research article a new iterative method for linear systems...

Research ArticleA New Iterative Method for Linear Systems from XFEM

Jianping Zhao12 Yanren Hou2 Haibiao Zheng2 and Bo Tang3

1 College of Mathematics and System Sciences Xinjiang University Urumqi 830046 China2 School of Mathematics and Statistics Center for Computational Geosciences Xirsquoan Jiaotong University Xirsquoan 710049 China3 School of Mathematics and Computer Hubei University of Arts and Science Xiangyang Hubei 441053 China

Correspondence should be addressed to Haibiao Zheng hbzhengxjtueducn

Received 12 September 2013 Accepted 16 December 2013 Published 14 January 2014

Academic Editor Trung NguyenThoi

Copyright copy 2014 Jianping Zhao et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The extend finite element method (XFEM) is popular in structural mechanics when dealing with the problem of the crackeddomains XFEM ends up with a linear system However XFEM usually leads to nonsymmetric and ill-conditioned stiff matrixIn this paper we take the linear elastostatics governing equations as the model problem We propose a new iterative method tosolve the linear equations Here we separate two variables119880 and Enr so that we change the problems into solving the smaller scaleequations iteratively The new program can be easily applied Finally numerical examples show that the proposed method is moreefficient than common methods we compare the 119871

2-error and the CPU time in whole process Furthermore the new XFEM can

be applied and optimized in many other problems

1 Introduction

Many physical phenomena can be modeled by partial dif-ferential equations with singularities and interfaces Muchattention has been focused on the development of interfaceproblem in fluid dynamics andmaterial scienceThe standardfinite difference and finite element methods may not be suc-cessful in giving satisfactory numerical results for such prob-lems Hence many newmethods have been developed Someof them are developed with the modifications in the standardmethods so that they can deal with the discontinuitiesand the singularities Peskin developed immersed boundarymethod (IBM) [1] basically to model blood flow in theheart Immersed interface method (IIM) [2ndash6] was basicallydeveloped because of the need for a second-order accuratefinite difference method to solve linear elliptic parabolicpartial differential equations in which an integral equationmay not be available IIM will give a nonsymmetric systemof equations The extended finite element method (XFEM)[7 8] enables the accurate approximation of solutions thatinvolve jumps kinks singularities and other locally nons-mooth features within elementsThis is achieved by enrichingthe polynomial approximation space of the classical finite

element method The generalized finite element method(GEFM)XFEM [9ndash11] has shown its potential in a varietyof applications that involve nonsmooth solutions near inter-faces Among them are the simulation of cracks shear bandsdislocations solidification and multifield problems A largenumber of examples can be found in the real world wherefield quantities change rapidly over length scales that aresmall with respect to the observed domain For the mod-eling of such phenomena the resulting solutions typicallyinvolve discontinuities singularities high gradients or othernonsmooth properties For the approximation of nonsmoothsolutions one of the approaches is to enrich a polynomialapproximation space such that the nonsmooth solutionscan be modeled independently of the mesh Methods thatextend polynomial approximation spaces are called ldquoenrichedmethodsrdquo in this work The enrichment can be achieved byadding special shape functions (which are customized soas to capture jumps singularities etc) to the polynomialapproximation space The total stiffness matrix consists offour parts the FEM part the enrich nodes part and anothertwo parts are the connecting parts between the real nodes andenrich nodes Compared with FEM the matrix size increases119874(119873) while the FEM stiff matrix size is119874(1198732)(119873 sim 119874(ℎ

minus1))

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 367802 8 pageshttpdxdoiorg1011552014367802

2 Mathematical Problems in Engineering

But engineering calculations in XFEM always encounteredthe problem of solving ill-conditioned matrix It made theXFEM widely

In this paper according to the characteristics of thestiffness matrix we use an iterative method to avoid dealingwith complicated stiff matrix First we can use the result oflinear equation from standard FEM The result can be usedto be the initial value for iteration Then we decompose theunknown values to two parts each part is governed by twolinear equations whose dimension is smaller than the wholestiffmatrixThe new programm can be easily applied Finallynumerical examples show that the proposed method ismore efficient than commonmethods Furthermore the newmethod works well for some problems while the commonmethod (such as LU) fails The new iterative method can beapplied to many other problems

Theoutline of this paper is as follows Section 2 introducesthe XFEM Subsequently the main part of this paper isSection 3 thatwe describe a new efficient and feasiblemethodWe also analyze the time complexity of the algorithm InSection 4 we briefly mention details on numerical examplesA final conclusion is drawn in Section 5

2 The Extended Finite Element Method

The basic mathematical foundation of the partition of unityfinite element method (PUFEM) was discussed in [12 13]They illustrated that PUFEM can be used to employ thestructure of the differential equation under consideration toconstruct effective and robust methods The global solutionof PUFEMhas been the theoretical basis of the local partitionof unity finite elementmethod to be called later the extendedfinite element method

The first effort for developing the extended finite elementmethod can be traced back to 1999 [7] presented a minimalenmeshing finite element method for crack growth Theyadded discontinuous enrichment functions to the finiteelement approximation to account for the presence of thecrackThe method allowed the crack to be arbitrarily alignedwithin the mesh though it required enmeshing for severelycurved cracks The method was improved and called extendFinite Element Method (XFEM) in [8 9 11 14ndash16]

In mathematics relevant theoretical parts of doing arealmost same in [17ndash19] Of course XFEM is still evolvingcurrently and keeping the new extension about (such as fluid-structure interaction) mathematics stiff matrix remainedprone to the sick

Consider a domain Ω in R119899 which is discredited by a setof elementsΩ

119890such that

⋃

119890isin120576

Ω119890= Ω (1)

and a set of nodes

T = ⋃

119868isinN

119909119868 (2)

The total ordering of the element basis is given by 120576 andthe total ordering of the nodes is given by N The standardfinite element basis is defined as follows

Bstd

= ⋃

119868isinN

119873119868(119909) (3)

where 119873119868(119909) is the finite element basis or shape function

of node 119868 119873119868(119909) is assumed to be of compact support and

piecewise continuously differentiable Typically Bstd spansthe space of piecewise continuous polynomials of a specificorder The XFEM aims to alleviate the burden associatedwithmesh generation for problemswith voids and interfacesIt does not require the finite element mesh to conform tointernal boundaries The essence of the XFEM lies in subdi-viding a model into two distinct parts mesh generation forthe domain (excluding internal boundaries) and enrichmentof the finite-element approximation by additional functionsA partitioning of a typical domain into its enriched andunenriched subdomains is shown in Figure 2(a) We definedthe subdomain where the enrichment is applied as Ω

Enr

This regionrsquos feature is that the enrichment is dominant Theelements coveringΩEnr are the enriched elements so

⋃

119890isin120576Enr

Ω119890= Ω

Enr (4)

where 120576Enr is the ordering of the subset of elements which

are to be enriched Any node 119868 for which the support of119873119868(119909) overlaps or is contained in Ω

Enr is enriched node andthe set is defined by the ordering NEnr NEnr

sub N For anenrichment function 120595(119909) the enriched basis for the localpartition of unity method is

BEnr

= Bstd

oplus ⋃

119868isinNEnr

119873119868(119909) 120595 (119909) (5)

The approximation enriched with a local partition of unitythe field 119906(119909) is given by the sum of a standard finite elementapproximation and a linear combination of the enriched basisfunctions

119906ℎ(119909) = sum

119868isinN

119873std119868

(119909) 119906119868+ sum

119869isinNEnr

119873Enr119869

(119909) 120595 (119909) 119902119869 (6)

where 119873std119868(119909) and 119873

Enr119869

(119909) are the shape functions for thestandard part and the partition of unity respectively differentorder interpolates may be used for the standard and enrichedshape functions

For example we consider the linear elastostatics govern-ing equations

nabla sdot 120590 + b = 0 in Ω

120590 = C 120576

120576 = nabla119904u

(7)

where 120590 is cauchy stress and b is the body force per unitvolumeThe first function is called equilibrium equationThesecond function is the equilibrium equation that describes

Mathematical Problems in Engineering 3

the relation of strain and displacement The third equationis for Hookersquos law The parameter 119862 implies two parameters120583 = 119866 = 119864(2(1+])) and shear elasticity120582 = 119864](1+])(1minus2])(Lame constants are 120582 120583 and Poissonrsquos ratio is ]) Consider

u = u on Γ119906

120590 sdot n = t on Γ119905

120590 sdot n119894ℎ= 0 on Γ

119894

ℎ(119894 = 1 2 119898)

[120590 sdot n119894119868] = 0 on Γ

119894

119868(119894 = 1 2 119899)

(8)

The boundary Γ is composed of the sets Γ119906 Γ119905 Γ119894ℎ and Γ119894

119868such

that

Γ = Γ119906cup Γ119905cup Γ119894

ℎcup Γ119894

119868 (9)

where 119899 is the unit outward normal to Ω and 119906 and t areprescribed displacements and traction respectivelyTheweakformulation of (12) is well known and reads as follows find119906 isin 119881 such that

intΩ

120590 (u) 120576 (k) 119889Ω = intΩ

b sdot k119889Ω + intΓ119905

t sdot k119889Γ forallV isin 1198810

(10)

Here 119881 = 1198671(Ω) and 119881

0= V isin 119881 V = 0 on Γ

119906

For the finite element discretization let 120591ℎbe a regular

rectangular element of domain Ω and define the meshparameter ℎ = max

119879isin120591ℎ

diam(119879)Consider the Bubnov-Galerkin implementation for the

XFEM in two-dimensional linear elasticity In the XFEMfinite-dimensional subspaces 119881

ℎsub 119881 and 119881

ℎ

0sub 1198810are

used as the approximating trial and test spaces The finitedimensional space 119881ℎ is given by

119881ℎ= V isin 119881 V|119879 isin 119876

119896(119879) forall119879 isin 120591

ℎ (11)

The weak form of the discrete problem can be stated asfollows find 119906

ℎisin 119881ℎ such that

intΩℎ

120590 (uℎ) 120576 (kℎ) 119889Ω

= intΩℎ

b sdot kℎ119889Ω + intΓℎ

119905

t sdot kℎ119889Γ forallVℎ isin 119881ℎ

0sub 1198810

(12)

where 119881ℎ

sub 119881 and 119881ℎ

0sub 1198810 In a Bubnov-Galkerkin

procedure the trial function 119906ℎ and the test function Vℎ are

represented as linear combinations of the same shape func-tions The trial and test functions are

uℎ (x) = sum

119868

120601119868(x) u119868+sum

119869

120601119869120595 (120593 (x)) a

119869

kℎ (x) = sum

119868

120601119868(x) k119868+sum

119869

120601119869120595 (120593 (x)) b

119869

(13)

where 119868 denotes the set of all nodes in the mesh and 119869 = 119895 isin

119868 119874120576x119895cup Γ = 0(120576 le ℎ) denotes the set of nodes near the

interface 120601119868(x) are the finite-element shape functions 120593(x) is

the level set function and120595(120593(x)) is the enrichment functionfor material interface

There are several kinds of enrichment functions suchas abs-enrichment function step-enrichment function andRamp-enrichment function The choice of function is basedon the behavior of the solution near the interface

The Ramp function is defined as

120595Ramp = 1 120593 gt 0

1 minus 2120593 120593 le 0(14)

This enrichment function yields only continuous solutionsThe advantage is that it automatically satisfies the continuitycondition [119906] = 0 and does not require the use of Lagrangemultipliers

The step-enrichment function is defined as

120595Step = 1 120593 gt 0

minus1 120593 le 0(15)

This enrichment function can yield a continuous or discon-tinuous solution across the interface but requires Lagrangemultipliers to apply the Dirichlet jump condition

On substituting the trial and test functions form (17)and using the arbitrariness of nodal variations the followingdiscrete system of linear equations is obtained

Kd = F (16)

where

K119868119869= intΩℎ

B119879119868CB119869119889Ω (17)

F119868= intΓℎ

119905

120601119868t 119889Γ + int

Ωℎ

120601119868b 119889Ω (18)

120601119868equiv 120601119868for a finite-element displacement degree of freedom

and 120601119868equiv 120601119868120595 for an enriched degree of freedom In the above

equations C is the constitutive matrix for an isotropic linearelastic material

119862 = (

120582 + 2119866 1 0

1 120582 + 2119866 0

0 0 119866

) (19)

and the matrix B119868is given by

B119868= (

120601119868119909

0

0 120601119868119910

120601119868119910

120601119868119909

) (20)

We can get the integral terms and thematrices computedfrom the integral terms are block matrices of the form

119870 = (119870119880119880

119870119880119860

119870119860119880

119870119860119860) 119865 = (

119865119880

119865119860)

119889 = (119880

Enr)

(21)

The next step is solving the linear equations system


3 The Program for Solving the Equation

We need to solve the following equations

(11987211

119899times11989911987212

119899times119898

11987221

119898times11989911987222

119898times119898

)(119880119899times1

Enr119898times1

) = (1198771

119899times1

1198772

119898times1

) (22)

Here 119899 sim 119874(1198732) and 119898 sim 119874(119873) the block matrix 119872

11 issame in FEM and (119872

21)1015840

= 11987212 are sparse matrix

We must point out that the matrix 11987212 has up to 119898

2

nonzero elements We can use the modified program in (22)can be converted to two sub-problems to solve (11987222)minus1 andsolve 119880 by 119899-order linear equations Now we list the detailedsteps

Algorithm 1 (i) We use the ordinary method to solve

11987211119880(0)

= 1198771 (23)

the results for the FEM equation and set the result as 119880(0)119864(0)

= 0 The matrix 11987211 is block tridiagonal symmetric

sparse matrix There are a lot of methods to solve thisequation efficiently

(ii) Solve the small scale equation

11987222Enr(119894+1) = 119877

2minus11987221119880(119894) 119894 = 0 1 (24)

The matrix11987222 is a sparse tridiagonal where the dimen-sion is much smaller than 119872

11 This equation can be solvedquicker than (i)

(iii) Then we use (25) to solve 119880(119894+1)

11987211119880(119894+1)

= 1198771minus11987212Enr(119894+1) 119894 = 0 1 (25)

the computation time complexity is the same with the firstone

(iv) If the norm 119880(119894+1)

minus 119880(119894) le 119908 or 119894 gt 119868max (giving

the maximum iterative times) output the result 119880 = 119880(119894+1)

Enr = Enr(119894+1) Else return to step (ii)In this part we can use the common FEM solver such as

LU or other useful solvers We do not change the stiffnessmatrix 119872

11 and 11987222 these two matrixes are symmetric

positive definite and banded We can compute easily for(23) (24) (25) The matrix conditions is not changedK(11987211) = 119874(ℎ

minus2) [17] First of all the store can choose a

one-dimensional We can choose a one dimensional storagewhen we use triangular decomposition for the matrix119872

Remark 2 If the matrix11987211 and 11986022 are not tridiagonal we

often use LU or GMRES method [20] to solve (23) (24) and(25)

The iteration has the following advantages

(1) Other than the half bandwidth of zero elements donot have storage nor involved in the calculation theamount of storage crunch the impact of roundingerror is relatively small even if it is ill-conditionedmatrix we often obtain a higher accuracy

(2) The operator only uses FEM linear equation solver itis a simple calculation

(3) The original matrix deposit triangular matrix anddiagonal matrix deposit area can overlap thus avoid-ing to take up a lot of the middle unit of work savingmemory

(4) This solution can be extended to block direct solution(5) With the iterative method it is easy to build and test

the general program and you can more accuratelyestimate the computation time the user is relativelyeasy to grasp the algorithm process

Also if we using in [21] to solve the equation ((2119873)2+

4119873) equations use LU solver which need 119874((2119873)2+ 4119873)

3

Multiplication here if the number of iterations is 119868 we useup to 119874((2119873)

6+ (4119873)

3)multiplications

4 Numerical Test

In this section we report two experiments to verify theefficacy and accuracy of the new iterative method in XFEMWe use Matlab and modify software [22] to demonstrateAlgorithm 1 For the convenience of presentation we intro-duce the following notations

SFEMmeans the standard finite element methodXFEMmeans the extended finite element methodℎmeans the mesh sizeGMRESmeans the generalizedminimal residual iter-ative method1198712-error means 119906ℎ minus 119906

0

119864ℎmeans the 119871

2-error when the mesh size equals ℎ

iterations mean the iterative times in the algorithmorders mean the numerical error order by XFEM

orders =log (119864

ℎ1

119864ℎ2

)

log (ℎ1ℎ2) (26)

Here ℎ1le ℎ2 First we list Example 1 for one-dimensional

problems

Example 1 The exact solution is

119906 =

minus(119909 minus 05)

2

21198962

minus(119909 minus 05) (119896

1minus 1198962)

41198962(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω1


2

21198961

minus(119909 minus 05) (119896

1minus 1198962)

41198961(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω2

(27)

HereΩ1= [0 05] andΩ

2= (05 1] the coefficient 119896 = 1

if119909 isin Ω1 119896 = 10 if119909 isin Ω

2 And119865 is point force from the righ

119865 = 10 We use Figure 1 to describe the 1198712error change when

we use different ℎ We can easily see that when the step issmaller than 180 thewhole computing time has economizedone order of magnitude in Example 1


10minus2

10minus3

10minus4

102 103 104

Tim

e

1h

ErrL2 for LUErrL2 for iteration

(a)

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

101 102 103 104

Erro

r L2

1h

ErrL2 for LUErrL1 for iterative method

(b)

Figure 1 (a) The comparison of time by iteration and LU with different mesh (b) The comparison of 1198712-error by iteration and LU with

different mesh

1080604020minus02minus04minus06minus08minus1

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

y

x

(a)

0 200 400 600 800 1000

0

100

200

300

400

500

600

700

800

900

1000

nz = 17803

(b)

Figure 2 (a) All nodes and enrichment nodes (in red color)119873 = 20 (b) distribution map of nonzero elements in whole matrix119872119873 = 20

Example 2 For the two-dimensional solid test we use thesame enrichment function and shape function to compute(4) We define Ω = [0 1] times [0 1] with the material interface

Γ 1199092+ 1199102= 1198862= 01608 (28)

Let Youngrsquos modulus and Poissonrsquos ratio in

Ω1= (119909 119910) isin Ω | 119909

2+ 1199102le 0404

2 (29)

be 1198641= 100 ]

1= 0 and let those in

Ω2= (119909 119910) isin Ω | 119909

2+ 1199102le1

4 (30)

be 1198642= 10 ]

2= 0 but area force in domain 119891

119909= 119891119910= 0


Table 1 DOF sparsity of matrix and condition number of linear equation by FEM and XFEM for Example 2

ℎ DOFFEM DOFXFEM condsFEM condsXFEM Sparsity of matrix ()1

20882 1018 17627 times 10

4790617 times 10

4 1721

403362 3626 375035 times 10

4320917 times 10

5 047741

6012800 13120 102381 times 10

5130023 times 10

6 02221

8012800 13120 102381 times 10

5130023 times 10

6 0131

10025600 26240 407606 times 10

5523974 times 10

6 00829

Table 2 The CPU time for different solvers in Example 2

ℎ FEMLU XFEMiteration XFEMLU (s) XFEMGMRES (s)1

2012166 13129 26019 21449

1

4057133 65198 73678 89494

1

60178348 206117 272810 266332

1

80543007 590695 704340 707011

1

10013789 14592 1651256 1665808

Table 3 The comparison of 1198712-error with different solvers for Example 2

ℎ FEMLU XFEMLU (s) XFEMGMRES XFEMiteration (s)1

2042101 times 10

minus264292 times 10



minus3

1

4021032 times 10




minus3

1

6014089 times 10




minus4

1

8010317 times 10




minus4

1

10082802 times 10




minus4

We denote 119903 = radic1199092 + 1199102 and tan 120579 = 119910119909 and the exactsolution is

119906 =119886

8120583(119903

119886(120581 + 1) cos 120579 + 2

119886

119903((1 + 120581) cos 120579 + cos 3120579)

minus 21198863

1199033cos 3120579)

V =119886

8120583(119903

119886(120581 minus 3) sin 120579 + 2

119886

119903((1 minus 120581) sin 120579 + sin 3120579)

minus 21198863

1199033sin 3120579)

(31)

Platewith a circular hole Figure 2(b) shows the consistentrectangular subdivision and labels the nodeswhich need to beextended The CPU time we counted is the total time for theentire programm running Table 1 shows the linear equationscale and the condition numbers of FEM and XFEM thedensity of sparse matrix show the matrix is sparse It can beeasily find the condition number of XFEM is bigger than thatof FEM The CPU times are also compared with LU solversand GMRES in Table 2 The cheapest is LU by FEM and thenew iterative method is the second cheapest however theerror for LU by FEM is the biggest in all methods Table 3shows an 119871

2-error comparison between the new iterative

method and other solvers And the numerical 1198712-error is

nearly 119874(ℎ2) 1198712-error order and iterations are given in

Table 4


Table 4 1198712-error order CPU time and iterations by XFEM with the new iterative method for Example 2

ℎ ErrorXFEM Orders CPUXFEM (s) Iterations1

2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion

In this paper we discuss how to use XFEM for the stiffmatrix ill-conditioned and came to the following conclusionsFirstly we found why some of problems cannot be computedby XFEM the stiff matrix is not then symmetric and ill-conditioned So a lot of storage space is wasted The timeof computing XFEM equation is longer which cannot becomputed in certain subdivision Secondly we use XFEMfor solving two-dimensional solid test when solving theXFEM equations the iterative method and the standardfinite element equations solver are adopted so that wechange the problems into solving the smaller scale equationsand standard finite element stiffness matrix need not bechanged Finally numerical examples show that the proposedmethod is more efficient than the common method We canuse this programm in many other interface problems laterBesides we also consider combining the stable GFEM withour iterative method in the future And it is believed thatthe XFEM can be widely used in more problems in fluiddynamics and material science The calculation module insome software can be improved for users

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The author also thanks the anonymous reviewers for theircomments in improving the paper This research was sup-ported by the special funds for NSFC (1127313 1117126911201369 and 61163027) and the PhD Programs Founda-tion of the Ministry of Education of China (Grant no20110201110027) and it was partially subsidized by the Fun-damental Research Funds for the Central Universities (Grantnos 08142003 and 08143007)

References

[1] C S Peskin ldquoFlow patterns around heart valves A numericalmethodrdquo Journal of Computational Physics vol 10 no 2 pp252ndash271 1972

[2] R J LeVeque and Z L Li ldquoThe immersed interface method forelliptic equations with discontinuous coefficients and singularsourcesrdquo SIAM Journal on Numerical Analysis vol 31 no 4 pp1019ndash1044 1994

[3] Z Li K Ito andM-C Lai ldquoAn augmented approach for Stokesequations with a discontinuous viscosity and singular forcesrdquoComputers amp Fluids vol 36 no 3 pp 622ndash635 2007

[4] A L Fogelson and J P Keener ldquoImmersed interface methodsfor Neumann and related problems in two and three dimen-sionsrdquo SIAM Journal on Scientific Computing vol 22 no 5 pp1630ndash1654 2000

[5] Z Li ldquoThe immersed interface method using a finite elementformulationrdquo Applied Numerical Mathematics vol 27 no 3 pp253ndash267 1998

[6] H Huang and Z Li ldquoConvergence analysis of the immersedinterface methodrdquo IMA Journal of Numerical Analysis vol 19no 4 pp 583ndash608 1999

[7] NMoes J Dolbow andT Belytschko ldquoA finite elementmethodfor crack growth without remeshingrdquo International Journal forNumerical Methods in Engineering vol 46 no 1 pp 131ndash1501999

[8] N Sukumar D L Chopp N Moes and T Belytschko ldquoMod-eling holes and inclusions by level sets in the extended finite-element methodrdquo Computer Methods in Applied Mechanics andEngineering vol 190 no 46-47 pp 6183ndash6200 2001

[9] J Chessa H Wang and T Belytschko ldquoOn the constructionof blending elements for local partition of unity enrichedfinite elementsrdquo International Journal for Numerical Methods inEngineering vol 57 no 7 pp 1015ndash1038 2003

[10] T-P Fries and T Belytschko ldquoThe extendedgeneralized finiteelement method an overview of the method and its applica-tionsrdquo International Journal for Numerical Methods in Engineer-ing vol 84 no 3 pp 253ndash304 2010

[11] T Belytschko R Gracie and G Ventura ldquoA review ofextendedgeneralized finite elementmethods formaterial mod-elingrdquo Modelling and Simulation in Materials Science andEngineering vol 17 pp 1ndash24 2009

[12] I Babuska and J M Melenk ldquoThe partition of unity methodrdquoInternational Journal for Numerical Methods in Engineering vol40 no 4 pp 727ndash758 1997

[13] J M Melenk and I Babuska ldquoThe partition of unity finiteelement method basic theory and applicationsrdquo ComputerMethods in AppliedMechanics and Engineering vol 139 no 1ndash4pp 289ndash314 1996

[14] T-P Fries and T Belytschko ldquoThe intrinsic XFEM a methodfor arbitrary discontinuities without additional unknownsrdquo


International Journal for Numerical Methods in Engineering vol68 no 13 pp 1358ndash1385 2006

[15] J-Y Wu ldquoUnified analysis of enriched finite elements for mod-eling cohesive cracksrdquo Computer Methods in Applied Mechanicsand Engineering vol 200 no 45-46 pp 3031ndash3050 2011

[16] H Sauerland and T-P Fries ldquoThe extended finite elementmethod for two-phase and free-surface flows a systematicstudyrdquo Journal of Computational Physics vol 230 no 9 pp3369ndash3390 2011

[17] I Babuska and U Banerjee ldquoStable generalized finite elementmethod (SGFEM)rdquo Computer Methods in Applied Mechanicsand Engineering vol 201ndash204 pp 91ndash111 2012

[18] T Strouboulis I Babuska and K Copps ldquoThe design andanalysis of the generalized finite element methodrdquo ComputerMethods in Applied Mechanics and Engineering vol 181 no 1ndash3pp 43ndash69 2000

[19] T Strouboulis K Copps and I Babuska ldquoThe generalizedfinite element method an example of its implementationand illustration of its performancerdquo International Journal forNumerical Methods in Engineering vol 47 no 8 pp 1401ndash14172000

[20] Y Saad and M H Schultz ldquoGMRES a generalized minimalresidual algorithm for solving nonsymmetric linear systemsrdquoSIAM Journal on Scientific and Statistical Computing vol 7 no3 pp 856ndash869 1986

[21] G Meurant ldquoA review on the inverse of symmetric tridiagonaland block tridiagonal matricesrdquo SIAM Journal on Matrix Anal-ysis and Applications vol 13 no 3 pp 707ndash728 1992

[22] Copyright (c) Thomas-Peter Fries RWTH Aachen University2010

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Stochastic AnalysisInternational Journal of


But engineering calculations in XFEM always encounteredthe problem of solving ill-conditioned matrix It made theXFEM widely

In this paper according to the characteristics of thestiffness matrix we use an iterative method to avoid dealingwith complicated stiff matrix First we can use the result oflinear equation from standard FEM The result can be usedto be the initial value for iteration Then we decompose theunknown values to two parts each part is governed by twolinear equations whose dimension is smaller than the wholestiffmatrixThe new programm can be easily applied Finallynumerical examples show that the proposed method ismore efficient than commonmethods Furthermore the newmethod works well for some problems while the commonmethod (such as LU) fails The new iterative method can beapplied to many other problems

Theoutline of this paper is as follows Section 2 introducesthe XFEM Subsequently the main part of this paper isSection 3 thatwe describe a new efficient and feasiblemethodWe also analyze the time complexity of the algorithm InSection 4 we briefly mention details on numerical examplesA final conclusion is drawn in Section 5

2 The Extended Finite Element Method

The basic mathematical foundation of the partition of unityfinite element method (PUFEM) was discussed in [12 13]They illustrated that PUFEM can be used to employ thestructure of the differential equation under consideration toconstruct effective and robust methods The global solutionof PUFEMhas been the theoretical basis of the local partitionof unity finite elementmethod to be called later the extendedfinite element method

The first effort for developing the extended finite elementmethod can be traced back to 1999 [7] presented a minimalenmeshing finite element method for crack growth Theyadded discontinuous enrichment functions to the finiteelement approximation to account for the presence of thecrackThe method allowed the crack to be arbitrarily alignedwithin the mesh though it required enmeshing for severelycurved cracks The method was improved and called extendFinite Element Method (XFEM) in [8 9 11 14ndash16]

In mathematics relevant theoretical parts of doing arealmost same in [17ndash19] Of course XFEM is still evolvingcurrently and keeping the new extension about (such as fluid-structure interaction) mathematics stiff matrix remainedprone to the sick

Consider a domain Ω in R119899 which is discredited by a setof elementsΩ

119890such that

⋃

119890isin120576

Ω119890= Ω (1)

and a set of nodes

T = ⋃

119868isinN

119909119868 (2)

The total ordering of the element basis is given by 120576 andthe total ordering of the nodes is given by N The standardfinite element basis is defined as follows

Bstd

= ⋃

119868isinN

119873119868(119909) (3)

where 119873119868(119909) is the finite element basis or shape function

of node 119868 119873119868(119909) is assumed to be of compact support and

piecewise continuously differentiable Typically Bstd spansthe space of piecewise continuous polynomials of a specificorder The XFEM aims to alleviate the burden associatedwithmesh generation for problemswith voids and interfacesIt does not require the finite element mesh to conform tointernal boundaries The essence of the XFEM lies in subdi-viding a model into two distinct parts mesh generation forthe domain (excluding internal boundaries) and enrichmentof the finite-element approximation by additional functionsA partitioning of a typical domain into its enriched andunenriched subdomains is shown in Figure 2(a) We definedthe subdomain where the enrichment is applied as Ω

Enr

This regionrsquos feature is that the enrichment is dominant Theelements coveringΩEnr are the enriched elements so

⋃

119890isin120576Enr

Ω119890= Ω

Enr (4)

where 120576Enr is the ordering of the subset of elements which

are to be enriched Any node 119868 for which the support of119873119868(119909) overlaps or is contained in Ω

Enr is enriched node andthe set is defined by the ordering NEnr NEnr

sub N For anenrichment function 120595(119909) the enriched basis for the localpartition of unity method is

BEnr

= Bstd

oplus ⋃

119868isinNEnr

119873119868(119909) 120595 (119909) (5)

The approximation enriched with a local partition of unitythe field 119906(119909) is given by the sum of a standard finite elementapproximation and a linear combination of the enriched basisfunctions

119906ℎ(119909) = sum

119868isinN

119873std119868

(119909) 119906119868+ sum

119869isinNEnr

119873Enr119869

(119909) 120595 (119909) 119902119869 (6)

where 119873std119868(119909) and 119873

Enr119869

(119909) are the shape functions for thestandard part and the partition of unity respectively differentorder interpolates may be used for the standard and enrichedshape functions

For example we consider the linear elastostatics govern-ing equations

nabla sdot 120590 + b = 0 in Ω

120590 = C 120576

120576 = nabla119904u

(7)

where 120590 is cauchy stress and b is the body force per unitvolumeThe first function is called equilibrium equationThesecond function is the equilibrium equation that describes



u = u on Γ119906

120590 sdot n = t on Γ119905

120590 sdot n119894ℎ= 0 on Γ

119894

ℎ(119894 = 1 2 119898)

[120590 sdot n119894119868] = 0 on Γ

119894

119868(119894 = 1 2 119899)

(8)


119868such

that

Γ = Γ119906cup Γ119905cup Γ119894

ℎcup Γ119894

119868 (9)


intΩ

120590 (u) 120576 (k) 119889Ω = intΩ

b sdot k119889Ω + intΓ119905


(10)

Here 119881 = 1198671(Ω) and 119881

0= V isin 119881 V = 0 on Γ

119906



119879isin120591ℎ



ℎsub 119881 and 119881

ℎ

0sub 1198810are


119881ℎ= V isin 119881 V|119879 isin 119876

119896(119879) forall119879 isin 120591

ℎ (11)



intΩℎ

120590 (uℎ) 120576 (kℎ) 119889Ω

= intΩℎ


119905


0sub 1198810

(12)

where 119881ℎ

sub 119881 and 119881ℎ




uℎ (x) = sum

119868

120601119868(x) u119868+sum

119869

120601119869120595 (120593 (x)) a

119869

kℎ (x) = sum

119868

120601119868(x) k119868+sum

119869

120601119869120595 (120593 (x)) b

119869

(13)







120595Ramp = 1 120593 gt 0

1 minus 2120593 120593 le 0(14)



120595Step = 1 120593 gt 0

minus1 120593 le 0(15)



Kd = F (16)

where

K119868119869= intΩℎ

B119879119868CB119869119889Ω (17)

F119868= intΓℎ

119905

120601119868t 119889Γ + int

Ωℎ

120601119868b 119889Ω (18)




119862 = (

120582 + 2119866 1 0

1 120582 + 2119866 0

0 0 119866

) (19)


B119868= (

120601119868119909

0

0 120601119868119910

120601119868119910

120601119868119909

) (20)


119870 = (119870119880119880

119870119880119860

119870119860119880

119870119860119860) 119865 = (

119865119880

119865119860)

119889 = (119880

Enr)

(21)





(11987211

119899times11989911987212

119899times119898

11987221

119898times11989911987222

119898times119898

)(119880119899times1

Enr119898times1

) = (1198771

119899times1

1198772

119898times1

) (22)



21)1015840



2



11987211119880(0)

= 1198771 (23)





11987222Enr(119894+1) = 119877

2minus11987221119880(119894) 119894 = 0 1 (24)




11987211119880(119894+1)

= 1198771minus11987212Enr(119894+1) 119894 = 0 1 (25)


(iv) If the norm 119880(119894+1)



















3


6+ (4119873)

3)multiplications

4 Numerical Test



0

119864ℎmeans the 119871



orders =log (119864

ℎ1

119864ℎ2

)

log (ℎ1ℎ2) (26)


problems


119906 =


2

21198962

minus(119909 minus 05) (119896

1minus 1198962)

41198962(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω1


2

21198961

minus(119909 minus 05) (119896

1minus 1198962)

41198961(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω2

(27)


2= (05 1] the coefficient 119896 = 1

if119909 isin Ω1 119896 = 10 if119909 isin Ω





10minus2

10minus3

10minus4

102 103 104

Tim

e

1h


(a)

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

101 102 103 104

Erro

r L2

1h


(b)


different mesh


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

y

x

(a)

0 200 400 600 800 1000

0

100

200

300

400

500

600

700

800

900

1000

nz = 17803

(b)



Γ 1199092+ 1199102= 1198862= 01608 (28)


Ω1= (119909 119910) isin Ω | 119909

2+ 1199102le 0404

2 (29)

be 1198641= 100 ]


Ω2= (119909 119910) isin Ω | 119909

2+ 1199102le1

4 (30)

be 1198642= 10 ]


119909= 119891119910= 0




20882 1018 17627 times 10

4790617 times 10

4 1721

403362 3626 375035 times 10

4320917 times 10

5 047741

6012800 13120 102381 times 10

5130023 times 10

6 02221

8012800 13120 102381 times 10

5130023 times 10

6 0131

10025600 26240 407606 times 10

5523974 times 10

6 00829



2012166 13129 26019 21449

1

4057133 65198 73678 89494

1

60178348 206117 272810 266332

1

80543007 590695 704340 707011

1

10013789 14592 1651256 1665808



2042101 times 10




minus3

1

4021032 times 10




minus3

1

6014089 times 10




minus4

1

8010317 times 10




minus4

1

10082802 times 10




minus4


119906 =119886

8120583(119903

119886(120581 + 1) cos 120579 + 2

119886

119903((1 + 120581) cos 120579 + cos 3120579)

minus 21198863

1199033cos 3120579)

V =119886

8120583(119903

119886(120581 minus 3) sin 120579 + 2

119886

119903((1 minus 120581) sin 120579 + sin 3120579)

minus 21198863

1199033sin 3120579)

(31)





Table 4




2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion




Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











u = u on Γ119906

120590 sdot n = t on Γ119905

120590 sdot n119894ℎ= 0 on Γ

119894

ℎ(119894 = 1 2 119898)

[120590 sdot n119894119868] = 0 on Γ

119894

119868(119894 = 1 2 119899)

(8)


119868such

that

Γ = Γ119906cup Γ119905cup Γ119894

ℎcup Γ119894

119868 (9)


intΩ

120590 (u) 120576 (k) 119889Ω = intΩ

b sdot k119889Ω + intΓ119905


(10)

Here 119881 = 1198671(Ω) and 119881

0= V isin 119881 V = 0 on Γ

119906



119879isin120591ℎ



ℎsub 119881 and 119881

ℎ

0sub 1198810are


119881ℎ= V isin 119881 V|119879 isin 119876

119896(119879) forall119879 isin 120591

ℎ (11)



intΩℎ

120590 (uℎ) 120576 (kℎ) 119889Ω

= intΩℎ


119905


0sub 1198810

(12)

where 119881ℎ

sub 119881 and 119881ℎ




uℎ (x) = sum

119868

120601119868(x) u119868+sum

119869

120601119869120595 (120593 (x)) a

119869

kℎ (x) = sum

119868

120601119868(x) k119868+sum

119869

120601119869120595 (120593 (x)) b

119869

(13)







120595Ramp = 1 120593 gt 0

1 minus 2120593 120593 le 0(14)



120595Step = 1 120593 gt 0

minus1 120593 le 0(15)



Kd = F (16)

where

K119868119869= intΩℎ

B119879119868CB119869119889Ω (17)

F119868= intΓℎ

119905

120601119868t 119889Γ + int

Ωℎ

120601119868b 119889Ω (18)




119862 = (

120582 + 2119866 1 0

1 120582 + 2119866 0

0 0 119866

) (19)


B119868= (

120601119868119909

0

0 120601119868119910

120601119868119910

120601119868119909

) (20)


119870 = (119870119880119880

119870119880119860

119870119860119880

119870119860119860) 119865 = (

119865119880

119865119860)

119889 = (119880

Enr)

(21)





(11987211

119899times11989911987212

119899times119898

11987221

119898times11989911987222

119898times119898

)(119880119899times1

Enr119898times1

) = (1198771

119899times1

1198772

119898times1

) (22)



21)1015840



2



11987211119880(0)

= 1198771 (23)





11987222Enr(119894+1) = 119877

2minus11987221119880(119894) 119894 = 0 1 (24)




11987211119880(119894+1)

= 1198771minus11987212Enr(119894+1) 119894 = 0 1 (25)


(iv) If the norm 119880(119894+1)



















3


6+ (4119873)

3)multiplications

4 Numerical Test



0

119864ℎmeans the 119871



orders =log (119864

ℎ1

119864ℎ2

)

log (ℎ1ℎ2) (26)


problems


119906 =


2

21198962

minus(119909 minus 05) (119896

1minus 1198962)

41198962(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω1


2

21198961

minus(119909 minus 05) (119896

1minus 1198962)

41198961(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω2

(27)


2= (05 1] the coefficient 119896 = 1

if119909 isin Ω1 119896 = 10 if119909 isin Ω





10minus2

10minus3

10minus4

102 103 104

Tim

e

1h


(a)

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

101 102 103 104

Erro

r L2

1h


(b)


different mesh


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

y

x

(a)

0 200 400 600 800 1000

0

100

200

300

400

500

600

700

800

900

1000

nz = 17803

(b)



Γ 1199092+ 1199102= 1198862= 01608 (28)


Ω1= (119909 119910) isin Ω | 119909

2+ 1199102le 0404

2 (29)

be 1198641= 100 ]


Ω2= (119909 119910) isin Ω | 119909

2+ 1199102le1

4 (30)

be 1198642= 10 ]


119909= 119891119910= 0




20882 1018 17627 times 10

4790617 times 10

4 1721

403362 3626 375035 times 10

4320917 times 10

5 047741

6012800 13120 102381 times 10

5130023 times 10

6 02221

8012800 13120 102381 times 10

5130023 times 10

6 0131

10025600 26240 407606 times 10

5523974 times 10

6 00829



2012166 13129 26019 21449

1

4057133 65198 73678 89494

1

60178348 206117 272810 266332

1

80543007 590695 704340 707011

1

10013789 14592 1651256 1665808



2042101 times 10




minus3

1

4021032 times 10




minus3

1

6014089 times 10




minus4

1

8010317 times 10




minus4

1

10082802 times 10




minus4


119906 =119886

8120583(119903

119886(120581 + 1) cos 120579 + 2

119886

119903((1 + 120581) cos 120579 + cos 3120579)

minus 21198863

1199033cos 3120579)

V =119886

8120583(119903

119886(120581 minus 3) sin 120579 + 2

119886

119903((1 minus 120581) sin 120579 + sin 3120579)

minus 21198863

1199033sin 3120579)

(31)





Table 4




2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion




Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












(11987211

119899times11989911987212

119899times119898

11987221

119898times11989911987222

119898times119898

)(119880119899times1

Enr119898times1

) = (1198771

119899times1

1198772

119898times1

) (22)



21)1015840



2



11987211119880(0)

= 1198771 (23)





11987222Enr(119894+1) = 119877

2minus11987221119880(119894) 119894 = 0 1 (24)




11987211119880(119894+1)

= 1198771minus11987212Enr(119894+1) 119894 = 0 1 (25)


(iv) If the norm 119880(119894+1)



















3


6+ (4119873)

3)multiplications

4 Numerical Test



0

119864ℎmeans the 119871



orders =log (119864

ℎ1

119864ℎ2

)

log (ℎ1ℎ2) (26)


problems


119906 =


2

21198962

minus(119909 minus 05) (119896

1minus 1198962)

41198962(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω1


2

21198961

minus(119909 minus 05) (119896

1minus 1198962)

41198961(1198962+ 1198961)

+1

4 (1198962+ 1198961)

if119909 isin Ω2

(27)


2= (05 1] the coefficient 119896 = 1

if119909 isin Ω1 119896 = 10 if119909 isin Ω





10minus2

10minus3

10minus4

102 103 104

Tim

e

1h


(a)

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

101 102 103 104

Erro

r L2

1h


(b)


different mesh


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

y

x

(a)

0 200 400 600 800 1000

0

100

200

300

400

500

600

700

800

900

1000

nz = 17803

(b)



Γ 1199092+ 1199102= 1198862= 01608 (28)


Ω1= (119909 119910) isin Ω | 119909

2+ 1199102le 0404

2 (29)

be 1198641= 100 ]


Ω2= (119909 119910) isin Ω | 119909

2+ 1199102le1

4 (30)

be 1198642= 10 ]


119909= 119891119910= 0




20882 1018 17627 times 10

4790617 times 10

4 1721

403362 3626 375035 times 10

4320917 times 10

5 047741

6012800 13120 102381 times 10

5130023 times 10

6 02221

8012800 13120 102381 times 10

5130023 times 10

6 0131

10025600 26240 407606 times 10

5523974 times 10

6 00829



2012166 13129 26019 21449

1

4057133 65198 73678 89494

1

60178348 206117 272810 266332

1

80543007 590695 704340 707011

1

10013789 14592 1651256 1665808



2042101 times 10




minus3

1

4021032 times 10




minus3

1

6014089 times 10




minus4

1

8010317 times 10




minus4

1

10082802 times 10




minus4


119906 =119886

8120583(119903

119886(120581 + 1) cos 120579 + 2

119886

119903((1 + 120581) cos 120579 + cos 3120579)

minus 21198863

1199033cos 3120579)

V =119886

8120583(119903

119886(120581 minus 3) sin 120579 + 2

119886

119903((1 minus 120581) sin 120579 + sin 3120579)

minus 21198863

1199033sin 3120579)

(31)





Table 4




2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion




Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










10minus2

10minus3

10minus4

102 103 104

Tim

e

1h


(a)

10minus3

10minus4

10minus5

10minus6

10minus7

10minus8

10minus9

10minus10

101 102 103 104

Erro

r L2

1h


(b)


different mesh


1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

y

x

(a)

0 200 400 600 800 1000

0

100

200

300

400

500

600

700

800

900

1000

nz = 17803

(b)



Γ 1199092+ 1199102= 1198862= 01608 (28)


Ω1= (119909 119910) isin Ω | 119909

2+ 1199102le 0404

2 (29)

be 1198641= 100 ]


Ω2= (119909 119910) isin Ω | 119909

2+ 1199102le1

4 (30)

be 1198642= 10 ]


119909= 119891119910= 0




20882 1018 17627 times 10

4790617 times 10

4 1721

403362 3626 375035 times 10

4320917 times 10

5 047741

6012800 13120 102381 times 10

5130023 times 10

6 02221

8012800 13120 102381 times 10

5130023 times 10

6 0131

10025600 26240 407606 times 10

5523974 times 10

6 00829



2012166 13129 26019 21449

1

4057133 65198 73678 89494

1

60178348 206117 272810 266332

1

80543007 590695 704340 707011

1

10013789 14592 1651256 1665808



2042101 times 10




minus3

1

4021032 times 10




minus3

1

6014089 times 10




minus4

1

8010317 times 10




minus4

1

10082802 times 10




minus4


119906 =119886

8120583(119903

119886(120581 + 1) cos 120579 + 2

119886

119903((1 + 120581) cos 120579 + cos 3120579)

minus 21198863

1199033cos 3120579)

V =119886

8120583(119903

119886(120581 minus 3) sin 120579 + 2

119886

119903((1 minus 120581) sin 120579 + sin 3120579)

minus 21198863

1199033sin 3120579)

(31)





Table 4




2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion




Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












20882 1018 17627 times 10

4790617 times 10

4 1721

403362 3626 375035 times 10

4320917 times 10

5 047741

6012800 13120 102381 times 10

5130023 times 10

6 02221

8012800 13120 102381 times 10

5130023 times 10

6 0131

10025600 26240 407606 times 10

5523974 times 10

6 00829



2012166 13129 26019 21449

1

4057133 65198 73678 89494

1

60178348 206117 272810 266332

1

80543007 590695 704340 707011

1

10013789 14592 1651256 1665808



2042101 times 10




minus3

1

4021032 times 10




minus3

1

6014089 times 10




minus4

1

8010317 times 10




minus4

1

10082802 times 10




minus4


119906 =119886

8120583(119903

119886(120581 + 1) cos 120579 + 2

119886

119903((1 + 120581) cos 120579 + cos 3120579)

minus 21198863

1199033cos 3120579)

V =119886

8120583(119903

119886(120581 minus 3) sin 120579 + 2

119886

119903((1 minus 120581) sin 120579 + sin 3120579)

minus 21198863

1199033sin 3120579)

(31)





Table 4




2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion




Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












2063754 times 10

minus3 1313 24

1

4017092 times 10

minus3 18992 652 191

6077554 times 10

minus4 19489 2061 191

8041733 times 10

minus4 21541 5907 191

10027295 times 10

minus4 19028 14592 18

5 Conclusion




Acknowledgments


References
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article a new iterative method for linear systems...

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