Applied Mathematics and Computation 166 (2005) 373–384
www.elsevier.com/locate/amc
Meshless collocation method based onDirichlet–Neumann substructure iteration
Yong Duan *, Yong-ji Tan
Institute of mathematics, Fudan University, Shanghai 200433, PR China
Abstract
In this paper, we will combine the Dirichlet–Neumann substructure iteration with so-
called meshless collocation method for PDE using RBF. The convergence of this
method and numerical examples are given to show the efficiency.
� 2004 Elsevier Inc. All rights reserved.
Keywords: Meshless; RBF; Domain decomposition; Parallel
1. Introduction
Since 1990s 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 for its �stiff�matrix being sparse, which can remarkably decrease the numerical complexity.Both Galerkin method and collocation are considered by some authors. Galer-
kin method has some difficulty to be applied. The theoretical analysis of this
method is given by Wendland in [1], the improvement of this method is given
0096-3003/$ - see front matter � 2004 Elsevier Inc. All rights reserved.
doi:10.1016/j.amc.2004.04.094
* Corresponding author.
E-mail address: [email protected] (Y. Duan).
374 Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384
by Cai in [2] for the Laplace equation with essential boundary condition. There
are also some attempt to solve PDE based on regions partitioned into subdo-
main using meshless Galerkin method [3]. Comparing with the Galerkin
method, the collocation method using RBF is excited in application. Many
numerical examples have shown that this method is very efficient, see, for
example, [4–7]. In views of the theory about this method, Schaback has gotthe error bound and convergent order in [8–10]. The feasibility of this method
has also been proven in [8,9].
Sometimes, when the problem is too �big�, there is a nature consideration to
change the origin problem into subproblems, named domain decomposition
method (DDM). Many authors had done some research about this method,
for example, Bjørstad [11], Bramble [12], Bernardi [13], Quarteroni [14]. The
DDM before are based on FEM. The FEM has to spend a lot of time on tech-
nical details concerning the mesh. Since, in contrast with FEM, the construc-tion 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 don�t need
to handle such problems like the mesh�s generation. This charming behavior
will dramatically decrease the computational time.
In this paper, the algorithm called meshless collocation based on Dirichlet–
Neumann iteration will be discussed. In Section 2, a brief script of the mathe-
matical foundations of DDM will is given. We refer the reader to [14] fordetails. We will mainly concentrate on the meshless (exactly, meshless colloca-
tion using RBF) numerical realization in solving PDE based on the computa-
tional domain partitioned into 2-subdomains. The convergence analysis of this
method is given in Section 3. Two numerical examples will be given in Section 4
to test this method.
2. Meshless collocation based on Dirichlet–Neumann iteration
2.1. Dirichlet–Neumann iteration
Consider:
Du ¼ f in X;
u ¼ 0 on oX;
�ð1Þ
where X is a d-dimensional domain, with a Lipschitz boundary oX, whoseouter unit normal direction is denoted by n*, f is a given function of L2(X),D ¼
Pdj¼1DjDj is the Laplace operator and Dj denotes the partial derivative
with respect to xj, j = 1, . . .,d. Firstly, we assume that X is partitioned into
two non-overlapping subdomains X1 and X2, and denote by C ¼ X1 \ X2.
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384 375
We indicate by ui the restriction to Xi, i = 1,2 of the solution u to (1), and ni bythe normal direction on oXi \ C, oriented outward. Set n = n1.
The Poisson�s problem (1) can be reformulated in the equivalent multido-
main form:
Du1 ¼ f in X1;
u1 ¼ 0 on oX1 \ oX;
u1 ¼ u2 on C;ou2on ¼ ou1
on on C;
u2 ¼ 0 on oX2 \ oX;
Du2 ¼ f in X2:
8>>>>>>>><>>>>>>>>:ð2Þ
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:
DHik ¼ 0 in Xi;
Hik ¼ 0 on oXi \ oX;
Hik ¼ k on C
8><>: ð3Þ
and
Du�i ¼ f in Xi;
u�i ¼ 0 on oXi \ oX;
u�i ¼ 0 on C
8><>: ð4Þ
for i = 1,2. Hik is the harmonic extension of k into Xi. Then k satisfies the Stek-lov–Poincare interface equation
Sk ¼ v on C; ð5Þ
where v ¼ ou�2
on �ou�
1
on . S is the Steklov–Poincare operator,
Sg ¼ oH1gon
� oH2gon
� Sig ¼ oHigon
8g 2 K;
where K is the trace space of H1(X) on C.Dirichlet–Neumann iterative scheme reads: given k0, solve for each kP 0:
Dukþ11 ¼ f in X1;
ukþ11 ¼ 0 on oX1 \ oX;
ukþ11 ¼ kk on C;
8><>: ð6Þ
then
376 Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384
Dukþ12 ¼ f in X2;
ukþ12 ¼ 0 on oX2 \ oX;
oukþ12
on ¼ oukþ11
on on C;
8><>: ð7Þ
with
kkþ1 ¼ hukþ12jC þ ð1� hÞkk; ð8Þ
h being a positive acceleration parameter. This method is equivalent to the so
called Richardson procedure for the Steklov–Poincare equation (5) with the
operator S2 as a preconditioner, i,e,
kkþ1 ¼ kk þ hS�12 ð�Skk þ vÞ: ð9Þ
It is worthy being pointed out that we may replace ukþ11 by uk1 at the second
step for D–N iteration in views of parallel implementation.
2.2. Algorithm based on meshless collocation using RBF
In this part, we give the presentation of RBF approximation theory for the
domain decomposition method. There are many kinds of RBF available, for
example:
Gaussians
exp(�ck Æ k2) c > 0Multiquadrics
(c2 + k Æ k2)b b 2 N, c50 Thin-plate splines k Æ kblogk Æ k b 2 2NThe RBF introduced above are global RBF. One of the most popular classes
of CS-RBFs is the one introduced by Wendland [15]. Another is given by Wu
[16]. 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 d2
� �þ jþ 1 and j = 0,1,2,3.
Wendland�s function reads:
/l;0ðrÞ¼: ð1� rÞlþ;
/l;1ðrÞ¼: ð1� rÞlþ1
þ ½ðlþ 1Þr þ 1�;/l;2ðrÞ¼
: ð1� rÞlþ2
þ ½ðl2 þ 4lþ 3Þr2 þ ð3lþ 6Þr þ 3�;/l;3ðrÞ¼
: ð1� rÞlþ3
þ ½ðl3 þ 9l2 þ 32lþ 15Þr3 þ ð6l2 þ 36lþ 45Þr2
þ ð15lþ 45Þr þ 15�:
Here and throughout, G denotes equality up to a constant factor. For exam-
ple, the functions
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384 377
/2;0 ¼: ð1� rÞ2þ;
/3;1ðrÞ¼: ð1� rÞ4þ½4r þ 1�;
/4;2ðrÞ¼: ð1� rÞ6þ½35r2 þ 18r þ 3�;
/5;3ðrÞ¼: ð1� rÞ8þ½32r3 þ 25r2 þ 8r þ 1�;
which are in C0, C2, C4, C6, respectively, and strictly positive definite in R3.
Wu�s CS-RBF reads:
/ðrÞ ¼ ð1� rÞ6þð5r5 þ 30r4 þ 72r3 þ 82r2 þ 36r þ 6Þ;
which is in C6. Here and throughout r = kxk is Euclidean norm, with x 2 Rd. In
order to solve the problem (1), we define the finite dimensional space for a
given discrete centers X � X
V N ¼ spanfð/ðkx� xjkÞg j ¼ 1; 2; . . . ;N :
Denoting by Xi, Xoi, XC the centers in Xi, oXi \ oX, C, by Ni, Noi
, NC the num-
ber of Xi, Xoi, XC by h = supx2Xminxi2Xkx�xik the density of the distinct cen-
ters X = {x1,x2, . . .,xN}. We use subscript h to substitute the correspondingform of each operator after that we had finished the CS-RBF approximation.
Noted that, for the split problem (2), we just need the centers in Xi to solve the
Dirichlet problem in X1 and the Neumann problem in X2, then the discrete
form of Eq. (2) reads:
Du1;hðX 1Þ ¼ f ðX 1Þ;
u1;hðX @1Þ ¼ 0;
u1;hðX CÞ ¼ u2;hðX CÞ;ou2;hon ðX CÞ ¼ ou1;h
on ðX CÞ
u2;hðX @2Þ ¼ 0;
Du2;hðX 2Þ ¼ f ðX 2Þ:
8>>>>>>>>>>><>>>>>>>>>>>:ð10Þ
The discrete form of Eq. (3) reads:
DHi;hkðX 1Þ ¼ 0;
Hi;hkðX @1Þ ¼ 0;
Hi;hkðX CÞ ¼ k:
8>><>>: ð11Þ
The corresponding Dirichlet–Neumann iterative method is:
Let k0C 2 RNC be the initial guess of the value of the u2 on XC. For kP 0
solve:
378 Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384
ukþ11;h 2 V 1;h;
Dukþ11;h ðX 1Þ ¼ f ðX 1Þ;
ukþ11;h ðX @1Þ ¼ 0;
ukþ11;h ðX CÞ ¼ kkC;
8>>>>>><>>>>>>:ð12Þ
ukþ12;h 2 V 2;h;
Dukþ12;h ðX 2Þ ¼ f ðX 2Þ;
ukþ12;h ðX @2Þ ¼ 0;
oukþ12;h
on ðX CÞ ¼oukþ1
1;h
on ðX CÞ;
8>>>>>><>>>>>>:ð13Þ
then update
kkþ1 ¼ hukþ12;h ðX CÞ þ ð1� hÞkk on C: ð14Þ
The corresponding Steklov–Poincare operator is
Si;hk ¼ oHi;hkon
;
hSi;hk; gi ¼ZXi
rHi;hkrHi;hg 8k; g 2 RNC :
The discrete form of the preconditioned Richardson iteration reads:
kkþ1 ¼ kk þ hS�12;hð�Shk
k þ vÞ:
3. Convergence analysis
As anticipated in the previous content, typical domain decomposition pre-
conditioners are expressed in terms of the sum of operators, each related to
a certain subdomain Xi of the computational domain X � Rd, In this section
we will prove that the iterative solution of (12)–(14) converge to the solution
of Eq. (10).
Firstly, we introduce some notation. For a given RBF {/i} [ {wi}, where /i,wi are the basis function in X1, X2, which are the same ones when the center is
on C, denoted by g1 the vector ðo/ion ðX CÞÞ, g2 the vector ðowi
on ðX CÞÞ. The matrix
A1 ¼ ðD/iðX 1ÞÞ; A2 ¼ ð/iðX @1Þ; A3 ¼ ð/iðX CÞ;B1 ¼ ðDwiðX 2ÞÞ; B2 ¼ ðwiðX @2Þ; B3 ¼ ðwiðX CÞ:
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384 379
Then matrix A and B is nonsingular according to Schaback [8,9]. Having set
A ¼ ðA1;A2;A3ÞT; B ¼ ðB1;B2;B3ÞT;where T is the transpose of a matrix.
Theorem 3.1. For given discrete centers and basis function, there exists
constants C1, C2, which may dependent on the density h and the basis functions,
such that:
C1kgk6 kHi;hgkH1ðXiÞ 6C2kgk 8g 2 RNC i ¼ 1; 2: ð15Þ
Proof. For i = 1 and a given g 2 RNC , we can get
AC ¼ ~g
by solving Eq. (12) using collocation method. Where C is the coefficients of ba-
sis function /i, ~g is the vector (0,g)T. So C ¼ A�1~g.
kCk2 ¼ ½C;C� ¼ ½A�1~g;A�1~g� ¼ ½~g; ðA�1ÞTA�1~g�¼: k~gk ¼ kgk;
where [ Æ , Æ ] is the Euclidean inner scale. Notes the definition of Hi;hg, it is easyto find that kH1;hgkH1ðXiÞ ¼ CTMC, with
M ¼ZXi
/i/j þXdk¼1
o/i
oxk
o/j
oxk
! !:
Then kH1;hgkH1ðXiÞ¼: kCk because that matrix M is positive defined. It is similar
for the case i = 2. This finish the proof. h
Remark 3.1. Here, the definition of the Steklov–Poincare follow the one in
weak form. It is hard to explain if we restrict us in the �strong� discrete spacebecause of the lack of the information about the dual space of the trace space
on C. The method to solve or discuss the relative problem are collocation yet.
Remark 3.2. It is easy to find that both Si,h are continuous, coercive in RNC , so
is S.
Remark 3.3. It can be observed that the constants in Theorem 3.1 are relative
to the condition number of matrix A, B and M.This indicates that this method
is not optimal.
380 Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384
Recall the theorem [14]:
Theorem 3.2. Suppose that (a) Q2 is continuous and coercive; that is,
(a)1 there exists b2 > 0 such that
hQ2g; li6 b2kgkXklkX 8g; l 2 X ;
(a)2 there exists a2 > 0 such that
hQ2g; giP a2kgk2X 8g 2 X ;
(b) Q1 is continuous; that is, there exists b1 > 0 such that
hQ1g; li6 b1kgkXklkX 8g; l 2 X ;
(c) there exists a constant j* > 0 such that
hQ2g;Q�12 QgiP j�kgk2X 8g 2 X :
Then for any given k0 in X and for any h satisfying 0 < h < hmax, with hmax ¼j�a2
2
b2ðb1þb2Þ2the sequence kkþ1 ¼ kk þ hQ�1
2 ðG� QkkÞ converges in X to the solution
of problem Qk = G, where X is some Hilbert space.
Remark 3.4. If the operator Q2 is symmetry, one just need to check whether Qis positively defined.
Theorem 3.3. If the RBF satisfy the condition in Theorem 3.1, the iterative solu-
tion of (12)–(14) converge to the solution of (10).
Proof. Assuming that kn ! k in RNC by using Theorem 3.2. Note the iteration
on XC, it can be yielded:
un2;hðX CÞ ! k ðn ! þ1Þ
from Eq. (14). Then
un1;hðX CÞ ! k ðn ! þ1Þ
from Eq. (12). Assuming that um1;h, un1;h with coefficient Cm, Cn are the solution
of Eq. (12) at m iterative step and n iterative step respectively, we can yield:
kCm � Cnk6C1kkm � knk;
because that um1;h � un1;h satisfy the Eq. (11) with km–kn on C. Then Cn is Cauchy
sequence in RN. This yield:
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384 381
Cn ! C0 ðn ! þ1Þ:
The iterative solution in X1 is given bygu1;h ¼P
iC0;i/i. It is similar to obtainthe solution in X2, denoted by gu2;h . Then
Dgu1;hðX 1Þ ¼ f ðX 1Þ;gu1;hðX @1Þ ¼ 0;
gu1;hðX CÞ ¼gu2;hðX CÞ;
ofu2;hon ðX CÞ ¼ ofu1;h
on ðX CÞ;gu2;hðX @2Þ ¼ 0;
Dgu2;hðX 2Þ ¼ f ðX 2Þ;
8>>>>>>>>>>>>><>>>>>>>>>>>>>:ð16Þ
because that the matrix operators A, B are bounded. h
4. Numerical examples
In this section, two numerical examples are given to test the feasibility of this
method. Letting X = [�1,1;�1,1], C = {(x,y)jy = 0}, X1 = [�1,1;�1,0],
X2 = [�1,1;0,1], the density of discrete centers is h = 0.1. Taken the uniformly
centers as xi ¼ �1þ i�110
� 0:1, yj ¼ �1þ j�1
10� 0:1, i, j = 1, . . ., 21, TPS-RBF as
the basis function. The pointwise relative error at the discrete centers will be
shown in the following tables.
ðL2erroriÞ
2 ¼RXiðeu � uexactÞ2RXiðuexactÞ2
are the relative error of L2-norm in Xi. ~u is the approximate solution.
Example 1. Let
f1ðx; yÞ ¼ exðy2 � 1Þðx2 þ 4xþ 1Þ þ 2exðx2 � 1Þ
the exact solution of this problem is
u1ðx; yÞ ¼ ðx2 � 1Þðy2 � 1Þex:
The tables (Table 1 for X1, Table 2 for X2) are obtained by means of the
method discussed before.
Table 1
L2error1 ¼ 0:0078
0.0330 0.0236 0.0128 0.0094 0.0086 0.0082 0.0063 0.0007 0.0130
0.0300 0.0221 0.0156 0.0123 0.0103 0.0090 0.0076 0.0058 0.0031
0.0211 0.0176 0.0141 0.0118 0.0101 0.0089 0.0078 0.0068 0.0060
0.0153 0.0143 0.0124 0.0108 0.0096 0.0085 0.0077 0.0069 0.0064
0.0130 0.0125 0.0112 0.0100 0.0089 0.0081 0.0073 0.0067 0.0062
0.0124 0.0115 0.0103 0.0093 0.0084 0.0076 0.0070 0.0064 0.0059
0.0120 0.0109 0.0097 0.0087 0.0079 0.0072 0.0066 0.0061 0.0056
0.0113 0.0103 0.0091 0.0082 0.0075 0.0068 0.0062 0.0058 0.0053
0.0106 0.0097 0.0087 0.0078 0.0071 0.0065 0.0059 0.0055 0.0050
0.0102 0.0094 0.0083 0.0075 0.0068 0.0062 0.0057 0.0052 0.0049
0.0101 0.0091 0.0081 0.0072 0.0065 0.0059 0.0054 0.0051 0.0049
0.0101 0.0090 0.0079 0.0070 0.0062 0.0056 0.0052 0.0048 0.0047
0.0101 0.0090 0.0077 0.0068 0.0060 0.0054 0.0049 0.0046 0.0045
0.0101 0.0089 0.0076 0.0066 0.0058 0.0052 0.0047 0.0044 0.0042
0.0104 0.0090 0.0076 0.0065 0.0056 0.0050 0.0045 0.0043 0.0041
0.0110 0.0093 0.0075 0.0063 0.0054 0.0048 0.0044 0.0042 0.0043
0.0117 0.0095 0.0075 0.0063 0.0053 0.0047 0.0043 0.0042 0.0046
0.0119 0.0097 0.0075 0.0063 0.0053 0.0046 0.0042 0.0041 0.0046
0.0120 0.0104 0.0086 0.0076 0.0065 0.0055 0.0048 0.0046 0.0047
Table 2
L2error2 ¼ 0:0079
0.0704 0.0383 0.0216 0.0169 0.0136 0.0114 0.0107 0.0082 0.0034
0.0293 0.0241 0.0188 0.0152 0.0129 0.0112 0.0098 0.0083 0.0061
0.0196 0.0186 0.0161 0.0137 0.0118 0.0104 0.0092 0.0083 0.0076
0.0200 0.0168 0.0142 0.0123 0.0108 0.0096 0.0086 0.0078 0.0072
0.0174 0.0148 0.0126 0.0111 0.0099 0.0088 0.0080 0.0072 0.0065
0.0139 0.0127 0.0113 0.0101 0.0090 0.0082 0.0074 0.0068 0.0061
0.0123 0.0114 0.0102 0.0092 0.0083 0.0076 0.0069 0.0064 0.0058
0.0115 0.0105 0.0095 0.0085 0.0077 0.0071 0.0065 0.0060 0.0056
0.0108 0.0099 0.0089 0.0080 0.0073 0.0066 0.0061 0.0056 0.0052
0.0104 0.0094 0.0084 0.0076 0.0068 0.0062 0.0057 0.0053 0.0049
0.0101 0.0091 0.0081 0.0072 0.0065 0.0059 0.0054 0.0051 0.0047
0.0099 0.0089 0.0078 0.0069 0.0062 0.0056 0.0052 0.0048 0.0047
0.0097 0.0087 0.0076 0.0067 0.0060 0.0054 0.0050 0.0047 0.0046
0.0098 0.0087 0.0075 0.0065 0.0058 0.0052 0.0048 0.0045 0.0044
0.0102 0.0088 0.0074 0.0064 0.0056 0.0050 0.0046 0.0043 0.0042
0.0104 0.0089 0.0074 0.0063 0.0054 0.0048 0.0044 0.0043 0.0045
0.0107 0.0091 0.0075 0.0063 0.0054 0.0048 0.0044 0.0043 0.0045
0.0125 0.0099 0.0078 0.0065 0.0056 0.0049 0.0045 0.0044 0.0047
0.0182 0.0123 0.0094 0.0083 0.0071 0.0062 0.0059 0.0059 0.0056
382 Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384
Example 2. Setting
f2ðx; yÞ ¼ �4ðx2 þ y2Þ sinðx2 � 1Þ sinðy2 � 1Þ þ 2 sinðy2 þ x2 � 2Þ;
Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384 383
whose exact solution is
u2ðx; yÞ ¼ sinðx2 � 1Þ sinðy2 � 1Þ:The tables (Table 3 for X1, Table 4 for X2) are obtained by means of the
method discussed before.
Table 3
L2error1 ¼ 0:0047
0.0062 0.0055 0.0033 0.0022 0.0015 0.0011 0.0012 0.0013 0.0004
0.0092 0.0069 0.0047 0.0036 0.0028 0.0023 0.0020 0.0019 0.0019
0.0079 0.0064 0.0049 0.0040 0.0032 0.0027 0.0024 0.0022 0.0022
0.0066 0.0060 0.0050 0.0041 0.0035 0.0030 0.0026 0.0024 0.0023
0.0064 0.0059 0.0050 0.0043 0.0036 0.0031 0.0028 0.0025 0.0022
0.0068 0.0061 0.0051 0.0044 0.0038 0.0033 0.0029 0.0026 0.0023
0.0072 0.0062 0.0052 0.0045 0.0039 0.0034 0.0030 0.0026 0.0024
0.0071 0.0062 0.0053 0.0046 0.0039 0.0034 0.0030 0.0027 0.0024
0.0068 0.0061 0.0053 0.0046 0.0040 0.0035 0.0031 0.0027 0.0024
0.0066 0.0061 0.0053 0.0046 0.0040 0.0035 0.0031 0.0027 0.0023
0.0067 0.0061 0.0053 0.0046 0.0040 0.0035 0.0031 0.0027 0.0023
0.0071 0.0062 0.0053 0.0046 0.0039 0.0034 0.0030 0.0027 0.0024
0.0071 0.0062 0.0052 0.0045 0.0039 0.0034 0.0030 0.0026 0.0024
0.0068 0.0061 0.0052 0.0044 0.0038 0.0033 0.0029 0.0026 0.0024
0.0064 0.0059 0.0051 0.0043 0.0037 0.0032 0.0028 0.0025 0.0024
0.0067 0.0061 0.0050 0.0042 0.0036 0.0031 0.0027 0.0025 0.0025
0.0081 0.0065 0.0050 0.0040 0.0033 0.0029 0.0025 0.0024 0.0025
0.0093 0.0069 0.0048 0.0037 0.0030 0.0026 0.0023 0.0021 0.0021
0.0058 0.0052 0.0032 0.0025 0.0020 0.0018 0.0018 0.0017 0.0004
Table 4
L2error2 ¼ 0:0052
0.0195 0.0095 0.0059 0.0057 0.0048 0.0033 0.0028 0.0030 0.0022
0.0095 0.0077 0.0060 0.0050 0.0042 0.0035 0.0030 0.0028 0.0029
0.0079 0.0071 0.0059 0.0049 0.0041 0.0035 0.0031 0.0029 0.0030
0.0089 0.0072 0.0059 0.0049 0.0042 0.0036 0.0032 0.0029 0.0027
0.0084 0.0070 0.0058 0.0049 0.0042 0.0036 0.0032 0.0028 0.0025
0.0075 0.0067 0.0057 0.0049 0.0042 0.0037 0.0032 0.0029 0.0025
0.0071 0.0065 0.0056 0.0049 0.0043 0.0037 0.0032 0.0029 0.0026
0.0071 0.0064 0.0056 0.0049 0.0043 0.0038 0.0033 0.0030 0.0027
0.0074 0.0065 0.0057 0.0049 0.0043 0.0038 0.0033 0.0029 0.0026
0.0076 0.0066 0.0057 0.0049 0.0043 0.0038 0.0033 0.0029 0.0025
0.0074 0.0066 0.0057 0.0049 0.0043 0.0038 0.0033 0.0029 0.0025
0.0072 0.0065 0.0056 0.0049 0.0043 0.0038 0.0033 0.0030 0.0027
0.0073 0.0065 0.0056 0.0049 0.0043 0.0037 0.0033 0.0029 0.0027
0.0076 0.0067 0.0057 0.0049 0.0042 0.0037 0.0032 0.0029 0.0026
0.0083 0.0069 0.0057 0.0049 0.0042 0.0036 0.0032 0.0028 0.0025
0.0087 0.0071 0.0057 0.0048 0.0041 0.0035 0.0031 0.0028 0.0026
0.0075 0.0068 0.0057 0.0048 0.0040 0.0034 0.0030 0.0028 0.0028
0.0091 0.0075 0.0058 0.0048 0.0040 0.0033 0.0029 0.0026 0.0026
0.0193 0.0093 0.0056 0.0054 0.0044 0.0030 0.0025 0.0027 0.0020
384 Y. Duan, Y.-j. Tan / Appl. Math. Comput. 166 (2005) 373–384
5. Conclusion
Some others RBF are advised, for example, the Sobolev spline or multiqua-
dric function. For multiquadric 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. At last we must pointout that the extension of this method to many sub-domains is direct.
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