Research Article A Numerical Approach to Static Deflection ...downloads.hindawi.com/journals/jam/2013/136358.pdf · A Numerical Approach to Static Deflection Analysis of an Infinite

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 136358 10 pageshttpdxdoiorg1011552013136358

Research ArticleA Numerical Approach to Static Deflection Analysisof an Infinite Beam on a Nonlinear Elastic FoundationOne-Way Spring Model

Jinsoo Park Hyeree Bai and T S Jang

Department of Naval Architecture and Ocean Engineering Pusan National UniversityBusan 609-735 Republic of Korea

Correspondence should be addressed to T S Jang taekpusanackr

Received 28 February 2013 Accepted 26 March 2013

Academic Editor Giuseppe Marino

Copyright copy 2013 Jinsoo Park 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

A numerical procedure proposed by Jang et al (2011) is applied for the numerical analyzing of static deflection of an infinitebeam on a nonlinear elastic foundation And one-way spring model is used for the modeling of fully nonlinear elastic foundationThe nonlinear procedure involves Greenrsquos function technique and an iterative method using the pseudo spring coefficient Theworkability of the numerical procedure is demonstrated through showing the validity of the solution and the convergence test withsome external loads

1 Introduction

Accurate modeling of nonlinear deflection of an infinitebeam on a nonlinear elastic foundation is crucial for materialand structural engineering The research can be applied tostrength analysis and practical engineering design applica-tion say to curved plate manufacturing Therefore manytheoretical and experimental studies have been carried outon the nonlinear modeling of an infinite beam on a nonlinearelastic foundation

The closed-form solutions for the static and dynamicresponse of a uniform beam resting on a linear elastic foun-dation can be found in several references [1ndash3] Timoshenko[4] Kenney [5] Saito and Murakami [6] and Fryba [7]formulated a closed-form solution using Greenrsquos functionapproach based on a linear assumption Beaufait andHoadley[8] Massalas [9] Lakshmanan [10] and Hui [11] proposedthe static dynamic and elastic stability analysis of a beamresting on a nonlinear elastic foundation And there aremanyresearches concerning the linear elastic foundation [12ndash15]Among the references Beaufait and Hoadley [8] approxi-mated the relationship of the stress-strain curve to be hyper-bolic but they approximated the bilinear curve to handlethe nonlinear problem The applied nonlinear foundation is

active only when the beam is pressing against the foundationand it is assumed to be inactive in the regions where the beamhas been displaced away from the foundation Soldatos andSelvadurai [16] also applied the hyperbolic-type nonlinearelastic foundation to analyze finite or infinite beams Leeet al [17ndash19] developed the exact and semiexact analysisof a nonuniform beam with general elastic end-restraintsKuo and Lee [20] derived the static deflection of a generalelastically end-restrained nonuniform beam on a nonlinearelastic foundation under axial and transverse forces

Recently Jang et al [21] proposed a new method forassessing the nonlinear deflection of an infinite beam ona nonlinear elastic foundation They approach the highnonlinear problems using Greenrsquos function technique withan iterative method Jang and Sung [22] proposed a newfunctional iterative method for static beam deflection whichhas a variable cross-section Choi and Jang [23] proved theexistence and uniqueness of the nonlinear deflections of aninfinite beam resting on a nonlinear elastic foundation usingthe Banach fixed point theorem Jang [24] also proposed anew iterative method for the large deflection of an infinitebeam resting on an elastic foundation based on the v Karmanapproximation of geometrical nonlinearity From the existing

2 Journal of Applied Mathematics

0Deflection 119906

Non

linea

r spr

ing

forc

e119891(119906)

One wayTwo way

Figure 1 Nonlinear elastic foundation model a nonlinear spring model (one way) and a conventional one (two way)

119891(119906)

119906

119909

Euler-Bernoulli beam

119906(119909)

119908(119909)

Separable

Figure 2 An infinite beam on a nonlinear elastic foundation a nonlinear spring model

literature a number of studies have analyzed a beam onan elastic foundation however they just use linear plus anonlinear term of spring force that is linear-cubic modelAnd they are related to the static analysis of nonuniformbeams which is resting on a nonlinear elastic foundation andthe recovered solution is not accurate or hasmany limits Fewstudies have fully adopted the nonlinear elastic foundationmodel whose spring force is based on one-way springmodelas shown in Figure 1 In the real world at the steady state thesoil or foundation would not be raised or they are separable(in Figures 1 and 2) A nonlinear spring force exists when theinfinite beam deflects downward but does not exist in case ofthe other cases

Although there are many researches fully nonlinearelastic foundation was not considered Beaufait and Hoadley[8] and Soldatos and Selvadurai [16] approximated the stress-strain relationship as a bilinear curve In this paper one-wayspring model is successfully used to examine the real non-linear elastic foundation and the nonlinear iterative methodproposed by Jang et al [21] is applied Some numerical exper-iments are carried out to report the accuracy of the methodand the convergence of the solution is investigated accordingto several physical properties of the system

2 Mathematical Modeling

21 Euler-Bernoullirsquos Beam on a Nonlinear Elastic FoundationANonlinear SpringModel In this paper the nonlinear springforce is fully analyzed by the one-way spring model instead ofthe conventional mathematical form of the two-way springmodel [25]

The well-known classical Euler-Bernoullirsquos beam the-ory is considered for the solution procedure which is a

simplification of elasticitywhich provides ameans of calculat-ing the load-carrying and deflection characteristics of beamsThe governing equation for the linear deflection of an infinitebeam on an elastic foundation that satisfies the fourth-orderdifferential equation is as follows (the weight of the beam isneglected)

1198641198681198894119906

1198891199094+ 119891 (119906) = 119908 (119909) (1)

And the reaction force 119891(119906)119891 (119906) = 119896 sdot 119906 + 119873 (119906) (2)

and 119864 119868 119896 119873(119906) and 119908(119909) are Youngrsquos modulus the massmoment of inertia a linear spring coefficient a nonlinear partof spring force and external load respectively

The boundary

119906119889119906

1198891199091198892119906

1198891199092 and 119889

3119906

1198891199093997888rarr 0 as |119909| 997888rarr infin (3)

Therefore (1) and (3) together form a well-defined boundaryvalue problem Timoshenko [4] Kenney [5] Saito andMurakami [6] and Fryba [7] derived the general linearsolutions neglecting the nonlinear part 119873(119906) in (2) asfollows

119906 (119909) = int

infin

minusinfin

119866 (119909 120585 119896) sdot 119908 (120585) 119889120585 (4)

where Greenrsquos function 119866 can be defined as

119866 (119909 120585 119896) =120572

2119896119890minus120572|120585minus119909|radic2

times sin(1205721003816100381610038161003816120585 minus 119909

1003816100381610038161003816

radic2

+120587

4) 120572 =

4radic119896119864119868

(5)

Journal of Applied Mathematics 3

minus10 minus5 0 5 10

minus05

0

05

1

119909119906(119909)

Case aCase bCase c

Figure 3 Exact solutions in Table 1

minus20

minus10

0

10

20

119908(119909)

minus10 minus5 0 5 10119909

Case aCase bCase c

Figure 4 Applied loading conditions corresponding to the exact solution in Table 1

The loading condition is assumed to be localized so 119906(119909)

in (4) satisfies the boundary conditions in (3) Deriving thelinear solution in (4) a uniformly upward nonlinear springforce depends on the beam deflection 119906

In this study the main idea of the present study is pro-posed by Jang et al [21] The nonlinear spring model is usedfor the formulation of a realistic nonlinear elastic foundationA pseudo linear spring coefficient 119896

119901and a (real) spring

force 119891(119906) are used Therefore the fourth-order differentialequation in (1) is equivalent to the following equation

1198641198681198894119906

1198891199094+ 119896119901119906 + 119891 (119906) = 119908 (119909) + 119896

119901119906 (6)

where the nonlinear spring force119891 depends on the deflection119906

119891 (119906) = 119896 sdot 119906 + 119873 (119906) for 119906 ge 00 for 119906 lt 0

(7)

or

1198641198681198894119906

1198891199094+ 119896119901119906 = 119908 (119909) + 119896

119901119906 minus 119891 (119906) equiv 120601 (8)

Table 1 Three cases of the exact solution

Case Exact solution 119906(119909)a 119890

minus1199092

b Sin119909 sdot 119890minus1199092

c Sin 119909 sdot 119890minus11990924

In (8) 119896119901119906 is the pseudolinear spring force term and is finally

compensated so it does not affect the nonlinear solutionTherefore (4) shows that (8) must be equivalent to the

following equation

119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 120601 (120585) 119889120585 (9)

where Greenrsquos function with pseudo linear spring coefficientcan be expressed as

119866(119909 120585 119896119901) =

120573


times sin(1205731003816100381610038161003816120585 minus 119909

1003816100381610038161003816

radic2

+120587

4) 120573 =

4radic119896119901119864119868

(10)



0

02

04

06

08

1

119909

119906(119909)

ExactNumerical

(a)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus05

0

05

(c)

Figure 5 Numerical solutions compared to the exact solutions

Regarding (8) the nonlinear relation for 119906 can be derived asfollows

119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906 (120585)] 119889120585

(11)

where the function119870 can be written as

119870 (119906) = 119896119901sdot 119906 minus 119891 (119906) (12)

Equation (11) is a nonlinear Fredholm integral equation of thesecond kind for 119906

22 Iterative Procedure From (11) the nonlinear iterativeprocedure is applied [21ndash24]

119906119899+1

(119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906

119899(120585)] 119889120585

(13)

where119870(119906) satisfies (7) and (12)

To examine the nonlinear iterative procedure in (13) bysimulation it should be discretized as follows

119906119899+1

(119909) =

Nsum

119895=1

119882119895119866 (119909 120585

119895 119896119901) sdot 119908 (120585

119895)

+ 119866 (119909 120585 119896119901) sdot 119870 [119906

119899(120585119895)]

119895 = 0 1 2 119873

(14)

where 119882119895denotes the weights for the integration rule The

number 119873 in the summation of (14) denotes the totalsegments of the interval (minus119877 119877) and 119877 is a sufficiently largevalue satisfying (3)

3 Numerical Experiments

In this section numerical experiments are performed todetermine the validity of the iterative method This studyassumes a nonlinear spring force 119891(119906) in (7) and examineshow accurate the solution converges to an exact solutionTheconvergence of themethod is investigated with some externalloading conditions


0

05

1

119899 = 1

119899 = 2

119899 = 3

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

(a)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

5 50minus

05

0

minus05

(c)

Figure 6 Convergence behaviors of the iterative solutions

31 Nonlinear Spring Model For simplicity the present studyconsiders an infinite beam on a nonlinear elastic foundationwhose spring force is derived as follows

119891 (119906) = 119896 sdot 119906 + 120574 sdot 119906

3 for 119906 ge 0

0 for 119906 lt 0(15)

where119873(119906) in (7) is chosen immediately as a cubic form

119873(119906) = 120574 sdot 1199063 (16)

32 Comparison with Exact Solution To determine if theiterative method converges to an exact solution the 3 casesof exact solutions listed in Table 1 are first assumed and areillustrated in Figure 3 These cases are chosen to be infinitelydifferentiable and satisfy the conditions in (3) The followingare assumed 119864 = 119868 = 119896 = 1 119896

119901= 3 and 120574 = 02 and the

initial guess of the deflection is 1199060= 0 Simpsonrsquos integration

rule is applied to the numerical integration Three cases ofexternal loads shown in Figure 4 are obtained by substitut-ing the exact solutions in Table 1 to (6) Under the load-ing condition the nonlinear iterative method in Section 2

is applied to obtain the solution Figure 5 compares the exactand numerical solutions and Figure 6 shows the convergencebehavior of the solutions in Figure 5 The errors of the solu-tions at the 119899th iteration are defined as follows

Error (119899) =1003817100381710038171003817119906exact minus 119906119899

100381710038171003817100381721003817100381710038171003817119906exact

10038171003817100381710038172

where 1199112equiv (

119873

sum

119894=1

10038161003816100381610038161199111198941003816100381610038161003816

2

)

12

(17)Three cases of Error (119899) are plotted as a function of theiteration number in Figure 7 The solutions converge at 200ndash300 iterations

33 Convergence of the Procedure The accuracy of theapplied iterative method for the nonlinear spring model isproven in Section 32 In this section 2 cases of loadingconditions are investigated to show the convergence of thesolutions

The locally distributed rectangular-type loadings inFigure 8 are taken

119908normal (single) (119909) = 1 |119909| le 1

0 otherwise(18)


100

101

102

103

0

02

04

06

The number of iterations (119899)

Erro

r (119899

)

(a)

100

101

102

103


Erro

r (119899

)

minus01

0

01

02

03

04

05

(b)

100

101

102

103

0

02

04

06


Erro

r (119899

)

(c)

Figure 7 Errors between the exact solutions and iterative solutions

minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal

(single)(119909)

Figure 8 Applied loading

Figure 9 presents the convergence behavior of the solutionsaccording to the 3 cases of 119896

119901 119864 = 119868 = 119896 = 1

and 120574 = 02 As specified in Section 21 119896119901does not

affect the converged iterative solutions but the convergencecharacteristics changed The solution converged to a steadystate faster when 119896

119901is small

In Figure 10 two cases of converged solutions are com-pared the numerical results using the one-way and two-wayspring models The loading condition is 119908normal (single)(119909) in(18)They are different only when the beamdeflected upwardthat is the deflections near 119909 = plusmn5 are larger when the beamis separable (or free) from the foundation whereas the one-way spring model is not Of course at 119909 = 0 the solutions areequivalent because the physical system is the same in Figure 1

for 119906 gt 0 Therefore Figure 10 shows the validity of theapplied iterative method for the nonlinear spring model

Another loading condition in Figure 11 is taken as follows

119908normal (119909) = 1 4 le |119909| le 5 1 le |119909| le 2

0 otherwise(19)

Other numerical experiments are also conducted accordingto the 3 cases of a linear spring coefficient 119896 whereas theproperties (119864 = 119868 = 119896 = 1 and 120574 = 02) are fixed andthe solution converged (Figure 12) Figure 13 compares thesolutions from the one-way and two-way spring models 119864 =

119868 = 119896 = 1 120574 = 02 and 119896119901= 3


minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)

Figure 9 Convergence behavior of the applied iterative method 119864 = 119868 = 119896 = 1 120574 = 02 (a) 119896119901= 3 (b) 119896

119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)

Figure 10 Validity of the solution


minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)

Figure 11 Applied loading conditions

119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)

Figure 12 Convergence behavior of the solutions 119864 = 119868 = 1 120574 = 02 and 119896119901= 3 (a) 119896 = 1 (b) 119896 = 15 and (c) 119896 = 2

Finally the validity and accuracy of the applied iterativeprocedure are investigated The nonlinear spring force isconsidered and the iterative procedure is applied successfullyfor the solution Convergence of the deflections according tothe external loading conditions is observed for the effects ofthe pseudo 119896

119901and real 119896

4 Conclusion

In this work we succeeded in applying the numerical methodproposed by Jang et al [21] to find the static deflection ofan infinite beam on a full nonlinear elastic foundation Forthat one-way spring model is considered for the formulation


minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)

minus15 minus10 minus5 0 5 10 15minus003

minus002

minus001

0

001

002

(b)

Figure 13 Validity of the solution in Figure 12(a)

of the nonlinear elastic one Since the problem concernsone-way spring force the governing equation for the staticbeam deflection is transformed as in (11) For the solution aniterative procedure is applied for the calculation of the highnonlinearity Some numerical experiments are carried out forshowing the validity and the fast convergence of the appliednumerical method And we can also find that the resultsconverge to the solutions fast for certain external loads

Acknowledgment

This work was supported by a 2-year research grant of PusanNational University

References

[1] M Hetenyi Beams on Elastic Foundation The University ofMichigan Press Ann Arbor Mich USA 1946

[2] C Miranda and K Nair ldquoFinite beams on elastic foundationrdquoJournal of the Structural Division vol 92 pp 131ndash142 1966

[3] B Y Ting ldquoFinite beams on elastic foundation with restraintsrdquoJournal of the Structural Division vol 108 no 3 pp 611ndash6211982

[4] S Timoshenko ldquoMethod of analysis of statistical and dynamicalstress in railrdquo in Proceeding of the International Congress ofApplied Mechanics pp 407ndash418 Zurich Switzerland 1926

[5] J T Kenney ldquoSteady-state vibrations of beam on elastic founda-tion for moving loadrdquo Journal of Applied Mechanics vol 21 pp359ndash364 1954

[6] H Saito and TMurakami ldquoVibrations of an infinite beam on anelastic foundation with consideration of mass of a foundationrdquoThe Japan Society of Mechanical Engineers vol 12 pp 200ndash2051969

[7] L Fryba ldquoInfinitely long beam on elastic foundation undermoving loadrdquo Aplikace Matematiky vol 2 pp 105ndash132 1957

[8] F W Beaufait and P W Hoadley ldquoAnalysis of elastic beams onnonlinear foundationsrdquo Computers amp Structures vol 12 no 5pp 669ndash676 1980

[9] C Massalas ldquoFundamental frequency of vibration of a beam ona non-linear elastic foundationrdquo Journal of Sound andVibrationvol 54 no 4 pp 613ndash615 1977

[10] M Lakshmanan ldquoComment on the fundamental frequency ofvibration of a beam on a non-linear elastic foundationrdquo Journalof Sound and Vibration vol 58 no 3 pp 455ndash457 1978

[11] D Hui ldquoPostbuckling behavior of infinite beams on elasticfoundations using Koiterrsquos improved theoryrdquo International Jour-nal of Non-Linear Mechanics vol 23 no 2 pp 113ndash123 1988

[12] D Z Yankelevsky and M Eisenberger ldquoAnalysis of a beam col-umn on elastic foundationrdquoComputers amp Structures vol 23 no3 pp 351ndash356 1986

[13] M Eisenberger and J Clastornik ldquoBeams on variable two-parameter elastic foundationrdquo Journal of EngineeringMechanicsvol 113 no 10 pp l454ndash1466 1987

[14] D Karamanlidis and V Prakash ldquoExact transfer and stiffnessmatrices for a beamcolumn resting on a two-parameter foun-dationrdquo Computer Methods in Applied Mechanics and Engineer-ing vol 72 no 1 pp 77ndash89 1989

[15] S N Sirosh A Ghali and A G Razaqpur ldquoA general finite ele-ment for beams or beam-columns with or without an elasticfoundationrdquo International Journal for Numerical Methods inEngineering vol 28 no 5 pp 1061ndash1076 1989

[16] K P Soldatos and A P S Selvadurai ldquoFlexure of beams rest-ing on hyperbolic elastic foundationsrdquo International Journal ofSolids amp Structures vol 21 no 4 pp 373ndash388 1985

[17] S Y Lee H Y Ke and Y H Kuo ldquoExact static deflection ofa non-uniform bernoulli-euler beam with general elastic endrestraintsrdquoComputersampStructures vol 36 no 1 pp 91ndash97 1990

[18] S Y Lee and H Y Ke ldquoFree vibrations of a non-uniformbeam with general elastically restrained boundary conditionsrdquoJournal of Sound andVibration vol 136 no 3 pp 425ndash437 1990

[19] S Y Lee H Y Ke and Y H Kuo ldquoAnalysis of non-uniformbeam vibrationrdquo Journal of Sound and Vibration vol 142 no1 pp 15ndash29 1990

[20] Y H Kuo and S Y Lee ldquoDeflection of nonuniform beams rest-ing on a nonlinear elastic foundationrdquo Computers amp Structuresvol 51 no 5 pp 513ndash519 1994

[21] T S Jang H S Baek and J K Paik ldquoA newmethod for the non-linear deflection analysis of an infinite beam resting on a non-linear elastic foundationrdquo International Journal of Non-LinearMechanics vol 46 no 1 pp 339ndash346 2011

[22] T S Jang and H G Sung ldquoA new semi-analytical method forthe non-linear static analysis of an infinite beam on a non-linear


elastic foundation a general approach to a variable beam cross-sectionrdquo International Journal of Non-Linear Mechanics vol 47pp 132ndash139 2012

[23] S W Choi and T S Jang ldquoExistence and uniqueness of non-linear deflections of an infinite beam resting on a non-uniformnonlinear elastic foundationrdquo Boundary Value Problems vol2012 article 5 2012

[24] T S Jang ldquoA new semi-analytical approach to large deflectionsof BernoullindashEuler-v Karman beams on a linear elastic founda-tion nonlinear analysis of infinite beamsrdquo International Journalof Mechanical Sciences vol 66 pp 22ndash32 2013

[25] M D Greenberg Foundations of Applied Mathematics Pren-tice-Hall New Jersey NJ USA 1978

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Stochastic AnalysisInternational Journal of


0Deflection 119906

Non

linea

r spr

ing

forc

e119891(119906)

One wayTwo way

Figure 1 Nonlinear elastic foundation model a nonlinear spring model (one way) and a conventional one (two way)

119891(119906)

119906

119909

Euler-Bernoulli beam

119906(119909)

119908(119909)

Separable

Figure 2 An infinite beam on a nonlinear elastic foundation a nonlinear spring model

literature a number of studies have analyzed a beam onan elastic foundation however they just use linear plus anonlinear term of spring force that is linear-cubic modelAnd they are related to the static analysis of nonuniformbeams which is resting on a nonlinear elastic foundation andthe recovered solution is not accurate or hasmany limits Fewstudies have fully adopted the nonlinear elastic foundationmodel whose spring force is based on one-way springmodelas shown in Figure 1 In the real world at the steady state thesoil or foundation would not be raised or they are separable(in Figures 1 and 2) A nonlinear spring force exists when theinfinite beam deflects downward but does not exist in case ofthe other cases

Although there are many researches fully nonlinearelastic foundation was not considered Beaufait and Hoadley[8] and Soldatos and Selvadurai [16] approximated the stress-strain relationship as a bilinear curve In this paper one-wayspring model is successfully used to examine the real non-linear elastic foundation and the nonlinear iterative methodproposed by Jang et al [21] is applied Some numerical exper-iments are carried out to report the accuracy of the methodand the convergence of the solution is investigated accordingto several physical properties of the system

2 Mathematical Modeling

21 Euler-Bernoullirsquos Beam on a Nonlinear Elastic FoundationANonlinear SpringModel In this paper the nonlinear springforce is fully analyzed by the one-way spring model instead ofthe conventional mathematical form of the two-way springmodel [25]

The well-known classical Euler-Bernoullirsquos beam the-ory is considered for the solution procedure which is a

simplification of elasticitywhich provides ameans of calculat-ing the load-carrying and deflection characteristics of beamsThe governing equation for the linear deflection of an infinitebeam on an elastic foundation that satisfies the fourth-orderdifferential equation is as follows (the weight of the beam isneglected)

1198641198681198894119906

1198891199094+ 119891 (119906) = 119908 (119909) (1)

And the reaction force 119891(119906)119891 (119906) = 119896 sdot 119906 + 119873 (119906) (2)

and 119864 119868 119896 119873(119906) and 119908(119909) are Youngrsquos modulus the massmoment of inertia a linear spring coefficient a nonlinear partof spring force and external load respectively

The boundary

119906119889119906

1198891199091198892119906

1198891199092 and 119889

3119906

1198891199093997888rarr 0 as |119909| 997888rarr infin (3)

Therefore (1) and (3) together form a well-defined boundaryvalue problem Timoshenko [4] Kenney [5] Saito andMurakami [6] and Fryba [7] derived the general linearsolutions neglecting the nonlinear part 119873(119906) in (2) asfollows

119906 (119909) = int

infin

minusinfin

119866 (119909 120585 119896) sdot 119908 (120585) 119889120585 (4)

where Greenrsquos function 119866 can be defined as

119866 (119909 120585 119896) =120572


times sin(1205721003816100381610038161003816120585 minus 119909

1003816100381610038161003816

radic2

+120587

4) 120572 =

4radic119896119864119868

(5)



minus05

0

05

1

119909119906(119909)

Case aCase bCase c


minus20

minus10

0

10

20

119908(119909)

minus10 minus5 0 5 10119909

Case aCase bCase c







1198641198681198894119906

1198891199094+ 119896119901119906 + 119891 (119906) = 119908 (119909) + 119896

119901119906 (6)


119891 (119906) = 119896 sdot 119906 + 119873 (119906) for 119906 ge 00 for 119906 lt 0

(7)

or

1198641198681198894119906

1198891199094+ 119896119901119906 = 119908 (119909) + 119896

119901119906 minus 119891 (119906) equiv 120601 (8)



minus1199092

b Sin119909 sdot 119890minus1199092

c Sin 119909 sdot 119890minus11990924



following equation

119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 120601 (120585) 119889120585 (9)


119866(119909 120585 119896119901) =

120573


times sin(1205731003816100381610038161003816120585 minus 119909

1003816100381610038161003816

radic2

+120587

4) 120573 =

4radic119896119901119864119868

(10)



0

02

04

06

08

1

119909

119906(119909)

ExactNumerical

(a)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus05

0

05

(c)



119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906 (120585)] 119889120585

(11)


119870 (119906) = 119896119901sdot 119906 minus 119891 (119906) (12)



119906119899+1

(119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906

119899(120585)] 119889120585

(13)



119906119899+1

(119909) =

Nsum

119895=1

119882119895119866 (119909 120585

119895 119896119901) sdot 119908 (120585

119895)

+ 119866 (119909 120585 119896119901) sdot 119870 [119906

119899(120585119895)]

119895 = 0 1 2 119873

(14)






0

05

1

119899 = 1

119899 = 2

119899 = 3

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

(a)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

5 50minus

05

0

minus05

(c)



119891 (119906) = 119896 sdot 119906 + 120574 sdot 119906

3 for 119906 ge 0

0 for 119906 lt 0(15)


119873(119906) = 120574 sdot 1199063 (16)


119901= 3 and 120574 = 02 and the





100381710038171003817100381721003817100381710038171003817119906exact

10038171003817100381710038172


119873

sum

119894=1

10038161003816100381610038161199111198941003816100381610038161003816

2

)

12




119908normal (single) (119909) = 1 |119909| le 1

0 otherwise(18)


100

101

102

103

0

02

04

06


Erro

r (119899

)

(a)

100

101

102

103


Erro

r (119899

)

minus01

0

01

02

03

04

05

(b)

100

101

102

103

0

02

04

06


Erro

r (119899

)

(c)


minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal

(single)(119909)



119901 119864 = 119868 = 119896 = 1



119901is small




119908normal (119909) = 1 4 le |119909| le 5 1 le |119909| le 2

0 otherwise(19)


119868 = 119896 = 1 120574 = 02 and 119896119901= 3


minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)


119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)



minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











minus05

0

05

1

119909119906(119909)

Case aCase bCase c


minus20

minus10

0

10

20

119908(119909)

minus10 minus5 0 5 10119909

Case aCase bCase c







1198641198681198894119906

1198891199094+ 119896119901119906 + 119891 (119906) = 119908 (119909) + 119896

119901119906 (6)


119891 (119906) = 119896 sdot 119906 + 119873 (119906) for 119906 ge 00 for 119906 lt 0

(7)

or

1198641198681198894119906

1198891199094+ 119896119901119906 = 119908 (119909) + 119896

119901119906 minus 119891 (119906) equiv 120601 (8)



minus1199092

b Sin119909 sdot 119890minus1199092

c Sin 119909 sdot 119890minus11990924



following equation

119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 120601 (120585) 119889120585 (9)


119866(119909 120585 119896119901) =

120573


times sin(1205731003816100381610038161003816120585 minus 119909

1003816100381610038161003816

radic2

+120587

4) 120573 =

4radic119896119901119864119868

(10)



0

02

04

06

08

1

119909

119906(119909)

ExactNumerical

(a)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus05

0

05

(c)



119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906 (120585)] 119889120585

(11)


119870 (119906) = 119896119901sdot 119906 minus 119891 (119906) (12)



119906119899+1

(119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906

119899(120585)] 119889120585

(13)



119906119899+1

(119909) =

Nsum

119895=1

119882119895119866 (119909 120585

119895 119896119901) sdot 119908 (120585

119895)

+ 119866 (119909 120585 119896119901) sdot 119870 [119906

119899(120585119895)]

119895 = 0 1 2 119873

(14)






0

05

1

119899 = 1

119899 = 2

119899 = 3

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

(a)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

5 50minus

05

0

minus05

(c)



119891 (119906) = 119896 sdot 119906 + 120574 sdot 119906

3 for 119906 ge 0

0 for 119906 lt 0(15)


119873(119906) = 120574 sdot 1199063 (16)


119901= 3 and 120574 = 02 and the





100381710038171003817100381721003817100381710038171003817119906exact

10038171003817100381710038172


119873

sum

119894=1

10038161003816100381610038161199111198941003816100381610038161003816

2

)

12




119908normal (single) (119909) = 1 |119909| le 1

0 otherwise(18)


100

101

102

103

0

02

04

06


Erro

r (119899

)

(a)

100

101

102

103


Erro

r (119899

)

minus01

0

01

02

03

04

05

(b)

100

101

102

103

0

02

04

06


Erro

r (119899

)

(c)


minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal

(single)(119909)



119901 119864 = 119868 = 119896 = 1



119901is small




119908normal (119909) = 1 4 le |119909| le 5 1 le |119909| le 2

0 otherwise(19)


119868 = 119896 = 1 120574 = 02 and 119896119901= 3


minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)


119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)



minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











0

02

04

06

08

1

119909

119906(119909)

ExactNumerical

(a)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

ExactNumerical

minus10 minus5 0 5 10119909

119906(119909)

minus05

0

05

(c)



119906 (119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906 (120585)] 119889120585

(11)


119870 (119906) = 119896119901sdot 119906 minus 119891 (119906) (12)



119906119899+1

(119909) = int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119908 (120585) 119889120585

+ int

infin

minusinfin

119866(119909 120585 119896119901) sdot 119870 [119906

119899(120585)] 119889120585

(13)



119906119899+1

(119909) =

Nsum

119895=1

119882119895119866 (119909 120585

119895 119896119901) sdot 119908 (120585

119895)

+ 119866 (119909 120585 119896119901) sdot 119870 [119906

119899(120585119895)]

119895 = 0 1 2 119873

(14)






0

05

1

119899 = 1

119899 = 2

119899 = 3

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

(a)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

5 50minus

05

0

minus05

(c)



119891 (119906) = 119896 sdot 119906 + 120574 sdot 119906

3 for 119906 ge 0

0 for 119906 lt 0(15)


119873(119906) = 120574 sdot 1199063 (16)


119901= 3 and 120574 = 02 and the





100381710038171003817100381721003817100381710038171003817119906exact

10038171003817100381710038172


119873

sum

119894=1

10038161003816100381610038161199111198941003816100381610038161003816

2

)

12




119908normal (single) (119909) = 1 |119909| le 1

0 otherwise(18)


100

101

102

103

0

02

04

06


Erro

r (119899

)

(a)

100

101

102

103


Erro

r (119899

)

minus01

0

01

02

03

04

05

(b)

100

101

102

103

0

02

04

06


Erro

r (119899

)

(c)


minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal

(single)(119909)



119901 119864 = 119868 = 119896 = 1



119901is small




119908normal (119909) = 1 4 le |119909| le 5 1 le |119909| le 2

0 otherwise(19)


119868 = 119896 = 1 120574 = 02 and 119896119901= 3


minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)


119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)



minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

05

1

119899 = 1

119899 = 2

119899 = 3

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

(a)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

minus04

minus02

0

02

04

(b)

119899 = 1

119899 = 5

119899 = 10

119899 = 300

Exact solution

minus10 minus5 0 5 10119909

119906(119909)

5 50minus

05

0

minus05

(c)



119891 (119906) = 119896 sdot 119906 + 120574 sdot 119906

3 for 119906 ge 0

0 for 119906 lt 0(15)


119873(119906) = 120574 sdot 1199063 (16)


119901= 3 and 120574 = 02 and the





100381710038171003817100381721003817100381710038171003817119906exact

10038171003817100381710038172


119873

sum

119894=1

10038161003816100381610038161199111198941003816100381610038161003816

2

)

12




119908normal (single) (119909) = 1 |119909| le 1

0 otherwise(18)


100

101

102

103

0

02

04

06


Erro

r (119899

)

(a)

100

101

102

103


Erro

r (119899

)

minus01

0

01

02

03

04

05

(b)

100

101

102

103

0

02

04

06


Erro

r (119899

)

(c)


minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal

(single)(119909)



119901 119864 = 119868 = 119896 = 1



119901is small




119908normal (119909) = 1 4 le |119909| le 5 1 le |119909| le 2

0 otherwise(19)


119868 = 119896 = 1 120574 = 02 and 119896119901= 3


minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)


119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)



minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










100

101

102

103

0

02

04

06


Erro

r (119899

)

(a)

100

101

102

103


Erro

r (119899

)

minus01

0

01

02

03

04

05

(b)

100

101

102

103

0

02

04

06


Erro

r (119899

)

(c)


minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal

(single)(119909)



119901 119864 = 119868 = 119896 = 1



119901is small




119908normal (119909) = 1 4 le |119909| le 5 1 le |119909| le 2

0 otherwise(19)


119868 = 119896 = 1 120574 = 02 and 119896119901= 3


minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)


119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)



minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










minus02

0

02

04

06

08

119899 = 1

119899 = 2

119899 = 5

119899 = 10

119899 = 30

minus20 minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

(a)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 30

119899 = 60

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(b)

minus02

0

02

04

06

08

119899 = 1

119899 = 3

119899 = 10

119899 = 40

119899 = 90

minus10 0 10 20119909

minus15 minus5 5 15

119906(119909)

minus20

(c)


119901= 6 and (c) 119896

119901= 10

minus20 minus10 0 10 20119909

119906(119909)

One wayTwo way

0

02

04

06

08

minus02

(a)

One wayTwo way

minus5 0 5

minus004

minus002

0

002

004

006

119909

119906(119909)

(b)



minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










minus10 minus5 0 5 10119909

minus1

0

1

2

3

119908no

rmal(119909)


119899 = 1

119899 = 2

119899 = 4

119899 = 20

119899 = 40

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(a)

119899 = 1

119899 = 2

119899 = 3

119899 = 10

119899 = 20

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(b)

119899 = 1

119899 = 2

119899 = 3

119899 = 5

119899 = 10

minus10minus20 0 10 20119909

minus15 minus5 5 15

05

04

03

02

01

0

119906(119909)

(c)



119901and real 119896

4 Conclusion



minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










minus20 minus10 0 10 20119909

One wayTwo way

119906(119909)

0

02

04

06

(a)

One wayTwo way

119909

119906(119909)


minus002

minus001

0

001

002

(b)



Acknowledgment


References



































Volume 2014




Journal of











Journal of


Function Spaces






Algebra





















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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