. For a subset B R n and a nonnegatve nteger r, let P r (B) be the space of all polynomals defned on B wth degree not greater than r, and Q r (B) the space of all polynomals wth degree n each varable not greater than r.
Nonconformng Fnte Elements for Ellptc Problems 217 2. A Class of Rectangular Fnte Elements Let m be a postve nteger. Ths secton s devoted to the rectangular fnte element for the 2m-th-oder ellptc boundary value problem n n-dmenson. Let T be an n-rectangle wth each edge parallel to some coordnate axs respectvely. Then there exst n postve numbers h 1,h 2,,h n, such that, { } T = x = (x 1,x 2,,x n ) T x = x 0 +ξ h, 1 ξ 1, 1 n, (2.1) where x 0 s the center pont of T. Defne ξ = 1 h (x x 0 ), 1 n, (2.2) and set ξ = (ξ 1,ξ 2,,ξ n ) T. Denote 2 n vertces of T by a j, 1 j 2 n, and (ξ 1,ξ 2,,ξ n ) T correspondng to a j by Ξ j = (ξ 1j,ξ 2j,,ξ nj ) T. For ξ = (ξ 1,ξ 2,,ξ n ) T and mult-ndex α = (α 1,α 2,,α n ), defne Set ϕ j = 1 2 n ξ α = n =1 ξ α, α! = n α!. =1 n (1+ξ j ξ ), 1 j 2 n, (2.3) =1 P T,m = span Then ϕ j, 1 j 2 n, form a bass of Q 1 (T), and P T,m = span { } ϕ j ξ 2α 1 j 2 n, α < m. (2.4) { } pξ 2α p Q1 (T), α < m. (2.5) The rectangular fnte element of order m s defned by the trple (T,P T,D T ) as follows, 1. T s the n-rectangle descrbed by (2.1); 2. P T = P T,m ; 3. the components of D T (v) for any v C m 1 (T) are α v(a j ), α < m, 1 j 2 n. Lemma 2.1. For the rectangular fnte element of order m, D T s P T -unsolvent and P 2m 1 (T) P T. set Proof. It s obvous that the dmensons of D T and P T are all 2 n C m 1 n+m 1. For 1 j 2n, ϕ j,α = Ξα j α!2 α (ξ2 1 1)α1 (ξ 2 2 1)α2 (ξ 2 n 1)αn ϕ j, α < m. (2.6)
218 B.R. GAO, S. ZHANG AND M. WANG Wrte the partal dervatve wth respect to ξ as α ξ = ξ α1 α. 1 ξαn n It can be verfed that { 1, β = α and j = k, β ξ ϕ j,α(a k ) = 0, otherwse, 1 j, k 2 n, β α < m. (2.7) Defne when α = m 1, and ψ j,α = ϕ j,α (2.8) ψ j,α = ϕ j,α 2 n m 1 k=1 β = α +1 β ξ ϕ j,α(a k )ψ k,β (2.9) when α < m 1. Then β ξ ψ j,α(a k ) = { 1, β = α and j = k, 0, otherwse, 1 j,k 2 n, α, β < m. (2.10) Therefore, h α1 1 hαn n ψ j,α (1 j 2 n, α < m) are the bass functons correspondng to the degrees of freedom snce ξ α = hα1 1 hαn n α. Thus, we obtan that D T s P T -unsolvent. Now we show that P 2m 1 (T) P T,m. Let p P 2m 1 (T), then p can be wrtten as p = C α x α, α 2m 1 wth C α constants. For term C α x α wth α 2m 1, defne β and γ by α 2, α s even, { 0, α s even, β = γ α 1 = 1 n. 1, α, α s odd, s odd, 2 Then α = 2β+γ, so that C α x α = C α x γ x 2β P T,m by (2.5) and the fact that x γ Q 1 (T) and β < m. Consequently, p P T,m. For the rectangular fnte element of order m, the correspondng fnte element spaces V h and V h0 are defned as follows. V h = {v L 2 (Ω) v T P T,m, T T h, α v, α < m, are contnuous at all vertces of elements n Th}. V h0 = {v V h α v, α < m, vansh at all vertces of elements n Th whch are belongng to Ω }. Remark 2.1. The rectangular fnte element of order m s just the (2m 1)-th-oder Hermte nterpolaton element when n = 1, the n-lnear fnte element when m = 1 and the Adn element when m = 2. For the 2m-th-oder problems, the rectangular fnte element of order m s conformng when m = 1 or n = 1, otherwse t s nonconformng. The rectangular fnte element of order m can be vewed as the natural and reasonable generalzatons of the one dmensonal (2m 1)-th-oder Hermte nterpolaton element to hgher dmensons or the n- lnear fnte element to hgher order problems. Ths generalzaton shows that the conformng elements and the nonconformng elements are n same category.
Nonconformng Fnte Elements for Ellptc Problems 219 Now let Π T,m be the correspondng nterpolaton operator to the rectangular fnte element of order m. Lemma 2.2. For the rectangular fnte element of order m, ( ) β p Π T,1 β p dx = 0, 1 n, β < m, p P T,m. (2.11) x T Proof. Let p P T,m, 1 n and β < m. We know by (2.4) that β p s a lnear combnaton of the followng functons, F j,α = β( (ξ 2 1 1) α1 (ξ 2 2 1) α2 (ξ 2 n 1) αn ϕ j ), 1 j 2 n, α < m. For 1 j 2 n and α < m, set f = dβ (ξ dξ β 2 1)α, g = dβ dξ β (ξ (ξ 2 1)α ). Then F j,α can be wrtten as the sum of such terms that each term has two factors, one s f or g, and another s ndependent of component ξ. Defne { G (T) = v C v (T) dx = 0, v(a j ) = 0, 1 j 2 n}. ξ T a) β < α. In ths case, df dξ s just the Legendre polynomal of ξ or ts ntegral. Hence 1 1 df dξ dξ = 0. On the other hand, f vanshes when ξ = ±1. Then f G (T). b) β α and α 1. In ths case, f Q 1 (T). c) β α 2. In ths case, we have f = (2α dβ 1 dξ β 1 ξ (ξ 2 1)α 1) = dβ 2 dξ β 2 d β 2 =C 1 dξ β 2 (2α (ξ 2 1) α 1 +4α (α 1)ξ 2 (ξ 2 1) α 2) (ξ 2 1) α 1 d β 2 +C 2 (ξ 2 1) α 2, dξ β 2 where C 1 and C 2 are constants. Repeatng the same argument, we can read f as the lnear combnaton of terms satsfyng case a) or case b). Then f G (T)+Q 1 (T). d) For g, we have g = 1 2(α +1) d β+1 dξ β+1 (ξ 2 1)α+1. Then g G (T)+Q 1 (T) by the dscusson from case a) to case c). Fnally, we conclude that β p G (T)+Q 1 (T). Therefore, β p Π T,1 β p G (T)+Q 1 (T). Snce β p Π T,1 β p vanshes at the vertces of T, we have that β p Π T,1 β p G (T), and (2.11) s proved.
220 B.R. GAO, S. ZHANG AND M. WANG 3. C 0 Nonconformng Plate Elements In ths secton, we wll focus on the plate bendng problem, and consder the nonconformng fnte element. Let n = 2. Gven a trangle T, ts vertces are denoted by a, 1 3. The sde of T opposte to a s denoted by F, ts unt outer normal by ν F and ts measure by F, 1 3. Let λ 1,λ 2,λ 3 be the barycentrc coordnates of T. Denote ( ) q 1 = 2 5(λ 1 λ 2 1 2λ 2λ 3 ) 1 λ 1 λ 2 λ 3, ( ) q 2 = 2 5(λ 2 λ 2 2 2λ 1λ 3 ) 1 λ 1 λ 2 λ 3, (3.1) ( ) q 3 = 2 5(λ 3 λ 2 3 2λ 1 λ 2 ) 1 λ 1 λ 2 λ 3. Set It s obvous that so the dmenson of P + 3 P + 3 (T) = P 3(T)+span{ q 1, q 2, q 3 }. (3.2) (T) s at most twelve. q 1 + q 2 + q 3 = 6λ 1 λ 2 λ 3, The new plate element s defned by (T,P T,D T ) wth 1. T s a trangle; 2. P T = P + 3 (T); 3. the components of D T (v) for any C 1 (T) are: 1 v v(a j ), ds, 1 j 3, F j F j ν Fj (a j a ) T v(a ), 1 j 3, (3.3) where s the gradent operator. Defne, for 1 j k 3, q = q λ, p = 3λ 2 2λ3 + 1 l 3 l λ T λ l λ l q l, p j = λ 2 λ j +10λ (λ j λ k )λ 1 λ 2 λ 3. Let δ j be the Kronecker delta. It can be verfed that q, p, p j P 3 + (T), and q (a k ) = 0, (a l a k ) T 1 q q (a k ) = 0, ds = δ k, F k F k p (a k ) = δ k, (a l a k ) T 1 p p (a k ) = 0, ds = 0, F k F k p j (a k ) = 0, (a l a k ) T 1 p j p j (a k ) = δ k δ jl, ds = 0, F k F k (3.4) (3.5)
Nonconformng Fnte Elements for Ellptc Problems 221 when 1 j 3 and 1 k l 3. Hence q, p and p j are the nodal bass functons wth respect to the degrees of freedom. Therefore, the dmenson of P + 3 (T) s 12 and D T s P T -unsolvent. One can verfy that F k λ l q ds = F k 2 δ k, when 1 3 and 1 k l 3, and that p j F 0, 1 3 1 3 F k λ p j ds = when 1 j 3, 1 k 3 and 1 l k, 3. Gven p P 3 + (T), t can be wrtten as p = p(a )p + 1 p df q + F ν F F F k λ l p ds = 0. (3.6) F k λ l p j ds = F k 12 νt F k λ j, (3.7) 1 j 3 (a j a ) T p(a )p j. Then for 1 j k 3, t can be computed by (3.6), (3.7) and the above equalty that 1 p λ j ds = 1 ( p (a j ) p ) (a k ) + 1 p ds. (3.8) F F ν F 12 ν F ν F 2 F F ν F Gven any edge F of T Th, denote ts unt outer normal by ν F. For any v L 2 (Ω) wth v T H 1 ( T), T Th, we defne the jump of v across F as follows: f F = T T for some other T Th, and [v] F = v T v T, [v] F = v T, f F = T Ω. For the new element, defne the correspondng fnte element spaces V h and V h0 as follows. V h = {v L 2 (Ω) v T P 3 + (T), T T h, v and v are contnuous at all vertces of elements n Th, and for any edge F of T wth F Ω the ntegral average of νf T[ v] F over F s zero}; and V h0 = {v V h v and v vansh at all vertces belongng to Ω, and for any edge F of T wth F Ω the ntegral average of ν F v over F s zero}. We clam that V h C 0 ( Ω) and V h0 C0 0(Ω). Let v h V h, F be a common edge of T,T Th. By the defnton, [v h ] F s n P 3 (F), and t and ts drectve dervatve along F are zero at two endponts of F. Hence [v h ] F 0, that s, v h C 0 ( Ω). Smlarly, we can show that v h C0 0(Ω) when v h V h0. By (3.8), the defntons of V h and V h0 and the fact that V h C 0 ( Ω) and V h0 C0(Ω), 0 we obtan the followng lemma. Lemma 3.1. If F s a common edge of dstnct T,T Th, then p[ v h ] F ds = 0, p P 1 (F), v h V h. (3.9) F
222 B.R. GAO, S. ZHANG AND M. WANG If an edge F of T Th s on Ω then p[ v h ] F ds = 0, p P 1 (F), v h V h0. (3.10) F From the new plate element gven above, we can deduce a new Zenkewcz-type element. For 1 3, defne φ (v) = 1 v ds 1 v (a j ), v C 1 (T). (3.11) F ν F 2 ν F F 1 j 3,j Set P z T = { } p P 3 + (T) φ1 (p) = φ 2 (p) = φ 3 (p) = 0. (3.12) Observng the fact that P 2 (T) P + 3 (T) and φ 1(p) = φ 2 (p) = φ 3 (p) = 0, p P 2 (T), we have P 2 (T) P z T. The new Zenkewcz-type element s defned by (T,P T,D T ) wth 1. T s a trangle; 2. P T = PT z; 3. the components of D T (v) for any v C 1 (T) are: { v(aj ), 1 j 3, (a j a ) T v(a ), 1 j 3. (3.13) It s easy to verfy that D T s P T -unsolvent. For the new Zenkewcz-type element, the correspondng fnte element spaces Vh z and V h0 z are defned as follows. Vh z = {v L2 (Ω) v T PT z, T T h, v and v are contnuous at all vertces of elements n Th}, Vh0 z = {v V h z v and v vansh at all vertces belongng to Ω}. The dfference between the new Zenkewcz-type element here and the two dmensonal one proposed n [14] s ther shape functon spaces. The consstent term of the element here s of order O(h 2 ), whle the consstent term of the element gven n [14] s of order O(h). 4. Convergence Analyss Let f L 2 (Ω). We take the followng boundary value problem as example to show the convergent result: ( 1) m m u = f, n Ω, u Ω = u = = m 1 u (4.1) Ω ν Ω ν m 1 = 0, where ν = (ν 1,ν 2,,ν n ) T s the unt outer normal to Ω and s the standard Laplacan operator. Defne m u m v a(u,v) = dx. (4.2) x j1 x jm x j1 x jm 1 j 1,,j m n Ω
Nonconformng Fnte Elements for Ellptc Problems 223 Then the weak form of problem (4.1) s: fnd u H m 0 (Ω) such that a(u,v) = (f,v), v H m 0 (Ω). (4.3) For nonnegatve nteger s and Th, defne { H s (T h) = v L 2 (Ω) v T H s (T), T T h }. For v,w H m (Th), defne a h (v,w) = T Th 1 j T 1,,j m n m v x j1 x jm m w x j1 x jm dx. (4.4) The fnte element method for problem (4.3) s: fnd u h V h0 such that a h (u h,v h ) = (f,v h ), v h V h0. (4.5) We ntroduce the followng mesh dependent norm s,h and sem-norm s,h : ( 1/2, v s,h = v s,t) 2 T Th ( ) v H s 1/2, (T h). v s,h = v 2 s,t T Th For the nonconformng elements, the basc mathematcal theory has been establshed(see [5, 7,10,13,18]). We can use them to gve the convergence analyss of our new elements. For the fnte elements gven n prevous two sectons, one can verfy the followng statements by ther constructons, Lemmas 2.1, 2.2 and 3.1: They all have the approxmablty. They all have the superapproxmaton. They all have the weak contnuty. They all pass the patch test. They all pass the generalzed patch test. Then by the result n [10] or by the one n [13] we can obtan the followng theorems. Theorem 4.1. Assume that m,n 1. Let V h0 be the fnte element space correspondng to the rectangular fnte element of order m, and let u and u h be the solutons of problems (4.3) and (4.5) respectvely. Then lm u u h m,h = 0. (4.6) h 0 Theorem 4.2. Assume that m = n = 2. Let V h0 be the fnte element space correspondng to the new plate element or new Zenkewcz-type element, and let u and u h be the solutons of problems (4.3) and (4.5) respectvely. Then lm u u h 2,h = 0. (4.7) h 0
224 B.R. GAO, S. ZHANG AND M. WANG By the result n [13], we know that the error of the rectangular fnte element of order m and the new Zenkewcz-type are all order O(h). For the new plate element, we can obtan the followng theorem by Lemma 3.1 and the usual technque dealng wth the consstent term. Theorem 4.3. Assume that m = n = 2. Let V h0 be the fnte element space correspondng to the new plate element, and let u and u h be the solutons of problems (4.3) and (4.5) respectvely. Then there exsts a constant C ndependent of h such that when u H 4 (Ω). In addton, when Ω s convex. u u h 2,h Ch 2 u 4,Ω, (4.8) u u h 1,Ω Ch 3 u 4,Ω, (4.9) Remark 4.1. Let k 1. The fnte element space V h correspondng to the rectangular element of order k s a subspace of H 1 (Ω). Hence the element s convergent wth order O(h 2k 1 ) by Lemma 2.1 when t s appled to solvng the second-oder problems. In general, the rectangular element of order k s a convergent nonconformng element for the 2m-th-oder problem when k m, whch can be shown by Lemmas 2.1 and 2.2. In ths stuaton, the fnte element space V h0 should be defned accordngly. 5. Numercal Examples In ths secton, we gve some numercal results of the new plate element. Now let m = n = 2, Ω = (0,1) (0,1) and defne u 1 (x) = x 2 1(x 1 1) 2 x 2 2(x 2 1) 2, u 2 (x) = ( sn(πx 1 )sn(πx 2 ) ) 2, u 3 (x) = e x1+x2. log( Π h u u h 2,h ) 4 2 0 u 2 (x)=(sn(πx 1 )sn(πx 2 )) 2 slope of whch s 1.1928. u 3 (x)=e x 1 +x 2 slope of whch s 1.3389. u 1 (x)=x 1 2 (x1 1) 2 x 2 2 (x2 1) 2 slope of whch s 1.2202. 2 4 6 8 10 12 1 2 3 4 5 6 7 8 9 log(h 2 ) Fg. 5.1. The error: Π h u u h 2,h
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