Research Article Prestress Force Identification for

Research ArticlePrestress Force Identification for ExternallyPrestressed Concrete Beam Based on FrequencyEquation and Measured Frequencies

Luning Shi Haoxiang He and Weiming Yan

Beijing Laboratory of Earthquake Engineering and Structure Retrofit Beijing University of Technology Beijing 100124 China

Correspondence should be addressed to Haoxiang He hhx7856163com

Received 29 March 2014 Accepted 14 May 2014 Published 29 May 2014

Academic Editor Qingsong Xu

Copyright copy 2014 Luning Shi 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 prestress force identification method for externally prestressed concrete uniform beam based on the frequency equation and themeasured frequencies is developed For the purpose of the prestress force identification accuracy we first look for the appropriatemethod to solve the free vibration equation of externally prestressed concrete beam and then combine the measured frequencieswith frequency equation to identify the prestress force To obtain the exact solution of the free vibration equation of multispanexternally prestressed concrete beam an analytical model of externally prestressed concrete beam is set up based on the Bernoulli-Euler beam theory and the function relation between prestress variation and vibration displacement is built The multispanexternally prestressed concrete beam is taken as themultiple single-span beamswhichmustmeet the bendingmoment and rotationangle boundary conditions the free vibration equation is solved using sublevel simultaneous method and the semi-analyticalsolution of the free vibration equation which considered the influence of prestress on section rigidity and beam length is obtainedTaking simply supported concrete beam and two-span concrete beamwith external tendons as examples frequency function curvesare obtained with the measured frequencies into it and the prestress force can be identified using the abscissa of the crosspoint offrequency functions Identification value of the prestress force is in good agreement with the test resultsThemethod can accuratelyidentify prestress force of externally prestressed concrete beam and trace the trend of effective prestress force

1 Introduction

Externally prestressed concrete structure is broadly applied inthe highway bridges urban bridges and railway bridges withthe development of external prestress technology In designand construction process of externally prestressed concretebridge the prestress force is often determined according tothe theory formula [1] But in the actual construction processmany factors such as relaxation of steel shrinkage and creepof concrete and ambient temperature can lead to the changeof the prestress force and the prestress force can show obviouschange when the concrete beam has the cracks or failureTherefore in order to effectively control the operating stateand the bearing capacity of bridges it is very important toidentify the prestress force of externally prestressed concretebridge The existing method which has good accuracy is

to install force sensors in the prestressed concrete beam tomonitor the change of the prestress forceThe disadvantage ofthis approach is that the sensor is expensive and the accuracyof the force sensor will decrease with the increase of agein services Above all it is necessary to find a simple andeffective method to identify the prestress force In recentyears scholars did a lot of research on identification ofprestress force and obtained some results

Lu and Law [2] presented a method for the identificationof prestress force of a prestressed concrete bridge deck usingthe measured structural dynamic responses and the prestressforce is identified using a sensitivity-based finite elementmodel updating method in the inverse analysis Law and Lu[3] also studied the time-domain response of a prestressedEuler-Bernoulli beam under external excitation based onmodal superposition and the prestress force is identified in

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 840937 13 pageshttpdxdoiorg1011552014840937

2 Mathematical Problems in Engineering

the time domain by a system identification approach Liet al [4] carried out numerical simulations to identify themagnitude of prestress force in a highway bridge by makinguse of the dynamic responses from moving vehicular loadsbased on dynamic response sensitivity-based finite elementmodel updating Law et al [5] developed a new method ofprestress identification using the wavelet-based method inwhich the approximation of the measured response is used toform the identification equation Bu and Wang [6] presenteda BP neural networkmethod to identify the effective prestressfor a simply supported PRC beam bridge based on modalfrequencies and dynamic responses of the bridge Abrahamet al [7] investigated the feasibility of using damage locationalgorithm technique for detecting loss of prestress in aprestressed concrete bridge Kim et al [8] studied a vibration-based method to detect prestress loss in beam-type PSCbridges by monitoring changes in a few natural frequenciesXuan et al [9] evaluated the prestress loss quantitativelyin the steel-strand reinforced structures by an optical fiber-sensor based monitoring technique However the prestressforce and prestress loss cannot be estimated directly simplyand accurately unless the beam has been instrumented atthe time of construction Several researchers also studied thedynamic behavior of prestressed beam with external tendonsand predicted the relation between the modal frequency andthe given prestress force Miyamoto et al [10] studied theeffect of the prestressing force introduced by the externaltendons on the vibration characteristics of a composite girderwith the results of dynamic tests and derived the formula forcalculating the natural frequency of a composite girder basedon a vibration equation Hamed and Frostig [11] presentedthe effect of the magnitude of the prestressing force on thenatural frequencies of prestressed beams with bonded andunbonded tendons Saiidi et al [12 13] reported a studyon modal frequency due to the prestress force with labora-tory test results The above researchers only considered theprestressing effect on dynamic characteristics of the simplysupported beam Very few works have been presented on theeffect of prestressing on the dynamic responses of a beam andidentification of prestress force directly or indirectly

The exact solution of the free vibration equation ofmultispan externally prestressed concrete uniform beam isobtained in this paper An inverse problem to identify theprestress force based on the frequency equation and themeasured frequencies is then presented taking the prestressforce as an unknown parameter in the frequency functionsThe prestress force identification method is suited to theexternally prestressed concrete uniform beam Firstly basedonMiyamoto et alrsquos study [10] the function relation betweenprestress variation and vibration displacement of multispanexternally prestressed concrete beam is built according to thebasic principle of the force methodThemultispan externallyprestressed concrete beam is considered as the multiplesingle-span beams which must meet the bending momentand rotation angle boundary conditions The free vibrationequations of multispan externally prestressed concrete beamby using sublevel simultaneous method which can simplifythe solution of dynamic equations are given and the semi-analytical solution of the free vibration equations which

x

dx

NN

uu(x t)

Figure 1 Analysis model of vibration system

considered the influence of prestress on section rigidity andbeam length is obtained Then frequency functions whichare obtained by frequency equation are used to identify theprestress force by the appropriate method Two dynamic testsof externally prestressed concrete beam in the laboratory aresubmitted to illustrate the effectiveness and robustness ofthe proposed method At last the effect of the error of themeasured frequencies on identification of the prestress forceis studied in the proposed method

2 Vibration Equation of Multispan ExternallyPrestressed Concrete Beam

21 Vibration Equation of Externally Prestressed Simply Sup-ported Beam An externally prestressed simply supportedbeam is shown in Figure 1 It is assumed that the prestressforce 119873 has no prestressing loss along the beam length andthe beam bending must meet the plane section assumptionThe vibration equation of this simply supported beam can beexpressed as follows

1205972

1205971199092[119864119868

1205972119906 (119909 119905)

1205971199092] + 119873

119909

1205972119906 (119909 119905)

1205971199092minus 119867

1205972(Δ119873)

1205971199092

+ 1198981205972119906 (119909 119905)

1205971199052= 0

(1)

where 119864119868 is the flexural rigidity of the beam119898 is the mass ofthe beam per unit length 120583(119909 119905) is the transverse deflection119873119909is the horizontal component of the prestress force119873119867 is

the equivalent eccentricity of the external tendons andΔ119873 isthe variation of the prestress force due to flexural vibrationBecause eccentricity of external tendons in different positionson the beam is not the same the equivalent eccentricity 119867can be calculated according to the principle of equal area inthe bending moment diagram

22 Vibration Equation of Multispan Externally PrestressedBeam A multispan externally prestressed continuous beamwhich has 119899 spans is shown in Figure 2 and the 119894th span ofthe beam is taken as the study subject The rotation angleand bending moment of the beam end at point 119894 are 120579

119894119894+1

and 119872119894119894+1

and the rotation angle and bending moment ofthe beam end at point 119894 + 1 are 120579

119894+1119894and119872

119894+1119894 respectively

Mathematical Problems in Engineering 3

xNi

u120579ii+1 120579i+1i

i + 1

Mi+1Mi

ui(x t)

Figure 2 Analysis model of the 119894th span of the beam

According to (1) the free vibration equation of the 119894th span ofthe beam can be written as follows

1205972

1205971199092[119864119868

1205972119906119894(119909 119905)

1205971199092] + 119873

119909119894

1205972119906119894(119909 119905)

1205971199092minus 119867119894

1205972(Δ119873119894)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(2)

where 120583119894(119909 119905) is the transverse deflection of the 119894th span119867

119894is

the equivalent eccentricity of the external tendons of the 119894thspanΔ119873

119894is the variation of the prestress force due to flexural

vibration of the 119894th span and119873119909119894is the horizontal component

of the prestress force119873119894 of the 119894th spanThe rotation angle and bending moment at both ends

of the 119894th span of the beam need to satisfy the followingboundary conditions

120579119894119894minus1

= 120579119894119894+1

119872119894119894minus1

= 119872119894119894+1

120579119894+1119894

= 120579119894+1119894+2

119872119894+1119894

= 119872119894+1119894+2

(3)

The first and the last span of multispan externally pre-stressed concrete beam must meet the boundary conditions

11987212= 0 120579

21= 12057923 119872

21= 11987223

119872119899+1119899

= 0 120579119899119899+1

= 120579119899119899minus1

119872119899119899+1

= 119872119899119899minus1

(4)

Obviously the free vibration equation of multispan exter-nally prestressed concrete beam can be considered to be thefree vibration equations of multiple single-span externallyprestressed beams which must satisfy the rotation angleand bending moment boundary conditions as shown in (3)and (4) In order to solve the vibration equations relationsbetween prestress variation Δ119873 and vibration displacement119906(119909 119905) should be defined firstly

23 Relations between Prestress Variation and Vibration Dis-placement Prestress force would change as the vibration dis-placement during the free vibration of multispan externallyprestressed concrete beam the free vibration of the beamis considered in small deformation condition so relationsbetween prestress variation Δ119873 and vibration displacement119906(119909 119905) can approximatively be seen as a linear relationshipon the geometric deformation [10 14] Assume that thereis a concentrated force 119865 on the midspan of the beamto get the relations between prestress variation Δ119873 andconcentrated force 119865 then to obtain the relations betweenvibration displacement 119906(119909 119905) and concentrated force 119865 andat last to find the relationship between prestress variationΔ119873and vibration displacement 119906(119909 119905) by variable replacing

The side span (119894 = 1 119899) of the multispan externallyprestressed concrete beam can be simplified approximatelyas the structure shown in Figure 3(a) Concentrated force119865 acts on the midspan of the side span beam modelthe prestress variation Δ119873 and bending moment on thesupport are identified as the unknown forces 119883

1and 119883

2

and the basic system can be generated after removing theredundant constraints The bending moment diagrams withthe unknown forces 119883

1= 1 and 119883

2= 1 and concentrated

force 119865 acting on the beam model are shown in Figure 3(a)The deformation compatibility equations can be written asfollows

120575111198831+ 120575121198832+ Δ1119865= 0

120575211198831+ 120575221198832+ Δ2119865= 0

(5)

where 120575119894119895= sumint(119872

119894119872119895119864119868)119889119909 + sumint(119873

119894119873119895119864119860)119889119909 Δ

119894119865=

sumint(119872119894119872119865119864119868)119889119909 + sumint(119873

119894119873119865119864119860)119889119909 119894 = 1 2 119895 = 1 2

Equation (5) can be rewritten as follows

Δ119873 =12057512Δ2119865minus 12057522Δ1119865

1205751112057522minus 1205751212057521

(6)

The vertical displacement 120583119865on the midspan of the side

span beam model can be expressed as follows

120583119865=71198651198713

768119864119868 (7)

Substituting (6) into (7) we can get the following

120583119865=

7 (1205751112057522minus 1205751212057521) 1198651198713

768 (12057512Δ2119865minus 12057522Δ1119865) 119864119868

Δ119873 (8)

When concentrated force 119865 acts on the midspan of thebeam model the vertical displacement 120583

119865can be produced

at the midspan and external tendons can produce internalforce which will produce the prestress variation Δ119873 At thesame time internal forcewill lead to the vertical displacement120583Δ119873

which has the opposite direction of the 120583119865 The vertical

displacement 120583Δ119873

can be written as follows

120583Δ119873

=12057522Δ1119865minus 12057512Δ2119865

12057522119864119868119865

Δ119873 (9)

The vertical displacement 120583 which is caused by theconcentrated force 119865 can be calculated as follows

120583 = 120583119865minus 120583Δ119873 (10)

Substituting (8) and (9) into (10) we can obtain

Δ119873 = 120601120583 (11)


NN

F

F

MF

X1

X2

M1

M2

N1

(a) The side span model

N

F

F

N

MF

X1

X2X3

M1

M2

M3

N1

(b) The middle span model

Figure 3 The analysis model and bending moment diagram

where

120601 = (119864119868)

times (7 (1205751112057522minus 1205751212057521) 1198651198713

768 (12057512Δ2119865minus 12057522Δ1119865)minus(12057522Δ1119865minus 12057512Δ2119865

12057522119865

)

minus1

(12)

Themiddle span (2 ⩽ 119894 ⩽ 119899minus1) of themultispan externallyprestressed concrete beam can be simplified as the structurewhich is shown in Figure 3(b) Concentrated force 119865 acts onthe midspan of the middle span beam model and unknownforces are 119883

1 1198832 and 119883

3 The deformation compatibility

equations can be expressed as follows

120575111198831+ 120575121198832+ 120575131198833+ Δ1119865= 0

120575211198831+ 120575221198832+ 120575231198833+ Δ2119865= 0

120575311198831+ 120575321198832+ 120575331198833+ Δ3119865= 0

(13)

where 120575119894119895and Δ

119894119865can be calculated by (5) and (13) can be

rewritten as follows

Δ119873 =119863Δ

1198630

(14)

where

119863Δ=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1205751212057513Δ1119865

1205752212057523Δ2119865

1205753212057533Δ3119865

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198630=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120575111205751212057513

120575211205752212057523

120575311205753212057533

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(15)

The vertical displacement 120583119865on the midspan of the

middle span model can be expressed as follows

120583119865=

1198651198713

192119864119868 (16)

The vertical displacement 120583Δ119873

caused by internal forcecan be written as follows

120583Δ119873

=(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119864119868119865

Δ119873 (17)

Substituting (16) and (17) into (10) the relationship betweenprestress variationΔ119873 and vibration displacement 119906(119909 119905) canbe expressed as in (11) But the coefficient 120601 can be written asfollows

120601 = (119864119868)

times (11986511987131198630

192119863Δ

minus(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119865

)

minus1

(18)

24 Equivalent Eccentricity119867 The equivalent eccentricity119867can be computed according to the principle which is thatthe areas of the bending moment diagram are equal [10 14]As shown in Figure 3(b) the bending moment of the middlespan caused by external tendons can be written as follows

119872119873= (119872

1minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198723)Δ119873

(19)

The area of the bending moment diagram is

119878119872119873

= (1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)Δ119873

(20)


where 119878119872119873

1198781198721

1198781198722

and 1198781198723

are the areas of the bendingmoment diagram which are shown in Figure 3(b)The equiv-alent eccentricity 119867 can be written as 119867 = 119878

119872119873(Δ119873 times 119871)

where 119871 is span length Equation (20) can be written asfollows

119867 =1

119871(1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)

(21)

Similarly the equivalent eccentricity 119867 of the side spancan be written as follows

119867 =1

119871(1198781198721minus12057512

12057522

1198781198722) (22)

Substituting (11) into (2) we can get

1198641198681205974119906119894(119909 119905)

1205971199094+ (119873119909119894minus 119867119894120601119894)1205972119906119894(119909 119905)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(23)

Equation (23) is the free vibration equation of the multispanexternally prestressed concrete beam and the section rigidityand beam length can be modified as follows

Kim et al [8] considered that the total rigidity ofprestressed beam 119864119868 is the sum of the flexural stiffness ofreinforced concrete beam 119864

119888119868119888and the flexural stiffness of

the prestressed steel 119864119904119868119904and took the prestressed steel as the

cable which is fixed at both ends of the beam According tothe principle that the natural frequency of the cable is equalto that of the beam we can obtain

119864119904119868119904= 119873(

119871119894

119899120587)

2

(24)

The total rigidity of prestressed beam can be written asfollows

119864119868 = 119864119888119868119888+ 119873(

119871119894

119899120587)

2

(25)

where 119871119894is the beam length of the 119894th span and 119899 is the modal

orderThe prestress force on the cross section can be regarded as

an axial force and a moment and the beam length will changeunder the axial force [15] The actual beam length of the 119894thspan can be written as follows

1198711015840

119894= (1 minus

119873119909119894

119864119860)119871119894 (26)

where 1198711015840119894is the actual beam length of the 119894th spanThe section

rigidity and beam length in (23) can be corrected accordingto (25) and (26) before solving it

3 Frequency Equation of Multispan ExternallyPrestressed Concrete Beam

31 To Solve the Vibration Equation Xiong et al [14 16]utilized Dirac function to establish vibration equation ofexternally prestressed continuous beam and this method isnot suitable for the solution of the vibration equation of three-span and more than three-span externally prestressed con-tinuous beam This paper translates the vibration equationof the multispan externally prestressed concrete beam intovibration equations of multi-single-span beams which mustsatisfy the rotation angle and bending moment conditionsAccording to (23) the vibration equation of 119894th single-spanbeam can be simplified as follows

1205974119906119894(119909 119905)

1205971199094+119873119909119894minus 119867119894120601119894

119864119868

1205972119906119894(119909 119905)

1205971199092+119898

119864119868

1205972119906119894(119909 119905)

1205971199052= 0

(27)

For any mode of vibration the lateral deflection 120583119894(119909 119905)

may be written in the form [17]

119906119894(119909 119905) = 120601

119894(119909) 119884 (119905) (28)

where 120601119894(119909) is the modal deflection and 119884(119905) is a harmonic

function of time 119905 Then substitution of (28) into (27) yields

1206011015840101584010158401015840

119894(119909) + 119892

212060110158401015840

119894(119909) minus 119886

4120601119894(119909) = 0 (29)

11988410158401015840(119905) + 120596

2119884 (119905) = 0 (30)

where1205962 = 11988641198641198681198981198922119894= (119873119909119894minus119867119894120601119894)119864119868 Equation (29) is the

fourth order constant coefficient differential equation and theassumption that the solution of (29) is Φ

119894(119909) = 119866119890

119904119909 Takingit into (29) we can get

11990412= plusmn119894ℎ119894 119904

34= plusmn119894119899119894 (31)

where ℎ119894

= radic(1198864 + (119892119894

44))12+ (119892119894

22) 119899119894

=

radic(1198864 + (119892119894

44))12minus (119892119894

22)The general solution of (29) can be written as follows

120601119894(119909) = 119860 sin (ℎ

119894119909) + 119861 cos (ℎ

119894119909)

+ 119862 sinh (119899119894119909) + 119863 cosh (119899

119894119909)

(32)

where119860 119861 119862 and119863 are constants which can be obtained byrotation angle and bending moment boundary conditions

32 To Solve Modal Equation As shown in Figure 2 thedisplacement and bending moment at the ends of the 119894thsingle-span beam can be written as follows

120601 (0) = 0 12060110158401015840(0) = minus

119872119894119894+1

119864119868

120601 (119871119894) = 0 120601

10158401015840(119871119894) = minus

119872119894+1119894

119864119868

(33)


Using (32) and its second partial derivative constants 119860 119861119862 and119863 can be obtained as follows

119860 =119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)

119864119868 (ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)

119861 =119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

119862 =cosh (119899

119894119871119894)119872119894119894+1

minus119872119894+1119894

119864119868 (ℎ2

119894+ 1198992

119894) sinh (119899

119894119871119894)

119863 =minus119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

(34)

Taking the values of constants into (32) model functions canbe derived

33 To Solve Frequency Equation According to (32) theangle equation of the 119894th single-span beam can be written asfollows

120579119894(119909) = 120578

119894[119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)] cos (ℎ

119894119909)

minus 120595119894[119872119894+1119894

minus cosh (119899119894119871119894)119872119894119894+1

] cosh (119899119894119909)

minus 120578119894119872119894119894+1

sin (ℎ119894119871119894) sin (ℎ

119894119909)

minus 120595119894119872119894119894+1

sinh (119899119894119871119894) sinh (119899

119894119909)

(35)

where 120578119894= ℎ119894119864119868(ℎ

2

119894+ 1198992

119894) sin(ℎ

119894119871119894) 120595119894= 119899119894119864119868(ℎ

2

119894+

1198992

119894) sinh(ℎ

119894119871119894)

For the 119894th support which is shown in Figure 2 theequation 119872

119894= 119872119894119894+1

= 119872119894119894minus1

always stands up and theangles on both sides of the 119894th support can be rewritten asfollows

120579119894119894+1

= [120595119894cosh (ℎ

119894119871119894) minus 120578119894cos (ℎ

119894119871119894)]119872119894

minus (120595119894minus 120578119894)119872119894+1

120579119894119894minus1

= (120595119894minus1minus 120578119894minus1)119872119894minus1

minus [120595119894minus1

cosh (119899119894minus1119871119894minus1) minus 120578119894minus1

cos (ℎ119894minus1119871119894minus1)]119872119894

(36)

The angles on both sides of the 119894th support must be equal(120579119894119894minus1

= 120579119894119894+1

2 ⩽ 119894 ⩽ 119899) so we can get

119883119894minus1119872119894minus1+ (119884119894minus1+ 119884119894)119872119894+ 119883119894119872119894+1= 0 (37)

where119883119894= 120595119894minus 120578119894 119884119894= 120578119894cos(ℎ119894119871119894) minus 120595119894cosh(119899

119894119871119894)

Equation (37) has 119899+1 unknowns and 119899minus1 equations andtaken (37) into matrix form

ΩM = 0 (38)

whereM = [11987211198722 119872

119899+1]119879

Ω =[[[

[

11988311198841+ 1198842

1198832

0

1198832

1198842+ 1198843

1198833

sdot sdot sdot sdot sdot sdot sdot sdot sdot

0 119883119899minus1

119884119899minus1

+ 119884119899119883119899

]]]

]

(39)

The bending moment within the first and last span beamends needs to satisfy that119872

1= 0 and119872

119899+1= 0 Equation

(38) can be simplified as follows

Ω0M0 = 0 (40)

whereM0 = [11987221198723 119872119899]119879

Ω0 =

[[[[[

[

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

]]]]]

]

(41)

Equation (40) must have a nonzero solution according to thephysical meaning of the formula so the frequency equationof the 119899 span externally prestressed concrete beam can bewritten as follows

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

= 0 (42)

The initial 119899 roots can be obtained by calculating (42) whichare the essential 119899 order frequencies of the 119899 span externallyprestressed concrete beam

4 Method for Prestress Force Identification

41 Identification from Frequency Equation and MeasuredFrequencies In order to identify the prestress force accordingto the frequency equation (42) the frequency function 119865(119873)can be defined as follows

119865 (119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(43)

where119873 is the prestress force which is the independent vari-able of the frequency function 119865(119873) Taking the measuredfrequency into the frequency function 119865(119873) and assumingthat the frequency function is equal to zero (119865(119873) = 0)the prestress force can be obtained by solving the formula119865(119873) = 0 The 119894th order measured frequency 119891

119894is taken into

frequency function 119865(119873) We can rewrite it as follows

119865119894(119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(44)


Table 1 Measured frequencies and identified prestress force of the single-span beam

Measured frequency (Hz) Prestress force (kN)Mode 1 Mode 2 Test value Identification value Error ()37549 138049 0 mdash mdash37634 138087 3095 3270 56537703 138123 5131 5324 37637819 138170 9027 9622 65937912 138215 12036 12627 491Note Error denotes the (identified value minus test value)test value times 100 the same meaning in the Table 3

Table 2 The corrected result of material parameters and frequencies

Material parameter Frequency result (HZ) of (42) Error ()Elastic modulus (Mpa) Density (kgm3) Mode 1 Mode 2 Mode 1 Mode 2

Uncorrected 2800 times 104 2500 38347 143998 minus212 minus431Corrected 2851 times 104 2583 37545 138079 001 minus002Note Error denotes the (mode with corrected or uncorrected material parameter minus test mode)test mode times 100 the same meaning in the Table 4

We can obtain 119894 frequency functions such as 1198651(119873)

1198652(119873) 119865

119894(119873) if there are 119894 measured frequencies The

prestress force can be identified by looking for the zerosof the frequency functions if the measured frequencies areaccurate enough Actually there are inevitable errors in themeasured frequencies and the true prestress force will appearnear the zero of the frequency function If we still identifiedthe prestress force at the zero of the frequency function theerrors of prestress force could be larger In order to obtainmore accurate results the finite order measured frequenciesare taken into the frequency function 119865(119873) and we can getthe relationship equations about the true prestress force119873 asfollows

1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)

Graphs of (45) must have the intersection and the valueof the independent variable at the intersection is the prestressforce which needs to be identified Concrete steps for theprestress force identification method will be given withexamples in Section 4

5 Examples and Discussion

51 Prestress Force Identification in a Single-Span Beam Asimply supported concrete beam with external tendons isstudied The length of the beam is 26m and the height andthe width of the section are 015m and 012m respectivelyThe concrete grade is C35 there is an external tendon 71205931199045

012

26

015

005

Figure 4 The single-span beam

within the beam the cross sectional area of the externaltendon is 139mm2 and eccentricity of the external tendon is0125m Schematic diagram of the single-span concrete beamwith external tendons is shown in Figure 4

The biggest tensile force of the external tendon is 120 kNaccording to principles that the tensile force cannot exceed75 tensile strength and the eccentric compressive concretebeam cannot cause cracks under the prestress force Thestretching device is a hydraulic jack and pull-press sensorat the beam end The tensile force in each tensioning stageis measured by the pull-press sensor The external tendonwas tensioned using multilevel tensioning method and thevibration signal of the beam using hammer to stimulate it wascollected by the acceleration sensor at each tensioning stageThe photos of test are shown in Figure 5 When the test iscompleted acceleration signals are analyzed by the methodsof digital signal processing including FFTThe first two orderfrequencies of the single-span beam in each tensioning stageare obtained and shown in Table 1

The values ofmaterial parameters such as elastic modulusand density cannot directly use the standard value because ofthe manufacturing error and material difference The valuesof material parameters must be corrected before identifyingthe prestress force On the basis of the external tendon layoutcompleted and no tensioning the values of material param-eters are corrected by using the frequency equation (42)and the measured frequencies After the completion of thecorrection the first two order frequencies with the modified


Table 3 Measured frequencies and identified prestress force of the two-span beam

Measured frequency (Hz) Prestress force (kN)Mode 1 Mode 2 Mode 3 Test value Identification value Error ()29809 45443 114421 0 mdash mdash30141 46062 116129 7152 7531 53030279 46316 116836 10136 10563 42130466 46659 117790 13971 14927 68430606 46918 118508 17176 18143 563

material parameter by frequency equation (42) are fit with themeasured frequencies which are shown in Table 2

Obviously the corrected results of the first two orderfrequencies have a good agreement with the measured fre-quencies at the external tendon layout completed and notensioning stage

The frequency function 119865119894(119873) of the simply supported

concrete beam with external straight tendon according to(44) can be written as follows

119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)

where 119891119894is the 119894th frequency of the test beam 119890 is the

eccentricity of the external tendon and 119903 is the radius ofgyration The frequency function 119865

119894(119873) can be rewritten as

1198651(119873) and 119865

2(119873) when 119894 = 1 and 119894 = 2 (to identify the

prestress force using the 1st and 2nd measured frequencies)The abscissa value of the intersection of the frequencyfunctions 119865

1(119873) and 119865

2(119873) is the prestress force which needs

to be identified Graphs of frequency functions 1198651(119873) and

1198652(119873) are shown in Figure 6 and the identified prestress force

and error are shown in Table 2Figure 6 shows that graphs of frequency functions 119865

1(119873)

and 1198652(119873) do meet in one point on every tensioning state

and the intersection of the frequency functions 1198651(119873) and

1198652(119873) seems close to the function zero which match with

theoretical analysis in Section 41 Table 2 shows that theidentified prestress force is slightly larger than the true valueand the maximum error is 659 This shows that the newmethod is available and can reflect the change trend of theprestress force in the beam

52 Prestress Force Identification in a Two-Span Beam Atwo-span concrete beam with external tendons is studiedThe height of the beam is 036m the width of the beam is017m concrete grade is C20 and the span length is 43m +

43mThere is an external tendon 71205931199045 within the beam thecross sectional area of the external tendon is 139mm2 andthe biggest tensile force of the external tendon is 180 kN Test

(a)

(b)

Figure 5 Photos of test

method and procedure of the two-span beam are the sameas the single-span beam Schematic diagram of the two-spanconcrete beam with external tendons is shown in Figure 7Thefirst three order frequencies of the two-span beam in eachtensioning stage are obtained and shown in Table 3

Frequency equation of the two-span concrete beam withexternal tendons based on (38) can be obtained as follows

2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)

where (47) is not corrected by (25) and (26) The valuesof material parameters of the two-span concrete beam withexternal tendons must be corrected based on the first threeordermeasured frequencies after (47) is corrected by (25) and(26) and the correction method is the same as the single-span beam in Section 51 The corrected values of materialparameters are shown in Table 4



Material parameter Frequency result (HZ) of (47) Error ()Elastic modulus (Mpa) Density (kgm3) Mode 1 Mode 2 Mode 3 Mode 1 Mode 2 Mode 3

Uncorrected 2550 times 10 2500 30494 47271 119897 230 402 479Corrected 2639 times 10 2591 29832 45527 114797 008 018 033

0 3276 50 100 150minus04

minus02

minus0006

Prestressing force (kN)

F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)

0 50 100 150Prestressing force (kN)

F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)

0 50 150Prestressing force (kN)

F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)

Figure 6 Graphs of frequency functions 1198651(119873) and 119865

2(119873)

05

036

017

005

005

43 4305 0630 0505 30

Figure 7 The two-span beam


50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in

Figure 8 Graphs of frequency functions 1198651(119873) 119865

2(119873) and 119865

3(119873)

The frequency function119865119894(119873) of two-span concrete beam

with external tendons can be presented according to (44)The frequency function 119865

119894(119873) can be rewritten as 119865

1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to

identify the prestress force using the first three measuredfrequencies) The abscissa value of the intersection of thefrequency functions119865

1(119873)119865

2(119873) and119865

3(119873) is the identified

prestress force Because there are always errors with themeasured frequencies graphs of frequency functions 119865

1(119873)

1198652(119873) and 119865

3(119873) cannot be accurate in one point Actually

graphs of any two frequency functions can meet in onepoint and the three frequency functions will have threeintersections The prestress force of the two-span concrete

beam will be identified based on the first three measuredfrequencies which will have higher accuracy than the resultbased upon the first two measured frequencies The trueprestress force will appear in the triangle with the threeintersections According to the geometric relationship of thetriangle the identified prestress force can be acquired by theabscissa value of the triangle of gravity Graphs of frequencyfunctions119865

1(119873)119865

2(119873) and119865

3(119873) are shown in Figure 8 and

the identified prestress force and error are shown in Table 4Figure 8 shows that graphs of frequency functions 119865

1(119873)

1198652(119873) and 119865

3(119873) have three intersections and frequency

functions value at the triangle of gravity is closed to thefunctions zero which match with theoretical analysis above


Table 5 The error analysis results

Frequency error Prestress force (kN)minus3 3 minus3 3 Theoretical value Identification value Error ()Mode 1 Mode 1 Mode 2 Mode 20 0 1 0

60

6515 8580 0 0 1 5485 minus8581 0 0 0 5859 minus2350 1 0 0 6143 2381 0 0 1 5344 minus10930 1 1 0 6659 10981 0 1 0 6375 6250 1 0 1 5629 minus6180 0 1 0

90


120

1252 4330 0 0 1 11431 minus4741 0 0 0 11857 minus1190 1 0 0 12143 1191 0 0 1 11338 minus5520 1 1 0 12667 5561 0 1 0 12377 3140 1 0 1 11663 minus281Note ldquo0rdquo stands for the error which does not appear in the certain calculating condition and ldquo1rdquo stands for the error which appears in the certain calculatingcondition

The identified prestress force appears near functions zeroTable 4 shows that the identified prestress force of the two-span concrete beam with external tendons is also larger thanthe true value and themaximumerror is 689 Identificationmethod for the prestress force based on the frequency equa-tion and themeasured frequencies can effectively identify theprestress force

53 Effect of Measured Frequency Errors Structural dynamicresponses can be collected using the acceleration sensorin practical engineering and the natural frequencies canbe obtained based on spectrum transformation with theacceleration data The low order natural frequencies usuallyhave higher precision than the high order natural frequenciesIf test environment is relatively stable and the test beam doesnot show cracks and plastic deformation in the tensioningstage the change of the measured frequencies reflects thatthe prestress force has effects on dynamic characteristics ofthe structureThe prestress force can be effectively recognizedbased on the measured frequencies and frequency functionsof externally prestressed concrete beam which is presentedin this paper Natural frequencies which are obtained by thesignal processing technology are relatively accurate but there

is a certain error between the measured results and the truevalues which is caused by the testmethod and data processingmethod It is necessary to study that the identified result ofthe prestress force is affected by the error of the measuredfrequencies

Taking the single-span beam which is shown inSection 51 as an example the first two order frequenciesunder different prestress force (119873 = 60 kN 90 kN and120 kN) can be obtained by (42) and then assuming that thefrequencies which are obtained by (42) have a maximumerror of plusmn3The frequencies and the error can be combinedto different calculating conditions Frequencies with differenterrors in different calculating conditions are plugged intofrequency functions (45) The graphs of frequency functionswith the first two order frequencies will have the intersectionand the identified prestress force can be modified with theintersection which is shown in Section 5 The error analysisresults under different calculating conditions are shown inTable 5

Table 5 shows that the identified prestress force has agreater difference with the different frequency and the errorcombination in the same tensioning stage which illustratethat the error of natural frequency has significant effect on


the accuracy of prestress force identification and the moreaccurate the measured frequencies are the higher precisionthe identified prestress force has In different tensioning stageand the same frequency error the smaller the prestress forceis the more significant effect by the frequency error theidentified results have and the influence of frequency error forprestressed force identification will wane with the increase oftensioning force The identified results are affected more sig-nificantly by the error of higher order frequency Apparentlythere exists nonlinear relationship between natural frequencyand prestressed force Above all when the test environmentis relatively stable and beam does not show cracks and plasticdeformation in the tensioning stage the natural frequenciesare obtained more accurately using the right test methodand data processing method then the prestress force can beidentified based on frequency equation and the measuredfrequencies and the identification accuracy for the prestressforce depends on the accuracy of the measured frequenciesIn the long-term bridge health monitoring the dynamicresponse of the bridge can be collected by sensors theinfluence of environmental factors and external incentivesis eliminated in signals and the measured frequencies areobtained by spectrum transformation The prestress force ofthe bridge can be identified based on the frequency equationand the measured frequencies and the change trend of theprestress force can be reflected

6 Conclusion

In this study a new method to identify the prestress force inexternally prestressed concrete beam based on the frequencyequation and the measured frequencies is proposed Theeffectiveness of the prestress force identification methodis demonstrated by the single-span externally prestressedconcrete beam and two-span externally prestressed concretebeam tests Taking the single-span beam as an examplethe influencing regularities of the error of the measuredfrequencies on the identified results are analyzed by numericcalculation The free vibration equation of multispan exter-nally prestressed concrete beam is solved using sublevelsimultaneous method and the multispan externally pre-stressed concrete beam is taken as the multiple single-spanbeams which must meet the bending moment and rotationangle boundary conditions The function relation betweenprestress variation and vibration displacement is built and theformula of equivalent eccentricity119867 is presented In the long-term bridge healthmonitoring themeasured frequencies canbe obtained by practical signal processingThe prestress forceof the bridge can be identified based on the new identifiedmethod and the change trend of the prestress force can bereflected

Conflict of Interests

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

Acknowledgments

This work is partially supported by Natural Science Founda-tion of China under Grant nos 51378039 and 51108009 andFoundation of Beijing Lab of Earthquake Engineering andStructural Retrofit under Grant no 2013TS02

References

[1] American Association of State Highway Transportation Offi-cials Bridge Design Specifications AASHTO Washington DCUSA 4th edition 2007

[2] Z R Lu and S S Law ldquoIdentification of prestress force frommeasured structural responsesrdquoMechanical Systems and SignalProcessing vol 20 no 8 pp 2186ndash2199 2006

[3] S S Law and Z R Lu ldquoTime domain responses of a prestressedbeam and prestress identificationrdquo Journal of Sound and Vibra-tion vol 288 no 4-5 pp 1011ndash1025 2005

[4] H Li Z Lv and J Liu ldquoAssessment of prestress force in bridgesusing structural dynamic responses under moving vehiclesrdquoMathematical Problems in Engineering vol 2013 Article ID435939 9 pages 2013

[5] S S Law S Q Wu and Z Y Shi ldquoMoving load and prestressidentification using wavelet-based methodrdquo Journal of AppliedMechanics Transactions ASME vol 75 no 2 Article ID 0210147 pages 2008

[6] J Q Bu and H Y Wang ldquoEffective prestress identification fora simply-supported PRC beam bridge by BP neural networkmethodrdquo Journal of Vibration and Shock vol 30 no 12 pp 155ndash159 2011

[7] M A Abraham S Park and N Stubbs ldquoLoss of prestress pre-diction based on nondestructive damage location algorithmsrdquoin Smart Structures and Materials vol 2446 of Proceedings ofSPIE pp 60ndash67 March 1995

[8] J-T Kim Y-S Ryu and C-B Yun ldquoVibration-based methodto detect prestress-loss in beam-type bridgesrdquo in Smart SystemsandNondestructive Evaluation for Civil Infrastructures vol 5057of Proceedings of SPIE pp 559ndash568 March 2003

[9] F Z Xuan H Tang and S T Tu ldquoIn situ monitoring onprestress losses in the reinforced structure with fiber-optic sen-sorsrdquo Measurement Journal of the International MeasurementConfederation vol 42 no 1 pp 107ndash111 2009

[10] A Miyamoto K Tei H Nakamura and J W Bull ldquoBehavior ofprestressed beam strengthened with external tendonsrdquo Journalof Structural Engineering vol 126 no 9 pp 1033ndash1044 2000

[11] E Hamed and Y Frostig ldquoNatural frequencies of bonded andunbonded prestressed beams-prestress force effectsrdquo Journal ofSound and Vibration vol 295 no 1-2 pp 28ndash39 2006

[12] M Saiidi B Douglas and S Feng ldquoPrestress force effect onvibration frequency of concrete bridgesrdquo Journal of StructuralEngineering vol 120 no 7 pp 2233ndash2241 1994

[13] M Saiidi B Douglas and S Feng ldquoPrestress force effect onvibration frequency of concrete bridgesrdquo Journal of StructuralEngineering vol 122 no 4 p 460 1996

[14] X Y Xiong F Gao and Y Li ldquoAnalysis on vibration behaviorof externally prestressed concrete continuous beamrdquo Journal ofVibration and Shock vol 30 no 6 pp 105ndash109 2011

[15] S Timoshenko Strength of Materials D Van Nostrand Com-pany Inc New York NY USA 1940


[16] R J Hosking S A Husain and F Milinazzo ldquoNatural flexuralvibrations of a continuous beam on discrete elastic supportsrdquoJournal of Sound and Vibration vol 272 no 1-2 pp 169ndash1852004

[17] RW Clough and J PenzienDynamics of Structure Computersand Structures Inc New York NY USA

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Stochastic AnalysisInternational Journal of


the time domain by a system identification approach Liet al [4] carried out numerical simulations to identify themagnitude of prestress force in a highway bridge by makinguse of the dynamic responses from moving vehicular loadsbased on dynamic response sensitivity-based finite elementmodel updating Law et al [5] developed a new method ofprestress identification using the wavelet-based method inwhich the approximation of the measured response is used toform the identification equation Bu and Wang [6] presenteda BP neural networkmethod to identify the effective prestressfor a simply supported PRC beam bridge based on modalfrequencies and dynamic responses of the bridge Abrahamet al [7] investigated the feasibility of using damage locationalgorithm technique for detecting loss of prestress in aprestressed concrete bridge Kim et al [8] studied a vibration-based method to detect prestress loss in beam-type PSCbridges by monitoring changes in a few natural frequenciesXuan et al [9] evaluated the prestress loss quantitativelyin the steel-strand reinforced structures by an optical fiber-sensor based monitoring technique However the prestressforce and prestress loss cannot be estimated directly simplyand accurately unless the beam has been instrumented atthe time of construction Several researchers also studied thedynamic behavior of prestressed beam with external tendonsand predicted the relation between the modal frequency andthe given prestress force Miyamoto et al [10] studied theeffect of the prestressing force introduced by the externaltendons on the vibration characteristics of a composite girderwith the results of dynamic tests and derived the formula forcalculating the natural frequency of a composite girder basedon a vibration equation Hamed and Frostig [11] presentedthe effect of the magnitude of the prestressing force on thenatural frequencies of prestressed beams with bonded andunbonded tendons Saiidi et al [12 13] reported a studyon modal frequency due to the prestress force with labora-tory test results The above researchers only considered theprestressing effect on dynamic characteristics of the simplysupported beam Very few works have been presented on theeffect of prestressing on the dynamic responses of a beam andidentification of prestress force directly or indirectly

The exact solution of the free vibration equation ofmultispan externally prestressed concrete uniform beam isobtained in this paper An inverse problem to identify theprestress force based on the frequency equation and themeasured frequencies is then presented taking the prestressforce as an unknown parameter in the frequency functionsThe prestress force identification method is suited to theexternally prestressed concrete uniform beam Firstly basedonMiyamoto et alrsquos study [10] the function relation betweenprestress variation and vibration displacement of multispanexternally prestressed concrete beam is built according to thebasic principle of the force methodThemultispan externallyprestressed concrete beam is considered as the multiplesingle-span beams which must meet the bending momentand rotation angle boundary conditions The free vibrationequations of multispan externally prestressed concrete beamby using sublevel simultaneous method which can simplifythe solution of dynamic equations are given and the semi-analytical solution of the free vibration equations which

x

dx

NN

uu(x t)

Figure 1 Analysis model of vibration system

considered the influence of prestress on section rigidity andbeam length is obtained Then frequency functions whichare obtained by frequency equation are used to identify theprestress force by the appropriate method Two dynamic testsof externally prestressed concrete beam in the laboratory aresubmitted to illustrate the effectiveness and robustness ofthe proposed method At last the effect of the error of themeasured frequencies on identification of the prestress forceis studied in the proposed method

2 Vibration Equation of Multispan ExternallyPrestressed Concrete Beam

21 Vibration Equation of Externally Prestressed Simply Sup-ported Beam An externally prestressed simply supportedbeam is shown in Figure 1 It is assumed that the prestressforce 119873 has no prestressing loss along the beam length andthe beam bending must meet the plane section assumptionThe vibration equation of this simply supported beam can beexpressed as follows

1205972

1205971199092[119864119868

1205972119906 (119909 119905)

1205971199092] + 119873

119909

1205972119906 (119909 119905)

1205971199092minus 119867

1205972(Δ119873)

1205971199092

+ 1198981205972119906 (119909 119905)

1205971199052= 0

(1)

where 119864119868 is the flexural rigidity of the beam119898 is the mass ofthe beam per unit length 120583(119909 119905) is the transverse deflection119873119909is the horizontal component of the prestress force119873119867 is

the equivalent eccentricity of the external tendons andΔ119873 isthe variation of the prestress force due to flexural vibrationBecause eccentricity of external tendons in different positionson the beam is not the same the equivalent eccentricity 119867can be calculated according to the principle of equal area inthe bending moment diagram

22 Vibration Equation of Multispan Externally PrestressedBeam A multispan externally prestressed continuous beamwhich has 119899 spans is shown in Figure 2 and the 119894th span ofthe beam is taken as the study subject The rotation angleand bending moment of the beam end at point 119894 are 120579

119894119894+1

and 119872119894119894+1

and the rotation angle and bending moment ofthe beam end at point 119894 + 1 are 120579

119894+1119894and119872

119894+1119894 respectively


xNi

u120579ii+1 120579i+1i

i + 1

Mi+1Mi

ui(x t)



1205972

1205971199092[119864119868

1205972119906119894(119909 119905)

1205971199092] + 119873

119909119894

1205972119906119894(119909 119905)

1205971199092minus 119867119894

1205972(Δ119873119894)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(2)


119894is






120579119894119894minus1

= 120579119894119894+1

119872119894119894minus1

= 119872119894119894+1

120579119894+1119894

= 120579119894+1119894+2

119872119894+1119894

= 119872119894+1119894+2

(3)


11987212= 0 120579

21= 12057923 119872

21= 11987223

119872119899+1119899

= 0 120579119899119899+1

= 120579119899119899minus1

119872119899119899+1

= 119872119899119899minus1

(4)




1and 119883

2


1= 1 and 119883



120575111198831+ 120575121198832+ Δ1119865= 0

120575211198831+ 120575221198832+ Δ2119865= 0

(5)

where 120575119894119895= sumint(119872

119894119872119895119864119868)119889119909 + sumint(119873

119894119873119895119864119860)119889119909 Δ

119894119865=

sumint(119872119894119872119865119864119868)119889119909 + sumint(119873

119894119873119865119864119860)119889119909 119894 = 1 2 119895 = 1 2


Δ119873 =12057512Δ2119865minus 12057522Δ1119865

1205751112057522minus 1205751212057521

(6)



120583119865=71198651198713

768119864119868 (7)


120583119865=

7 (1205751112057522minus 1205751212057521) 1198651198713

768 (12057512Δ2119865minus 12057522Δ1119865) 119864119868

Δ119873 (8)







120583Δ119873

=12057522Δ1119865minus 12057512Δ2119865

12057522119864119868119865

Δ119873 (9)


120583 = 120583119865minus 120583Δ119873 (10)


Δ119873 = 120601120583 (11)


NN

F

F

MF

X1

X2

M1

M2

N1


N

F

F

N

MF

X1

X2X3

M1

M2

M3

N1



where

120601 = (119864119868)

times (7 (1205751112057522minus 1205751212057521) 1198651198713


12057522119865

)

minus1

(12)


1 1198832 and 119883



120575111198831+ 120575121198832+ 120575131198833+ Δ1119865= 0

120575211198831+ 120575221198832+ 120575231198833+ Δ2119865= 0

120575311198831+ 120575321198832+ 120575331198833+ Δ3119865= 0

(13)

where 120575119894119895and Δ



Δ119873 =119863Δ

1198630

(14)

where

119863Δ=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1205751212057513Δ1119865

1205752212057523Δ2119865

1205753212057533Δ3119865

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198630=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120575111205751212057513

120575211205752212057523

120575311205753212057533

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(15)



120583119865=

1198651198713

192119864119868 (16)



120583Δ119873

=(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119864119868119865

Δ119873 (17)


120601 = (119864119868)

times (11986511987131198630

192119863Δ

minus(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119865

)

minus1

(18)


119872119873= (119872

1minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198723)Δ119873

(19)


119878119872119873

= (1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)Δ119873

(20)


where 119878119872119873

1198781198721

1198781198722

and 1198781198723


119872119873(Δ119873 times 119871)


119867 =1

119871(1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)

(21)


119867 =1

119871(1198781198721minus12057512

12057522

1198781198722) (22)


1198641198681205974119906119894(119909 119905)

1205971199094+ (119873119909119894minus 119867119894120601119894)1205972119906119894(119909 119905)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(23)






119864119904119868119904= 119873(

119871119894

119899120587)

2

(24)


119864119868 = 119864119888119868119888+ 119873(

119871119894

119899120587)

2

(25)




1198711015840

119894= (1 minus

119873119909119894

119864119860)119871119894 (26)





1205974119906119894(119909 119905)

1205971199094+119873119909119894minus 119867119894120601119894

119864119868

1205972119906119894(119909 119905)

1205971199092+119898

119864119868

1205972119906119894(119909 119905)

1205971199052= 0

(27)



119906119894(119909 119905) = 120601

119894(119909) 119884 (119905) (28)



1206011015840101584010158401015840

119894(119909) + 119892

212060110158401015840

119894(119909) minus 119886

4120601119894(119909) = 0 (29)

11988410158401015840(119905) + 120596

2119884 (119905) = 0 (30)



119894(119909) = 119866119890


11990412= plusmn119894ℎ119894 119904

34= plusmn119894119899119894 (31)

where ℎ119894

= radic(1198864 + (119892119894

44))12+ (119892119894

22) 119899119894

=

radic(1198864 + (119892119894

44))12minus (119892119894


120601119894(119909) = 119860 sin (ℎ

119894119909) + 119861 cos (ℎ

119894119909)

+ 119862 sinh (119899119894119909) + 119863 cosh (119899

119894119909)

(32)



120601 (0) = 0 12060110158401015840(0) = minus

119872119894119894+1

119864119868

120601 (119871119894) = 0 120601

10158401015840(119871119894) = minus

119872119894+1119894

119864119868

(33)



119860 =119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)

119864119868 (ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)

119861 =119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

119862 =cosh (119899

119894119871119894)119872119894119894+1

minus119872119894+1119894

119864119868 (ℎ2

119894+ 1198992

119894) sinh (119899

119894119871119894)

119863 =minus119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

(34)



120579119894(119909) = 120578

119894[119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)] cos (ℎ

119894119909)

minus 120595119894[119872119894+1119894

minus cosh (119899119894119871119894)119872119894119894+1

] cosh (119899119894119909)

minus 120578119894119872119894119894+1

sin (ℎ119894119871119894) sin (ℎ

119894119909)

minus 120595119894119872119894119894+1

sinh (119899119894119871119894) sinh (119899

119894119909)

(35)

where 120578119894= ℎ119894119864119868(ℎ

2

119894+ 1198992

119894) sin(ℎ

119894119871119894) 120595119894= 119899119894119864119868(ℎ

2

119894+

1198992

119894) sinh(ℎ

119894119871119894)


119894= 119872119894119894+1

= 119872119894119894minus1


120579119894119894+1

= [120595119894cosh (ℎ

119894119871119894) minus 120578119894cos (ℎ

119894119871119894)]119872119894

minus (120595119894minus 120578119894)119872119894+1

120579119894119894minus1




cos (ℎ119894minus1119871119894minus1)]119872119894

(36)


= 120579119894119894+1

2 ⩽ 119894 ⩽ 119899) so we can get



119894119871119894)


ΩM = 0 (38)

whereM = [11987211198722 119872

119899+1]119879

Ω =[[[

[

11988311198841+ 1198842

1198832

0

1198832

1198842+ 1198843

1198833


0 119883119899minus1

119884119899minus1

+ 119884119899119883119899

]]]

]

(39)


1= 0 and119872



Ω0M0 = 0 (40)

whereM0 = [11987221198723 119872119899]119879

Ω0 =

[[[[[

[

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

]]]]]

]

(41)


1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

= 0 (42)




119865 (119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(43)


119894is taken into


119865119894(119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(44)








1198652(119873) 119865



1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)




012

26

015

005












119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










xNi

u120579ii+1 120579i+1i

i + 1

Mi+1Mi

ui(x t)



1205972

1205971199092[119864119868

1205972119906119894(119909 119905)

1205971199092] + 119873

119909119894

1205972119906119894(119909 119905)

1205971199092minus 119867119894

1205972(Δ119873119894)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(2)


119894is






120579119894119894minus1

= 120579119894119894+1

119872119894119894minus1

= 119872119894119894+1

120579119894+1119894

= 120579119894+1119894+2

119872119894+1119894

= 119872119894+1119894+2

(3)


11987212= 0 120579

21= 12057923 119872

21= 11987223

119872119899+1119899

= 0 120579119899119899+1

= 120579119899119899minus1

119872119899119899+1

= 119872119899119899minus1

(4)




1and 119883

2


1= 1 and 119883



120575111198831+ 120575121198832+ Δ1119865= 0

120575211198831+ 120575221198832+ Δ2119865= 0

(5)

where 120575119894119895= sumint(119872

119894119872119895119864119868)119889119909 + sumint(119873

119894119873119895119864119860)119889119909 Δ

119894119865=

sumint(119872119894119872119865119864119868)119889119909 + sumint(119873

119894119873119865119864119860)119889119909 119894 = 1 2 119895 = 1 2


Δ119873 =12057512Δ2119865minus 12057522Δ1119865

1205751112057522minus 1205751212057521

(6)



120583119865=71198651198713

768119864119868 (7)


120583119865=

7 (1205751112057522minus 1205751212057521) 1198651198713

768 (12057512Δ2119865minus 12057522Δ1119865) 119864119868

Δ119873 (8)







120583Δ119873

=12057522Δ1119865minus 12057512Δ2119865

12057522119864119868119865

Δ119873 (9)


120583 = 120583119865minus 120583Δ119873 (10)


Δ119873 = 120601120583 (11)


NN

F

F

MF

X1

X2

M1

M2

N1


N

F

F

N

MF

X1

X2X3

M1

M2

M3

N1



where

120601 = (119864119868)

times (7 (1205751112057522minus 1205751212057521) 1198651198713


12057522119865

)

minus1

(12)


1 1198832 and 119883



120575111198831+ 120575121198832+ 120575131198833+ Δ1119865= 0

120575211198831+ 120575221198832+ 120575231198833+ Δ2119865= 0

120575311198831+ 120575321198832+ 120575331198833+ Δ3119865= 0

(13)

where 120575119894119895and Δ



Δ119873 =119863Δ

1198630

(14)

where

119863Δ=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1205751212057513Δ1119865

1205752212057523Δ2119865

1205753212057533Δ3119865

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198630=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120575111205751212057513

120575211205752212057523

120575311205753212057533

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(15)



120583119865=

1198651198713

192119864119868 (16)



120583Δ119873

=(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119864119868119865

Δ119873 (17)


120601 = (119864119868)

times (11986511987131198630

192119863Δ

minus(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119865

)

minus1

(18)


119872119873= (119872

1minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198723)Δ119873

(19)


119878119872119873

= (1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)Δ119873

(20)


where 119878119872119873

1198781198721

1198781198722

and 1198781198723


119872119873(Δ119873 times 119871)


119867 =1

119871(1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)

(21)


119867 =1

119871(1198781198721minus12057512

12057522

1198781198722) (22)


1198641198681205974119906119894(119909 119905)

1205971199094+ (119873119909119894minus 119867119894120601119894)1205972119906119894(119909 119905)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(23)






119864119904119868119904= 119873(

119871119894

119899120587)

2

(24)


119864119868 = 119864119888119868119888+ 119873(

119871119894

119899120587)

2

(25)




1198711015840

119894= (1 minus

119873119909119894

119864119860)119871119894 (26)





1205974119906119894(119909 119905)

1205971199094+119873119909119894minus 119867119894120601119894

119864119868

1205972119906119894(119909 119905)

1205971199092+119898

119864119868

1205972119906119894(119909 119905)

1205971199052= 0

(27)



119906119894(119909 119905) = 120601

119894(119909) 119884 (119905) (28)



1206011015840101584010158401015840

119894(119909) + 119892

212060110158401015840

119894(119909) minus 119886

4120601119894(119909) = 0 (29)

11988410158401015840(119905) + 120596

2119884 (119905) = 0 (30)



119894(119909) = 119866119890


11990412= plusmn119894ℎ119894 119904

34= plusmn119894119899119894 (31)

where ℎ119894

= radic(1198864 + (119892119894

44))12+ (119892119894

22) 119899119894

=

radic(1198864 + (119892119894

44))12minus (119892119894


120601119894(119909) = 119860 sin (ℎ

119894119909) + 119861 cos (ℎ

119894119909)

+ 119862 sinh (119899119894119909) + 119863 cosh (119899

119894119909)

(32)



120601 (0) = 0 12060110158401015840(0) = minus

119872119894119894+1

119864119868

120601 (119871119894) = 0 120601

10158401015840(119871119894) = minus

119872119894+1119894

119864119868

(33)



119860 =119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)

119864119868 (ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)

119861 =119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

119862 =cosh (119899

119894119871119894)119872119894119894+1

minus119872119894+1119894

119864119868 (ℎ2

119894+ 1198992

119894) sinh (119899

119894119871119894)

119863 =minus119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

(34)



120579119894(119909) = 120578

119894[119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)] cos (ℎ

119894119909)

minus 120595119894[119872119894+1119894

minus cosh (119899119894119871119894)119872119894119894+1

] cosh (119899119894119909)

minus 120578119894119872119894119894+1

sin (ℎ119894119871119894) sin (ℎ

119894119909)

minus 120595119894119872119894119894+1

sinh (119899119894119871119894) sinh (119899

119894119909)

(35)

where 120578119894= ℎ119894119864119868(ℎ

2

119894+ 1198992

119894) sin(ℎ

119894119871119894) 120595119894= 119899119894119864119868(ℎ

2

119894+

1198992

119894) sinh(ℎ

119894119871119894)


119894= 119872119894119894+1

= 119872119894119894minus1


120579119894119894+1

= [120595119894cosh (ℎ

119894119871119894) minus 120578119894cos (ℎ

119894119871119894)]119872119894

minus (120595119894minus 120578119894)119872119894+1

120579119894119894minus1




cos (ℎ119894minus1119871119894minus1)]119872119894

(36)


= 120579119894119894+1

2 ⩽ 119894 ⩽ 119899) so we can get



119894119871119894)


ΩM = 0 (38)

whereM = [11987211198722 119872

119899+1]119879

Ω =[[[

[

11988311198841+ 1198842

1198832

0

1198832

1198842+ 1198843

1198833


0 119883119899minus1

119884119899minus1

+ 119884119899119883119899

]]]

]

(39)


1= 0 and119872



Ω0M0 = 0 (40)

whereM0 = [11987221198723 119872119899]119879

Ω0 =

[[[[[

[

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

]]]]]

]

(41)


1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

= 0 (42)




119865 (119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(43)


119894is taken into


119865119894(119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(44)








1198652(119873) 119865



1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)




012

26

015

005












119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










NN

F

F

MF

X1

X2

M1

M2

N1


N

F

F

N

MF

X1

X2X3

M1

M2

M3

N1



where

120601 = (119864119868)

times (7 (1205751112057522minus 1205751212057521) 1198651198713


12057522119865

)

minus1

(12)


1 1198832 and 119883



120575111198831+ 120575121198832+ 120575131198833+ Δ1119865= 0

120575211198831+ 120575221198832+ 120575231198833+ Δ2119865= 0

120575311198831+ 120575321198832+ 120575331198833+ Δ3119865= 0

(13)

where 120575119894119895and Δ



Δ119873 =119863Δ

1198630

(14)

where

119863Δ=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1205751212057513Δ1119865

1205752212057523Δ2119865

1205753212057533Δ3119865

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198630=

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

120575111205751212057513

120575211205752212057523

120575311205753212057533

100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(15)



120583119865=

1198651198713

192119864119868 (16)



120583Δ119873

=(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119864119868119865

Δ119873 (17)


120601 = (119864119868)

times (11986511987131198630

192119863Δ

minus(12057522+ 12057523) Δ1119865minus (12057512+ 12057513) Δ2119865

(12057522+ 12057523) 119865

)

minus1

(18)


119872119873= (119872

1minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198723)Δ119873

(19)


119878119872119873

= (1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722

minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)Δ119873

(20)


where 119878119872119873

1198781198721

1198781198722

and 1198781198723


119872119873(Δ119873 times 119871)


119867 =1

119871(1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)

(21)


119867 =1

119871(1198781198721minus12057512

12057522

1198781198722) (22)


1198641198681205974119906119894(119909 119905)

1205971199094+ (119873119909119894minus 119867119894120601119894)1205972119906119894(119909 119905)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(23)






119864119904119868119904= 119873(

119871119894

119899120587)

2

(24)


119864119868 = 119864119888119868119888+ 119873(

119871119894

119899120587)

2

(25)




1198711015840

119894= (1 minus

119873119909119894

119864119860)119871119894 (26)





1205974119906119894(119909 119905)

1205971199094+119873119909119894minus 119867119894120601119894

119864119868

1205972119906119894(119909 119905)

1205971199092+119898

119864119868

1205972119906119894(119909 119905)

1205971199052= 0

(27)



119906119894(119909 119905) = 120601

119894(119909) 119884 (119905) (28)



1206011015840101584010158401015840

119894(119909) + 119892

212060110158401015840

119894(119909) minus 119886

4120601119894(119909) = 0 (29)

11988410158401015840(119905) + 120596

2119884 (119905) = 0 (30)



119894(119909) = 119866119890


11990412= plusmn119894ℎ119894 119904

34= plusmn119894119899119894 (31)

where ℎ119894

= radic(1198864 + (119892119894

44))12+ (119892119894

22) 119899119894

=

radic(1198864 + (119892119894

44))12minus (119892119894


120601119894(119909) = 119860 sin (ℎ

119894119909) + 119861 cos (ℎ

119894119909)

+ 119862 sinh (119899119894119909) + 119863 cosh (119899

119894119909)

(32)



120601 (0) = 0 12060110158401015840(0) = minus

119872119894119894+1

119864119868

120601 (119871119894) = 0 120601

10158401015840(119871119894) = minus

119872119894+1119894

119864119868

(33)



119860 =119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)

119864119868 (ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)

119861 =119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

119862 =cosh (119899

119894119871119894)119872119894119894+1

minus119872119894+1119894

119864119868 (ℎ2

119894+ 1198992

119894) sinh (119899

119894119871119894)

119863 =minus119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

(34)



120579119894(119909) = 120578

119894[119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)] cos (ℎ

119894119909)

minus 120595119894[119872119894+1119894

minus cosh (119899119894119871119894)119872119894119894+1

] cosh (119899119894119909)

minus 120578119894119872119894119894+1

sin (ℎ119894119871119894) sin (ℎ

119894119909)

minus 120595119894119872119894119894+1

sinh (119899119894119871119894) sinh (119899

119894119909)

(35)

where 120578119894= ℎ119894119864119868(ℎ

2

119894+ 1198992

119894) sin(ℎ

119894119871119894) 120595119894= 119899119894119864119868(ℎ

2

119894+

1198992

119894) sinh(ℎ

119894119871119894)


119894= 119872119894119894+1

= 119872119894119894minus1


120579119894119894+1

= [120595119894cosh (ℎ

119894119871119894) minus 120578119894cos (ℎ

119894119871119894)]119872119894

minus (120595119894minus 120578119894)119872119894+1

120579119894119894minus1




cos (ℎ119894minus1119871119894minus1)]119872119894

(36)


= 120579119894119894+1

2 ⩽ 119894 ⩽ 119899) so we can get



119894119871119894)


ΩM = 0 (38)

whereM = [11987211198722 119872

119899+1]119879

Ω =[[[

[

11988311198841+ 1198842

1198832

0

1198832

1198842+ 1198843

1198833


0 119883119899minus1

119884119899minus1

+ 119884119899119883119899

]]]

]

(39)


1= 0 and119872



Ω0M0 = 0 (40)

whereM0 = [11987221198723 119872119899]119879

Ω0 =

[[[[[

[

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

]]]]]

]

(41)


1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

= 0 (42)




119865 (119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(43)


119894is taken into


119865119894(119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(44)








1198652(119873) 119865



1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)




012

26

015

005












119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where 119878119872119873

1198781198721

1198781198722

and 1198781198723


119872119873(Δ119873 times 119871)


119867 =1

119871(1198781198721minus1205751212057522minus 1205751312057523

1205752

22minus 1205752

23

1198781198722minus1205751312057522minus 1205751212057523

1205752

22minus 1205752

23

1198781198723)

(21)


119867 =1

119871(1198781198721minus12057512

12057522

1198781198722) (22)


1198641198681205974119906119894(119909 119905)

1205971199094+ (119873119909119894minus 119867119894120601119894)1205972119906119894(119909 119905)

1205971199092

+ 1198981205972119906119894(119909 119905)

1205971199052= 0

(23)






119864119904119868119904= 119873(

119871119894

119899120587)

2

(24)


119864119868 = 119864119888119868119888+ 119873(

119871119894

119899120587)

2

(25)




1198711015840

119894= (1 minus

119873119909119894

119864119860)119871119894 (26)





1205974119906119894(119909 119905)

1205971199094+119873119909119894minus 119867119894120601119894

119864119868

1205972119906119894(119909 119905)

1205971199092+119898

119864119868

1205972119906119894(119909 119905)

1205971199052= 0

(27)



119906119894(119909 119905) = 120601

119894(119909) 119884 (119905) (28)



1206011015840101584010158401015840

119894(119909) + 119892

212060110158401015840

119894(119909) minus 119886

4120601119894(119909) = 0 (29)

11988410158401015840(119905) + 120596

2119884 (119905) = 0 (30)



119894(119909) = 119866119890


11990412= plusmn119894ℎ119894 119904

34= plusmn119894119899119894 (31)

where ℎ119894

= radic(1198864 + (119892119894

44))12+ (119892119894

22) 119899119894

=

radic(1198864 + (119892119894

44))12minus (119892119894


120601119894(119909) = 119860 sin (ℎ

119894119909) + 119861 cos (ℎ

119894119909)

+ 119862 sinh (119899119894119909) + 119863 cosh (119899

119894119909)

(32)



120601 (0) = 0 12060110158401015840(0) = minus

119872119894119894+1

119864119868

120601 (119871119894) = 0 120601

10158401015840(119871119894) = minus

119872119894+1119894

119864119868

(33)



119860 =119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)

119864119868 (ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)

119861 =119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

119862 =cosh (119899

119894119871119894)119872119894119894+1

minus119872119894+1119894

119864119868 (ℎ2

119894+ 1198992

119894) sinh (119899

119894119871119894)

119863 =minus119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

(34)



120579119894(119909) = 120578

119894[119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)] cos (ℎ

119894119909)

minus 120595119894[119872119894+1119894

minus cosh (119899119894119871119894)119872119894119894+1

] cosh (119899119894119909)

minus 120578119894119872119894119894+1

sin (ℎ119894119871119894) sin (ℎ

119894119909)

minus 120595119894119872119894119894+1

sinh (119899119894119871119894) sinh (119899

119894119909)

(35)

where 120578119894= ℎ119894119864119868(ℎ

2

119894+ 1198992

119894) sin(ℎ

119894119871119894) 120595119894= 119899119894119864119868(ℎ

2

119894+

1198992

119894) sinh(ℎ

119894119871119894)


119894= 119872119894119894+1

= 119872119894119894minus1


120579119894119894+1

= [120595119894cosh (ℎ

119894119871119894) minus 120578119894cos (ℎ

119894119871119894)]119872119894

minus (120595119894minus 120578119894)119872119894+1

120579119894119894minus1




cos (ℎ119894minus1119871119894minus1)]119872119894

(36)


= 120579119894119894+1

2 ⩽ 119894 ⩽ 119899) so we can get



119894119871119894)


ΩM = 0 (38)

whereM = [11987211198722 119872

119899+1]119879

Ω =[[[

[

11988311198841+ 1198842

1198832

0

1198832

1198842+ 1198843

1198833


0 119883119899minus1

119884119899minus1

+ 119884119899119883119899

]]]

]

(39)


1= 0 and119872



Ω0M0 = 0 (40)

whereM0 = [11987221198723 119872119899]119879

Ω0 =

[[[[[

[

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

]]]]]

]

(41)


1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

= 0 (42)




119865 (119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(43)


119894is taken into


119865119894(119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(44)








1198652(119873) 119865



1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)




012

26

015

005












119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119860 =119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)

119864119868 (ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)

119861 =119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

119862 =cosh (119899

119894119871119894)119872119894119894+1

minus119872119894+1119894

119864119868 (ℎ2

119894+ 1198992

119894) sinh (119899

119894119871119894)

119863 =minus119872119894119894+1

119864119868 (ℎ2

119894+ 1198992

119894)

(34)



120579119894(119909) = 120578

119894[119872119894+1119894

minus119872119894119894+1

cos (ℎ119894119871119894)] cos (ℎ

119894119909)

minus 120595119894[119872119894+1119894

minus cosh (119899119894119871119894)119872119894119894+1

] cosh (119899119894119909)

minus 120578119894119872119894119894+1

sin (ℎ119894119871119894) sin (ℎ

119894119909)

minus 120595119894119872119894119894+1

sinh (119899119894119871119894) sinh (119899

119894119909)

(35)

where 120578119894= ℎ119894119864119868(ℎ

2

119894+ 1198992

119894) sin(ℎ

119894119871119894) 120595119894= 119899119894119864119868(ℎ

2

119894+

1198992

119894) sinh(ℎ

119894119871119894)


119894= 119872119894119894+1

= 119872119894119894minus1


120579119894119894+1

= [120595119894cosh (ℎ

119894119871119894) minus 120578119894cos (ℎ

119894119871119894)]119872119894

minus (120595119894minus 120578119894)119872119894+1

120579119894119894minus1




cos (ℎ119894minus1119871119894minus1)]119872119894

(36)


= 120579119894119894+1

2 ⩽ 119894 ⩽ 119899) so we can get



119894119871119894)


ΩM = 0 (38)

whereM = [11987211198722 119872

119899+1]119879

Ω =[[[

[

11988311198841+ 1198842

1198832

0

1198832

1198842+ 1198843

1198833


0 119883119899minus1

119884119899minus1

+ 119884119899119883119899

]]]

]

(39)


1= 0 and119872



Ω0M0 = 0 (40)

whereM0 = [11987221198723 119872119899]119879

Ω0 =

[[[[[

[

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

]]]]]

]

(41)


1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

= 0 (42)




119865 (119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(43)


119894is taken into


119865119894(119873) =

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

1198841+ 1198842

1198832

1198832

1198842+ 1198843

1198833


119883119899minus2

119884119899minus2

+ 119884119899minus1

119883119899minus1

119883119899minus1

119884119899minus1

+ 119884119899

1003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

(44)








1198652(119873) 119865



1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)




012

26

015

005












119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















1198652(119873) 119865



1198651(119873) asymp 0

1198652(119873) asymp 0

1198653(119873) asymp 0

119865119899(119873) asymp 0

(45)




012

26

015

005












119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















119865119894(119873)

= 119891119894minus

119894119864119860

2 (119864119860 minus 119873) 119871

radic119864119868

119898+119873

119898(119871

119894120587)

2

timesradic[119894120587119864119860

(119864119860 minus 119873) 119871]

2

minus119873

119864119868 + 119873(119871119894120587)2+

241198902

1198712 (1198902 + 41199032)

(46)




1198651(119873) and 119865



1(119873) and 119865





1(119873)







(a)

(b)




2

sum

119894=1

ℎ119894cos (ℎ

119894119871119894)

(ℎ2

119894+ 1198992

119894) sin (ℎ

119894119871119894)minus

119899119894cosh (119899

119894119871119894)

(ℎ2

119894+ 1198992

119894) sinh (ℎ

119894119871119894)= 0 (47)






0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













0 3276 50 100 150minus04

minus02

minus0006


F(N

)

F1(N)F2(N)

12627minus01

00022

01

02

03

04

F(N

)


F1(N)F2(N)

9622minus02

minus00043

01

02

03

F(N

)


F1(N)F2(N)

5324minus03

minus02

00037

01

02


F1(N)F2(N)

F(N

)


2(119873)

05

036

017

005

005

43 4305 0630 0505 30



50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










50 60 70 753 80 90 100minus6

minus4

minus2

0

2

4

6


F1(N)F2(N)F3(N)

F(N

)

753

0


F(N

)

Zoom-in

60 70 80 90 100 1056 120minus14

minus10

minus6

minus2

0

4


F1(N)F2(N)F3(N)

F(N

)

1056333

0


F(N

)

Zoom

-in90 110 130 1492 160 180

minus30

minus20

minus10

0

10


F1(N)F2(N)F3(N)

F(N

)

1492

0


F(N

)

Zoom

-in

120 140 160 1814 200minus35

minus25

minus15

minus5

0

5


F1(N)F2(N)F3(N)

F(N

)

1814

0


F(N

)

Zoom-in


2(119873) and 119865

3(119873)




1(119873)

1198652(119873) and 119865

3(119873) when 119894 = 1 119894 = 2 and 119894 = 3 (to


1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865




1(119873)119865

2(119873) and119865



1(119873)

1198652(119873) and 119865






60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












60


90


120









6 Conclusion




Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











6 Conclusion




Acknowledgments


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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