Modied Morley Element Method for a ourth Order Elliptic Singular Perturbation Problem Λ Wang Ming LMAM, School of Mathematical Science, Peking University Jinchao u School of Mathematical Science, Peking University, and Department of Mathematics, Pennsylvania State University and Hu Yucheng School of Mathematical Science, Peking University Abstract. his paper proposes a modied Morley element method for a fourth order elliptic singular perturbation problem. he method also uses Morley element or rectangle Morley element, but linear or bilinear approximation of nite element functions is used in the lower part of the bilinear form. It is shown that the modied method converges uniformly in the perturbation parameter. Keywords: Morley element, singular perturbation problem AMS Subject Classication (1991): 65N30. 1. Introduction Let Ω be a bounded polygonal domain of R 2. Denote the boundary of Ω by @Ω. or f 2 L 2 (Ω), we consider the following boundary value problem of fourth order elliptic singular perturbation equation: 8 < : " 2 2u u = f; in Ω; uj @Ω = @u =0 @ν @Ω (1.1) where ν =(ν 1 ;ν 2 ) > is the unit outer normal to @Ω, is the standard Laplacian operator and " is a real small parameter with 0 < "» 1. When "! 0 the differential equation formally degenerates to Poisson equation. o overcome the C 1 difcult, it is prefer to using nonconforming nite element to solve problem (1.1). Since the differential equation degenerates to Poisson equation as "! 0, C 0 nonconforming elements seem better to be used. An C 0 nonconforming nite element was proposed in [4], and its uniform convergence in " was shown. Λ he work of the rst author was supported by the National Natural Science oundation of China (10571006). he work of the second author was supported by National Science oundation DMS-0209479 and DMS-0215392 and the Changjiang Professorship through Peking University 1
It is known that Morley element is not C 0 element and it is divergent for Poisson equation (see [6]). When Morley element is applied to solve problem (1.1), it fails when "! 0 (see [4]). On the other hand, we have noticed the remark in the end of paper [4]: the best result uniformly in " seems to be order of O(h 1=2 )for any nite element method to problem (1.1). Here h is the mesh size. Since Morley element has the least number of element degrees of freedom, we prefer to use a method which still uses the degrees of freedom of Morley element to solve problem (1.1). In this paper, we will propose a modied Morley element method for problem (1.1). he method also uses Morley element, but the linear approximation of nite element functions is used in the part of the bilinear form corresponding to the second order differential term. he modied method degenerates to the conforming linear element method for Poisson equation when " = 0, and this is consistent with the degenerate case of problem (1.1). We will show that the modied method converges uniformly in perturbation parameter ". he modied rectangle Morley element method is also considered in this paper. he paper is organized as follows. he rest of this section lists some preliminaries. Section 2 gives the detail descriptions of the modied Morley element method. Section 3 shows the uniform convergence of the method. he last section gives some numerical results. or nonnegativeinteger s, H s (Ω), k k s;ω and j j s;ω denote the usual Sobolev space, norm and semi-norm respectively. Let H0(Ω)be s the closure of C 1 0 (Ω)in H s (Ω)with respect to the norm k k s;ω and ( ; )denote the inner product of L 2 (Ω). Dene a(v; w) = b(v; w) = 2 Ω i;j=1 Ω i=1 @ 2 v he weak form of problem (1.1)is: nd u 2 H 2 0 (Ω)such that 2 @ 2 w ; 8v; w 2 H 2 (Ω): (1.2) @v @w ; 8v; w 2 H 1 (Ω): (1.3) " 2 a(u; v)+b(u; v) =(f; v); 8v 2 H 2 0 (Ω): (1.4) Let u 0 be the solution of following boundary value problem: ( u 0 = f; in Ω; u 0 j @Ω =0 he following lemma is shown in paper [4]. Lemma 1.1 If Ω is convex, then there exists a constant C independent of " such that for all f 2 L 2 (Ω). (1.5) juj 2;Ω + "juj 3;Ω» C" 1 2 kfk0;ω (1.6) ju u 0 j 1;Ω» C" 1 2 kfk0;ω (1.7) 2. Modied Morley Element Method or a subset B ρ R 2 and r a nonnegative integer, let P r (B)be the space of all polynomials with degree not greater than r. 2
Morley Element Given a triangle, its three vertices is denoted by a j,1» j» 3: he edge of opposite a j is denoted by j,1» j» 3. Denote the measures of and i by j j and j i j respectively. Morley element can be described by (;P ; Φ )with 1) is a triangle. 2) P = P 2 ( ). 3) Φ is the vector of degrees of freedom whose components are: for v 2 C 1 ( ). v(a j ); Rectangle Morley Element 1 j j j j @v @ν ds; 1» j» 3 Given a rectangle, its four vertices and edges are denoted by a j and j, 1» j» 4, respectively. Rectangle Morley element can be described by (;P ; Φ )with 1) is a rectangle with its edges parallel to some coordinate axes respectively. 2) P = P 2 ( )+spanf x 3 1 ;x3 2g. 3) Φ is the vector of degrees of freedom whose components are: for v 2 C 1 ( ). v(a j ); 1 j j j j @v @ν ds; 1» j» 4 he degrees of freedom of these two elements are shown in ig. 1. 6 Y 3 ff -? Morley element? Rectangle Morley element ig. 1 he Morley element and its convergence for biharmonic equations can be found in [1-3,5], while the rectangle Morley element in [7]. or mesh size h, take h a triangulation of Ω. or Morley element h consists of triangles, otherwise h consists of rectangles with their edges parallel to some coordinate axes respectively. or each 2 h, leth be the diameter of the smallest disk containing and ρ be 3
the diameter of the largest disk contained in. Let fhg be a family of triangulations with h! 0. hroughout the paper, we assume that fhg is quasi-uniform, namely it satised that h» h» ρ, 8 2 h for a positive constant independent ofh. or each h, letv h and V h0 be the corresponding nite element spaces associated with Morley element or with rectangle Morley element for the discretization of H 2 (Ω)and H0 2 (Ω) respectively. his denes two family of nite element spaces fv h g and fv h0 g. It is known that V h 6ρ H 2 (Ω)and V h0 6ρ H0 2 (Ω). Let Π h be the interpolation operator corresponding to h and Morley element or the rectangular Morley element. We dene 2 @ 2 v @ 2 w a h (v; w) = ; v; w 2 H 2 (Ω)+ V h (2.1) b h (v; w) = i;j=1 2 i=1 @v @w ; v; w 2 H 1 (Ω)+ V h (2.2) he standard nite element method for problem (1.4)corresponding to Morley element or to rectangle Morley element is: nd u h 2 V h0 such that " 2 a h (u h ;v h )+b h (u h ;v h )=(f; v h ); 8v h 2 V h0 : (2.3) or Morley element, let Π 1 h be the interpolation operator corresponding to linear conforming element for second order partial differential equation and h. or the rectangle Morley element let Π 1 h be the bilinear interpolation operator. We consider the following modied Morley element method: to nd u h 2 V h0 such that " 2 a h (u h ;v h )+b h (Π 1 h u h; Π 1 h v h)=(f;π 1 h v h); 8v h 2 V h0 (2.4) Problem (2.4)has unique solution when " >0. When " = 0, the problem degenerates to b h (Π 1 h u h; Π 1 h v h)=(f;π 1 h v h); 8v h 2 V h0 (2.5) Although the solution of problem (2.5) is not unique yet, Π 1 h u h is uniquely determined. Actually, Π 1 h u h is the exact nite element solution of linear or bilinear conforming elemet for problem (1.5). Hence the modied Morley element method seems to give a more natural way to solve problem (1.1). We introduce the following mesh dependent norm k k m;h and semi-norm j j m;h : 8 >< >: kvk m;h = jvj m;h = kvk 2 m; 1=2; jvj 2 m; 3. Convergence Analysis 1=2; 8v 2 V h + H m (Ω): In this section, we discuss the convergence properties of the modied Morley element methods in previous section. Let u and u h be the solutions of problem (1.4)and (2.4)respectively. 4
Lemma 3.1 here exists a constant C independent of h and " such that 8v h 2 V h0 when u 2 H 3 (Ω). j b h (Π 1 h u; Π1 h v h)+( u; Π 1 h v h)j»chjuj 2;Ω jπ 1 h v hj 1;h (3.1) Proof. Let v h 2 V h0. hen Π 1 h v h 2 H 1 0 (Ω)and j a h (u; v h ) ( 2u; Π 1 h v h)j»chjuj 3;Ω jv h j 2;h (3.2) j b h (Π 1 h u; Π1 h v h)+( u; Π 1 h v h)j = j b h (u Π 1 h u; Π1 h v h)j: By the interpolation theory and Schwarz inequality we obtain (3.1). Now take f 2 H 1 (Ω). Given 2 h and an edge of, let P 0 be the orthogonal projection operator from L 2 ( )to P 0 ( ). @ Let i; j 2 f1; 2g. It is known that the integral average of @xj v h on is continuous through and vanishes when ρ @Ω. hen Green formula gives f @2 v h + @f @v h = = @ ρ@ = ρ@ f @v h ν i ds = ρ@ f @vh P 0 @v h ν i ds f @v h ν i ds (f P 0 f) @vh P 0 @v h ν i ds rom Schwarz inequality and the interpolation theory we have ρ@ (f P 0 f) @vh P 0 @v h» ρ@» C ν i ds fl fl kf P 0 fk 0;fl @v h P 0 @v h hjfj 1; jv h j 2;» Chjfj 1;Ω jv h j 2;h : fl flfl 0; Consequently, we obtain that 8f 2 H 1 (Ω); 8v h 2 V h0, i; j 2f1; 2g, f @2 v h + @f @v h» Chjfj 1;Ω jv h j 2;h : (3.3) We obtain the conclusion of the lemma from (3.3), the interpolation theory and the 5
following equality, a h (u; v h ) ( 2u; Π 1 h v h) = 2 i=1 + 2 i=1 + @ u @(Π 1 h v h v h ) 1»i6=j»2 1»i6=j»2 u @2 v h @x 2 i @ 2 u @ 2 u @ 2 v h @x 2 i @x 2 j + @ u @v h @ 2 v h + @3 u @x 2 i + @3 u @v h @x 2 i @x : j @v h (3.4) rom lemma 3.1, we have heorem 3.1 here exists a constant C independent of h and " such that when u 2 H 3 (Ω). Proof. Let w h =Π h u, then "ku u h k 2;h + ku Π 1 h u hk 1;Ω» Ch("juj 3;Ω + juj 2;Ω ) (3.5) "ku u h k 2;h + ku Π 1 h u hk 1;Ω» "ku w h k 2;h + ku Π 1 h w hk 1;Ω +"ku h w h k 2;h + kπ 1 h(u h w h )k 1;Ω (3.6) Set v h = u h w h. rom (2.4)and (1.1), we have " 2 a h (v h ;v h )+b h (Π 1 h v h; Π 1 h v h) rom the interpolation theory, (3.1)and (3.2), Since we obtain that =" 2 a h (u w h ;v h )+b h (Π 1 h(u w h ); Π 1 h v h) + " ( 2u; 2 Π 1 h v h) a h (u; v h ) ( u; Π 1 h v h)+b h (Π 1 h u; Π1 h v h) " 2 a h (v h ;v h )+b h (Π 1 h v h; Π 1 h v h)» Ch("juj 3;Ω + juj 2;Ω )("jv h j 2;h + jπ 1 h v hj 1;Ω ): " 2 kv h k 2 2;h + kπ 1 h v hk 2 1;Ω» C " 2 a h (v h ;v h )+b h (Π 1 h v h; Π 1 h v h) "ku h w h k 2;h + kπ 1 h(u h w h )k 1;Ω» Ch("juj 3;Ω + juj 2;Ω ): (3.7) he theorem follows from the interpolation theory, (3.6)and (3.7). 6
heorem 3.2 If Ω is convex, then there exists a constant C independent of h and " such that "ku u h k 2;h + ku Π 1 h u hk 1;Ω» Ch 1=2 kfk 0;Ω : (3.8) Proof. rom the interpolation theory, ku Π h uk 2 2;h» Cjuj 2;Ω ku Π h uk 2;h» Chjuj 2;Ω juj 3;Ω By lemma 1.1, we have "ku Π h uk 2;h» Ch 1=2 kfk 0;Ω : (3.9) Similar to (4.4)in [4], we can show that kv Π 1 h vk2» Chjvj 1;Ω 1;Ωjvj 2;Ω ; 8v 2 H 2 0 (Ω): (3.10) Using (3.10), we obtain ku u 0 Π 1 h(u u 0 )k 2 1;Ω» Chju u 0 j 1;Ω ju u 0 j 2;Ω and by theinterpolation theory, ku 0 Π 1 h u0 k 1;Ω» Chju 0 j 2;Ω : rom lemma 1.1 and the following inequalities ku 0 k 2;Ω» Ckfk 0;Ω we have ku Π 1 h uk 1;Ω»ku u 0 Π 1 h(u u 0 )k 1;Ω + ku 0 Π 1 h u0 k 1;Ω Set v h = u h Π h u. hen Π 1 h v h 2 H 1 0 (Ω)and rom (3.11)and Schwarz inequality, ku Π 1 h uk 1;Ω» Ch 1=2 kfk 0;Ω : (3.11) j b h (Π 1 h u; Π1 h v h)+( u; Π 1 h v h)j = j b h (u Π 1 h u; Π1 h v h)j: j b h (Π 1 h u; Π1 h v h)+( u; Π 1 h v h)j»ch 1=2 kfk 0;Ω kπ 1 h v hk 1;Ω : (3.12) Now take f 2 H 1 (Ω)and i; j 2f1; 2g. rom the proof of lemma 3.1, we have f @2 v h + @f @v h» ρ@ fl fl kf P 0 fk 0;fl @v h P 0 @v h fl flfl 0; : Since kf P 0 fk 0;» 2kfk 0;» 2kfk 0;@» Ckfk 1=2 0; kfk1=2 1; (3.13) 7
we have, by the interpolation theory f @2 v h + @f @v h If "» h, then by Green formula @f @(Π 1 h v h v h ) =» Ch 1=2 kfk 1=2 @ 0;Ω kfk1=2 f @(Π1 h v h v h ) ν i ds f @2 (Π 1 h v h v h ) @x 2 i rom Schwarz inequality, the interpolation thoery and (3.13), we obtain Hence when "» h " 2 @f @(Π 1 h v h v h )» + @f @(Π 1 h v h v h )» Ch " 1=2 2 kfk 1=2 kfk 0;@ fl flfl @(Π 1 h v h v h ) kfk 0; jπ 1 h v h v h j 2; 1;Ω jv hj 2;h : (3.14) fl flfl 0;@»C h 1=2 kfk 1=2 0;Ω kfk1=2 1;Ω + kfk 0;Ω jvh j 2;h : 0;Ω kfk1=2 When ">h,by Schwarz inequality and the interpolation theory we have, " 2 1;Ω + "3=2 kfk 0;Ω jv h j 2;h : (3.15) @f @(Π 1 h v h v h )» Ch" 2 jfj 1;Ω jv h j 2;h» Ch 1=2 " 5=2 jfj 1;Ω jv h j 2;h : (3.16) rom lemma 1.1, (3.4), (3.14), (3.15) and (3.16) we obtain " 2 j a h (u; v h ) ( 2u; Π 1 h v h)j»c"h 1=2 kfk 0;Ω jv h j 2;h : (3.17) Combining (3.9), (3.11), (3.12), (3.17) and the proof of theorem 3.1, we obtain the theorem. 4. Numerical Results In this section, we will show some numerical results of the modied Morley element methods. We will use the same example used in [4] for comparison. Let Ω = [0; 1] [0; 1] and u(x) = sin(ßx 1 )sin(ßx 2 ) 2. or " 0, set f = " 2 2u u. hen u is the solution of problem (1.1) when ">0, and is the solution of problem (1.5) when " = 0. or the rectangle Morley element, Ω is divided into h h squares, and for 8
Morley element, each square is further divided into two triangles by the diagonal with a negative slash. Dene kv h k ";h = " 2 a h (v h ;v h )+b h (Π 1 h v h; Π 1 h v h) 1=2 ; 8vh 2 V h0 : Different values of " and h are chosen to demonstrate the behaviors of the following relative error of two modied Morley element methods, E ";h = kui h u hk ";h ku I h k ";h (4.1) where u h is the solution of problem (2.4) and u I h denote the interpolant of u by Morley element or rectangular Morley element. Let g = 2u, then u is the solution of the following boundary value problem of biharmonic equation, 8 < : 2u = g; in Ω; uj @Ω = @u =0 @ν @Ω (4.2) or comparison, we also consider the error of nite element solution to problem (4.2). Let ~u h 2 V h0 be the solution of the following problem, In this situation, the relative error ~ Eh is presented by a h (~u h ;v h )=(g; Π 1 h v h); 8v h 2 V h0 : (4.3) ~E 2 h = a h(u I h ~u h;u I h ~u h) a h (u I h ;ui h ) (4.4) or the modied Morley element method and the modied rectangular Morley element method, E ";h and ~ Eh, corresponding some " and h, are listed in able 1 and able 2 respectively. " n h 2 3 2 4 2 5 2 6 0 0.0576 0.0145 0.0036 0.0009 2 10 0.0577 0.0146 0.0037 0.0009 2 8 0.0586 0.0156 0.0046 0.0017 2 6 0.0713 0.0266 0.0119 0.0057 2 4 0.1653 0.0832 0.0418 0.0210 2 2 0.3404 0.1749 0.0885 0.0451 2 0 0.3869 0.1979 0.1000 0.0512 Biharmonic 0.3908 0.1998 0.1010 0.0517 able 1. Modied Morley Element Method 9
" n h 2 3 2 4 2 5 2 6 0 0.0205 0.0046 0.0011 0.0002 2 10 0.0205 0.0047 0.0011 0.0003 2 8 0.0211 0.0052 0.0016 0.0006 2 6 0.0285 0.0107 0.0049 0.0024 2 4 0.0757 0.0360 0.0179 0.0091 2 2 0.1568 0.0770 0.0392 0.0211 2 0 0.1774 0.0875 0.0449 0.0246 Biharmonic 0.1791 0.0884 0.0453 0.0249 able 2. Modied Rectangular Morley Element Method rom able 1 we see that the modied Morley element method, unlike Morley element method (see able 1 in [4]), converges for all " 2 [0; 1]. More precisely, the result shows that E ";h tends to E 0;h as " approaches 0, while E ";h is linear with respect to h as well as ~ Eh is when " is large. hat is, the modied Morley element method behaves in the way that the conforming linear element does for Poisson equation when " is small, while it likes Morley element for the biharmoni equation when " is large. We can get the similar discussion about the modied rectangular Morley element method from able 2. References 1. Ciarlet P C, he inite Element Method for Elliptic Problems, North-Holland, Amsterdam, New York, 1978. 2. Lascaux P and Lesaint P, Some nonconforming nite elements for the plate bending problem, RAIRO Anal. Numer., R-1 (1985), 9-53. 3. Morley L S D, he triangular equilibrium element in the solution of plate bending problems, Aero. Quart., 19 (1968), 149-169. 4. Nilssen K, ai -C and Winther R, A robust nonconforming H 2 -element, Math. Comp., 70 (2001), 489-505. 5. Strang G and ix G J, An Analysis of the inite Element Method, Prentice-Hall, Englewood Cliffs, 1973. 6. Wang Ming, On the necessity and sufciency of the patch test for convergence of nonconforming nite elements, SIAM JNumer Anal, 39, 2(2002), 363-384. 7. hang Hongqing and Wang Ming, he Mathematical heory of inite Elements, Science Press, Beijing, 1991. 10