A Novel Fractional Tikhonov Regularization Coupled with an ...MathematicalProblemsinEngineering 5.5....

Research ArticleA Novel Fractional Tikhonov Regularization Coupled withan Improved Super-Memory Gradient Method and Applicationto Dynamic Force Identification Problems

NengjianWang1 Chunping Ren 1 and Chunsheng Liu2

1College of Mechanical and Electrical Engineering Harbin Engineering University Harbin 150001 China2School of Mechanical Engineering Heilongjiang University of Science and Technology Harbin 150022 China

Correspondence should be addressed to Chunping Ren renchunpinsinacom

Received 8 January 2018 Revised 27 March 2018 Accepted 15 April 2018 Published 2 July 2018

Academic Editor Kishin Sadarangani

Copyright copy 2018 NengjianWang et alThis 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

This paper presents a novel inverse technique to provide a stable optimal solution for the ill-posed dynamic force identificationproblems Due to ill-posedness of the inverse problems conventional numerical approach for solutions results in arbitrarily largeerrors in solution However in the field of engineering mathematics there are famous mathematical algorithms to tackle the ill-posed problem which are known as regularization techniques In the current study a novel fractional Tikhonov regularization(NFTR) method is proposed to perform an effective inverse identification then the smoothing functional of the ill-posed problemprocessed by the proposedmethod is regarded as an optimization problem and finally a stable optimal solution is obtained by usingan improved super-memory gradient (ISMG) method The result of the present method is compared with that of the standard TRmethod and FTR method new findings can be obtained that is the present method can improve accuracy and stability of theinverse identification problem robustness is stronger and the rate of convergence is faster The applicability and efficiency of thepresent method in this paper are demonstrated through a mathematical example and an engineering example

1 Introduction

In this paper we present a novel regularization technique tosolve the following linear operator equations of the form

Cx = y (1)

where x y isin R푛 and C isin R푛times푛 represent the identifiedvector the measured vector and a compact linear operatorrespectively

Under these circumstances a generalized solution of (1)can be obtained

x = Cdaggery (2)

where Cdagger is the MoorendashPenrose pseudoinverse of C which isusually an ill-conditioned operator as for example it couldbe if det119862 = 0 which causes (2) to be a typical ill-posedproblem see [1 2]

Ill-posed problem exists widely in many fields of scienceand has important application in engineering [3] Howeverin many cases of linear equation which appears in practicalengineering problem generally it is clear that the measuredvector y is rarely obtained exactly Conversely noises are con-tained in y owing to the measurement error ConsequentlywhenC is severely ill-conditioned and y is disturbed throughnoise we will be confronted with a problem in which thegeneralized solution of an ill-posed linear equation may beseriously deviated from the exact solution [4ndash6]

In mathematical theory there are several typical regu-larization methods which can solve the ill-posed problemin practical engineering applications For example the stan-dard Tikhonov regularization (TR) method as a traditionaltechnique has been widely adopted for overcoming theill-posed nature (or numerical instability) such as systemparameter identification load source identification problemand so on [7ndash10] T S Jang et al [11] applied the standardTR method to identify the loading source of infinite beams

HindawiMathematical Problems in EngineeringVolume 2018 Article ID 4790950 16 pageshttpsdoiorg10115520184790950

2 Mathematical Problems in Engineering

on an elastic foundation from given information of verticaldeflection of infinite beams J Gao et al [12] utilized thestandard TR method for identifying the unknown loadsource VMelicher et al [13] identified the piecewise constantparameter using the standard TR method In addition manyworks are mainly concerned with the identification of loadposition and load number in nonlinear system using thestandard TR method [14ndash16] However the standard TRmethod is not completely perfect there are some unavoidablelimitations and disadvantages (i) the standard TR methodmay determine approximate solutions that are too smooth(ii) the approximate solution may lack many details that thedesired exact solution might possess therefore it is very nec-essary to discuss other regularizationmethods to improve thelimitations and disadvantages of the regularization methodmentioned above

In the past few years fractional Tikhonov regularization(FTR) method was gradually introduced to deal with theill-posed problems mentioned above [17] Furthermore theFTR method has been applied to many fields and obtainedsome remarkable achievements [18ndash20] M E Hochstenbachet al [21] used the FTR method to obtain approximatesolution of higher quality which is better than the standardTR method in solving linear system equation with a severeill-posedness In [22] we can see that the FTR method issignificantly better than the standard TR method which wasproved by several numerical examples It can be found from[23] that the FTR method can obtain more approximationsolutions than the standard TRmethod which was discussedthrough several computed examples J Zhang et al [24]illustrated that the FTR method was more suitable fortackling partly textured images problems Hence the FTRmethod mentioned above has many advantages to a certainextent which may not only solve the shortcomings of thestandard TRmethod but also be able to provide a useful helpfor obtaining the effective solution of the ill-posed problemHowever FTR method is also not very perfect which hasgreat limitations for solving the ill-posed problem First thesolutions of the larger singular values can not be retainedwell Then it is not completely effective for the large-scaleill-posed problem Thus it is very necessary to discuss aneffective technique to tackle the limitations of the problemencountered

Considering the limitations of TR method and FTRmethod in a certain case we propose a novel fractionalTikhonov regularization (NFTR) method to offer a stablesolution for the ill-posed problems in practical engineeringapplications In our work the ill-posed problem processed bythe proposed method is redefined as a class of unconstrainedoptimization problems to solve the following minimumproblem

minxisinR119899

119891 (x) = Λ|NFTR (3)

In which 119891(x) R푛 997888rarr R is a continuously differentiablefunction andΛ|NFTR denotes the smoothing functional usingthe NFTR method

In fact there are many optimization algorithms thatcan improve the efficiency of solving (3) like conjugate

gradient (CG) method memory gradient (MG) methodsuper-memory gradient (SMG) method and so on [25ndash27]It is generally known that CG method does not achieveglobal convergence for common functions under inexact linesearches [28] whereas MG method requires dealing withthe trust region subproblem [29] Although SMG methodis superior to CG and MG method in some cases itis also not globally convergent [30] Therefore it is verymeaningful to find a global convergence algorithm based onthe SMG method Hence in this paper an improved super-memory gradient (ISMG) method is introduced to deal withunconstrained optimization problems of (3)

This article is focused on a novel stable inverse techniquedeveloped to provide an optimal solution for the ill-posedproblems in practical engineering Compared with our pre-vious works some highlights and advances of the proposedtechnique are derived in this article

First of all the article is different from our previousstudies and the highlights of this article are describedas follows (i) a novel fractional Tikhonov regularization(NFTR) coupled with an improved super-memory gradi-ent (ISMG) method is proposed (ii) solving the ill-posedproblem procedure is regarded as a class of unconstrainedoptimization problems (iii) the detailed information of thesolution is clearly enhanced (iv) the technique we proposecan effectively overcome the smoothness of the solution ofthe ill-posed problem (v) global convergence and stability ofthe solution are proved

Then some advances of the proposed technique aredescribed as follows (i) a stable optimal solution for the ill-posed dynamic force identification problems can be obtained(ii) the proposed method yields approximate solutions ofhigher quality than the standard TR method and FTRmethod (iii) the technique proposed does not need any apriori information on the model of the inverse identificationonly the measured response is sufficient for the presentmethod (iv) the applicability of the present technique isexamined through a mathematical example and an engineer-ing example

In this paper we propose a novel fractional Tikhonovregularization (NFTR) coupled with an improved super-memory gradient (ISMG) method to solve the ill-poseddynamic force identification problems in engineering appli-cations The remainder of this paper is formulated as followsThe inverse problem model is described in Section 2 InSection 3 we present the NFTR method for solving the ill-posed problem In Section 4 ISMG method is introducedto deal with an optimization problem The convergenceproperties of the present method are proved in Section 5 InSection 6 we discuss the application of a mathematical andan engineering example Section 7 shows some meaningfulfindings and new conclusions

2 Description of Inverse Problem Model

The model of inverse problems encountered in science andengineering can be uniformly expressed by a Fredholmintegral equation of the first kind [31ndash33]

Mathematical Problems in Engineering 3

intb

ac (119905 minus 120591) x (120591) d120591 = y (119905) (4)

where x(t) c(t minus 120591) and y(t) represent the identified objectthe kernel function and the measured response respectively

The discretization form of (4) is processed based on therectangular formula

푛sum푖=1

c (119905푘 minus 120591푖) x (120591푖) ΔT = y (119905푘) (5)

in which ΔT is time interval we consider that

x푖 = x (120591푖) y푘 = y (119905푘)

c푘minus푖 = c (119905푘 minus 120591푖) (6)

Then we have

x = (x1 x2 x3 sdot sdot sdot x푛)T y = (1199101 y2 y3 sdot sdot sdot y푛)T C = (c푘minus푖) ΔT

(7)

Hence (5) can be simply written as

Cx = y (8)

The vector 119910 represents available data which is contaminatedby an error 119890 The error may stem from measurementinaccuracies or discretization Thus

y = y + 119890 (9)

where 119910 is the unknown error-free vector associated with 119910We will assume the unavailable error-free system

Cx = y (10)

to be consistent and denote its solution of minimal Euclideannorm by 119909 We would like to determine an approximationof 119909 by computing a suitable approximate solution of (10)Due to the ill-conditioning of the matrix 119862 and the error 119890 in119910 the solution of the least-squares problem (10) of minimalEuclidean norm is typically a poor approximation of 119909 and asmall perturbation in 119910 may give rise to an arbitrarily largeperturbation in 119909 or even make the problem unsolvableMoreover the right-hand side function that is available inapplications represents data that is contaminated by noiseThus instead of 119910 the error-contaminated function 119910 isavailable [34 35] Therefore it is very necessary to find aneffective method to solve the ill-posed problem

3 The Regularization Method for SolvingThe Ill-Posed Problem

In some instances we all know that some traditional numer-ical methods are not able to solve (8) effectively However

the regularization method as a way to deal with the ill-posed problem has been used by mathematical workersand engineering technicians [36ndash38] In the past few yearsthe standard Tikhonov regularization (TR) and fractionalTikhonov regularization (FTR) method have been success-fully applied to many engineering fields Although manymeaningful achievements have been made there are stillmany limitations Hence in this work considering thedisadvantages of the standard TR method and FTR methoda novel fractional Tikhonov regularization (NFTR) methodis discussed

31 The Standard Tikhonov Regularization (TR) Method Thestandard Tikhonov regularization (TR) method is a popularapproach to determine an approximation of x see eg [39ndash41] for properties and applications This method replacesthe minimization problem (8) by a penalized least-squaresproblem of the following form [42]

min119891 (x) = 1003817100381710038171003817y minus Cx10038171003817100381710038172 + 10038171003817100381710038171003817A휇x100381710038171003817100381710038172 (11)

in which A휇 = 120583I and A휇 120583 and I represent theregularization matrix the regularization parameter and theunit matrix respectively

32 Fractional Tikhonov Regularization (FTR) Method It iswell known that Tikhonov regularization in standard formtypically determines a regularized solution that is too smoothie many details of the desired solution are not representedby the regularized solution This shortcoming led Klann andRamlau to introduce the fractional Tikhonov regularizationmethod [17] Subsequently another approach also referred toas fractional Tikhonov regularization (FTR) was investigatedin [21] the model of which is described as follows

min119891 (x) = 1003817100381710038171003817y minus Cx10038171003817100381710038172W + 1205832 x22 (12)

Then we know

x2W = xTWx (13)

In which W is defined with the aid of the MoorendashPenrosepseudoinverse of 119862119862푇

W = (CCT)(훼minus1)2 (14)

where 120572 denotes fractional order and 120572 isin (0 1) 119862T denotesthe transpose of 119862 It is obvious that FTR method is differentthan that of TR method it mainly considers the influenceof fractional order on the identified objects which thusmay be more suitable for solving the ill-posed identificationproblem However FTR method is not perfect which hasgreat limitations for solving the ill-posed problem First thesolutions of the larger singular values can not be retainedwell Then it is not completely effective for solving the large-scale ill-posed problem Thus it is very necessary to findan effective method to solve the limitations of problem weencountered


33 A Novel Fractional Tikhonov Regularization (NFTR)Method From Sections 31 and 32 on the one hand wecan find that the regular solution is obtained by using thestandard TR method that may be too smooth and maylose some details of the solution we expected On theother hand the solutions of the larger singular values cannot be retained perfectly using the FTR method for deal-ing with some practical complex engineering applicationsHence in order to solve the limitations of TR methodand FTR method in a certain case in this paper a novelfractional Tikhonov regularization (NFTR) method is pro-posed The model of the proposed method is described asfollows

min119891 (x) = 1003817100381710038171003817y minus Cx10038171003817100381710038172M + 10038171003817100381710038171003817S휇x1003817100381710038171003817100381722 (15)

in which

x2M = 12xTMxM = (C119862T)(훼minus1)2 S휇 = ΓN휇V

T(16)

where S휇 N휇 VT Γ and M represent the regularizationmatrix the diagonal matrix right column orthogonal matrixof singular value decomposition of C adjustment coefficientmatrix and symmetric positive semidefinite matrix respec-tively In this work Γ is arranged as the identity matrix thatis Γ = 119868푛 = diag (1 1 1) in which N휇 can be describedas follows

N휇 =[[[[[[[

max 1205832 minus 12059012 0

max 1205832 minus 12059022 0

d

max 1205832 minus 120590푛2 0

]]]]]]] (17)

Hence the NFTR method corresponding to the normalequation can be expressed in the following form

((CTC)(훼+1)2 + ST휇S휇) x = (CTC)(훼+1)2CTy (18)

where 119878T휇 denotes the transpose of 119878휇Introduce the singular value decomposition (SVD) [43]

C = UΣVT = 푛sum푖=1

119906푖120590푖V푇푖 (19)

in which U = (1199061 1199062 1199063 sdot sdot sdot 119906푛) isin R푛times푛 andV = (V1 V2 V3 sdot sdot sdot V푛) isin R푛times푛 are orthogonal matrices

and

Σ = diag (1205901 1205902 1205903 sdot sdot sdot 120590푛) isin R푛times푛 (20)

The singular values are ordered according to

1205901 ge 1205902 ge 1205903 ge sdot sdot sdot ge 120590푟 gt 120590푟+1 = = 120590푛 = 0 (21)

in which the index 119903 is the rank of 119862 see eg [43] fordiscussions on properties and the computation of the SVD

Combining (18) and (19) then we have

((ΣTΣ)(훼+1)2 + S2휇)VTx = (ΣT)훼 UTy (22)

Hence we can obtain the regular solution x푦푠 of (22)

x푦푠 = V((ΣTΣ)(훼+1)2 + S2휇)minus1 (ΣT)훼 UTy (23)

Meanwhile the filter factor of TR FTR and NFTR methodcan be expressed as follows

Ω푇푅 (120590푖) = 1205902푖1205902

푖 + 1205832

Ω퐹푇푅 (120590푖) = 120590훼+1푖120590훼+1

푖 + 1205832

Ω푁퐹푇푅 (120590푖) =1 1 le 119894 le 119896120590훼+1푖1205832

119896 lt 119894 le 119899

(24)

34 Stability Analysis In the paper we may determine asuitable regular parameter 120583 gt 0 by the discrepancy principle[44 45] we choose 120583 gt 0 so that

10038171003817100381710038171003817Cxys minus y100381710038171003817100381710038172 = 120596ℓ (25)

where ℓ is equivalent to the noise level120575 which satisfies 1198902 leℓ = 120575 119890 is the noise vector 120596 is a user-supplied constantindependent of ℓ and 120596 gt 1

Using (23) and (24) we have

10038171003817100381710038171003817xys1003817100381710038171003817100381722 =푛sum

푖=1

Ω2푁퐹푇푅 (120590푖) (119906푇

푖 y)21205902푖

= 푘sum푖=1

(119906푇푖 y)21205902푖

+ 푛sum푖=푘+1

1205902훼+2푖 (119906푇

푖 y)212058341205902푖

(26)


Finding the partial derivative of 120583 and 120572 in (26) respectivelythen

120597120597120583 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = minus4 푛sum푖=푘+1

1205902훼푖 (119906푇

푖 y)21205835

120597120597120572 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = 2 푛sum푖=푘+1

1205902훼푖 ln120590 (119906푇

푖 y)21205834

(27)

We can find from (27) that 120583 997888rarr xys22 is a monotonousdecreasing function When C lt 1 then ln120590 lt 0 so we canobtain 120572 997888rarr xys22 also a monotonous decreasing function

However it is clearly seen from (27) that the regular-ization parameter becomes large and the norm of regularsolution becomes small Due to 120572 997888rarr xys22 being amonotonous decreasing function so reducing 120572 can increasethe norm of regular solution according to 120572 lt 1Theorem 1 For ℓ gt 0 the regularization parameter 120583 = 120583(ℓ)meets (25) and then one has

119889120583119889ℓ gt 0 (28)

Proof The following form can be found

10038171003817100381710038171003817Cxys minus y푑1003817100381710038171003817100381722 =푛sum

푖=1

(1 minus Ω푁퐹푇푅 (120590푖))2 (119906푇푖 y)2 (29)

Then the following form is obtained by substituting (24) and(25) into (29)

1205962ℓ (120583)2 = 푛sum푖=1

(1 minus 120590훼+1푖1205832

)2 (119906푇푖 y)2 (30)

Offering the derivative of 12058321205962ℓ (120583) 119889ℓ119889120583 = 2 푛sum

푖=1

(1 minus 120590훼+1푖1205832

) 120590훼+1푖1205833

(119906푇푖 y)2 (31)

We can see from the previous formula that 119889ℓ119889120583 gt 0 then119889120583119889ℓ gt 0 (32)

Hence the conclusion is proved

In this paper the NFTR method is of order optimalwhen the regularization parameter is chosen according tothe discrepancy principle and the proof process is shown asfollows

Let 119860T119860 = (119862T119862)minus훾 119878T휇119878휇 = 120583119868 then (18) can bedescribed as a normal equation

(119860T119860 + 120583 (119860T119860)minus훾) xys = 119860Ty (33)

where 119860T and 119878T휇 denote the transpose of 119860 and 119878T휇 120574 = (120572 minus1)2

For our analysis we assume a Holder-type smoothnessassumption ie that the minimal norm solution 119909푦푠 of theerror-free problem (1) satisfies a smoothness condition of theform for some constant 119911

x isin 119877((119862T119862)휗) (34)

With x휗 fl (sum푛ge1 120590minus2휗푖 |⟨x 119906푛⟩|2)12 le 119911

Here (120590푛 119906푛 V푛)푛ge1 is the singular system of the operator119862 120599 is a parameter which controls the a priori assumptionson the regularity of the domain set

Theorem 2 ([17 Theorem 31]) Let 119910 isin range(119862) and 119910 minus119910푑 le 120575 Ω푁퐹푇푅(120590) is a regularizing filter and assume that for0 le 120599 le 120599lowast

sup0lt휎le휎1

120590minus1 1003816100381610038161003816Ω푁퐹푇푅 (120590)1003816100381610038161003816 le 120581120583minus휛(35)

and

sup0lt휎le휎1

120590휗lowast 10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 le 120581휗lowast120583휛휗lowast (36)

where 120603 gt 0 and 120581 120581휗lowast are constants independent of 120575 Thedefinition of 120599lowast is the same as that of 120599 Then with the a prioriparameter choice

120583 = 119901(120575119911)(1휛)(휗+1) 119901 gt 0 fixed (37)

The NFTR method induced by the filter Ω푁퐹푇푅(120590) is orderoptimal for all 0 le 120599 le 120599lowast

Lemma 3 The regularizing filter Ω푁퐹푇푅(120590) from (24) withminus1 le 120572 le 1 fulfillsTheorem2with120603 = 1(120572+1) and120599lowast = 120572+1Proof The filter Ω푁퐹푇푅(120590) is continuous on (0infin) The reg-ularizing properties of the filter Ω푁퐹푇푅(120590) are easily verifiedOne sees that

lim휎㨀rarr0

120590minus1 1003816100381610038161003816Ω푁퐹푇푅 (120590)1003816100381610038161003816 = lim휎㨀rarrinfin

120590minus1 1003816100381610038161003816Ω푁퐹푇푅 (120590)1003816100381610038161003816 = 0lim휎㨀rarr0

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 120590휗 = 0

lim휎㨀rarrinfin

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590푖)1003816100381610038161003816 120590휗 =

infin 120599 gt 120572 + 11 120599 = 120572 + 10 120599 lt 120572 + 1

(38)

Hence as long as 120599 lt 120572 + 1 the suprema in Theorem 2are attained as local maximal which can be derived by simplecalculus One obtains

sup0lt휎le휎1

120590minus1 1003816100381610038161003816Ω푁퐹푇푅 (120590)1003816100381610038161003816 le 120581120583minus1(훼+1)(39)

and

sup0lt휎le휎1

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 120590휗lowast le 120581휗lowast120583(1(훼+1))휗 (40)

Consequently by Theorem 2 the NFTR method (15) isorder optimal for minus1 lt 120572 le 1 and 0 le 120599 le 120572 + 1


4 The Algorithm for Solving UnconstrainedOptimization Problem

In general it has always been known that the regular solutionof (15) may give rise to an arbitrarily large perturbation inactual solution However one of the highlights of this work isexpressed as follows (15) is deemed considered as an uncon-strained optimization problem and the optimal solution of(15) can be obtained by using some optimization algorithmsIn this paper an improved super-memory gradient (ISMG)method was introduced to solve objective function of (15)Most of the well-known iterative algorithms for solving (15)take the following form [46ndash49]

119909푘+1 = 119909푘 + ℎ푘119889푘 (41)

where 119889푘 is the search directionIn this paper a new search direction 119889푘 of ISMGmethod

is determined as follows

119889푘 =

minus119892푘 119896 lt 119898minus119892푘 + 푚sum

푖=1

120573푖푘119889푘minus푖 119896 ge 119898 + 1 (42)

where 119892푘 denotes the gradient of 119891(119909푘) at the point 119909푘 and119892푘 = nabla119891(119909푘) 120573푖푘 is defined as follows

120573푖푘 = 120588 1003817100381710038171003817119892푘

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 120588 isin (0 1119898) (43)

ℎ푘 of (41) is the step length which can be computed usingmodified nonmonotone line search technique ℎ푘 = 120573푚119896 and119898푘 satisfies the following inequalities

119891 (119909푘 + 120573푚119896119889푘) le 119891 (119909푘) + 120592120573푚119896 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (44)

where 120592 120573 isin (0 1) and 120578 gt 0Algorithm 4

Step 1 Given an initial point 1199090 isin 119877푛 120592 isin (0 1) 120573 isin (0 1)and 120578 gt 0 119898 is a given positive integer 120588 isin (0 1119898) and0 lt 120576 le 1 set 119896 = 0Step 2 Compute 119892푘 if 119892푘 le 120576 then terminate else go to thenext step

Step 3 Compute 119889푘 based on (42)

Step 4 Compute ℎ푘 based on (44)

Step 5 Set 119909푘+1 = 119909푘 + ℎ푘119889푘

Step 6 Set 119896 = 119896 + 1 and go to Step 2

Lemma 5 For any line search the search direction 119889푘 isdefined by (42) then one has

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (45)

Proof If 119896 le 119898 so we have

119892푇푘 119889푘 = 119892푇

푘 (minus119892푘) = minus 1003817100381710038171003817119892푘10038171003817100381710038172 (46)

and 120588 isin (0 1119898) so it has 0 lt 119898120588 lt 1 minus1 lt 119898120588 minus 1 lt 0Hence we obtain

119892푇푘 119889푘 = minus 1003817100381710038171003817119892푘

10038171003817100381710038172 le (119898120588 minus 1) 1003817100381710038171003817119892푘10038171003817100381710038172 (47)

If 119896 ge 119898+1 according to (42) and (43) we can obtain thefollowing

119892푇푘 119889푘 = 119892푇

푘 (minus119892푘 + 푚sum푖=1

120573푖푘119889푘minus푖)

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 sdot 119892

푇푘 119889푘minus푖

le minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 = (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172

(48)

Then we have

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (49)

Hence (45) is proved

Lemma 6 For any 119896 then one has1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (50)

Proof If 119896 le 119898 so we have1003817100381710038171003817119889푘

1003817100381710038171003817 = minus 1003817100381710038171003817119892푘1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘

1003817100381710038171003817 (51)


1003817100381710038171003817119889푘1003817100381710038171003817 =

1003817100381710038171003817100381710038171003817100381710038171003817minus119892푘 + 푚sum푖=1

120573푖푘119889푘minus푖

1003817100381710038171003817100381710038171003817100381710038171003817 le1003817100381710038171003817119892푘

1003817100381710038171003817 +푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= 1003817100381710038171003817119892푘1003817100381710038171003817 +

푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 = (119898120588 minus 1) 1003817100381710038171003817119892푘

1003817100381710038171003817 (52)

So we have1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (53)


It is clear that we can draw the following conclusionaccording to (45) and (50)

minus119892푇푘 119889푘1003817100381710038171003817119889푘10038171003817100381710038172 ge 1 minus 119898120588

(1 + 119898120588)2 (54)


5 The Convergence Properties Analysis forOptimal Algorithm

In this section we propose an algorithm and then we studythe global convergence property of this algorithm Firstly wemake the following two assumptions which have beenwidelyused in the literature to analyze the global convergence of thepresented method

Assumption 7 The objective function 119891(119909) is considered tobe bounded in the following circumstances

120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)

Assumption 8 The gradient 119892(119909) is Lipschitz continuousthere exists a constant 120585 gt 0 for any 119909 119910 isin 120595 we have

1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)

Lemma 9 Suppose that Assumptions 7 and 8 hold thenAlgorithm 4 is well-posed that is there exists the step lengthℎ푘 which makes (44) hold

Proof According to (44) we have

limℎ㨀rarr0

119891 (119909푘 + ℎ푘119889푘) minus 119891 (119909푘) minus 120592ℎ푘 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)

It indicates that the following formula is set up for whenthere exists ℎ耠

푘 gt 0 and forallℎ isin [0 ℎ耠푘]


10038171003817100381710038172) le 0 (58)


Lemma 10 Suppose that Assumptions 7 and 8 hold 120578 isin (0 1minusm120588) and there is a constant 120575 then

119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)

Proof ℎ = infforall푘ℎ푘 is defined we just need proof that ℎ gt 0First of all we prove it with a contradiction suppose that ℎ =0 holds and there are a series of subfamily ℎ푘 then

lim푘㨀rarr+prop

ℎ푘 = 0 (60)

Set ℎ = ℎ푘120573 and ℎ푘 is the maximum in 1 120573 1205732 sdot sdot sdot whichmeets (44) Thus ℎ = ℎ푘120573 gt ℎ푘 can not meet (44) Then

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)

Based on the mean value theorem then

119891 (119909푘 + ℎ119889푘) minus 119891 (119909푘) = 119892 (119909푘 + ℎ120577푘119889푘)푇 sdot ℎ119889푘 (63)

in which 120577푘 isin [0 1]Thus it follows from (62) and (63) that

119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)

From Assumption 8 and Cauchy Schwartz inequality wecan obtain

[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘

le 1003817100381710038171003817119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘

1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)

From (64) and (65) it follows that

(120592 minus 1) 119892푇푘 119889푘 lt [119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]2 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (66)

Then we have

ℎ ge 1 minus 120592120585 (minus119892푇푘 119889푘1003817100381710038171003817119889푘10038171003817100381710038172 ) (67)

Combining (67) and (54) we obtain

ℎ푘 = ℎ120573 ge 120573 (1 minus 120592)120585 sdot (1 minus 119898120588)(1 + 119898120588)2 gt 0 (68)

Obviously we clearly see that (68) and (60) are contradic-tory Hence we have

ℎ gt 0 (69)

Then combining (44) and (45) we can obtain

119891 (119909푘) minus 119891 (119909푘+1) ge minus120592ℎ푘 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)

in which set 120575 = 120592ℎ(1 minus 119898120588 minus 120578)Hence (59) is proved

According to lim푘㨀rarrinfin inf 119892푘 = 0 (see [50 51]) it is clearthat the sequence xk is contained in 120595 and there exists aconstant 119891lowast such that

lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)

forallx y isin [a b] suppose altxltyltb then it can satisfy thefollowing conditions a + x lt x b minus x gt y for x smallenough Because 119891(x) is a convex function so we obtain

119891 (a + x) minus 119891 (a)xle 119891 (y) minus 119891 (x)

y minus x

le 119891 (b) minus 119891 (b minus x)x

(72)


Letx 997888rarr 0+ we have

119891耠+ (a) le 119891 (y) minus 119891 (x)

y minus xle 119891耠

minus (b) (73)

Set K = max|119891耠+(a)| |119891耠

minus(b)| then100381610038161003816100381610038161003816100381610038161003816119891 (y) minus 119891 (x)

y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)

Hence we get1003816100381610038161003816119891 (x) minus 119891 (y)1003816100381610038161003816 le K 1003816100381610038161003816x minus y1003816100381610038161003816 (75)

Therefore according to the above discussion we can drawa conclusion that the objective function 119891 defined in (15) isdifferentiable and Lipschitz continuous

6 Examples and Discussions

In this section a mathematical example and an engineeringexample are investigated to clarify how to solve the solution ofthe ill-posed problem quickly and accurately by using a novelfractional Tikhonov regularization (NFTR) coupled with animproved super-memory gradient (ISMG) method

61 A Mathematical Example In this section our first exam-ple is a severely ill-posed Fredholm integral equation of thefirst kind given by

int휋

0exp (120591 cos (119905)) x (119905) d119905 = 2 sinh (120591)120591 0 le 120591 le 1205872 (76)

Set C(119905 120591) = exp(120591 cos(119905)) y(120591) = 2(sinh(120591)120591) thenint휋

0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)

in which the perturbation of the right-hand side is set asfollows

y (120591) = y (120591) + 119890 (78)

where 119890 = 120575 sdot (119910(120591)) sdot 119903119886119899119889 in which std(y(120591)) rand and120575 denote the standard deviation of y(120591) a random numberwithin [minus1 1] and the noise level respectively

Then (77) was discretized which can be obtained

120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)

in which 119905푖 = 120587119894119899 120591푘 = 1205871198962The problem of solving (76) is an ill-posed problem we

solve the approximate solution of (76) with the right-handside given a perturbation The true solution of (76) is x(119905) =sin(119905) then we need to calculate the identified solution x(119905)Here TR method FTR method and the present method inthis paper are applied to deal with the problem

In order to further discuss the quality of the differentmethods used and assess the degree of the approximation

of the identified result to the true result we provide thefollowing evaluation metrics

Relative error (RE)

RE = x minus xx (80)

Correlation coefficient (CC)

CC = sum푛푖=0 [x minus E (x)] [x minus E (x)]x minus E (x) sdot x minus E (x) (81)

The program processing is executed in the MATLABregularization tools package and the condition number of thematrix C is calculated by the MATLAB function cond it canbe found that cond(C) = 2sdot1037 whichmeans that thematrixis singular

In our experiments the noise vector 119890 is selected asa normal random distribution vector with zero mean andnormalized to an ideal noise level

= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)

In this experiment the regularization parameter 120583 is selectedby using the discrepancy principle in which 1198902 = 120575 = ℓ and120596 = 101 Other parameters are set to the following 120592 = 05120573 = 03 120578 = 02 119898 = 2 and 120576 = 10minus3 The computer stopsthe iteration processing when |119892푘| le 10minus3 is met120572lowastdenotes the optimal value of the fractional order 120572 isin0 01 02 10 that gives the most accurate identifiedresult of (76) when 120583 is selected using the discrepancyprinciple (1198902 = 120575 = ℓ and 120596 = 101)

In this section we employ the standard TR FTR and thepresent method in the paper to identify the results of x(119905)respectively

First of all when 119899 isin [20 30 40 50] and the noise level isset to 10 5 and 1 then RE and CC and iterative steps ofthe identified result are listed in Table 1

Table 1 shows that whether the noise level is set to 105 or 1 RE and iterative steps of the identified result withapplication of the above methods are seen to increase with 119899then CC is the opposite

Moreover Table 1 also shows that when 119899 isin[20 30 40 50] RE and iterative steps of the identifiedresult with application of the above methods are seen toincrease with the noise level then CC is the opposite

It can be seen clearly that the evaluation indicatorsusing the present method are associated with minimum REminimum iterative steps and maximum CC in comparisonto the standard TR method and FTR method

In a word Table 1 shows that the identified result of thepresentmethod in this paper is closer to the true solution thanthat of the other methods

Hence it can be seen clearly that the evaluation indicatorsusing the present method are associated with minimum REminimum iterative steps and maximum CC in comparisonto the other methods which demonstrates that the presentmethod in this paper is superior to the standard TR and FTRmethod


Table 1 The evaluation metrics comparisons of different methods

119899 Noise level () Methods RE CC Iterative steps

20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13

62 An Engineering Example In this section the presentmethod is applied to an engineering example of the identi-fication problem of dynamic force generated between conicalpick and coal-seam structure

621 Material Properties A schematic of conical pick andcoal-seam structure is shown in Figure 1 In this diagramthe size of the coal-seam structure specimen manufacturedartificially was set as 2000 mm times 800 mm times 1750 mm Thedensity of conical pick is 7800 kgm3 and the density ofcoal-seam structure is 160798 kgm3 the elastic modulus ofconical pick is 600MPa and the elastic modulus of coal-seamstructure is 32209 MPa the Poisson ratio of conical pickis 025 and the Poisson ratio of coal-seam structure is 025respectively [52ndash54]

622 Experimental Procedure In our current study theexperimental procedure was carried out on an experimentalsetup with rotatable and adjustable parameter which wasshown in Figure 2The experimental setup includes a test bedand a test system The test bed mainly includes the followingdevices motor reducer coupling cutting mechanism plat-form and coal-seam structure The test system includes theforce-measured device force sensor the signal amplifier andtheDaspV10 intelligent data acquisition and signal processingsystem [55]

It is clearly seen from Figure 3 that the conical pick isinstalled on the force-measured device the various surfacesof which are intercepting with the force sensor once thecoal-seam structure is hit by conical pick the deformationof a force sensor is converted into a signal which is called


X

Y

Z

Conical pick

Coal-seam structure

Figure 1 A schematic of conical pick and coal-seam structure

V10Dasp data intelligent system Coal-seam structure

Test bed

Motor Coupling Reducer

Platform

The cutting mechanism

Figure 2 Experimental procedure

the dynamic force then the dynamic force is converted by asignal amplifier finally recorded by V10Dasp data intelligentacquisition and signal processing system

623 The Selection of the Optimal Fractional Order In ourexperiment the installation angle of conical pick is 45∘ themaximum cutting thickness of conical pick is 20 mm thehauling speed is 08 rmin and the rotary arm speed is30 mmin In order to effectively identify dynamic forcegenerated between conical pick and coal-seam structure first

Conical pick

Sleeve

Holder

Force sensor

Force sensor Force sensor

Force sensor

Figure 3 The dynamic force-measured device

0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)

Figure 4 The measured displacement response

we provide the measured displacement response within 03seconds which is shown in Figure 4 Since the measureddisplacement response can be given by the experiment thenwe can use the present method in the paper to identifydynamic force

In this part we can study the effect of different fractionalorder on the results of dynamic force identification in orderto determine the most optimal fractional order

First of all all program codes were made in MATLAB70 and Windows 7 operating system On the one hand 119889푘

is obtained using (42) all parameters are set to the following120596 = 101 120592 = 05 120573 = 03 120578 = 02 119898 = 2 and 120576 = 10minus3 Theiteration processing stops when |119892푘| le 10minus3 is met

The identified results of dynamic force are achieved byusing the present method in this paper which are shown inFigures 5(a) 5(b) 5(c) 5(d) 5(e) 5(f) 5(g) 5(h) and 5(i)respectively RE and CC and iterative steps of the identifiedresults under different fractional order (120572) are listed inTable 2


0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)

Identified forceActual force

(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)Identified forceActual force

(i) 훼 = 09

Figure 5 The identified results under different fractional order

Table 2 The selection of the optimal fractional order

Evaluation metrics Fractional order 12057201 02 03 04 05 06 07 08 09

RE 14687 11069 06304 05295 01038 03119 02825 04811 03463CC 03410 03686 04649 05058 09422 06583 08200 08188 08199Iterative steps 60 53 47 44 9 29 22 39 32

Figure 5(a) illustrates that the identified force is gainedat 120572 = 01 and Figure 5(b) provides the identified force at120572 = 02 the identified force at 120572 = 03 can be seen fromFigure 5(c) Figure 5(d) gives the identified force at 120572 = 04the identified force at 120572 = 05 is shown in Figure 5(e)Figure 5(f) illustrates that the identified force is gained at120572 = 06 Figure 5(g) provides the identified force at 120572 = 07the identified force at 120572 = 08 can be seen from Figure 5(h)and Figure 5(i) gives the identified force at 120572 = 09

We can see from Figures 5(a) 5(b) 5(c) 5(d) 5(e) 5(f)5(g) 5(h) and 5(i) that the different fractional order canidentify the dynamic force generated between conical pickand coal-seam structure from the measured displacementresponse However the performance of dynamic force iden-tification results has a certain difference

Figures 5(a) and 5(b) illustrate that the identified forceis not ideal owing to the smaller fractional order On thecontrary when the fractional order is larger the identifiedforce is very seriously deviating from the actual force whichcan be seen from Figures 5(h) and 5(i) Although Figures5(c) and 5(d) can identify the contours of the load there isa big error with the actual load Also Figures 5(f) and 5(g)can also identify the contours of the load but not the idealsolution we want However we can find that Figure 5(e) is therelatively ideal result that we expect Hence within the scaleof experiment we know that the optimal fractional order ofdynamic force identification using the present method is 05

In order to confirm the optimal fractional order RE andCC and iterative steps are listed in Table 2 It is clearly seenfrom Table 2 that RE and iterative steps of Figures 5(a) 5(b)5(c) and 5(d) are larger than those of Figure 5(e) HoweverRE and iterative steps of Figure 5(e) are smaller than thoseof Figures 5(f) 5(g) 5(h) and 5(i) And the CC value is the

opposite Hence compared with the other fractional orderwe find from Table 2 that the fractional order 120572 = 05 has thesmallest RE the largest CC and the least iterative steps

Taking into account the above analysis we can draw aconclusion that the optimal fractional order (120572 = 05) usingthe present method makes it more successful to identifydynamic force generated between conical pick and coal-seamstructure based on the measured displacement response

624 Comparison of Dynamic Force Identification MethodsThe proposed method in this paper is compared with thestandard TR method and FTR method and the identifiedresults of the three methods are shown in Figures 6(a) 6(b)and 6(c) respectively

Figure 6(a) explains that the identified force is gainedusing the standardTRmethod at120572 = 10 Figure 6(b) explainsthat the identified force is given by FTR method at 120572lowast = 04and the identified result at 120572lowast = 05 is shown in Figure 6(c)using the present method in the paper

We can find from Figures 6(a) 6(b) and 6(c) that thethree methods can all identify dynamic force however thepresent method in the paper can successfully and effectivelyidentify the details of dynamic force in comparison to thestandard TR method and FTR method

In order to deeply distinguish the stability of the proposedmethod in this paper we give the filter functions curves ofthe three methods with their corresponding optimal regularparameters (120583 = 10minus2) shown in Figure 7 In additionthe evaluation metrics of the identified results through threemethods are listed in Table 3

We can clearly find from Table 3 that the present methodunder the fractional order 120572lowast = 05 has the smallest RE thelargest CC and the least iterative steps


Table 3 Comparison of different identification methods

Evaluation metrics The standard TR method (120572 = 10) FTR method (120572lowast = 04) Present method (120572lowast = 05)RE 02673 02033 01038CC 07098 07799 09422Iterative steps 22 15 9

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(a) The standard TR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method

Figure 6 The comparison of identified results under different methods

Figure 7 shows that the three curves basically coincidedin the small singular value part with the regular parametersTherefore the three methods can be used to modify smallsingular values to suppress uncertain noise However withregard to the parts of the middle and large singular value thethree filter functions Ω(120572 120590푖) of the present method in thepaper have relatively larger value in comparison to the stan-dard TR and FTR method In addition Table 3 demonstratesthat the identified result of the proposed method is betterthan that of the standard TR method and FTR method

Taking into account the above analysis and discussionwe can conclude that the filter functions Ω(120572 120590푖) of thepresent method in the paper can not only retain the details ofthe solution components corresponding to smaller singular

values but also resist the solution components correspondingto larger singular values In other words the present methodin the paper can reduce the ill-posedness of dynamic forceidentification and can offer a stable identified result

7 Conclusions

This paper presents a novel inverse technique to offer anoptimal solution for the ill-posed problems in mathematicaland practical engineering The inverse problem models areuniformly represented as a Fredholm integral equation ofthe first kind which has the ill-posedness to a certain extentA novel fractional Tikhonov regularization (NFTR) methodis proposed to perform an effective inverse identification


TR methodFTR methodPresent method

100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)

Figure 7The curve of the filter factor with the singular value underdifferent methods

The smoothing functional of the ill-posed problem processedby the proposed method is considered as an unconstrainedoptimization problem an improved super-memory gradient(ISMG) method is introduced to offer an optimal solution

A mathematical problem and an engineering exampleof the identification problem of dynamic force generatedbetween conical pick and coal-seam structure show thatthe result of the present method is compared with that ofthe standard TR and FTR methods new findings can beseen that is the present method has a better effect andstronger robustness which can not only improve accuracyand stability of the inverse identification problem but alsoaccelerate the rate of convergence and enhance the detailsTherefore we can make a clear conclusion that the presentmethod in this paper can reduce the ill-posedness of theinverse problems andprovide a unified optimization solutionIn addition there is reason to believe that the present methodcan offer a reference for future research

Data Availability

This section describes the data used in this research Readerscan find the source of data in the following ways (i)httpswwwhindawicomjournalsmpe20179125734 and(ii) httpsdoiorg104028wwwscientificnetAMM577196

Conflicts of Interest

The authors declare that they have no conflicts of interestregarding the publication of this article

Acknowledgments

This work was supported by the Chinese National NaturalScience Foundation (Contracts nos 51674106 and 51274091)

References

[1] T S Jang H Baek S L Hana and T Kinoshita ldquoIndirectmeasurement of the impulsive load to a nonlinear system fromdynamic responses inverse problem formulationrdquo MechanicalSystems and Signal Processing vol 24 no 6 pp 1665ndash1681 2010

[2] T Hein and B Hofmann ldquoOn the nature of ill-posedness of aninverse problemarising in option pricingrdquo Inverse Problems vol19 no 6 pp 1319ndash1338 2003

[3] T S Jang H Baek M C Kim and B Y Moon ldquoA newmethod for detecting the time-varying nonlinear damping innonlinear oscillation systems nonparametric identificationrdquoMathematical Problems in Engineering Art ID 749309 12 pages2011

[4] S-J Kim and S-K Lee ldquoExperimental identification for inverseproblem of a mechanical system with a non-minimum phasebased on singular value decompositionrdquo Journal of MechanicalScience and Technology vol 22 no 8 pp 1504ndash1509 2008

[5] N S Hoang and A G Ramm ldquoSolving ill-conditioned linearalgebraic systems by the dynamical systems methodrdquo InverseProblems in Science and Engineering vol 16 no 5 pp 617ndash6302008

[6] W Luo ldquoParameter identifiability of ship manoeuvring mod-eling using system identificationrdquo Mathematical Problems inEngineering vol 2016 Article ID 8909170 10 pages 2016

[7] A Beck and A Ben-Tal ldquoOn the solution of the Tikhonovregularization of the total least squares problemrdquo SIAM Journalon Optimization vol 17 no 1 pp 98ndash118 2006

[8] H Egger and H W Engl ldquoTikhonov regularization applied tothe inverse problem of option pricing convergence analysis andratesrdquo Inverse Problems vol 21 no 3 pp 1027ndash1045 2005

[9] S G Solodky and G L Myleiko ldquoAbout regularization ofseverely ill-posed problems by standard Tikhonovrsquos methodwith the balancing principlerdquo Mathematical Modelling andAnalysis vol 19 no 2 pp 199ndash215 2014

[10] S George and P Jidesh ldquoReconstruction of signals by standardTikhonov methodrdquo Applied Mathematical Sciences vol 5 no57-60 pp 2819ndash2829 2011

[11] T S Jang H G Sung S L Han and S H Kwon ldquoInversedetermination of the loading source of the infinite beam onelastic foundationrdquo Journal of Mechanical Science amp Technologyvol 22 no 12 pp 2350ndash2356 2008

[12] J Gao D Wang and J Peng ldquoA Tikhonov-type regularizationmethod for identifying the unknown source in the modifiedHelmholtz equationrdquo Mathematical Problems in Engineeringvol 2012 Article ID 878109 13 pages 2012

[13] V Melicher and V Vrabel ldquoOn a continuation approachin Tikhonov regularization and its application in piecewise-constant parameter identificationrdquo Inverse Problems vol 29 no11 pp 193ndash212 2013

[14] BWeber P Paultre and J Proulx ldquoStructural damage detectionusing nonlinear parameter identification with Tikhonov regu-larizationrdquo Structural Control andHealthMonitoring vol 14 no3 pp 406ndash427 2007

[15] V A Postnov ldquoUse of Tikhonovrsquos regularization method forsolving identification problem for elastic systemsrdquoMechanics ofSolids vol 45 no 1 pp 51ndash56 2010

[16] T A Johansen ldquoOn Tikhonov regularization bias and variancein nonlinear system identificationrdquo Automatica vol 33 no 3pp 441ndash446 1997


[17] D Gerth E Klann R Ramlau and L Reichel ldquoOn fractionalTikhonov regularizationrdquo Journal of Inverse and ILL-PosedProblems vol 23 no 6 pp 611ndash625 2015

[18] T K Huckle and M Sedlacek ldquoTikhonov-Phillips regulariza-tion with operator dependent seminormsrdquo Numerical Algo-rithms vol 60 no 2 pp 339ndash353 2012

[19] D Bianchi A Buccini M Donatelli and S Serra-CapizzanoldquoIterated fractional Tikhonov regularizationrdquo PAMM vol 15no 1 pp 581-582 2015

[20] D Bianchi andMDonatelli ldquoOn generalized iterated Tikhonovregularization with operator-dependent seminormsrdquo ElectronicTransactions on Numerical Analysis vol 47 pp 73ndash99 2017

[21] M E Hochstenbach and L Reichel ldquoFractional Tikhonov regu-larization for linear discrete ill-posed problemsrdquo BITNumericalMathematics vol 51 no 1 pp 197ndash215 2011

[22] R H Chan A Lanza S Morigi and F Sgallari ldquoAn adaptivestrategy for the restoration of textured images using fractionalorder regularizationrdquo Numerical Mathematics Theory Methodsand Applications vol 6 no 1 pp 276ndash296 2013

[23] M D Ruiz-Medina J M Angulo and V V Anh ldquoFractional-order regularization and wavelet approximation to the inverseestimation problem for random fieldsrdquo Journal of MultivariateAnalysis vol 85 no 1 pp 192ndash216 2003

[24] J Zhang Z Wei and L Xiao ldquoA fast adaptive reweightedresidual-feedback iterative algorithm for fractional-order totalvariation regularized multiplicative noise removal of partly-textured imagesrdquo Signal Processing vol 98 pp 381ndash395 2014

[25] A Y Al-Bayati and R S Muhammad ldquoNew scaled sufficientdescent conjugate gradient algorithm for solving unconstraintoptimization problemsrdquo Journal of Computer Science vol 6 no5 pp 511ndash518 2010

[26] W W Hager and H Zhang ldquoThe limited memory conjugategradient methodrdquo SIAM Journal on Optimization vol 23 no4 pp 2150ndash2168 2013

[27] Y J Wang C Y Wang and N H Xiu ldquoA family of supermem-ory gradient projection methods for constrained optimizationrdquoOptimization A Journal of Mathematical Programming andOperations Research vol 51 no 6 pp 889ndash905 2002

[28] K W Ng and A Rohanin ldquoSolving optimal control problem ofmonodomainmodel using hybrid conjugate gradientmethodsrdquoMathematical Problems in Engineering vol 2012 Article ID734070 14 pages 2012

[29] Y Narushima and H Yabe ldquoGlobal convergence of a memorygradient method for unconstrained optimizationrdquo Computa-tional optimization and applications vol 35 no 3 pp 325ndash3462006

[30] J-B Jian Y-F Zeng and C-M Tang ldquoA generalized super-memory gradient projection method of strongly sub-feasibledirections with strong convergence for nonlinear inequalityconstrained optimizationrdquo Computers amp Mathematics withApplications vol 54 no 4 pp 507ndash524 2007

[31] Y Jia Z Yang N Guo and L Wang ldquoRandom dynamic loadidentification based on error analysis and weighted total leastsquares methodrdquo Journal of Sound and Vibration vol 358 pp111ndash123 2015

[32] J Liu X Sun X Han C Jiang and D Yu ldquoDynamic load iden-tification for stochastic structures based on Gegenbauer poly-nomial approximation and regularization methodrdquoMechanicalSystems and Signal Processing vol 56 pp 35ndash54 2015

[33] T Aktosun and C van der Mee ldquoSolution of the inversescattering problem for the three-dimensional Schroedinger

equation using a Fredholm integral equationrdquo SIAM Journal onMathematical Analysis vol 22 no 3 pp 717ndash731 2006

[34] J N Franklin ldquoMinimum principles for ill-posed problemsrdquoSIAM Journal on Mathematical Analysis vol 9 no 4 pp 638ndash650 1978

[35] N N Abdelmalek ldquoAn algorithm for the solution of ill-posedlinear systems arising from the discretization of Fredholmintegral equation of the first kindrdquo Journal of MathematicalAnalysis and Applications vol 97 no 1 pp 95ndash111 1983

[36] G L Mazzieri and R D Spies ldquoRegularization methods for ill-posed problems in multiple Hilbert scalesrdquo Inverse Problemsvol 28 no 5 055005 30 pages 2012

[37] V A Morozov ldquoApplication of the regularization method tothe solution of an ill-posed problemrdquo Proceedings of the RoyalSociety of Medicine vol 18 no 4 pp 401ndash406 1965

[38] N H Tuan and D D Trong ldquoA simple regularization methodfor the ill-posed evolution equationrdquo Czechoslovak Mathemati-cal Journal vol 61(136) no 1 pp 85ndash95 2011

[39] F Stout and J H Kalivas ldquoTikhonov regularization in stan-dardized and general form for multivariate calibration withapplication towards removing unwanted spectral artifactsrdquoJournal of Chemometrics vol 20 no 1-2 pp 22ndash33 2006

[40] Y Wei P Xie and L Zhang ldquoTikhonov Regularization andRandomized GSVDrdquo Siam Journal on Matrix Analysis andApplications vol 37 no 2 pp 649ndash675 2016

[41] S COmowunmi andAA Susu ldquoApplication of TikhonovReg-ularization Technique to the Kinetic Data of an AutocatalyticReaction Pyrolysis of N-Eicosanerdquo Engineering Journal vol 03no 12 pp 1161ndash1170 2011

[42] M Z Nashed ldquoThe theory of Tikhonov regularization forfredholm equations of the first kind (C W Groetsch)rdquo SiamReview vol 28 no 1 pp 116ndash118 2006

[43] H Liu X Wang and C Lu ldquoRolling bearing fault diagnosisunder variable conditions using Hilbert-Huang transform andsingular value decompositionrdquoMathematical Problems in Engi-neering vol 2014 Article ID 765621 10 pages 2014

[44] G M Vainikko ldquoThe discrepancy principle for a class ofregularization methodsrdquo USSR Computational Mathematicsand Mathematical Physics vol 22 no 3 pp 1ndash19 1982

[45] V Albani A De Cezaro and J P Zubelli ldquoOn the choice of theTikhonov regularization parameter and the discretization levela discrepancy-based strategyrdquo Inverse Problems and Imagingvol 10 no 1 pp 1ndash25 2016

[46] C S Liu ldquoNovel algorithms based on the conjugate gradientmethod for inverting ill-conditioned matrices and a new regu-larization method to solve ill-posed linear systemsrdquo ComputerModeling in Engineering and Sciences vol 60 no 3 pp 279ndash3082010

[47] C-S Liu ldquoAn optimally generalized steepest-descent algorithmfor solving ill-posed linear systemsrdquo Journal of Applied Mathe-matics vol 2013 Article ID 154358 15 pages 2013

[48] C-S Liu ldquoModifications of steepest descentmethod and conju-gate gradient method against noise for ill-posed linear systemsrdquoCommunications in Numerical Analysis vol 2012 Article IDcna-00115 24 pages 2012

[49] G Zoutendijk ldquoNonlinear programming computationalmeth-odsrdquo in Integer and Nonlinear Programming pp 37ndash86 North-Holland Amsterdam The Netherlands 1970

[50] P Wolfe ldquoConvergence conditions for ascent methodsrdquo SIAMReview vol 11 pp 226ndash235 1969


[51] P Wolfe ldquoConvergence Conditions for Ascent Methods IISome Correctionsrdquo Siam Review vol 13 no 2 pp 226ndash2352006

[52] M P Baldwin and T J Dunkerton ldquoStratospheric harbingersof anomalous weather regimesrdquo Science vol 294 no 5542 pp581ndash584 2001

[53] C-S Liu C-P Ren and D-G Li ldquoReconstruction and deduc-tion of cutting coal and rock load spectrumonmodified discreteregularization algorithmrdquo Journal of China Coal Society vol 39no 5 pp 981ndash986 2014

[54] C S Liu C P Ren and F Han ldquoStudy on time-frequencyspectrum characteristic of dynamic cutting load based onwavelet regularizationrdquo Applied Mechanics and Materials vol577 pp 196ndash200 2014

[55] C RenNWang andC Liu ldquoIdentification of randomdynamicforce using an improvedmaximumentropy regularization com-bined with a novel conjugate gradientrdquoMathematical Problemsin Engineering vol 2017 Article ID 9125734 14 pages 2017

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Submit your manuscripts atwwwhindawicom


on an elastic foundation from given information of verticaldeflection of infinite beams J Gao et al [12] utilized thestandard TR method for identifying the unknown loadsource VMelicher et al [13] identified the piecewise constantparameter using the standard TR method In addition manyworks are mainly concerned with the identification of loadposition and load number in nonlinear system using thestandard TR method [14ndash16] However the standard TRmethod is not completely perfect there are some unavoidablelimitations and disadvantages (i) the standard TR methodmay determine approximate solutions that are too smooth(ii) the approximate solution may lack many details that thedesired exact solution might possess therefore it is very nec-essary to discuss other regularizationmethods to improve thelimitations and disadvantages of the regularization methodmentioned above

In the past few years fractional Tikhonov regularization(FTR) method was gradually introduced to deal with theill-posed problems mentioned above [17] Furthermore theFTR method has been applied to many fields and obtainedsome remarkable achievements [18ndash20] M E Hochstenbachet al [21] used the FTR method to obtain approximatesolution of higher quality which is better than the standardTR method in solving linear system equation with a severeill-posedness In [22] we can see that the FTR method issignificantly better than the standard TR method which wasproved by several numerical examples It can be found from[23] that the FTR method can obtain more approximationsolutions than the standard TRmethod which was discussedthrough several computed examples J Zhang et al [24]illustrated that the FTR method was more suitable fortackling partly textured images problems Hence the FTRmethod mentioned above has many advantages to a certainextent which may not only solve the shortcomings of thestandard TRmethod but also be able to provide a useful helpfor obtaining the effective solution of the ill-posed problemHowever FTR method is also not very perfect which hasgreat limitations for solving the ill-posed problem First thesolutions of the larger singular values can not be retainedwell Then it is not completely effective for the large-scaleill-posed problem Thus it is very necessary to discuss aneffective technique to tackle the limitations of the problemencountered

Considering the limitations of TR method and FTRmethod in a certain case we propose a novel fractionalTikhonov regularization (NFTR) method to offer a stablesolution for the ill-posed problems in practical engineeringapplications In our work the ill-posed problem processed bythe proposed method is redefined as a class of unconstrainedoptimization problems to solve the following minimumproblem

minxisinR119899

119891 (x) = Λ|NFTR (3)

In which 119891(x) R푛 997888rarr R is a continuously differentiablefunction andΛ|NFTR denotes the smoothing functional usingthe NFTR method

In fact there are many optimization algorithms thatcan improve the efficiency of solving (3) like conjugate

gradient (CG) method memory gradient (MG) methodsuper-memory gradient (SMG) method and so on [25ndash27]It is generally known that CG method does not achieveglobal convergence for common functions under inexact linesearches [28] whereas MG method requires dealing withthe trust region subproblem [29] Although SMG methodis superior to CG and MG method in some cases itis also not globally convergent [30] Therefore it is verymeaningful to find a global convergence algorithm based onthe SMG method Hence in this paper an improved super-memory gradient (ISMG) method is introduced to deal withunconstrained optimization problems of (3)

This article is focused on a novel stable inverse techniquedeveloped to provide an optimal solution for the ill-posedproblems in practical engineering Compared with our pre-vious works some highlights and advances of the proposedtechnique are derived in this article

First of all the article is different from our previousstudies and the highlights of this article are describedas follows (i) a novel fractional Tikhonov regularization(NFTR) coupled with an improved super-memory gradi-ent (ISMG) method is proposed (ii) solving the ill-posedproblem procedure is regarded as a class of unconstrainedoptimization problems (iii) the detailed information of thesolution is clearly enhanced (iv) the technique we proposecan effectively overcome the smoothness of the solution ofthe ill-posed problem (v) global convergence and stability ofthe solution are proved

Then some advances of the proposed technique aredescribed as follows (i) a stable optimal solution for the ill-posed dynamic force identification problems can be obtained(ii) the proposed method yields approximate solutions ofhigher quality than the standard TR method and FTRmethod (iii) the technique proposed does not need any apriori information on the model of the inverse identificationonly the measured response is sufficient for the presentmethod (iv) the applicability of the present technique isexamined through a mathematical example and an engineer-ing example

In this paper we propose a novel fractional Tikhonovregularization (NFTR) coupled with an improved super-memory gradient (ISMG) method to solve the ill-poseddynamic force identification problems in engineering appli-cations The remainder of this paper is formulated as followsThe inverse problem model is described in Section 2 InSection 3 we present the NFTR method for solving the ill-posed problem In Section 4 ISMG method is introducedto deal with an optimization problem The convergenceproperties of the present method are proved in Section 5 InSection 6 we discuss the application of a mathematical andan engineering example Section 7 shows some meaningfulfindings and new conclusions

2 Description of Inverse Problem Model

The model of inverse problems encountered in science andengineering can be uniformly expressed by a Fredholmintegral equation of the first kind [31ndash33]


intb

ac (119905 minus 120591) x (120591) d120591 = y (119905) (4)



푛sum푖=1

c (119905푘 minus 120591푖) x (120591푖) ΔT = y (119905푘) (5)


x푖 = x (120591푖) y푘 = y (119905푘)


Then we have


(7)


Cx = y (8)


y = y + 119890 (9)


Cx = y (10)









min119891 (x) = 1003817100381710038171003817y minus Cx10038171003817100381710038172W + 1205832 x22 (12)

Then we know

x2W = xTWx (13)







in which


T(16)


N휇 =[[[[[[[

max 1205832 minus 12059012 0

max 1205832 minus 12059022 0

d

max 1205832 minus 120590푛2 0

]]]]]]] (17)





119906푖120590푖V푇푖 (19)


and










Ω푇푅 (120590푖) = 1205902푖1205902

푖 + 1205832

Ω퐹푇푅 (120590푖) = 120590훼+1푖120590훼+1

푖 + 1205832

Ω푁퐹푇푅 (120590푖) =1 1 le 119894 le 119896120590훼+1푖1205832

119896 lt 119894 le 119899

(24)


10038171003817100381710038171003817Cxys minus y100381710038171003817100381710038172 = 120596ℓ (25)



10038171003817100381710038171003817xys1003817100381710038171003817100381722 =푛sum

푖=1

Ω2푁퐹푇푅 (120590푖) (119906푇

푖 y)21205902푖

= 푘sum푖=1

(119906푇푖 y)21205902푖

+ 푛sum푖=푘+1

1205902훼+2푖 (119906푇

푖 y)212058341205902푖

(26)



120597120597120583 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = minus4 푛sum푖=푘+1

1205902훼푖 (119906푇

푖 y)21205835

120597120597120572 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = 2 푛sum푖=푘+1

1205902훼푖 ln120590 (119906푇

푖 y)21205834

(27)



119889120583119889ℓ gt 0 (28)



푖=1

(1 minus Ω푁퐹푇푅 (120590푖))2 (119906푇푖 y)2 (29)


1205962ℓ (120583)2 = 푛sum푖=1

(1 minus 120590훼+1푖1205832

)2 (119906푇푖 y)2 (30)


푖=1

(1 minus 120590훼+1푖1205832

) 120590훼+1푖1205833

(119906푇푖 y)2 (31)





(119860T119860 + 120583 (119860T119860)minus훾) xys = 119860Ty (33)



x isin 119877((119862T119862)휗) (34)




sup0lt휎le휎1


and

sup0lt휎le휎1



120583 = 119901(120575119911)(1휛)(휗+1) 119901 gt 0 fixed (37)



lim휎㨀rarr0



10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 120590휗 = 0

lim휎㨀rarrinfin

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590푖)1003816100381610038161003816 120590휗 =

infin 120599 gt 120572 + 11 120599 = 120572 + 10 120599 lt 120572 + 1

(38)


sup0lt휎le휎1


and

sup0lt휎le휎1






119909푘+1 = 119909푘 + ℎ푘119889푘 (41)



119889푘 =


푖=1

120573푖푘119889푘minus푖 119896 ge 119898 + 1 (42)


120573푖푘 = 120588 1003817100381710038171003817119892푘

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 120588 isin (0 1119898) (43)


119891 (119909푘 + 120573푚119896119889푘) le 119891 (119909푘) + 120592120573푚119896 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (44)





Step 5 Set 119909푘+1 = 119909푘 + ℎ푘119889푘



119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (45)


119892푇푘 119889푘 = 119892푇

푘 (minus119892푘) = minus 1003817100381710038171003817119892푘10038171003817100381710038172 (46)


119892푇푘 119889푘 = minus 1003817100381710038171003817119892푘

10038171003817100381710038172 le (119898120588 minus 1) 1003817100381710038171003817119892푘10038171003817100381710038172 (47)


119892푇푘 119889푘 = 119892푇



= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 sdot 119892


le minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 = (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172

(48)

Then we have

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (49)



1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (50)


1003817100381710038171003817 = minus 1003817100381710038171003817119892푘1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘

1003817100381710038171003817 (51)


1003817100381710038171003817119889푘1003817100381710038171003817 =

1003817100381710038171003817100381710038171003817100381710038171003817minus119892푘 + 푚sum푖=1


1003817100381710038171003817100381710038171003817100381710038171003817 le1003817100381710038171003817119892푘

1003817100381710038171003817 +푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= 1003817100381710038171003817119892푘1003817100381710038171003817 +

푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 = (119898120588 minus 1) 1003817100381710038171003817119892푘

1003817100381710038171003817 (52)

So we have1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (53)




(1 + 119898120588)2 (54)





120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)


1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)



limℎ㨀rarr0


10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)




10038171003817100381710038172) le 0 (58)



119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)


lim푘㨀rarr+prop

ℎ푘 = 0 (60)


119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)




119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)


[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘


1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)



10038171003817100381710038172 (66)

Then we have





ℎ gt 0 (69)



10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)



lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)



y minus x


(72)



119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018









intb

ac (119905 minus 120591) x (120591) d120591 = y (119905) (4)



푛sum푖=1

c (119905푘 minus 120591푖) x (120591푖) ΔT = y (119905푘) (5)


x푖 = x (120591푖) y푘 = y (119905푘)


Then we have


(7)


Cx = y (8)


y = y + 119890 (9)


Cx = y (10)









min119891 (x) = 1003817100381710038171003817y minus Cx10038171003817100381710038172W + 1205832 x22 (12)

Then we know

x2W = xTWx (13)







in which


T(16)


N휇 =[[[[[[[

max 1205832 minus 12059012 0

max 1205832 minus 12059022 0

d

max 1205832 minus 120590푛2 0

]]]]]]] (17)





119906푖120590푖V푇푖 (19)


and










Ω푇푅 (120590푖) = 1205902푖1205902

푖 + 1205832

Ω퐹푇푅 (120590푖) = 120590훼+1푖120590훼+1

푖 + 1205832

Ω푁퐹푇푅 (120590푖) =1 1 le 119894 le 119896120590훼+1푖1205832

119896 lt 119894 le 119899

(24)


10038171003817100381710038171003817Cxys minus y100381710038171003817100381710038172 = 120596ℓ (25)



10038171003817100381710038171003817xys1003817100381710038171003817100381722 =푛sum

푖=1

Ω2푁퐹푇푅 (120590푖) (119906푇

푖 y)21205902푖

= 푘sum푖=1

(119906푇푖 y)21205902푖

+ 푛sum푖=푘+1

1205902훼+2푖 (119906푇

푖 y)212058341205902푖

(26)



120597120597120583 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = minus4 푛sum푖=푘+1

1205902훼푖 (119906푇

푖 y)21205835

120597120597120572 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = 2 푛sum푖=푘+1

1205902훼푖 ln120590 (119906푇

푖 y)21205834

(27)



119889120583119889ℓ gt 0 (28)



푖=1

(1 minus Ω푁퐹푇푅 (120590푖))2 (119906푇푖 y)2 (29)


1205962ℓ (120583)2 = 푛sum푖=1

(1 minus 120590훼+1푖1205832

)2 (119906푇푖 y)2 (30)


푖=1

(1 minus 120590훼+1푖1205832

) 120590훼+1푖1205833

(119906푇푖 y)2 (31)





(119860T119860 + 120583 (119860T119860)minus훾) xys = 119860Ty (33)



x isin 119877((119862T119862)휗) (34)




sup0lt휎le휎1


and

sup0lt휎le휎1



120583 = 119901(120575119911)(1휛)(휗+1) 119901 gt 0 fixed (37)



lim휎㨀rarr0



10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 120590휗 = 0

lim휎㨀rarrinfin

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590푖)1003816100381610038161003816 120590휗 =

infin 120599 gt 120572 + 11 120599 = 120572 + 10 120599 lt 120572 + 1

(38)


sup0lt휎le휎1


and

sup0lt휎le휎1






119909푘+1 = 119909푘 + ℎ푘119889푘 (41)



119889푘 =


푖=1

120573푖푘119889푘minus푖 119896 ge 119898 + 1 (42)


120573푖푘 = 120588 1003817100381710038171003817119892푘

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 120588 isin (0 1119898) (43)


119891 (119909푘 + 120573푚119896119889푘) le 119891 (119909푘) + 120592120573푚119896 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (44)





Step 5 Set 119909푘+1 = 119909푘 + ℎ푘119889푘



119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (45)


119892푇푘 119889푘 = 119892푇

푘 (minus119892푘) = minus 1003817100381710038171003817119892푘10038171003817100381710038172 (46)


119892푇푘 119889푘 = minus 1003817100381710038171003817119892푘

10038171003817100381710038172 le (119898120588 minus 1) 1003817100381710038171003817119892푘10038171003817100381710038172 (47)


119892푇푘 119889푘 = 119892푇



= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 sdot 119892


le minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 = (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172

(48)

Then we have

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (49)



1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (50)


1003817100381710038171003817 = minus 1003817100381710038171003817119892푘1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘

1003817100381710038171003817 (51)


1003817100381710038171003817119889푘1003817100381710038171003817 =

1003817100381710038171003817100381710038171003817100381710038171003817minus119892푘 + 푚sum푖=1


1003817100381710038171003817100381710038171003817100381710038171003817 le1003817100381710038171003817119892푘

1003817100381710038171003817 +푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= 1003817100381710038171003817119892푘1003817100381710038171003817 +

푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 = (119898120588 minus 1) 1003817100381710038171003817119892푘

1003817100381710038171003817 (52)

So we have1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (53)




(1 + 119898120588)2 (54)





120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)


1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)



limℎ㨀rarr0


10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)




10038171003817100381710038172) le 0 (58)



119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)


lim푘㨀rarr+prop

ℎ푘 = 0 (60)


119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)




119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)


[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘


1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)



10038171003817100381710038172 (66)

Then we have





ℎ gt 0 (69)



10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)



lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)



y minus x


(72)



119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018











in which


T(16)


N휇 =[[[[[[[

max 1205832 minus 12059012 0

max 1205832 minus 12059022 0

d

max 1205832 minus 120590푛2 0

]]]]]]] (17)





119906푖120590푖V푇푖 (19)


and










Ω푇푅 (120590푖) = 1205902푖1205902

푖 + 1205832

Ω퐹푇푅 (120590푖) = 120590훼+1푖120590훼+1

푖 + 1205832

Ω푁퐹푇푅 (120590푖) =1 1 le 119894 le 119896120590훼+1푖1205832

119896 lt 119894 le 119899

(24)


10038171003817100381710038171003817Cxys minus y100381710038171003817100381710038172 = 120596ℓ (25)



10038171003817100381710038171003817xys1003817100381710038171003817100381722 =푛sum

푖=1

Ω2푁퐹푇푅 (120590푖) (119906푇

푖 y)21205902푖

= 푘sum푖=1

(119906푇푖 y)21205902푖

+ 푛sum푖=푘+1

1205902훼+2푖 (119906푇

푖 y)212058341205902푖

(26)



120597120597120583 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = minus4 푛sum푖=푘+1

1205902훼푖 (119906푇

푖 y)21205835

120597120597120572 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = 2 푛sum푖=푘+1

1205902훼푖 ln120590 (119906푇

푖 y)21205834

(27)



119889120583119889ℓ gt 0 (28)



푖=1

(1 minus Ω푁퐹푇푅 (120590푖))2 (119906푇푖 y)2 (29)


1205962ℓ (120583)2 = 푛sum푖=1

(1 minus 120590훼+1푖1205832

)2 (119906푇푖 y)2 (30)


푖=1

(1 minus 120590훼+1푖1205832

) 120590훼+1푖1205833

(119906푇푖 y)2 (31)





(119860T119860 + 120583 (119860T119860)minus훾) xys = 119860Ty (33)



x isin 119877((119862T119862)휗) (34)




sup0lt휎le휎1


and

sup0lt휎le휎1



120583 = 119901(120575119911)(1휛)(휗+1) 119901 gt 0 fixed (37)



lim휎㨀rarr0



10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 120590휗 = 0

lim휎㨀rarrinfin

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590푖)1003816100381610038161003816 120590휗 =

infin 120599 gt 120572 + 11 120599 = 120572 + 10 120599 lt 120572 + 1

(38)


sup0lt휎le휎1


and

sup0lt휎le휎1






119909푘+1 = 119909푘 + ℎ푘119889푘 (41)



119889푘 =


푖=1

120573푖푘119889푘minus푖 119896 ge 119898 + 1 (42)


120573푖푘 = 120588 1003817100381710038171003817119892푘

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 120588 isin (0 1119898) (43)


119891 (119909푘 + 120573푚119896119889푘) le 119891 (119909푘) + 120592120573푚119896 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (44)





Step 5 Set 119909푘+1 = 119909푘 + ℎ푘119889푘



119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (45)


119892푇푘 119889푘 = 119892푇

푘 (minus119892푘) = minus 1003817100381710038171003817119892푘10038171003817100381710038172 (46)


119892푇푘 119889푘 = minus 1003817100381710038171003817119892푘

10038171003817100381710038172 le (119898120588 minus 1) 1003817100381710038171003817119892푘10038171003817100381710038172 (47)


119892푇푘 119889푘 = 119892푇



= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 sdot 119892


le minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 = (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172

(48)

Then we have

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (49)



1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (50)


1003817100381710038171003817 = minus 1003817100381710038171003817119892푘1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘

1003817100381710038171003817 (51)


1003817100381710038171003817119889푘1003817100381710038171003817 =

1003817100381710038171003817100381710038171003817100381710038171003817minus119892푘 + 푚sum푖=1


1003817100381710038171003817100381710038171003817100381710038171003817 le1003817100381710038171003817119892푘

1003817100381710038171003817 +푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= 1003817100381710038171003817119892푘1003817100381710038171003817 +

푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 = (119898120588 minus 1) 1003817100381710038171003817119892푘

1003817100381710038171003817 (52)

So we have1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (53)




(1 + 119898120588)2 (54)





120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)


1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)



limℎ㨀rarr0


10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)




10038171003817100381710038172) le 0 (58)



119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)


lim푘㨀rarr+prop

ℎ푘 = 0 (60)


119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)




119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)


[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘


1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)



10038171003817100381710038172 (66)

Then we have





ℎ gt 0 (69)



10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)



lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)



y minus x


(72)



119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018










120597120597120583 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = minus4 푛sum푖=푘+1

1205902훼푖 (119906푇

푖 y)21205835

120597120597120572 10038171003817100381710038171003817xys1003817100381710038171003817100381722 = 2 푛sum푖=푘+1

1205902훼푖 ln120590 (119906푇

푖 y)21205834

(27)



119889120583119889ℓ gt 0 (28)



푖=1

(1 minus Ω푁퐹푇푅 (120590푖))2 (119906푇푖 y)2 (29)


1205962ℓ (120583)2 = 푛sum푖=1

(1 minus 120590훼+1푖1205832

)2 (119906푇푖 y)2 (30)


푖=1

(1 minus 120590훼+1푖1205832

) 120590훼+1푖1205833

(119906푇푖 y)2 (31)





(119860T119860 + 120583 (119860T119860)minus훾) xys = 119860Ty (33)



x isin 119877((119862T119862)휗) (34)




sup0lt휎le휎1


and

sup0lt휎le휎1



120583 = 119901(120575119911)(1휛)(휗+1) 119901 gt 0 fixed (37)



lim휎㨀rarr0



10038161003816100381610038161 minus Ω푁퐹푇푅 (120590)1003816100381610038161003816 120590휗 = 0

lim휎㨀rarrinfin

10038161003816100381610038161 minus Ω푁퐹푇푅 (120590푖)1003816100381610038161003816 120590휗 =

infin 120599 gt 120572 + 11 120599 = 120572 + 10 120599 lt 120572 + 1

(38)


sup0lt휎le휎1


and

sup0lt휎le휎1






119909푘+1 = 119909푘 + ℎ푘119889푘 (41)



119889푘 =


푖=1

120573푖푘119889푘minus푖 119896 ge 119898 + 1 (42)


120573푖푘 = 120588 1003817100381710038171003817119892푘

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 120588 isin (0 1119898) (43)


119891 (119909푘 + 120573푚119896119889푘) le 119891 (119909푘) + 120592120573푚119896 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (44)





Step 5 Set 119909푘+1 = 119909푘 + ℎ푘119889푘



119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (45)


119892푇푘 119889푘 = 119892푇

푘 (minus119892푘) = minus 1003817100381710038171003817119892푘10038171003817100381710038172 (46)


119892푇푘 119889푘 = minus 1003817100381710038171003817119892푘

10038171003817100381710038172 le (119898120588 minus 1) 1003817100381710038171003817119892푘10038171003817100381710038172 (47)


119892푇푘 119889푘 = 119892푇



= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 sdot 119892


le minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 = (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172

(48)

Then we have

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (49)



1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (50)


1003817100381710038171003817 = minus 1003817100381710038171003817119892푘1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘

1003817100381710038171003817 (51)


1003817100381710038171003817119889푘1003817100381710038171003817 =

1003817100381710038171003817100381710038171003817100381710038171003817minus119892푘 + 푚sum푖=1


1003817100381710038171003817100381710038171003817100381710038171003817 le1003817100381710038171003817119892푘

1003817100381710038171003817 +푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= 1003817100381710038171003817119892푘1003817100381710038171003817 +

푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 = (119898120588 minus 1) 1003817100381710038171003817119892푘

1003817100381710038171003817 (52)

So we have1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (53)




(1 + 119898120588)2 (54)





120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)


1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)



limℎ㨀rarr0


10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)




10038171003817100381710038172) le 0 (58)



119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)


lim푘㨀rarr+prop

ℎ푘 = 0 (60)


119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)




119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)


[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘


1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)



10038171003817100381710038172 (66)

Then we have





ℎ gt 0 (69)



10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)



lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)



y minus x


(72)



119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018











119909푘+1 = 119909푘 + ℎ푘119889푘 (41)



119889푘 =


푖=1

120573푖푘119889푘minus푖 119896 ge 119898 + 1 (42)


120573푖푘 = 120588 1003817100381710038171003817119892푘

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 120588 isin (0 1119898) (43)


119891 (119909푘 + 120573푚119896119889푘) le 119891 (119909푘) + 120592120573푚119896 (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (44)





Step 5 Set 119909푘+1 = 119909푘 + ℎ푘119889푘



119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (45)


119892푇푘 119889푘 = 119892푇

푘 (minus119892푘) = minus 1003817100381710038171003817119892푘10038171003817100381710038172 (46)


119892푇푘 119889푘 = minus 1003817100381710038171003817119892푘

10038171003817100381710038172 le (119898120588 minus 1) 1003817100381710038171003817119892푘10038171003817100381710038172 (47)


119892푇푘 119889푘 = 119892푇



= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817 sdot 119892


le minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= minus 1003817100381710038171003817119892푘10038171003817100381710038172 +

푚sum푖=1

120588 1003817100381710038171003817119892푘10038171003817100381710038172 = (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172

(48)

Then we have

119892푇푘 119889푘 le (119898120588 minus 1) 1003817100381710038171003817119892푘

10038171003817100381710038172 (49)



1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (50)


1003817100381710038171003817 = minus 1003817100381710038171003817119892푘1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘

1003817100381710038171003817 (51)


1003817100381710038171003817119889푘1003817100381710038171003817 =

1003817100381710038171003817100381710038171003817100381710038171003817minus119892푘 + 푚sum푖=1


1003817100381710038171003817100381710038171003817100381710038171003817 le1003817100381710038171003817119892푘

1003817100381710038171003817 +푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 sdot 1003817100381710038171003817119889푘minus푖

10038171003817100381710038171003817100381710038171003817119889푘minus푖1003817100381710038171003817

= 1003817100381710038171003817119892푘1003817100381710038171003817 +

푚sum푖=1

120588 1003817100381710038171003817119892푘1003817100381710038171003817 = (119898120588 minus 1) 1003817100381710038171003817119892푘

1003817100381710038171003817 (52)

So we have1003817100381710038171003817119889푘

1003817100381710038171003817 le (1 + 119898120588) 1003817100381710038171003817119892푘1003817100381710038171003817 (53)




(1 + 119898120588)2 (54)





120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)


1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)



limℎ㨀rarr0


10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)




10038171003817100381710038172) le 0 (58)



119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)


lim푘㨀rarr+prop

ℎ푘 = 0 (60)


119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)




119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)


[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘


1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)



10038171003817100381710038172 (66)

Then we have





ℎ gt 0 (69)



10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)



lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)



y minus x


(72)



119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018












120595 = 119909 isin 119877푛 | 119891 (119909) le 119891 (1199090) (55)


1003817100381710038171003817119892 (119909) minus 119892 (119910)1003817100381710038171003817 le 120585 1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (56)



limℎ㨀rarr0


10038171003817100381710038172)ℎ푘

= 119892푇푘 119889푘 minus 120592 (119892푇

푘 119889푘 + 120578 1003817100381710038171003817119892푘10038171003817100381710038172)

= (1 minus 120592) 119892푇푘 119889푘 minus 120592120578 1003817100381710038171003817119892푘

10038171003817100381710038172 lt 0(57)




10038171003817100381710038172) le 0 (58)



119891 (119909푘) minus 119891 (119909푘+1) ge 120575 1003817100381710038171003817119892푘10038171003817100381710038172 (59)


lim푘㨀rarr+prop

ℎ푘 = 0 (60)


119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ (119892푇푘 119889푘 + 120578 1003817100381710038171003817119892푘

10038171003817100381710038172) (61)

Further we have

119891 (119909푘 + ℎ119889푘) gt 119891 (119909푘) + 120592ℎ119892푇푘 119889푘 (62)




119892 (119909푘 + ℎ120577푘119889푘)푇 119889푘 lt 120592119892푇푘 119889푘 (64)


[119892 (119909푘 + ℎ120577푘119889푘) minus 119892푘]푇 119889푘


1003817100381710038171003817le 120585 sdot 120577푘 sdot ℎ 1003817100381710038171003817119889푘

1003817100381710038171003817 sdot 1003817100381710038171003817119889푘1003817100381710038171003817 le 120585ℎ 1003817100381710038171003817119889푘

10038171003817100381710038172 (65)



10038171003817100381710038172 (66)

Then we have





ℎ gt 0 (69)



10038171003817100381710038172)ge 120592ℎ푘 [(1 minus 119898120588) 1003817100381710038171003817119892푘

10038171003817100381710038172 minus 120578 1003817100381710038171003817119892푘10038171003817100381710038172]

ge 120592ℎ (1 minus 119898120588 minus 120578) 1003817100381710038171003817119892푘10038171003817100381710038172

(70)



lim푘㨀rarrinfin

119891 (xk) = 119891lowast (71)



y minus x


(72)



119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018










119891耠+ (a) le 119891 (y) minus 119891 (x)


minus (b) (73)

Set K = max|119891耠+(a)| |119891耠


y minus x

100381610038161003816100381610038161003816100381610038161003816 le K (74)






int휋



0C (119905 120591) x (119905) d119905 = y (120591) 0 le 120591 le 1205872 (77)


y (120591) = y (120591) + 119890 (78)



120587119899nsumi=0

C (119905푖 120591푘) x (119905푖) = y (120591푘) 119896 = 0 1 119899 (79)





Relative error (RE)






= 1198902100381710038171003817100381711991010038171003817100381710038172 (82)












20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018











20

10The standard TR

01136 08950 335 01128 08960 301 01109 08966 2710

FTR01080 08994 25

5 01069 09218 221 01052 09268 2010

Present00483 09858 12

5 00368 09876 101 00224 09882 7

30

10The standard TR

01183 08946 405 01180 08958 381 01179 08964 3310

FTR01177 08969 30

5 01163 09200 261 01111 09213 2210

Present00591 09709 16

5 00503 09712 121 00407 09732 11

40

10The standard TR

01199 08870 485 01196 08876 461 01187 08881 4310

FTR01184 08889 38

5 01182 08894 321 01174 08899 2710

Present00699 09678 18

5 00653 09695 131 00618 09603 12

50

10The standard TR

01219 08776 525 01214 08780 501 01203 08787 4710

FTR01198 08790 44

5 01188 08799 341 01144 08799 3010

Present00706 09012 20

5 00701 09093 151 00681 09117 13






X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018









X

Y

Z

Conical pick

Coal-seam structure



Test bed


Platform





Conical pick

Sleeve

Holder

Force sensor


Force sensor


0 005 01 015 02 025 030

100

200

300

400

500

600

700

800

Time (s)

Disp

lace

men

t res

pons

e (m

m)








0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018









0 005 01 015 02 025 03minus6

minus4

minus2

0

2

4

6

8

Time (s)

Forc

e (kN

)


(a) 훼 = 01

0 005 01 015 02 025 03minus5

minus3

minus1

1

3

5

Time (s)

Forc

e (kN

)


(b) 훼 = 02

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(c) 훼 = 03

0 005 01 015 02 025 03minus2

minus1

0

1

2

3

4

Time (s)

Forc

e (kN

)


(d) 훼 = 04

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(e) 훼 = 05

0 005 01 015 02 025 03minus05

0

05

1

15

2

25

3

Time (s)

Forc

e (kN

)


(f) 훼 = 06

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(g) 훼 = 07

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(h) 훼 = 08

Figure 5 Continued


0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018









0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN


(i) 훼 = 09



















0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018











0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)



0 005 01 015 02 025 030

05

1

15

2

25

Time (s)Fo

rce (

kN)


(b) FTR method

0 005 01 015 02 025 030

05

1

15

2

25

Time (s)

Forc

e (kN

)


(c) Present method





7 Conclusions




100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018










100

10minus2

10minus4

10minus6

10minus8

100

10minus2

10minus4

10minus6

10minus8

10minus10

Ω(

)




Data Availability




Acknowledgments


References


































































Journal of












Journal of







Volume 2018






Volume 2018

























































Journal of












Journal of







Volume 2018






Volume 2018








A Novel Fractional Tikhonov Regularization Coupled with an ...MathematicalProblemsinEngineering 5.5....

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Transcript of A Novel Fractional Tikhonov Regularization Coupled with an ...MathematicalProblemsinEngineering 5.5....