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Meshless Galerkin method based on regions partitioned into subdomains
Applied Mathematics and Computation 162 (2005) 317–327 www.elsevier.com/locate/amc
Meshless Galerkin method based on regions partitioned into subdomains Yong Duan *, Yong-ji Tan Institute of Mathematics, Fudan University, Shanghai 200433, People’s Republic of China
Abstract In this paper, we will combine the domain decomposition method with so-called meshless Galerkin method for PDE using RBF. The convergence of this new method and numerical examples are given. Ó 2004 Elsevier Inc. All rights reserved. Keywords: Meshless; RBF; Domain decomposition; Parallel
1. Introduction The partition of elliptic problem into subproblems is a very nature idea and is very much in use in practices. Many peoples had done some research about this method, for example, [1–4]. Also are there some books about this method, see, for example, [5]. We must point out that the methods before are based on FEM. The FEM has to spend a lot of time on technical details concerning the mesh. Since 1990Õs last century, interpolation by radial basis functions (RBF) has become a powerful tool in multivariate approximation theory, especially since compactly supported radial basis functions (CS-RBF) are available. Since, in contrast to FEM, the construction of the finite dimensional subspace using RBF is independent of the current space dimension, it is possible to solve high dimensional problems as they occur in quantum mechanics and so on. Meshless methods do not need to handle such problems like the meshÕs
*
Corresponding author. E-mail address: [email protected] (Y. Duan).
0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.12.138
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Y. Duan, Y.-j. Tan / Appl. Math. Comput. 162 (2005) 317–327
generation. See [6] for an overview of general meshless methods and applications in engineering. In this paper, we do not give the details of the mathematical foundations of domain decomposition method (DDM). We refer the reader [5] for details. Which we will mainly concentrate on is the meshless numerical realization in solving PDE. About the RBF, we refer [7–11] to reader.
2. The mathematical foundation of DDM We consider: Du ¼ f u¼0
in X; on oX;
ð1Þ
where X is d-dimensional domain, with a Lipschitz boundary oX, whose outer unit P normal direction is denoted by n , f is a given function of L2 ðXÞ, d D ¼ j¼1 Dj Dj is the Laplace operator and Dj denotes the partial derivative with respect to xj , j ¼ 1; . . . ; d. Firstly, we assume that X is partitioned into two non-overlapping subdomains X1 and X2 , and denote by C ¼ X1 \ X2 . We indicate by ui the restriction to Xi , i ¼ 1; 2, of the solution u to (1.1), and ni by the normal direction on oXi \ C, oriented outward. Set n ¼ n1 . The Poisson problem (1) can be reformulated in the equivalent multidomain form: 8 Du1 ¼ f in X1 ; > > > > u on oX1 \ oX; > 1 ¼ 0 > < u1 ¼ u2 on C; ð2Þ ou2 1 ¼ ou on C; > > on > on > u ¼0 on oX2 \ oX; > > : 2 Du2 ¼ f in X2 : The third and fourth terms of Eq. (2) are the transmission condition for u1 and u2 on C. Let k denote the unknown value of u on C. We consider the two Dirichlet problems: 8 < DHi k ¼ 0 in Xi ; Hk¼0 on oXi \ oX; ð3Þ : i Hi k ¼ k on C and 8 < Dui ¼ f u ¼ 0 : i ui ¼ 0
in Xi ; on oXi \ oX; on C
ð4Þ
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319
for i ¼ 1; 2. Hi k is the harmonic extension of k into Xi . Then k satisfies the Steklov–Poincare interface equation Sk ¼ v
ð5Þ
on C;
where v¼
ou2 ou1 : on on
S is the Steklov–Poincare operator, Sg ¼
oH1 g oH2 g : on on
We introduce the following function spaces: Vi ¼ fvi 2 H 1 ðXi ÞjvijoX\oXi ¼0 g; Vi 0 ¼ H01 ðXi Þ. K ¼ fg 2 H 1=2 ðCÞjg ¼ vjC for a suitable v 2 V g. We can prove that the Poisson problem can be equivalently reformulated as: find ui 2 Vi such that 8 8v1 2 V10 ; a1 ðu1 ; v1 Þ ¼ ðf ; v1 ÞX1 > > < u1 ¼ u2 on C; ð6Þ a ðu ; v Þ ¼ ðf ; v Þ 8v2 2 V20 ; > 2 2 2 2 X2 > : a2 ðu2 ; R2 lÞ ¼ ðf ; R2 lÞX2 þ ðf ; R1 lÞX1 a1 ðu1 ; R1 lÞ 8l 2 K; where ai ðwi ; vi Þ ¼ ðrwi ; rvi ÞXi ¼
Z
rwi rvi ;
ð7Þ
Xi
Ri denotes any possible extension operator from K to Vi . The Dirichlet–Neumann iterative scheme reads: given k0 , solve for each k P 0: 8 ¼ f in X1 ; < Dukþ1 1 ð8Þ ukþ1 ¼ 0 on oX1 \ oX; : 1kþ1 u1 ¼ kk on C; then 8 kþ1 > < Du2 ¼ f kþ1 u2 ¼ 0 > oukþ1 : oukþ1 2 ¼ 1 on
on
in X2 ; on oX2 \ oX;
ð9Þ
on C;
with k kkþ1 ¼ hukþ1 2jC þ ð1 hÞk ;
ð10Þ
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h being a positive acceleration parameter. This method is equivalent to the socalled Richardson procedure for the Steklov–Poincare Eq. (5) with the operator S2 as a preconditioner, i.e., kkþ1 ¼ kk þ hS21 ðSkk þ vÞ:
ð11Þ
We must point out that you may replace ukþ1 by uk1 at the second step for D–N 1 iteration in view of parallel implementation. 3. Domain decomposition method at RBF level In this part, we give the presentation of RBF approximation theory for the domain decomposition method. We only concentrate on the CS-RBF and TPSRBF (thin plate spline). One of the most popular classes of CS-RBFs is the one introduced by Wendland [7]. Another is given by Wu and Schaback [10]. These functions are strictly positive definite in Rd for all d less than or equal to some fixed value d0 , and can be constructed to have any desired amount of smoothness 2j. For l ¼ ½d2 þ j þ 1 and j ¼ 0; 1; 2; 3. TPS-RBF reads: /ðrÞ ¼ r2m logðrÞ; m ¼ 1; 2; . . . WendlaudÕs function reads: : /l;0 ðrÞ¼ð1 rÞlþ ; : lþ1 /l;1 ðrÞ¼ð1 rÞþ ½ðl þ 1Þr þ 1; : 2 2 /l;2 ðrÞ¼ð1 rÞlþ2 þ ½ðl þ 4l þ 3Þr þ ð3l þ 6Þr þ 3; : lþ3 /l;3 ðrÞ¼ð1 rÞþ ½ðl3 þ 9l2 þ 32l þ 15Þr3 þ ð6l2 þ 36l þ 45Þr2 þ ð15l þ 45Þr þ 15: : Here ¼ denotes equality up to a constant factor. For example, the functions : 2 /2;0 ¼ð1 rÞþ ; : /3;1 ðrÞ¼ð1 rÞ4þ ½4r þ 1; : 6 /4;2 ðrÞ¼ð1 rÞþ ½35r2 þ 18r þ 3; : 8 /5;3 ðrÞ¼ð1 rÞþ ½32r3 þ 25r2 þ 8r þ 1; which are in C0 , C2 , C4 , C6 , respectively, and strictly positive definite in R3 . WuÕs CS-RBF reads: 6
/ðrÞ ¼ ð1 rÞþ ð5r5 þ 30r4 þ 72r3 þ 82r2 þ 36r þ 6Þ; which is in C6 . Here and throughout r ¼ kxk is Euclidean norm, with x 2 Rd . In order to solve the problem (1), we define the finite dimensional space VN ¼ spanf/ðkx xj kÞg;
j ¼ 1; 2; . . . ; N :
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321
Denoting by N1 , N2 , NC the corresponding number of knots in X1 , X2 , C. Suppose that suppf/ðkx xi kÞg \ suppf/ðkx xj kÞg ¼ ; if xi , xj are not in the same subdomain X1 or X2 and suppf/ðkx xi kÞg \ C ¼ ; for 8xj 2 X1 or X2 . Denoting by h ¼ supx2X minxi 2X kx xi k, the density of the distinct centers X ¼ fx1 ; x2 ; . . . ; xN g. We use subscript h to substitute the corresponding form of each operator after that we had finished the CS-RBF approximation. The weak form of Dirichlet–Neumann iterative substructuring method is: let k0C 2 RNC be the initial guess of the coefficients of the CS-RBFs, which center are on C. For k P 0 solve: 8 kþ1 > < u1;h 2 V1;h ; 0 8v1;h 2 V1;h ; a1 ðukþ1 ð12Þ 1;h ; v1;h Þ ¼ ðf ; v1;h ÞX1 P > N k : ukþ1 ¼ C on C; 1;h k¼1 kh /ðx xk Þ kþ1 ukþ1 2;h 2 V2;h : a2 ðu2;h ; v2;h Þ
¼ ðf ; v2;h ÞX2 þ ðf ; R1;h v2;hjC ÞX1 a1 ðukþ1 1;h ; R1;h v2;hjC Þ
8v2;h 2 V2;h ; ð13Þ
then update NC X
kþ1 kkþ1 h /ðx xk Þ ¼ hu2;hjC þ ð1 hÞ
k¼1
NC X
kkh /ðx xk Þ on C:
ð14Þ
k¼1
Assuming that ukþ1 1;h ¼ ukþ1 2;h ¼
N1 X
kkþ1 1;i /ðkx xi kÞ þ
NC X
i¼1
i¼1
N2 X
NC X
kkþ1 2;i /ðkx xi kÞ þ
i¼1
kkþ1 1 ,
kkC;i /ðkx xi kÞ; kkþ1 2;C;i /ðkx xi kÞ;
i¼1
kþ1 kþ1 denoting by the column vector consisted of kkþ1 1;i , k2;i , kC;i , kþ1 k2;C;i , respectively. By solving the Eqs. (12)–(14), we get the algebraic form for the Dirichlet–Neumann iterative procedure on the Stekelov–Poincare interface Eq. (11). ! 1 X X k kþ1 kC ¼ h vC kC þ ð1 hÞkkC ; ð15Þ 2
kkþ1 2 ,
kkþ1 C ,
kkþ1 2;C
1
where A2CC ¼ ðð/ðkx xi kÞ; /ðkx xj kÞÞX2 Þ;
xi ; xj
on C;
A1CC ¼ ðð/ðkx xi kÞ; /ðkx xj kÞÞX1 Þ;
xi ; xj
on C;
A11 ¼ a1 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi ; xj 2 X 1 ;
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A22 ¼ a2 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi ; xj 2 X 2 ;
A1C ¼ a1 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi 2 X1 ; xj on C;
A2C ¼ a2 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi 2 X2 ; xj on C;
AC2 ¼ transposeðA2C Þ AC1 ¼ transposeðA1C Þ; i ¼ 1; 2; R ¼ R1 þ R2 ; Ri ¼ AiCC ACi A1 ii AiC ; Z f ðxÞ/ðkx xi kÞ ; xi 2 X ; f ¼ ðf 1 ; f 2 ; f C Þtranspose ; f¼ X 1 vC ¼ f C AC1 A1 11 f 1 AC2 A22 f 2 :
4. Convergence analysis As anticipated in the previous content, typical domain decomposition preconditioned are expressed in terms of the sum of operators, each related to a certain subdomain Xi of the computational domain X Rd , d ¼ 2; 3. In this part we present some abstract theorems that are useful for proving the convergence of iteration-by-subdomain methods at RBF level. Theorem 4.1 (RBF-interpolant error estimate theorem [7]). Let X Rd be an open and bounded domain, having Lipschitz boundary. Denote by Su the interpolant on X ¼ fx1 ; x2 ; . . . ; xN g X to a function u H k ðXÞ with k > d=2. Then there exists a constant h0 > 0 such that for all X with h < h0 where h is the density of X , the estimate ku su kH j ðXÞ 6 Chkj kukH k ðXÞ is valid for 0 6 j 6 k. Theorem 4.2. Denoting by dk 6 1 the support radius of /ðx xk Þ, if the center xk is on line C Bdk ðxk Þ, then there exists constant C dependent on /ðrÞ such that kukH 2 ðBd
k
ðxk ÞÞ
6 Cd3=2 kukH 1=2 ðCÞ :
Proof. For simplicity, we do not consider the subscript k. Direct calculation shows: r 2 c 2 6 2 k/00 ðtÞkL2 ð0;1Þ ; /xx 2 d L ðBd ðxÞÞ d r 2 6 k/0 ðtÞk2L2 ð0;1Þ ; /x d L2 ðBd ðxÞÞ
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323
r 2 2 ¼ d2 kt/2 ðtÞkL2 ð0;1Þ ; / d L2 ðBd ðxÞÞ r 2 2 ¼ dk/2 ðatÞkL2 ð0;1Þ ; / d L2 ðCÞ where a is a constant dependent on the C. Hence r r 6 cd3=2 / ; / 2 d H ðBd ðxÞÞ d H 1=2 ðCÞ where c is a constant dependent on the RBF/ðrÞ. h Theorem 4.3 (RBF Uniform extension theorem). Let oX, oXi be piecewise c2 domains and locally convex near the corner. Assume that the family of CS-RBF with centers on C satisfy that its support radius are kh, k < 1. Then there exist two positive constants C1 and C2 , which depend on the relative size of X1 and X2 and k and the density h, such that C1 kgh kK 6 kHi;h gh k1;Xi 6 C2 kgh kK
8gh 2 Kh ; i ¼ 1; 2:
ð16Þ
Proof. By some choice, for example WuÕs function, we know that gh 2 H 3=2 ðCÞ and the naturally extension of gh in Xi Ri gh 2 H 2 ðXi Þ. Therefore Hi gh 2 H 2 ðXi Þ (see [13]), the following regularity estimate also holds: kHi gh k2;Xi 6 kRi gh k1;Xi ; but kHi;h gh k1;Xi 6 kHi;h gh Hi gh k1;Xi þ kHi gh k1;Xi : By using Theorems 4.1 and 4.2 and prior estimate, we can get: kHi;h gh k1;Xi 6 C2 kgh kK : The another part of this theorem is an easy Pcorollary of trace theorem. Letting Euclidean inner product about i;h as " # X k; g ¼ hSi;h k/C ; g/C i ¼ ai ðHi;h k/C ; Hi;h g/C Þ i;h
for 8k; g 2 RNC. /C is the vector consists of the RBF with centers on C. h By using of Theorem 4.3, we can easily find. Theorem 4.4. Under the assumption in Theorem 4.3, the discrete Steklov– Poincare operators Si;h are all bounded and coercive in Kh , with continuity constant and coerciveness constant dependent on RBF /ðrÞ and the density h.
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To finish our argument about the convergence of this meshless DDM based on RBF, we recall the theorem [5]: Theorem 4.5. Suppose that (a) Q2 is continuous and coercive; that is (a)1 there exists b2 > 0 such that hQ2 g; li 6 b2 kgkX klkX
8g; l 2 X ;
(a)2 there exists a2 > 0 such that 2
hQ2 g; gi P a2 kgkX
8g; l 2 X ;
(b) Q1 is continuous; that is, there exists b1 > 0 such that hQ1 g; li 6 b1 kgkX klkX
8g; l 2 X ;
(c) there exists a constant j > 0 such that 2
hQ2 g; Q1 2 Qgi P j kgkX
8g 2 X :
Then for any given k0 in X and for any h satisfying 0 < h < hmax , with j a2 k hmax ¼ b ðb þb2 Þ2 the sequence kkþ1 ¼ kk þ hQ1 2 ðG Qk Þ converges in X to the 2 1 2 solution of problem Qk ¼ G. Where, X is some Hilbert space. Remark. In the argument above, we find that the convergent rate of this method is dependent on the density of the centers. So it is not optimal. We must point out that we do not constrain ourselves in two-dimensional space in the proof of the theorem above. It is also right in high dimensional space. Following the argument above, the Neumann–Neumann iterative scheme is also convergent.
5. Numerical aspect For Eq. (1), we let f ðx; yÞ ¼ exþy ½ðx2 þ 3xÞðy 2 2yÞ þ ðy 2 þ 2y 2Þðx2 xÞ; X ¼ ½0; 1; 0; 2, the exact solution of these problem are u ¼ exþy ðx2 xÞðy 2 2yÞ: Let C : y ¼ 1. As we have mentioned before, we make two lines l1 : y ¼ 1 þ hh, l2 ¼ 1 hh to guarantee the conditions on the support of the CS-RBF. If the center is X1 , the true support is suppf/g \ fy 6 1 hhg. For the centers in X2 ,
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325
Table 1 x
du
x
du
x
du
(0.22,0.625) (0.24,0.625) (0.26,0.675) (0.28,0.625) (0.26,0.325) (0.24,0.425) (0.38,1.625) (0.52,1.625) (0.560,1.625) (0.58,1.650)
0.0272 0.0068 0.0576 0.0582 0.0321 0.0013 0.0089 0.0421 0.0330 0.0063
(0.22,0.650) (0.24,0.675) (0.26,0.725) (0.28,0.675) (0.22,0.375) (0.28,0.425) (0.40,1.625) (0.52,1.700) (0.560,1.650) (0.60,1.625)
0.0195 0.0281 0.0936 0.0824 0.0234 0.0467 0.0066 0.0674 0.0024 0.0130
(0.22,0.675) (0.24,0.725) (0.26,0.700) (0.22,0.325) (0.24,0.375) (0.22,0.600) (0.40,1.675) (0.56,1.6500) (0.58,1.625) (0.60,1.650)
0.0074 0.0643 0.0740 0.0131 0.0047 0.0013 0.0635 0.0083 0.0244 0.0179
the real support is suppf/g \ fy P 1 þ hhg. By selecting the centers ðxi ; yi Þ with xi ¼ i=10, yj ¼ j=10, i ¼ 0; 1; . . . ; 20 and the WuÕs function as our basis functions of the finite dimensional approximation of H01 ðXÞ, we obtain Table 1 (du, is the relative error). In the calculation for the problems above, we choose d is near 2.5 if the center is in Xi . Otherwise d ¼ 0:999. For accuracy, we should choose the acceleration parameter h in each iteration step to ensure the convergence or accelerate the convergent speed. The numerical result shows that it is satisfactory if the point is not near oX or C. But the result gets worse when the point be nearer to the boundary. We must point out that it is the defect of the CSRBF because that the finite approximation VN does not belong to H01 ðXÞ. To overcome this defect, we advise the following theorem [12]: Theorem 5.1. Consider the equation: Dua ¼ f in X; oua þ au ¼ 0 on oX; a on
ð17Þ
when a ! þ1 the solution ua converges to the solution u of Eq. (1). The estimate C kua ukH 1 ðXÞ 6 pffiffiffi kf kL2 ðXÞ ; a where C is a constant independent of a. By using Theorem 5.1, we will solve a Robin type problem in the Dirichlet domain X1 . Given: f1 ðx; yÞ ¼ ex ðy 2 1Þðx2 þ 4x þ 1Þ 2ex ðx2 1Þ; f2 ðx; yÞ ¼ 4ðx2 þ y 2 Þ sinðx2 1Þ sinðy 2 1Þ 2 sinðy 2 þ x2 2Þ; X ¼ ½1; 1; 1; 1, C ¼ fðx; yÞjy ¼ 0g, the exact solution is:
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Y. Duan, Y.-j. Tan / Appl. Math. Comput. 162 (2005) 317–327
Table 2 L2 error ¼ 0:0320 0.0059 0.0330 0.0315 0.0296 0.0241 0.0202 0.0176 10.0184 0.0256
0.0042 0.0330 0.0357 0.0332 0.0273 0.0193 0.0123 0.0079 0.0110
0.0137 0.0400 0.0438 0.0426 0.0352 0.0226 0.0101 0.0021 0.0043
0.0333 0.0527 0.0525 0.0611 0.0570 0.0324 0.0054 0.0089 0.0093
0.0185 0.0504 0.0451 0.0843 0.0749 0.0447 0.0146 0.0415 0.0478
0.0059 0.0330 0.0315 0.0296 0.0241 0.0202 0.0176 0.0184 0.0256
0.0042 0.0330 0.0357 0.0332 0.0273 0.0193 0.0123 0.0079 0.0110
0.0137 0.0400 0.0438 0.0426 0.0352 0.0226 0.0101 0.0021 0.0043
0.0333 0.0527 0.0525 0.0611 0.0570 0.0324 0.0054 0.0089 0.0093
0.0124 0.0170 0.0228 0.0255 0.0254 0.0210 0.0144 0.0056 0.0264
0.0480 0.0197 0.0256 0.0299 0.0331 0.0222 0.0114 0.0002 0.0740
0.0045 0.0128 0.0124 0.0123 0.0108 0.0111 0.0100 0.0098 0.0008
0.0076 0.0071 0.0116 0.0132 0.0131 0.0115 0.0084 0.0029 0.0126
0.0091 0.0087 0.0149 0.0173 0.0171 0.0146 0.0099 0.0021 0.0170
0.0124 0.0170 0.0228 0.0255 0.0254 0.0210 0.0144 0.0056 0.0264
Table 3 L2 error ¼ 0:0203 0.0045 0.0128 0.0124 0.0123 0.0108 0.0111 0.0100 0.0098 0.0008
0.0076 0.0071 0.0116 0.0132 0.0131 0.0115 0.0084 0.0029 0.0126
0.0091 0.0087 0.0149 0.0173 0.0171 0.0146 0.0099 0.0021 0.0170
u1 ðx; yÞ ¼ ðx2 1Þðy 2 1Þex ; u2 ðx; yÞ ¼ sinðx2 1Þ sinðy 2 1Þ; respectively. Taken the centers as xi ¼ 1 þ i1 0:2, yj ¼ 1 þ j1 0:2, 10 10 i; j ¼ 1; . . . ; 11. Tables 2 and 3, are the corresponding relative error (at the centers in X, i.e. i ¼ 2; . . . ; 10, j ¼ 2; . . . ; 10) in solving the Eq. (1) with f1 ðx; yÞ, f2 ðx; yÞ as the right side function by using TPS-RBF. L2 -error is the relative error of L2 -norm.
6. Conclusion Some others RBF are advised, for example, the Sobolev spline or multiquadric function. For multi-quadric function, the numerical result is very sensitive to the control parameter c. Considering the charming perspective of meshless method, we should spend concentration on this method. In the examples above, the resource of error also include the large condition number of the stiffness matrix and the numerical quadratures, which may be relative
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 162 (2005) 317–327
327
with the density h. At last we must point out that the extension of this method to many subdomains is direct.
References [1] P.E. Bjørstad, O.B. widlund, Iterative methods for the solution of elliptic problems on regions partitioned into substructures, SIAM J. Numer. Anal. 23 (1986) 1097–1120. [2] J.H. Bramble, J.E. Pasciak, A.H. Schatz, An iterative method for elliptic problems on regions partitioned into substructures, Math. Comput. 46 (1986) 361–369. [3] C. Bernardi, Y. Maday, A.T. Petera, A new nonconforming approach to domain decomposition: the mortar element method, in: Nonlinear Partial Differential Equations and Their Applications, in: H. Brezis, J.-L. Lions (Eds.), College de France Seminar, vol. XI, Longman, Harlow, 1994, pp. 13–51. [4] T.F. Chan, T.P. Mathew, Domain decomposition algorithms, Acta. Numer. (1994) 143–161. [5] A. Quarteroni, A. Valli, Domain Decomposition Method for Partial Differential Equation, Oxford University Press, England, 1999. [6] T. Belytschko, Y. Krongauze, D. Organ, M. Fleming, P. Krysl, Meshless methods: an overview and recent developments, Comput. Methods Appl. Mech. Eng. 139 (1996) 3–47, special issue on Meshless Methods. [7] H. Wendland, Meshless Galerkin methods using radial basis functions, Math. Comput. 68 (1999) 1521–1531. [8] H. Wendland, Sobolev-type error estimates for interpolation by radial basis functions, in: A. LeMehaute, C. Rabut, L.L. Schumaker (Eds.), Surface Fitting and Multiresolution Methods, Vanderbilt University Press, Nashville, 1997, pp. 337–344. [9] H. Wendland, Error estimates for interpolation by compactly supported radial basis functions of minimal degree, J. Approx. Theory 93 (1998) 258–272. [10] Z. Wu, R. Schaback, Local error estimates for radial basis function interpolation of scattered data, IMA J. Numer. Anal. (13) (1993) 13–27. [11] Z. Wu, Compactly supported positive definite radial functions, Adv. Comput. Math. 4 (1995) 283–292. [12] Zhi-jie Cai, Error estimate for elementfree Galerkin methods, preprint, 2003. [13] Ph.G. Ciarlet, J.L. Lions (Eds.), Basic Error Estimates for Elliptic Problems, Handbook of Numerical Analysis, vol. II. Finite Element Methods (part1), Elsevier, 1991.
Meshless Galerkin method based on regions partitioned into subdomains Yong Duan *, Yong-ji Tan Institute of Mathematics, Fudan University, Shanghai 200433, People’s Republic of China
Abstract In this paper, we will combine the domain decomposition method with so-called meshless Galerkin method for PDE using RBF. The convergence of this new method and numerical examples are given. Ó 2004 Elsevier Inc. All rights reserved. Keywords: Meshless; RBF; Domain decomposition; Parallel
1. Introduction The partition of elliptic problem into subproblems is a very nature idea and is very much in use in practices. Many peoples had done some research about this method, for example, [1–4]. Also are there some books about this method, see, for example, [5]. We must point out that the methods before are based on FEM. The FEM has to spend a lot of time on technical details concerning the mesh. Since 1990Õs last century, interpolation by radial basis functions (RBF) has become a powerful tool in multivariate approximation theory, especially since compactly supported radial basis functions (CS-RBF) are available. Since, in contrast to FEM, the construction of the finite dimensional subspace using RBF is independent of the current space dimension, it is possible to solve high dimensional problems as they occur in quantum mechanics and so on. Meshless methods do not need to handle such problems like the meshÕs
*
Corresponding author. E-mail address: [email protected] (Y. Duan).
0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.12.138
318
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 162 (2005) 317–327
generation. See [6] for an overview of general meshless methods and applications in engineering. In this paper, we do not give the details of the mathematical foundations of domain decomposition method (DDM). We refer the reader [5] for details. Which we will mainly concentrate on is the meshless numerical realization in solving PDE. About the RBF, we refer [7–11] to reader.
2. The mathematical foundation of DDM We consider: Du ¼ f u¼0
in X; on oX;
ð1Þ
where X is d-dimensional domain, with a Lipschitz boundary oX, whose outer unit P normal direction is denoted by n , f is a given function of L2 ðXÞ, d D ¼ j¼1 Dj Dj is the Laplace operator and Dj denotes the partial derivative with respect to xj , j ¼ 1; . . . ; d. Firstly, we assume that X is partitioned into two non-overlapping subdomains X1 and X2 , and denote by C ¼ X1 \ X2 . We indicate by ui the restriction to Xi , i ¼ 1; 2, of the solution u to (1.1), and ni by the normal direction on oXi \ C, oriented outward. Set n ¼ n1 . The Poisson problem (1) can be reformulated in the equivalent multidomain form: 8 Du1 ¼ f in X1 ; > > > > u on oX1 \ oX; > 1 ¼ 0 > < u1 ¼ u2 on C; ð2Þ ou2 1 ¼ ou on C; > > on > on > u ¼0 on oX2 \ oX; > > : 2 Du2 ¼ f in X2 : The third and fourth terms of Eq. (2) are the transmission condition for u1 and u2 on C. Let k denote the unknown value of u on C. We consider the two Dirichlet problems: 8 < DHi k ¼ 0 in Xi ; Hk¼0 on oXi \ oX; ð3Þ : i Hi k ¼ k on C and 8 < Dui ¼ f u ¼ 0 : i ui ¼ 0
in Xi ; on oXi \ oX; on C
ð4Þ
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319
for i ¼ 1; 2. Hi k is the harmonic extension of k into Xi . Then k satisfies the Steklov–Poincare interface equation Sk ¼ v
ð5Þ
on C;
where v¼
ou2 ou1 : on on
S is the Steklov–Poincare operator, Sg ¼
oH1 g oH2 g : on on
We introduce the following function spaces: Vi ¼ fvi 2 H 1 ðXi ÞjvijoX\oXi ¼0 g; Vi 0 ¼ H01 ðXi Þ. K ¼ fg 2 H 1=2 ðCÞjg ¼ vjC for a suitable v 2 V g. We can prove that the Poisson problem can be equivalently reformulated as: find ui 2 Vi such that 8 8v1 2 V10 ; a1 ðu1 ; v1 Þ ¼ ðf ; v1 ÞX1 > > < u1 ¼ u2 on C; ð6Þ a ðu ; v Þ ¼ ðf ; v Þ 8v2 2 V20 ; > 2 2 2 2 X2 > : a2 ðu2 ; R2 lÞ ¼ ðf ; R2 lÞX2 þ ðf ; R1 lÞX1 a1 ðu1 ; R1 lÞ 8l 2 K; where ai ðwi ; vi Þ ¼ ðrwi ; rvi ÞXi ¼
Z
rwi rvi ;
ð7Þ
Xi
Ri denotes any possible extension operator from K to Vi . The Dirichlet–Neumann iterative scheme reads: given k0 , solve for each k P 0: 8 ¼ f in X1 ; < Dukþ1 1 ð8Þ ukþ1 ¼ 0 on oX1 \ oX; : 1kþ1 u1 ¼ kk on C; then 8 kþ1 > < Du2 ¼ f kþ1 u2 ¼ 0 > oukþ1 : oukþ1 2 ¼ 1 on
on
in X2 ; on oX2 \ oX;
ð9Þ
on C;
with k kkþ1 ¼ hukþ1 2jC þ ð1 hÞk ;
ð10Þ
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h being a positive acceleration parameter. This method is equivalent to the socalled Richardson procedure for the Steklov–Poincare Eq. (5) with the operator S2 as a preconditioner, i.e., kkþ1 ¼ kk þ hS21 ðSkk þ vÞ:
ð11Þ
We must point out that you may replace ukþ1 by uk1 at the second step for D–N 1 iteration in view of parallel implementation. 3. Domain decomposition method at RBF level In this part, we give the presentation of RBF approximation theory for the domain decomposition method. We only concentrate on the CS-RBF and TPSRBF (thin plate spline). One of the most popular classes of CS-RBFs is the one introduced by Wendland [7]. Another is given by Wu and Schaback [10]. These functions are strictly positive definite in Rd for all d less than or equal to some fixed value d0 , and can be constructed to have any desired amount of smoothness 2j. For l ¼ ½d2 þ j þ 1 and j ¼ 0; 1; 2; 3. TPS-RBF reads: /ðrÞ ¼ r2m logðrÞ; m ¼ 1; 2; . . . WendlaudÕs function reads: : /l;0 ðrÞ¼ð1 rÞlþ ; : lþ1 /l;1 ðrÞ¼ð1 rÞþ ½ðl þ 1Þr þ 1; : 2 2 /l;2 ðrÞ¼ð1 rÞlþ2 þ ½ðl þ 4l þ 3Þr þ ð3l þ 6Þr þ 3; : lþ3 /l;3 ðrÞ¼ð1 rÞþ ½ðl3 þ 9l2 þ 32l þ 15Þr3 þ ð6l2 þ 36l þ 45Þr2 þ ð15l þ 45Þr þ 15: : Here ¼ denotes equality up to a constant factor. For example, the functions : 2 /2;0 ¼ð1 rÞþ ; : /3;1 ðrÞ¼ð1 rÞ4þ ½4r þ 1; : 6 /4;2 ðrÞ¼ð1 rÞþ ½35r2 þ 18r þ 3; : 8 /5;3 ðrÞ¼ð1 rÞþ ½32r3 þ 25r2 þ 8r þ 1; which are in C0 , C2 , C4 , C6 , respectively, and strictly positive definite in R3 . WuÕs CS-RBF reads: 6
/ðrÞ ¼ ð1 rÞþ ð5r5 þ 30r4 þ 72r3 þ 82r2 þ 36r þ 6Þ; which is in C6 . Here and throughout r ¼ kxk is Euclidean norm, with x 2 Rd . In order to solve the problem (1), we define the finite dimensional space VN ¼ spanf/ðkx xj kÞg;
j ¼ 1; 2; . . . ; N :
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321
Denoting by N1 , N2 , NC the corresponding number of knots in X1 , X2 , C. Suppose that suppf/ðkx xi kÞg \ suppf/ðkx xj kÞg ¼ ; if xi , xj are not in the same subdomain X1 or X2 and suppf/ðkx xi kÞg \ C ¼ ; for 8xj 2 X1 or X2 . Denoting by h ¼ supx2X minxi 2X kx xi k, the density of the distinct centers X ¼ fx1 ; x2 ; . . . ; xN g. We use subscript h to substitute the corresponding form of each operator after that we had finished the CS-RBF approximation. The weak form of Dirichlet–Neumann iterative substructuring method is: let k0C 2 RNC be the initial guess of the coefficients of the CS-RBFs, which center are on C. For k P 0 solve: 8 kþ1 > < u1;h 2 V1;h ; 0 8v1;h 2 V1;h ; a1 ðukþ1 ð12Þ 1;h ; v1;h Þ ¼ ðf ; v1;h ÞX1 P > N k : ukþ1 ¼ C on C; 1;h k¼1 kh /ðx xk Þ kþ1 ukþ1 2;h 2 V2;h : a2 ðu2;h ; v2;h Þ
¼ ðf ; v2;h ÞX2 þ ðf ; R1;h v2;hjC ÞX1 a1 ðukþ1 1;h ; R1;h v2;hjC Þ
8v2;h 2 V2;h ; ð13Þ
then update NC X
kþ1 kkþ1 h /ðx xk Þ ¼ hu2;hjC þ ð1 hÞ
k¼1
NC X
kkh /ðx xk Þ on C:
ð14Þ
k¼1
Assuming that ukþ1 1;h ¼ ukþ1 2;h ¼
N1 X
kkþ1 1;i /ðkx xi kÞ þ
NC X
i¼1
i¼1
N2 X
NC X
kkþ1 2;i /ðkx xi kÞ þ
i¼1
kkþ1 1 ,
kkC;i /ðkx xi kÞ; kkþ1 2;C;i /ðkx xi kÞ;
i¼1
kþ1 kþ1 denoting by the column vector consisted of kkþ1 1;i , k2;i , kC;i , kþ1 k2;C;i , respectively. By solving the Eqs. (12)–(14), we get the algebraic form for the Dirichlet–Neumann iterative procedure on the Stekelov–Poincare interface Eq. (11). ! 1 X X k kþ1 kC ¼ h vC kC þ ð1 hÞkkC ; ð15Þ 2
kkþ1 2 ,
kkþ1 C ,
kkþ1 2;C
1
where A2CC ¼ ðð/ðkx xi kÞ; /ðkx xj kÞÞX2 Þ;
xi ; xj
on C;
A1CC ¼ ðð/ðkx xi kÞ; /ðkx xj kÞÞX1 Þ;
xi ; xj
on C;
A11 ¼ a1 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi ; xj 2 X 1 ;
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A22 ¼ a2 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi ; xj 2 X 2 ;
A1C ¼ a1 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi 2 X1 ; xj on C;
A2C ¼ a2 ð/ðkx xi kÞ; /ðkx xj kÞÞ;
xi 2 X2 ; xj on C;
AC2 ¼ transposeðA2C Þ AC1 ¼ transposeðA1C Þ; i ¼ 1; 2; R ¼ R1 þ R2 ; Ri ¼ AiCC ACi A1 ii AiC ; Z f ðxÞ/ðkx xi kÞ ; xi 2 X ; f ¼ ðf 1 ; f 2 ; f C Þtranspose ; f¼ X 1 vC ¼ f C AC1 A1 11 f 1 AC2 A22 f 2 :
4. Convergence analysis As anticipated in the previous content, typical domain decomposition preconditioned are expressed in terms of the sum of operators, each related to a certain subdomain Xi of the computational domain X Rd , d ¼ 2; 3. In this part we present some abstract theorems that are useful for proving the convergence of iteration-by-subdomain methods at RBF level. Theorem 4.1 (RBF-interpolant error estimate theorem [7]). Let X Rd be an open and bounded domain, having Lipschitz boundary. Denote by Su the interpolant on X ¼ fx1 ; x2 ; . . . ; xN g X to a function u H k ðXÞ with k > d=2. Then there exists a constant h0 > 0 such that for all X with h < h0 where h is the density of X , the estimate ku su kH j ðXÞ 6 Chkj kukH k ðXÞ is valid for 0 6 j 6 k. Theorem 4.2. Denoting by dk 6 1 the support radius of /ðx xk Þ, if the center xk is on line C Bdk ðxk Þ, then there exists constant C dependent on /ðrÞ such that kukH 2 ðBd
k
ðxk ÞÞ
6 Cd3=2 kukH 1=2 ðCÞ :
Proof. For simplicity, we do not consider the subscript k. Direct calculation shows: r 2 c 2 6 2 k/00 ðtÞkL2 ð0;1Þ ; /xx 2 d L ðBd ðxÞÞ d r 2 6 k/0 ðtÞk2L2 ð0;1Þ ; /x d L2 ðBd ðxÞÞ
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323
r 2 2 ¼ d2 kt/2 ðtÞkL2 ð0;1Þ ; / d L2 ðBd ðxÞÞ r 2 2 ¼ dk/2 ðatÞkL2 ð0;1Þ ; / d L2 ðCÞ where a is a constant dependent on the C. Hence r r 6 cd3=2 / ; / 2 d H ðBd ðxÞÞ d H 1=2 ðCÞ where c is a constant dependent on the RBF/ðrÞ. h Theorem 4.3 (RBF Uniform extension theorem). Let oX, oXi be piecewise c2 domains and locally convex near the corner. Assume that the family of CS-RBF with centers on C satisfy that its support radius are kh, k < 1. Then there exist two positive constants C1 and C2 , which depend on the relative size of X1 and X2 and k and the density h, such that C1 kgh kK 6 kHi;h gh k1;Xi 6 C2 kgh kK
8gh 2 Kh ; i ¼ 1; 2:
ð16Þ
Proof. By some choice, for example WuÕs function, we know that gh 2 H 3=2 ðCÞ and the naturally extension of gh in Xi Ri gh 2 H 2 ðXi Þ. Therefore Hi gh 2 H 2 ðXi Þ (see [13]), the following regularity estimate also holds: kHi gh k2;Xi 6 kRi gh k1;Xi ; but kHi;h gh k1;Xi 6 kHi;h gh Hi gh k1;Xi þ kHi gh k1;Xi : By using Theorems 4.1 and 4.2 and prior estimate, we can get: kHi;h gh k1;Xi 6 C2 kgh kK : The another part of this theorem is an easy Pcorollary of trace theorem. Letting Euclidean inner product about i;h as " # X k; g ¼ hSi;h k/C ; g/C i ¼ ai ðHi;h k/C ; Hi;h g/C Þ i;h
for 8k; g 2 RNC. /C is the vector consists of the RBF with centers on C. h By using of Theorem 4.3, we can easily find. Theorem 4.4. Under the assumption in Theorem 4.3, the discrete Steklov– Poincare operators Si;h are all bounded and coercive in Kh , with continuity constant and coerciveness constant dependent on RBF /ðrÞ and the density h.
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To finish our argument about the convergence of this meshless DDM based on RBF, we recall the theorem [5]: Theorem 4.5. Suppose that (a) Q2 is continuous and coercive; that is (a)1 there exists b2 > 0 such that hQ2 g; li 6 b2 kgkX klkX
8g; l 2 X ;
(a)2 there exists a2 > 0 such that 2
hQ2 g; gi P a2 kgkX
8g; l 2 X ;
(b) Q1 is continuous; that is, there exists b1 > 0 such that hQ1 g; li 6 b1 kgkX klkX
8g; l 2 X ;
(c) there exists a constant j > 0 such that 2
hQ2 g; Q1 2 Qgi P j kgkX
8g 2 X :
Then for any given k0 in X and for any h satisfying 0 < h < hmax , with j a2 k hmax ¼ b ðb þb2 Þ2 the sequence kkþ1 ¼ kk þ hQ1 2 ðG Qk Þ converges in X to the 2 1 2 solution of problem Qk ¼ G. Where, X is some Hilbert space. Remark. In the argument above, we find that the convergent rate of this method is dependent on the density of the centers. So it is not optimal. We must point out that we do not constrain ourselves in two-dimensional space in the proof of the theorem above. It is also right in high dimensional space. Following the argument above, the Neumann–Neumann iterative scheme is also convergent.
5. Numerical aspect For Eq. (1), we let f ðx; yÞ ¼ exþy ½ðx2 þ 3xÞðy 2 2yÞ þ ðy 2 þ 2y 2Þðx2 xÞ; X ¼ ½0; 1; 0; 2, the exact solution of these problem are u ¼ exþy ðx2 xÞðy 2 2yÞ: Let C : y ¼ 1. As we have mentioned before, we make two lines l1 : y ¼ 1 þ hh, l2 ¼ 1 hh to guarantee the conditions on the support of the CS-RBF. If the center is X1 , the true support is suppf/g \ fy 6 1 hhg. For the centers in X2 ,
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325
Table 1 x
du
x
du
x
du
(0.22,0.625) (0.24,0.625) (0.26,0.675) (0.28,0.625) (0.26,0.325) (0.24,0.425) (0.38,1.625) (0.52,1.625) (0.560,1.625) (0.58,1.650)
0.0272 0.0068 0.0576 0.0582 0.0321 0.0013 0.0089 0.0421 0.0330 0.0063
(0.22,0.650) (0.24,0.675) (0.26,0.725) (0.28,0.675) (0.22,0.375) (0.28,0.425) (0.40,1.625) (0.52,1.700) (0.560,1.650) (0.60,1.625)
0.0195 0.0281 0.0936 0.0824 0.0234 0.0467 0.0066 0.0674 0.0024 0.0130
(0.22,0.675) (0.24,0.725) (0.26,0.700) (0.22,0.325) (0.24,0.375) (0.22,0.600) (0.40,1.675) (0.56,1.6500) (0.58,1.625) (0.60,1.650)
0.0074 0.0643 0.0740 0.0131 0.0047 0.0013 0.0635 0.0083 0.0244 0.0179
the real support is suppf/g \ fy P 1 þ hhg. By selecting the centers ðxi ; yi Þ with xi ¼ i=10, yj ¼ j=10, i ¼ 0; 1; . . . ; 20 and the WuÕs function as our basis functions of the finite dimensional approximation of H01 ðXÞ, we obtain Table 1 (du, is the relative error). In the calculation for the problems above, we choose d is near 2.5 if the center is in Xi . Otherwise d ¼ 0:999. For accuracy, we should choose the acceleration parameter h in each iteration step to ensure the convergence or accelerate the convergent speed. The numerical result shows that it is satisfactory if the point is not near oX or C. But the result gets worse when the point be nearer to the boundary. We must point out that it is the defect of the CSRBF because that the finite approximation VN does not belong to H01 ðXÞ. To overcome this defect, we advise the following theorem [12]: Theorem 5.1. Consider the equation: Dua ¼ f in X; oua þ au ¼ 0 on oX; a on
ð17Þ
when a ! þ1 the solution ua converges to the solution u of Eq. (1). The estimate C kua ukH 1 ðXÞ 6 pffiffiffi kf kL2 ðXÞ ; a where C is a constant independent of a. By using Theorem 5.1, we will solve a Robin type problem in the Dirichlet domain X1 . Given: f1 ðx; yÞ ¼ ex ðy 2 1Þðx2 þ 4x þ 1Þ 2ex ðx2 1Þ; f2 ðx; yÞ ¼ 4ðx2 þ y 2 Þ sinðx2 1Þ sinðy 2 1Þ 2 sinðy 2 þ x2 2Þ; X ¼ ½1; 1; 1; 1, C ¼ fðx; yÞjy ¼ 0g, the exact solution is:
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Table 2 L2 error ¼ 0:0320 0.0059 0.0330 0.0315 0.0296 0.0241 0.0202 0.0176 10.0184 0.0256
0.0042 0.0330 0.0357 0.0332 0.0273 0.0193 0.0123 0.0079 0.0110
0.0137 0.0400 0.0438 0.0426 0.0352 0.0226 0.0101 0.0021 0.0043
0.0333 0.0527 0.0525 0.0611 0.0570 0.0324 0.0054 0.0089 0.0093
0.0185 0.0504 0.0451 0.0843 0.0749 0.0447 0.0146 0.0415 0.0478
0.0059 0.0330 0.0315 0.0296 0.0241 0.0202 0.0176 0.0184 0.0256
0.0042 0.0330 0.0357 0.0332 0.0273 0.0193 0.0123 0.0079 0.0110
0.0137 0.0400 0.0438 0.0426 0.0352 0.0226 0.0101 0.0021 0.0043
0.0333 0.0527 0.0525 0.0611 0.0570 0.0324 0.0054 0.0089 0.0093
0.0124 0.0170 0.0228 0.0255 0.0254 0.0210 0.0144 0.0056 0.0264
0.0480 0.0197 0.0256 0.0299 0.0331 0.0222 0.0114 0.0002 0.0740
0.0045 0.0128 0.0124 0.0123 0.0108 0.0111 0.0100 0.0098 0.0008
0.0076 0.0071 0.0116 0.0132 0.0131 0.0115 0.0084 0.0029 0.0126
0.0091 0.0087 0.0149 0.0173 0.0171 0.0146 0.0099 0.0021 0.0170
0.0124 0.0170 0.0228 0.0255 0.0254 0.0210 0.0144 0.0056 0.0264
Table 3 L2 error ¼ 0:0203 0.0045 0.0128 0.0124 0.0123 0.0108 0.0111 0.0100 0.0098 0.0008
0.0076 0.0071 0.0116 0.0132 0.0131 0.0115 0.0084 0.0029 0.0126
0.0091 0.0087 0.0149 0.0173 0.0171 0.0146 0.0099 0.0021 0.0170
u1 ðx; yÞ ¼ ðx2 1Þðy 2 1Þex ; u2 ðx; yÞ ¼ sinðx2 1Þ sinðy 2 1Þ; respectively. Taken the centers as xi ¼ 1 þ i1 0:2, yj ¼ 1 þ j1 0:2, 10 10 i; j ¼ 1; . . . ; 11. Tables 2 and 3, are the corresponding relative error (at the centers in X, i.e. i ¼ 2; . . . ; 10, j ¼ 2; . . . ; 10) in solving the Eq. (1) with f1 ðx; yÞ, f2 ðx; yÞ as the right side function by using TPS-RBF. L2 -error is the relative error of L2 -norm.
6. Conclusion Some others RBF are advised, for example, the Sobolev spline or multiquadric function. For multi-quadric function, the numerical result is very sensitive to the control parameter c. Considering the charming perspective of meshless method, we should spend concentration on this method. In the examples above, the resource of error also include the large condition number of the stiffness matrix and the numerical quadratures, which may be relative
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327
with the density h. At last we must point out that the extension of this method to many subdomains is direct.
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