meshless collocation method based on dirichlet–neumann substructure iteration

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).

mailto:[email protected]

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


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 > 0
Multiquadrics
(c2 + k Æ k2)b b 2 N, c50 Thin-plate splines k Æ kblogk Æ k b 2 2N
The 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


/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:


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:


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.


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:


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


Example 2. Setting

f2ðx; yÞ ¼ �4ðx2 þ y2Þ sinðx2 � 1Þ sinðy2 � 1Þ þ 2 sinðy2 þ x2 � 2Þ;


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


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.

References

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meshless collocation method based on dirichlet–neumann substructure iteration

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