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Hindawi Publishing CorporationAdvances in Mathematical PhysicsVolume 2013 Article ID 485273 8 pageshttpdxdoiorg1011552013485273

Research ArticleA Point Source Identification Problem for a Time FractionalDiffusion Equation

Xiao-Mei Yang1 and Zhi-Liang Deng2

1 School of Mathematics Southwest Jiaotong University Chengdu 610031 China2 School of Mathematical Sciences University of Electronic Science and Technology of China Chengdu 610054 China

Correspondence should be addressed to Xiao-Mei Yang yangxiaomathhomeswjtueducn

Received 19 September 2013 Revised 20 October 2013 Accepted 21 October 2013

Academic Editor Ming Li

Copyright copy 2013 X-M Yang and Z-L Deng This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

An inverse source identification problem for a time fractional diffusion equation is discussedThe unknown heat source is supposedto be space dependent only Based on the use of Greenrsquos function an effective numerical algorithm is developed to recover both theintensities and locations of unknown point sources from final measurements Numerical results indicate that the proposed methodis efficient and accurate

1 Introduction

LetΩ be a bounded domain inR2 and let 120597Ω be the boundaryof Ω Consider the following time fractional diffusion pro-cess

0119863120574

119905119906 (119909 119905) minusL119906 (119909 119905) = 119891 (119909) 119909 isin Ω 119905 isin (0 119879)

119906 (119909 0) = 0 119909 isin Ω

B119906 (119909 119905) = 120583119906 (119909 119905) + 120573120597119906

120597V= 0 119909 isin 120597Ω 119905 isin (0 119879)

(1)

where L is the uniformly elliptic operator ] is the outwardnormal at the boundary 120597Ω and 120583 120573 are known constantswhich are not simultaneously zero Here

0119863120574

119905stands for the

Caputo fractional derivative operator of order 0 lt 120574 le 1

defined by

0119863120574

119905120595 (119905) =

1

Γ (1 minus 120574)int

119905

0

1205951015840(119904)

(119905 minus 119904)120574119889119904 0 lt 120574 lt 1

1205951015840(119905) 120574 = 1

(2)

where Γ(sdot) is the standard Γ-function and the prime denotesthe general derivative

From the last few decades fractional calculus grabbedgreat attention of not only mathematicians and engineers but

also many scientists from all fields (eg see [1ndash4]) Fractionaldiffusion equations describe anomalous diffusions on fractals(physical objects of fractional dimension like some amor-phous semiconductors or strongly porous materials see [56] and references therein) Indeed fractional derivativesprovide an excellent tool for the description of memory andhereditary properties of variousmaterials and processesThisis themain superiority of fractional derivatives in comparisonwith classical integer-order models in which such effects arein fact neglected For the detailed theory and applicationof fractional calculus one can refer to [1ndash4] and referencestherein Not only have differential equations of fractionalorder attracted peoplersquos attention but also theories and appli-cations related to physics and geometry of fractal dimensionhave been well studied (eg [7ndash11])

If the initial condition is nonhomogeneous that is119906(119909 0) = 120593(119909) we are always able to simplify the system(1) into two components that is 119906 = V + 119908 where V solvesthe homogeneous equation with nonhomogeneous initialcondition and 119908 satisfies the nonhomogeneous equationwith homogeneous initial condition As we know the initialvalueboundary value problem associated with V is well-posed and there exist many works on such forward problemfor example [12 13] In the following instead of nonhomo-geneous initial condition we only focus on the system (1)with homogeneous initial condition Ordinarily when 119891 is

2 Advances in Mathematical Physics

a known function we are asked to determine the solutionfunction 119906(119909 119905) so as to satisfy (1) So posed this is a directproblem However the source term 119891 is not always knownand has to be computed from some additional data Theadditional information is mainly the following the inte-riorboundary transient measurement values and the finalmeasurement values Here we suppose that the measureddata are given in final time 119905 = 119879 as 119906(119909 119879) + 120598 lowast rand (120598is the noise level)

For most classical partial differential equations the iden-tification and reconstruction of source functions from thefinal data or the partial boundary data are an inverse problemwith many applications (eg [14]) A number of articlesaddress the solvability problem of source term identificationFor parabolic-type differential equation please see [15ndash26]For elliptic-type differential equation one can refer to [27ndash29] though the source identification problem has been welldiscussed in the classic framework yet to the best of theauthorsrsquo knowledge there are rare researches in the aspect ofthe source identification problem associated with fractionaldifferential equation in spite of the physical and practicalimportance As indicated in [30ndash33] the source identifica-tion problem associated with the time fractional diffusionequation is also ill-posed That means the solution doesnot depend continuously on the given data and any smallperturbation in the given data may cause large change tothe solution In [33] when additional data is given on thepartial boundary the uniqueness in identifying a source termindependent of time 119905 is established for one-dimensional timefractional diffusion equation In [30] if the final time tem-perature distribution is known the existence and uniquenessresults are proved Murio and Mejıa [31] propose a mollifi-cation regularization technique to reconstruct the unknownforcing term 119891(119909 119905) In this paper we aim to deal with thespecial case that the sources andmeasurements are both pointlike The main focus will be placed on the recovery of bothintensities and locations of the unknown point source termFor this we propose a method based on the use of Greenrsquosfunction to solve the inverse source identification problems

The outline of the paper is as follows In Section 2we provide a brief sketch on the considered identificationproblem The reconstruction method by Greenrsquos function isthen given in Section 3 Numerical implementation of theproposed method is provided in Section 4 In Section 5 wesummarize the results

2 Statement of the Problem

In this paper we deal with the special case that the sourcefunction 119891(119909) is of the form

119891 (119909) =

119872

sum

1

120582119895120575 (119909 minus 120579

119895) 119895 = 1 2 119872 (3)

where 120579119895denotes the location of the point source and 120582

119895is

the intensity associated with each point source at 120579119895 Thereby

the temperature distribution 119906 = 119906(119909 119905) inside the domainΩis generated by 119891(119909) satisfying

0119863120574

119905119906 (119909 119905) minusL119906 (119909 119905) =

119872

sum

119895=1

120582119895120575 (119909 minus 120579

119895) 119909 isin Ω

119905 isin (0 119879)

119906 (119909 0) = 0 119909 isin Ω

B119906 (119909 119905) = 120583119906 (119909 119905) + 120573120597119906

120597]= 0 119909 isin 120597Ω

119905 isin (0 119879)

(4)

where 120575(sdot) is the Dirac delta function Meanwhile let119873 be anatural number and let 119909

119894119873

119894=1be a group of points inΩ Here

the points 119909119894 119894 = 1 119873 scattered in Ω are the collocation

points Our goal is to determine the strength sources 120582119895and

the locations 120579119895from user-input estimated position and the

set of final measurement data

119906120598(119909119894 119879) = 119906

119894119879+ 120592119894 119909119894isin Ω 119894 = 1 2 119873 (5)

where (120592119894)119873

119894=1denotes the Gaussian variable with mean zero

and variance 120598 This magnitude 120598 also represents the level ofnoises

Let us first suppose that the locations of the point sources120579119895 are given Under this assumption we come to the

problem of the recovery of the intensity 120582119895associated with

the point sources 120579119895from the119873 distinct final collocation data

119906120598(119909119894 119879) This recovery problem is ill-posed which prompts

us to use some regularization methodsConsequently we assume that the locations of the point

sources 120579119895are not known but an initial guess location 120579

119895is

given for each unknown point sourceMoreover wemake theassumption that each point source belongs to a distinct ballinside the domain that is

120579119895isin 119861 (120579

119895 120588119895) cap Ω 119895 = 1 2 119872

119861 (120579119895 120588119895) cap 119861 (120579

119896 120588119896) = 0 1 le 119895 lt 119896 le 119872

(6)

where 119861(120579 120588) denotes the ball centered at 120579 with radius120588 It should be pointed that if two or more point sourcesare concentrated in a sufficient small domain the proposedmethod in the following section will treat them as one pointsource

3 Methodology Based on Greenrsquos Function

In this section we discuss the identificationmethod based onGreenrsquos function Greenrsquos function119866(119909 119905 120579) can be defined as

Advances in Mathematical Physics 3

the solution of the following equations

0119863120574

119905119866 (119909 119905 120579) minusL119866 (119909 119905 120579) = 120575 (119909 minus 120579) 119909 120579 isin Ω

119905 isin (0 119879)

119866 (119909 0 120579) = 0 119909 120579 isin Ω

B119866 (119909 119905 120579) = 120583119866 (119909 119905 120579) + 120573120597119866

120597]= 0 119909 120579 isin 120597Ω

119905 isin (0 119879)

(7)

By applying Laplace transform technique we have that

119866 (119909 119905 120579) =

infin

sum

119899=1

119905120574119864120574120574+1

(minus1205822

119899119905120574) 120593119899(120579) 120593119899(119909) (8)

where 120593119899is the 119899th orthonormal eigenfunction and 120582

119899is the

corresponding eigenvalue to the Sturm-Liouville problem

L120593 + 1205822

119899120593 = 0

120583120593 + 120573120597120593

120597]= 0

(9)

and 119864120574120577(119911) is the Mittag-Leffler function defined by

119864120574120577(119911) =

infin

sum

119896=0

119911119896

Γ (120574119896 + 120577) (10)

For the details of Mittag-Leffler function one can refer to [2]Utilizing Greenrsquos function we then can write the solution

of (4) as

119906 (119909 119905) =

119872

sum

119895=1

120582119895119866(119909 119905 120579

119895) (11)

Therefore when the locations 120579119895 of point sources are

known once we obtain the final time measurement dataspecified in (5) we can solve the following linear algebraicequation

119872

sum

119895=1

120582119895119866(119909119894 119879 120579119895) = 119906120598

119894119879 119894 = 1 2 119873 (12)

to get the unknown values of the intensities 120582119895 Moreover

denoting 120582 = (120582119895) and 120603 = (119906

120576

119894119879) (12) can be rewritten as the

following matrix form

119860120582 = 120603 (13)

where 119860 is an119873 times119872matrix

119860119894119895= 119866 (119909

119894 119879 120579119895) (14)

Taking 119873 = 119872 then the system of (13) contains 119873 linearequations with119873 unknowns Subsequently if the matrix119860 isinvertible one simply has

120582 = 119860minus1120603 (15)

However due to the ill-posedness of the source identificationproblem the system of (13) is ill-conditioned and hence adirect solution as given by (15) will be either impossibleor will produce very inaccurate results To obtain stablesolutions to these kinds of ill-conditioning systems variousregularization techniques have been studied and appliedextensively [34] Here a standard Tikhonov regularizationtechnique is adopted to find the approximation solution ofthe matrix equation (13) By 120582

120572 we denote the Tikhonov

regularized solution defined to be theminimal element of thefollowing least square problem

min120582

119869120572(120582) = min

120582

119860120582 minus 1206032+ 12057221205822 (16)

where 120572 gt 0 is the regularization parameter and sdot denotesthe usual Euclidean norm It is well known [34] that theminimal element of 119869

120572can be written as

120582120572= (119860lowast119860 + 120572

2119868)minus1

119860lowast120603 (17)

where 119860lowast denotes the conjugate transposed matrix of 119860 and119868 denotes the identity matrix

Next assume that the locations 120579119895 of the point sources

are also unknown In such case we will get a nonlinear sys-temThe nonlinear system is not suitable or difficult for directnumerical computation In order to eliminate the difficultyin implementing the numerical computation we propose inthe following to linearize the nonlinear system

For the estimate locations 120579119895 we define the following

union set

Θ =

119872

⋃

119895=1

119861 (120579119895 120588119895) cap Ω (18)

and suppose that it contains all exact positions of the pointsources with proper radius 120588

119895 To linearize the nonlinear sys-

tem we take some additional collocation points 120585119897119899

119897=1from

the set Θ Assume that 120585119897119899

119897=1are uniformly distributed in Θ

On each point 120585119897 we put in a point source with intensity 120591

119897

Suppose that the temperature distribution generated by the 119899point sources 120585

119897119899

119897=1is equal to that generated by the119872 point

sources 120579119895119872

119895=1 Subsequently we have

0119863120574

119905119906 (119909 119905) minusL119906 (119909 119905) =

119899

sum

119897=1

120591119897120575 (119909 minus 120585

119897)

119909 isin Ω 119905 isin (0 119879)

119906 (119909 0) = 0 119909 isin Ω

120583119906 (119909 119905) + 120573120597119906

120597]= 0 119909 isin 120597Ω 119905 isin (0 119879)

(19)

with additional data (5) where 120591119897is the intensity at the

location 120585119897 By using the above proposed method to solve (19)

with (5) the intensity 120591119897of each point source 120585

119897can then be

obtained approximately Next we transform the intensities120591119897 sub 119861(120579

119895 120588119895) cap Ω back to a single source point as follows


the 119895th unknown source intensity 120582119895associated with each ball

119861(120579119895 120588119895) cap Ω is approximated by

119895as

119895= sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897 119895 = 1 2 119872 (20)

With the approximation intensity in hand we can start tolook at how to find the locations of the point sources Forevery point source 120579

119895 we use the weight sum of the location

coordinate 120585119897in the ball 119861(120579

119895 120588119895) to approximate the exact

location More specifically the approximation location 120579119895

corresponding to the intensity 120582119895is defined by

120579119895=

1

119895

sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897120585119897 119895 = 1 2 119872 (21)

4 Numerical Examples

In this section some numerical examples are given to verifythe effectiveness of the method proposed in Section 3 Inour computation we use the MATLAB code developed byHansen [35 36] for solving the ill-conditioned system (13) Tocompare the accuracy of the approximation we use the rootmean square (RMS) which is defined as

RMS = radic1

119872

119872

sum

119895=1

(120572119895

minus 120582119895)2

(22)

The noisy data 119906120598(119909119894 119879) at measurement points 119909

119894is obtained

by adding random noise to the exact data 119906(119909119894 119879) by

119906120598(119909119894 119879) = 119906 (119909

119894 119879) + 120598rand (119894) (23)

for 119909119894isin Ω where rand(119894) is a random number between

[minus1 1] The measurement points 119909119894 are equally distributed

in Ω In addition as we know for ill-posed problem theregularization parameter 120572 plays an important role and hencehas to be chosen appropriately In theory 120572 depends on somea priori knowledge of exact solution and noise level 120598 [34]However in practice the a priori knowledge and noise levelmay not always be known Therefore to compensate thislack of information for the noise level it is necessary for usto consider some error-free parameter choice rules Herewe adopt the 119871-curve criterion [35ndash37] to choose the regu-larization parameter

Example 1 Consider the following heat conduction problemon a semi-infinite stripe domain Ω = (119909 119910) | 0 le 119909 le 119897 119910 ge

0

119906119905(119909 119910 119905) minus Δ119906 = 119891 (119909 119910) 0 lt 119909 lt 119897

119910 gt 0 119905 isin (0 119879)

119906 (119909 119910 0) = 0 0 le 119909 le 119897 119910 ge 0

119906|119909=0

= 119906|119909=119897

= 0 119910 ge 0

119906|119910=0

= 0 0 le 119909 le 119897

(24)

Greenrsquos function is given by

119866 (119909 119910 119905 120585 120578) =2

119897radic120587119905119890minus(1199102+1205782)4119905 sinh

119910120578

2119905

times

infin

sum

119899=1

119890minus119899212058721199051198972

sin 119899120587120585

119897sin 119899120587119909

119897119867 (119905)

(25)

where119867(sdot) is the Heaviside function Without loss of gener-ality we take 119897 = 1

In this test we consider the case that the source function(3) contains five source points 120579

1198955

119895=1 The input source

locations 120579119895are randomly chosen such that

10038161003816100381610038161003816120579119895minus 120579119895

10038161003816100381610038161003816lt 120588119895 for 119895 = 1 2 3 4 5 (26)

where 120588119895are sufficient small to ensure 120579

119895isin Ω The value

of the parameters 1205881198955

119895=1in (18) used in this computations

is 01 When the noisy data are given in the final time 119879 =

1 we demonstrate the numerical performance under twonoise levels 120598 = 001 and 120598 = 0001 The computations areperformed by using a total of 119876 trial centers in each ballWe report the numerical results under different 119876 in Tables 1and 2 The displayed results show that the total number 119876 oftrial centers plays no role in the convergence of the schemeOnly a small number of trial centers are sufficient to approx-imate the unknown source function Therefore we onlyconsider the case when 100 trial centers are taken in each ballin subsequent examples

Example 2 In this example we consider the following inverseidentification problem on a square domainΩ = [0 1] times [0 1]

for 0 lt 120574 lt 1

0119863120574

119905119906 (119909 119910 119905) minus Δ119906 = 119891 (119909 119910) (119909 119910) isin Ω 119905 isin (0 119879)

119906 (119909 119910 0) = 0 (119909 119910) isin Ω

119906 (119909 119910 119905) = 0 (119909 119910) isin 120597Ω 119905 isin (0 119879)

(27)

By virtue of Laplacersquos transform [2 38] one can derive thecorresponding Greenrsquos function

119866 (119909 119910 119905 120585 120578) = 4

infin

sum

119899=1

119905120574119864120574120574+1

(minus11989921205872119905120574) sin (119899120587119909)

times sin (119899120587119910) sin (119899120587120585) sin (119899120587120578)

(28)

Firstly we see the robustness of the proposed algorithmabout the parameter 120574 For 120598 = 001 0001 120588

119895= 01 and

119879 = 1 we report the RMS in Table 3 under different 120574 forthree point sources located at

(03 03) (05 07) (07 03) (29)

with intensities 1 3 and 5The corresponding approximationlocations are given in Tables 4 and 5 The displayed results


Table 1 Example 1 numerical comparison for 120598 = 001 and119872 = 5

Exact 120582119895

11 1 21 minus5 minus7120579119895

(05 1) (05 5) (05 10) (05 15) (05 25)

119876 = 25119895

116978 10539 220974 minus51806 minus73625120579119895

(05196 09876) (05349 49513) (05173 100002) (05059 149989) (05129 250005)

119876 = 100119895

117866 09881 221760 minus51460 minus73026120579119895

(05226 09925) (05196 49688) (05188 100002) (05059 150016) (05102 250001)

119876 = 256119895

114531 10006 221702 minus51226 minus71781120579119895

(05136 09929) (05219 49689) (05190 100005) (05027 150023) (05046 249995)

119876 = 400119895

116968 10962 221453 minus50899 minus71799120579119895

(05213 09977) (05521 49829) (05186 100005) (05009 150027) (05047 250008)

119876 = 625119895

119726 10230 220421 minus51289 minus71438120579119895

(05274 09951) (05309 50000) (05171 100004) (05033 150033) (05028 249993)

119876 = 900119895

115229 10794 223220 minus49636 minus71709120579119895

(05163 09975) (05477 49907) (05209 100004) (04919 150005) (05035 250000)


Exact 120582119895

5 7 2 5 7120579119895

(03 1) (06 5) (05 10) (07 15) (02 25)

119876 = 25119895

49670 73015 18876 51820 69783120579119895

(03016 10113) (06133 50007) (04816 100010) (07097 149972) (01985 249974)

119876 = 100119895

49553 71001 19929 50736 69830120579119895

(03009 10117) (06057 49999) (04993 100079) (07061 150017) (01982 249966)

119876 = 256119895

49891 70950 20070 49055 67970120579119895

(03041 10112) (06054 49992) (05018 100075) (07003 150016) (02031 250043)

119876 = 400119895

49690 70538 19133 49015 67483120579119895

(03020 10114) (06041 49998) (04859 100036) (07002 150014) (02032 250023)

119876 = 625119895

49678 70433 19759 49636 66110120579119895

(03022 10115) (06036 50003) (04966 100061) (07025 150035) (02078 250015)

119876 = 900119895

49803 70981 20104 50513 66630120579119895

(03030 10103) (06052 49991) (05019 100072) (07047 150059) (02065 250024)

Table 3 Example 2 RMS under different 120598 and 120574 for119872 = 3

120598120574

00909 01818 02727 03636 04545 05455 06364 07273 08182 09091001 00231 00121 00388 00266 00137 00250 00184 00233 00822 005890001 00083 00045 00088 00080 00089 00114 00099 00091 00096 00070

Table 4 Example 2 the approximation locations for (03 03) (05 07) and (07 03)

120598120574

00909 01818 02727 03636 04545

001(02986 02844)(05017 07031)(07023 02998)

(02983 02987)(05013 06993)(07018 03014)

(03053 03013)(05022 06979)(07030 03061)

(02951 02875)(05005 07029)(07036 03016)

(03035 03049)(05009 06985)(07022 03030)

0001(03002 02937)(05010 07010)(07024 03018)

(03031 02956)(05002 07002)(07016 03016)

(03019 02951)(05006 07003)(07020 03017)

(03021 02950)(05006 07003)(07021 03019)

(02996 02947)(05008 07008)(07024 03018)


0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1

Exact locationApproximation location

x

y

(a)

0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(b)

Figure 1 Approximation for 120598 = 01 with 119879 = 01 (a) 1 (b)


120598120574

05455 06364 07273 08182 09091

001(03029 02955)(05021 07004)(07029 03036)

(03003 03007)(05007 06983)(07010 03029)

(03083 03026)(04974 06976)(06995 03008)

(02957 02992)(05050 06980)(07018 03088)

(02830 02881)(04980 07038)(07056 03000)

0001(03011 02940)(05010 07006)(07025 03022)

(03029 02946)(05006 07002)(07021 03021)

(03017 02943)(05009 07006)(07021 03017)

(02992 02954)(05004 07008)(07026 03019)

(03023 02972)(05005 06996)(07021 03024)

Table 6 Example 2 numerical comparison for 120598 = 001 using different 119879

Exact 120582119895

1 3 5120579119895

(03 03) (05 07) (07 03)

119879 = 01119895

10108 30096 50584120579119895

(02886 02917) (05007 07032) (07043 03034)

119879 = 1119895

10039 30035 48805120579119895

(03265 03068) (04966 06984) (06993 03001)

119879 = 2119895

08916 29666 50708120579119895

(02959 02993) (05046 06980) (07004 03093)

119879 = 5119895

08560 29201 51635120579119895

(02805 03130) (05134 06938) (07026 03100)

show that the change of the parameter 120574 has little effect onthe numerical computations which reflects that the proposedmethod is robust about 120574 On the other hand one can see thatfor smaller noise level 120598 we obtain better numerical effect

Secondly we also consider the effect of the final time 119879on the numerical precision Fixing 120574 = 05 and choosingparameter 120588

119895= 01 we report the numerical results in

Table 6 from which one can see that the accuracy of the

approximation decreases with respect to the increase of thenumber of 119879 Such phenomenon can be explained by thenature of ill-posed inverse source identification problem

Finally using the previous point source we plot the exactand approximation locations of source points in Figure 1 for120598 = 01 and 119879 = 01 1 The computational intensities are09326 29555 and 51065 for 119879 = 01 and 14365 19797and 58621 for 119879 = 1 respectively It can be seen that even for


high noise level 120598 = 01 the proposed method produces anacceptable numerical approximation

5 Conclusion

Based on the use of Greenrsquos function we propose in this paperan effective numerical method to recover both the intensitiesand locations of point sources for a time fractional diffusionprocess Some numerical results show that the proposedalgorithm provides an accurate and reliable scheme

Acknowledgments

This work is supported by the Fundamental ResearchFunds for the Central Universities (SWJTU11BR078ZYGX2011J104) and the NSF of China (no 11226040)

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006

[2] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press San DiegoCalif USA 1999

[3] J SabatierO PAgrawal and J A TMachadoAdvances in Frac-tional Calculus Theoretical Developments and Applications inPhysics and Engineering Springer Dordrecht The Netherlands2007

[4] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Gordon and Breach Science YverdonSwitzerland 1993

[5] V V Anh and N N Leonenko ldquoSpectral analysis of fractionalkinetic equations with random datardquo Journal of StatisticalPhysics vol 104 no 5-6 pp 1349ndash1387 2001

[6] R Metzler and J Klafter ldquoThe random walkrsquos guide to anoma-lous diffusion a fractional dynamics approachrdquo Physics Reportsvol 339 no 1 p 77 2000

[7] C Cattani A Ciancio and B Lods ldquoOn a mathematical modelof immune competitionrdquo Applied Mathematics Letters vol 19no 7 pp 678ndash683 2006

[8] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering vol 2010 Article ID 157264 26 pages2010

[9] M Li Y Q Chen J Y Li and W Zhao ldquoHolder scales of sealevelrdquo Mathematical Problems in Engineering vol 2012 ArticleID 863707 22 pages 2012

[10] M Li W Zhao and C Cattani ldquoDelay bound fractal trafficpasses through serversrdquoMathematical Problems in Engineeringvol 2013 Article ID 157636 15 pages 2013

[11] M Li and W Zhao ldquoOn 1119891 noiserdquo Mathematical Problems inEngineering vol 2012 Article ID 673648 23 pages 2012

[12] M M Khader ldquoOn the numerical solutions for the fractionaldiffusion equationrdquo Communications in Nonlinear Science andNumerical Simulation vol 16 no 6 pp 2535ndash2542 2011

[13] Y Luchko ldquoSome uniqueness and existence results for theinitial-boundary-value problems for the generalized time-fractional diffusion equationrdquo Computers amp Mathematics withApplications vol 59 no 5 pp 1766ndash1772 2010

[14] V Isakov Inverse Problems for Partial Differential Equations vol127 of Applied Mathematical Sciences Springer New York NYUSA 1998

[15] E C Baran and A G Fatullayev ldquoDetermination of anunknown source parameter in two-dimensional heat equationrdquoApplied Mathematics and Computation vol 159 no 3 pp 881ndash886 2004

[16] A de Cezaro and B T Johansson ldquoA note on uniqueness in theidentification of a spacewise dependent source anddiffusioncoefficient for the heat equationrdquo httparxivorgabs12107346

[17] A de Cezaro and F T de Cezaro ldquoUniqueness and regulariza-tion for unknown spacewise lower-order coefficient and sourcefor the heat type equationrdquo httparxivorgabs12107348

[18] S Drsquohaeyer B T Johansson and M Slodicka ldquoReconstructionof a spacewise-dependent heat source in a time-dependent heatdiffusion processrdquo IMA Journal of Applied Mathematics 2012

[19] V Isakov ldquoInverse parabolic problems with the final overdeter-minationrdquo Communications on Pure and Applied Mathematicsvol 44 no 2 pp 185ndash209 1991

[20] T Johansson and D Lesnic ldquoDetermination of a spacewisedependent heat sourcerdquo Journal of Computational and AppliedMathematics vol 209 no 1 pp 66ndash80 2007

[21] B T Johansson and D Lesnic ldquoA procedure for determining aspacewise dependent heat source and the initial temperaturerdquoApplicable Analysis vol 87 no 3 pp 265ndash276 2008

[22] I A Kaliev andMM Sabitova ldquoProblems of the determinationof the temperature and density of heat sources from theinitial and final temperaturesrdquo Journal of Applied and IndustrialMathematics vol 4 no 3 pp 332ndash339 2010

[23] G A Kriegsmann and W E Olmstead ldquoSource identificationfor the heat equationrdquo Applied Mathematics Letters vol 1 no 3pp 241ndash245 1988

[24] W Rundell ldquoThe determination of a parabolic equation frominitial and final datardquo Proceedings of the AmericanMathematicalSociety vol 99 no 4 pp 637ndash642 1987

[25] L Yan C-L Fu and F-L Yang ldquoThe method of fundamentalsolutions for the inverse heat source problemrdquo EngineeringAnalysis with Boundary Elements vol 32 no 3 pp 216ndash2222008

[26] L Yan F-L Yang and C-L Fu ldquoAmeshless method for solvingan inverse spacewise-dependent heat source problemrdquo Journalof Computational Physics vol 228 no 1 pp 123ndash136 2009

[27] Y C HonM Li and Y A Melnikov ldquoInverse source identifica-tion by Greenrsquos functionrdquo Engineering Analysis with BoundaryElements vol 34 no 4 pp 352ndash358 2010

[28] N F M Martins ldquoAn iterative shape reconstruction of sourcefunctions in a potential problem using the MFSrdquo InverseProblems in Science and Engineering vol 20 no 8 pp 1175ndash11932012

[29] L Ling Y C Hon and M Yamamoto ldquoInverse source identi-fication for Poisson equationrdquo Inverse Problems in Science andEngineering vol 13 no 4 pp 433ndash447 2005

[30] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linearheat equation involving fractional derivative in timerdquo AppliedMathematics and Computation vol 218 no 1 pp 163ndash170 2011

[31] D A Murio and C E Mejıa ldquoSource terms identificationfor time fractional diffusion equationrdquo Revista Colombiana deMatematicas vol 42 no 1 pp 25ndash46 2008

[32] J GWang Y B Zhou andTWei ldquoTwo regularizationmethodsto identify a space-dependent source for the time-fractional


diffusion equationrdquoAppliedNumericalMathematics vol 68 pp39ndash57 2013

[33] Y Zhang and X Xu ldquoInverse source problem for a fractionaldiffusion equationrdquo Inverse Problems vol 27 no 3 Article ID035010 12 pages 2011

[34] H W Engl M Hanke and A Neubauer Regularization ofInverse Problems vol 375 of Mathematics and its ApplicationsKluwer Academic Dordrecht The Netherlands 1996

[35] P C Hansen Rank-Deficient and Discrete Ill-Posed ProblemsSIAM Monographs on Mathematical Modeling and Computa-tion Society for Industrial and Applied Mathematics Philadel-phia Pa USA 1998

[36] P C Hansen ldquoRegularization tools a Matlab package foranalysis and solution of discrete ill-posed problemsrdquoNumericalAlgorithms vol 6 no 1-2 pp 1ndash35 1994

[37] P C Hansen and D P OrsquoLeary ldquoThe use of the 119871-curve in theregularization of discrete ill-posed problemsrdquo SIAM Journal onScientific Computing vol 14 no 6 pp 1487ndash1503 1993

[38] R Gorenflo and F Mainardi ldquoFractional calculus integral anddifferential equations of fractional orderrdquo in Fractals and Frac-tional Calculus in Continuum Mechanics A Carpinteri and FMainardi Eds pp 223ndash276 SpringerNewYorkNYUSA 1997

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Applied MathematicsJournal of


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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


a known function we are asked to determine the solutionfunction 119906(119909 119905) so as to satisfy (1) So posed this is a directproblem However the source term 119891 is not always knownand has to be computed from some additional data Theadditional information is mainly the following the inte-riorboundary transient measurement values and the finalmeasurement values Here we suppose that the measureddata are given in final time 119905 = 119879 as 119906(119909 119879) + 120598 lowast rand (120598is the noise level)

For most classical partial differential equations the iden-tification and reconstruction of source functions from thefinal data or the partial boundary data are an inverse problemwith many applications (eg [14]) A number of articlesaddress the solvability problem of source term identificationFor parabolic-type differential equation please see [15ndash26]For elliptic-type differential equation one can refer to [27ndash29] though the source identification problem has been welldiscussed in the classic framework yet to the best of theauthorsrsquo knowledge there are rare researches in the aspect ofthe source identification problem associated with fractionaldifferential equation in spite of the physical and practicalimportance As indicated in [30ndash33] the source identifica-tion problem associated with the time fractional diffusionequation is also ill-posed That means the solution doesnot depend continuously on the given data and any smallperturbation in the given data may cause large change tothe solution In [33] when additional data is given on thepartial boundary the uniqueness in identifying a source termindependent of time 119905 is established for one-dimensional timefractional diffusion equation In [30] if the final time tem-perature distribution is known the existence and uniquenessresults are proved Murio and Mejıa [31] propose a mollifi-cation regularization technique to reconstruct the unknownforcing term 119891(119909 119905) In this paper we aim to deal with thespecial case that the sources andmeasurements are both pointlike The main focus will be placed on the recovery of bothintensities and locations of the unknown point source termFor this we propose a method based on the use of Greenrsquosfunction to solve the inverse source identification problems

The outline of the paper is as follows In Section 2we provide a brief sketch on the considered identificationproblem The reconstruction method by Greenrsquos function isthen given in Section 3 Numerical implementation of theproposed method is provided in Section 4 In Section 5 wesummarize the results

2 Statement of the Problem

In this paper we deal with the special case that the sourcefunction 119891(119909) is of the form

119891 (119909) =

119872

sum

1

120582119895120575 (119909 minus 120579

119895) 119895 = 1 2 119872 (3)

where 120579119895denotes the location of the point source and 120582

119895is

the intensity associated with each point source at 120579119895 Thereby

the temperature distribution 119906 = 119906(119909 119905) inside the domainΩis generated by 119891(119909) satisfying

0119863120574

119905119906 (119909 119905) minusL119906 (119909 119905) =

119872

sum

119895=1

120582119895120575 (119909 minus 120579

119895) 119909 isin Ω

119905 isin (0 119879)

119906 (119909 0) = 0 119909 isin Ω

B119906 (119909 119905) = 120583119906 (119909 119905) + 120573120597119906

120597]= 0 119909 isin 120597Ω

119905 isin (0 119879)

(4)

where 120575(sdot) is the Dirac delta function Meanwhile let119873 be anatural number and let 119909

119894119873

119894=1be a group of points inΩ Here

the points 119909119894 119894 = 1 119873 scattered in Ω are the collocation

points Our goal is to determine the strength sources 120582119895and

the locations 120579119895from user-input estimated position and the

set of final measurement data

119906120598(119909119894 119879) = 119906

119894119879+ 120592119894 119909119894isin Ω 119894 = 1 2 119873 (5)

where (120592119894)119873

119894=1denotes the Gaussian variable with mean zero

and variance 120598 This magnitude 120598 also represents the level ofnoises

Let us first suppose that the locations of the point sources120579119895 are given Under this assumption we come to the

problem of the recovery of the intensity 120582119895associated with

the point sources 120579119895from the119873 distinct final collocation data

119906120598(119909119894 119879) This recovery problem is ill-posed which prompts

us to use some regularization methodsConsequently we assume that the locations of the point

sources 120579119895are not known but an initial guess location 120579

119895is

given for each unknown point sourceMoreover wemake theassumption that each point source belongs to a distinct ballinside the domain that is

120579119895isin 119861 (120579

119895 120588119895) cap Ω 119895 = 1 2 119872

119861 (120579119895 120588119895) cap 119861 (120579

119896 120588119896) = 0 1 le 119895 lt 119896 le 119872

(6)

where 119861(120579 120588) denotes the ball centered at 120579 with radius120588 It should be pointed that if two or more point sourcesare concentrated in a sufficient small domain the proposedmethod in the following section will treat them as one pointsource

3 Methodology Based on Greenrsquos Function

In this section we discuss the identificationmethod based onGreenrsquos function Greenrsquos function119866(119909 119905 120579) can be defined as



0119863120574


119905 isin (0 119879)

119866 (119909 0 120579) = 0 119909 120579 isin Ω

B119866 (119909 119905 120579) = 120583119866 (119909 119905 120579) + 120573120597119866

120597]= 0 119909 120579 isin 120597Ω

119905 isin (0 119879)

(7)


119866 (119909 119905 120579) =

infin

sum

119899=1

119905120574119864120574120574+1

(minus1205822

119899119905120574) 120593119899(120579) 120593119899(119909) (8)


119899is the


L120593 + 1205822

119899120593 = 0

120583120593 + 120573120597120593

120597]= 0

(9)


119864120574120577(119911) =

infin

sum

119896=0

119911119896

Γ (120574119896 + 120577) (10)


of (4) as

119906 (119909 119905) =

119872

sum

119895=1

120582119895119866(119909 119905 120579

119895) (11)



119872

sum

119895=1

120582119895119866(119909119894 119879 120579119895) = 119906120598

119894119879 119894 = 1 2 119873 (12)


denoting 120582 = (120582119895) and 120603 = (119906

120576



119860120582 = 120603 (13)


119860119894119895= 119866 (119909

119894 119879 120579119895) (14)


120582 = 119860minus1120603 (15)




min120582

119869120572(120582) = min

120582

119860120582 minus 1206032+ 12057221205822 (16)



120582120572= (119860lowast119860 + 120572

2119868)minus1

119860lowast120603 (17)





union set

Θ =

119872

⋃

119895=1

119861 (120579119895 120588119895) cap Ω (18)




119897=1from




119897


119897119899


sources 120579119895119872


0119863120574

119905119906 (119909 119905) minusL119906 (119909 119905) =

119899

sum

119897=1

120591119897120575 (119909 minus 120585

119897)

119909 isin Ω 119905 isin (0 119879)

119906 (119909 0) = 0 119909 isin Ω

120583119906 (119909 119905) + 120573120597119906

120597]= 0 119909 isin 120597Ω 119905 isin (0 119879)

(19)




119897can then be






119895as

119895= sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897 119895 = 1 2 119872 (20)







120579119895=

1

119895

sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897120585119897 119895 = 1 2 119872 (21)



RMS = radic1

119872

119872

sum

119895=1

(120572119895

minus 120582119895)2

(22)


119894is obtained


119906120598(119909119894 119879) = 119906 (119909

119894 119879) + 120598rand (119894) (23)





0

119906119905(119909 119910 119905) minus Δ119906 = 119891 (119909 119910) 0 lt 119909 lt 119897

119910 gt 0 119905 isin (0 119879)

119906 (119909 119910 0) = 0 0 le 119909 le 119897 119910 ge 0

119906|119909=0

= 119906|119909=119897

= 0 119910 ge 0

119906|119910=0

= 0 0 le 119909 le 119897

(24)


119866 (119909 119910 119905 120585 120578) =2

119897radic120587119905119890minus(1199102+1205782)4119905 sinh

119910120578

2119905

times

infin

sum

119899=1

119890minus119899212058721199051198972

sin 119899120587120585

119897sin 119899120587119909

119897119867 (119905)

(25)



1198955



10038161003816100381610038161003816120579119895minus 120579119895

10038161003816100381610038161003816lt 120588119895 for 119895 = 1 2 3 4 5 (26)








for 0 lt 120574 lt 1

0119863120574


119906 (119909 119910 0) = 0 (119909 119910) isin Ω

119906 (119909 119910 119905) = 0 (119909 119910) isin 120597Ω 119905 isin (0 119879)

(27)


119866 (119909 119910 119905 120585 120578) = 4

infin

sum

119899=1

119905120574119864120574120574+1

(minus11989921205872119905120574) sin (119899120587119909)

times sin (119899120587119910) sin (119899120587120585) sin (119899120587120578)

(28)


119895= 01 and


(03 03) (05 07) (07 03) (29)




Exact 120582119895

11 1 21 minus5 minus7120579119895

(05 1) (05 5) (05 10) (05 15) (05 25)

119876 = 25119895

116978 10539 220974 minus51806 minus73625120579119895

(05196 09876) (05349 49513) (05173 100002) (05059 149989) (05129 250005)

119876 = 100119895

117866 09881 221760 minus51460 minus73026120579119895

(05226 09925) (05196 49688) (05188 100002) (05059 150016) (05102 250001)

119876 = 256119895

114531 10006 221702 minus51226 minus71781120579119895

(05136 09929) (05219 49689) (05190 100005) (05027 150023) (05046 249995)

119876 = 400119895

116968 10962 221453 minus50899 minus71799120579119895

(05213 09977) (05521 49829) (05186 100005) (05009 150027) (05047 250008)

119876 = 625119895

119726 10230 220421 minus51289 minus71438120579119895

(05274 09951) (05309 50000) (05171 100004) (05033 150033) (05028 249993)

119876 = 900119895

115229 10794 223220 minus49636 minus71709120579119895

(05163 09975) (05477 49907) (05209 100004) (04919 150005) (05035 250000)


Exact 120582119895

5 7 2 5 7120579119895

(03 1) (06 5) (05 10) (07 15) (02 25)

119876 = 25119895

49670 73015 18876 51820 69783120579119895

(03016 10113) (06133 50007) (04816 100010) (07097 149972) (01985 249974)

119876 = 100119895

49553 71001 19929 50736 69830120579119895

(03009 10117) (06057 49999) (04993 100079) (07061 150017) (01982 249966)

119876 = 256119895

49891 70950 20070 49055 67970120579119895

(03041 10112) (06054 49992) (05018 100075) (07003 150016) (02031 250043)

119876 = 400119895

49690 70538 19133 49015 67483120579119895

(03020 10114) (06041 49998) (04859 100036) (07002 150014) (02032 250023)

119876 = 625119895

49678 70433 19759 49636 66110120579119895

(03022 10115) (06036 50003) (04966 100061) (07025 150035) (02078 250015)

119876 = 900119895

49803 70981 20104 50513 66630120579119895

(03030 10103) (06052 49991) (05019 100072) (07047 150059) (02065 250024)


120598120574

00909 01818 02727 03636 04545 05455 06364 07273 08182 09091001 00231 00121 00388 00266 00137 00250 00184 00233 00822 005890001 00083 00045 00088 00080 00089 00114 00099 00091 00096 00070


120598120574

00909 01818 02727 03636 04545

001(02986 02844)(05017 07031)(07023 02998)

(02983 02987)(05013 06993)(07018 03014)

(03053 03013)(05022 06979)(07030 03061)

(02951 02875)(05005 07029)(07036 03016)

(03035 03049)(05009 06985)(07022 03030)

0001(03002 02937)(05010 07010)(07024 03018)

(03031 02956)(05002 07002)(07016 03016)

(03019 02951)(05006 07003)(07020 03017)

(03021 02950)(05006 07003)(07021 03019)

(02996 02947)(05008 07008)(07024 03018)


0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(a)

0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(b)



120598120574

05455 06364 07273 08182 09091

001(03029 02955)(05021 07004)(07029 03036)

(03003 03007)(05007 06983)(07010 03029)

(03083 03026)(04974 06976)(06995 03008)

(02957 02992)(05050 06980)(07018 03088)

(02830 02881)(04980 07038)(07056 03000)

0001(03011 02940)(05010 07006)(07025 03022)

(03029 02946)(05006 07002)(07021 03021)

(03017 02943)(05009 07006)(07021 03017)

(02992 02954)(05004 07008)(07026 03019)

(03023 02972)(05005 06996)(07021 03024)


Exact 120582119895

1 3 5120579119895

(03 03) (05 07) (07 03)

119879 = 01119895

10108 30096 50584120579119895

(02886 02917) (05007 07032) (07043 03034)

119879 = 1119895

10039 30035 48805120579119895

(03265 03068) (04966 06984) (06993 03001)

119879 = 2119895

08916 29666 50708120579119895

(02959 02993) (05046 06980) (07004 03093)

119879 = 5119895

08560 29201 51635120579119895

(02805 03130) (05134 06938) (07026 03100)









5 Conclusion


Acknowledgments


References
















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











0119863120574


119905 isin (0 119879)

119866 (119909 0 120579) = 0 119909 120579 isin Ω

B119866 (119909 119905 120579) = 120583119866 (119909 119905 120579) + 120573120597119866

120597]= 0 119909 120579 isin 120597Ω

119905 isin (0 119879)

(7)


119866 (119909 119905 120579) =

infin

sum

119899=1

119905120574119864120574120574+1

(minus1205822

119899119905120574) 120593119899(120579) 120593119899(119909) (8)


119899is the


L120593 + 1205822

119899120593 = 0

120583120593 + 120573120597120593

120597]= 0

(9)


119864120574120577(119911) =

infin

sum

119896=0

119911119896

Γ (120574119896 + 120577) (10)


of (4) as

119906 (119909 119905) =

119872

sum

119895=1

120582119895119866(119909 119905 120579

119895) (11)



119872

sum

119895=1

120582119895119866(119909119894 119879 120579119895) = 119906120598

119894119879 119894 = 1 2 119873 (12)


denoting 120582 = (120582119895) and 120603 = (119906

120576



119860120582 = 120603 (13)


119860119894119895= 119866 (119909

119894 119879 120579119895) (14)


120582 = 119860minus1120603 (15)




min120582

119869120572(120582) = min

120582

119860120582 minus 1206032+ 12057221205822 (16)



120582120572= (119860lowast119860 + 120572

2119868)minus1

119860lowast120603 (17)





union set

Θ =

119872

⋃

119895=1

119861 (120579119895 120588119895) cap Ω (18)




119897=1from




119897


119897119899


sources 120579119895119872


0119863120574

119905119906 (119909 119905) minusL119906 (119909 119905) =

119899

sum

119897=1

120591119897120575 (119909 minus 120585

119897)

119909 isin Ω 119905 isin (0 119879)

119906 (119909 0) = 0 119909 isin Ω

120583119906 (119909 119905) + 120573120597119906

120597]= 0 119909 isin 120597Ω 119905 isin (0 119879)

(19)




119897can then be






119895as

119895= sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897 119895 = 1 2 119872 (20)







120579119895=

1

119895

sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897120585119897 119895 = 1 2 119872 (21)



RMS = radic1

119872

119872

sum

119895=1

(120572119895

minus 120582119895)2

(22)


119894is obtained


119906120598(119909119894 119879) = 119906 (119909

119894 119879) + 120598rand (119894) (23)





0

119906119905(119909 119910 119905) minus Δ119906 = 119891 (119909 119910) 0 lt 119909 lt 119897

119910 gt 0 119905 isin (0 119879)

119906 (119909 119910 0) = 0 0 le 119909 le 119897 119910 ge 0

119906|119909=0

= 119906|119909=119897

= 0 119910 ge 0

119906|119910=0

= 0 0 le 119909 le 119897

(24)


119866 (119909 119910 119905 120585 120578) =2

119897radic120587119905119890minus(1199102+1205782)4119905 sinh

119910120578

2119905

times

infin

sum

119899=1

119890minus119899212058721199051198972

sin 119899120587120585

119897sin 119899120587119909

119897119867 (119905)

(25)



1198955



10038161003816100381610038161003816120579119895minus 120579119895

10038161003816100381610038161003816lt 120588119895 for 119895 = 1 2 3 4 5 (26)








for 0 lt 120574 lt 1

0119863120574


119906 (119909 119910 0) = 0 (119909 119910) isin Ω

119906 (119909 119910 119905) = 0 (119909 119910) isin 120597Ω 119905 isin (0 119879)

(27)


119866 (119909 119910 119905 120585 120578) = 4

infin

sum

119899=1

119905120574119864120574120574+1

(minus11989921205872119905120574) sin (119899120587119909)

times sin (119899120587119910) sin (119899120587120585) sin (119899120587120578)

(28)


119895= 01 and


(03 03) (05 07) (07 03) (29)




Exact 120582119895

11 1 21 minus5 minus7120579119895

(05 1) (05 5) (05 10) (05 15) (05 25)

119876 = 25119895

116978 10539 220974 minus51806 minus73625120579119895

(05196 09876) (05349 49513) (05173 100002) (05059 149989) (05129 250005)

119876 = 100119895

117866 09881 221760 minus51460 minus73026120579119895

(05226 09925) (05196 49688) (05188 100002) (05059 150016) (05102 250001)

119876 = 256119895

114531 10006 221702 minus51226 minus71781120579119895

(05136 09929) (05219 49689) (05190 100005) (05027 150023) (05046 249995)

119876 = 400119895

116968 10962 221453 minus50899 minus71799120579119895

(05213 09977) (05521 49829) (05186 100005) (05009 150027) (05047 250008)

119876 = 625119895

119726 10230 220421 minus51289 minus71438120579119895

(05274 09951) (05309 50000) (05171 100004) (05033 150033) (05028 249993)

119876 = 900119895

115229 10794 223220 minus49636 minus71709120579119895

(05163 09975) (05477 49907) (05209 100004) (04919 150005) (05035 250000)


Exact 120582119895

5 7 2 5 7120579119895

(03 1) (06 5) (05 10) (07 15) (02 25)

119876 = 25119895

49670 73015 18876 51820 69783120579119895

(03016 10113) (06133 50007) (04816 100010) (07097 149972) (01985 249974)

119876 = 100119895

49553 71001 19929 50736 69830120579119895

(03009 10117) (06057 49999) (04993 100079) (07061 150017) (01982 249966)

119876 = 256119895

49891 70950 20070 49055 67970120579119895

(03041 10112) (06054 49992) (05018 100075) (07003 150016) (02031 250043)

119876 = 400119895

49690 70538 19133 49015 67483120579119895

(03020 10114) (06041 49998) (04859 100036) (07002 150014) (02032 250023)

119876 = 625119895

49678 70433 19759 49636 66110120579119895

(03022 10115) (06036 50003) (04966 100061) (07025 150035) (02078 250015)

119876 = 900119895

49803 70981 20104 50513 66630120579119895

(03030 10103) (06052 49991) (05019 100072) (07047 150059) (02065 250024)


120598120574

00909 01818 02727 03636 04545 05455 06364 07273 08182 09091001 00231 00121 00388 00266 00137 00250 00184 00233 00822 005890001 00083 00045 00088 00080 00089 00114 00099 00091 00096 00070


120598120574

00909 01818 02727 03636 04545

001(02986 02844)(05017 07031)(07023 02998)

(02983 02987)(05013 06993)(07018 03014)

(03053 03013)(05022 06979)(07030 03061)

(02951 02875)(05005 07029)(07036 03016)

(03035 03049)(05009 06985)(07022 03030)

0001(03002 02937)(05010 07010)(07024 03018)

(03031 02956)(05002 07002)(07016 03016)

(03019 02951)(05006 07003)(07020 03017)

(03021 02950)(05006 07003)(07021 03019)

(02996 02947)(05008 07008)(07024 03018)


0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(a)

0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(b)



120598120574

05455 06364 07273 08182 09091

001(03029 02955)(05021 07004)(07029 03036)

(03003 03007)(05007 06983)(07010 03029)

(03083 03026)(04974 06976)(06995 03008)

(02957 02992)(05050 06980)(07018 03088)

(02830 02881)(04980 07038)(07056 03000)

0001(03011 02940)(05010 07006)(07025 03022)

(03029 02946)(05006 07002)(07021 03021)

(03017 02943)(05009 07006)(07021 03017)

(02992 02954)(05004 07008)(07026 03019)

(03023 02972)(05005 06996)(07021 03024)


Exact 120582119895

1 3 5120579119895

(03 03) (05 07) (07 03)

119879 = 01119895

10108 30096 50584120579119895

(02886 02917) (05007 07032) (07043 03034)

119879 = 1119895

10039 30035 48805120579119895

(03265 03068) (04966 06984) (06993 03001)

119879 = 2119895

08916 29666 50708120579119895

(02959 02993) (05046 06980) (07004 03093)

119879 = 5119895

08560 29201 51635120579119895

(02805 03130) (05134 06938) (07026 03100)









5 Conclusion


Acknowledgments


References
















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119895as

119895= sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897 119895 = 1 2 119872 (20)







120579119895=

1

119895

sum

119897120585119897isin119861(120579119895 120588119895)capΩ

120591119897120585119897 119895 = 1 2 119872 (21)



RMS = radic1

119872

119872

sum

119895=1

(120572119895

minus 120582119895)2

(22)


119894is obtained


119906120598(119909119894 119879) = 119906 (119909

119894 119879) + 120598rand (119894) (23)





0

119906119905(119909 119910 119905) minus Δ119906 = 119891 (119909 119910) 0 lt 119909 lt 119897

119910 gt 0 119905 isin (0 119879)

119906 (119909 119910 0) = 0 0 le 119909 le 119897 119910 ge 0

119906|119909=0

= 119906|119909=119897

= 0 119910 ge 0

119906|119910=0

= 0 0 le 119909 le 119897

(24)


119866 (119909 119910 119905 120585 120578) =2

119897radic120587119905119890minus(1199102+1205782)4119905 sinh

119910120578

2119905

times

infin

sum

119899=1

119890minus119899212058721199051198972

sin 119899120587120585

119897sin 119899120587119909

119897119867 (119905)

(25)



1198955



10038161003816100381610038161003816120579119895minus 120579119895

10038161003816100381610038161003816lt 120588119895 for 119895 = 1 2 3 4 5 (26)








for 0 lt 120574 lt 1

0119863120574


119906 (119909 119910 0) = 0 (119909 119910) isin Ω

119906 (119909 119910 119905) = 0 (119909 119910) isin 120597Ω 119905 isin (0 119879)

(27)


119866 (119909 119910 119905 120585 120578) = 4

infin

sum

119899=1

119905120574119864120574120574+1

(minus11989921205872119905120574) sin (119899120587119909)

times sin (119899120587119910) sin (119899120587120585) sin (119899120587120578)

(28)


119895= 01 and


(03 03) (05 07) (07 03) (29)




Exact 120582119895

11 1 21 minus5 minus7120579119895

(05 1) (05 5) (05 10) (05 15) (05 25)

119876 = 25119895

116978 10539 220974 minus51806 minus73625120579119895

(05196 09876) (05349 49513) (05173 100002) (05059 149989) (05129 250005)

119876 = 100119895

117866 09881 221760 minus51460 minus73026120579119895

(05226 09925) (05196 49688) (05188 100002) (05059 150016) (05102 250001)

119876 = 256119895

114531 10006 221702 minus51226 minus71781120579119895

(05136 09929) (05219 49689) (05190 100005) (05027 150023) (05046 249995)

119876 = 400119895

116968 10962 221453 minus50899 minus71799120579119895

(05213 09977) (05521 49829) (05186 100005) (05009 150027) (05047 250008)

119876 = 625119895

119726 10230 220421 minus51289 minus71438120579119895

(05274 09951) (05309 50000) (05171 100004) (05033 150033) (05028 249993)

119876 = 900119895

115229 10794 223220 minus49636 minus71709120579119895

(05163 09975) (05477 49907) (05209 100004) (04919 150005) (05035 250000)


Exact 120582119895

5 7 2 5 7120579119895

(03 1) (06 5) (05 10) (07 15) (02 25)

119876 = 25119895

49670 73015 18876 51820 69783120579119895

(03016 10113) (06133 50007) (04816 100010) (07097 149972) (01985 249974)

119876 = 100119895

49553 71001 19929 50736 69830120579119895

(03009 10117) (06057 49999) (04993 100079) (07061 150017) (01982 249966)

119876 = 256119895

49891 70950 20070 49055 67970120579119895

(03041 10112) (06054 49992) (05018 100075) (07003 150016) (02031 250043)

119876 = 400119895

49690 70538 19133 49015 67483120579119895

(03020 10114) (06041 49998) (04859 100036) (07002 150014) (02032 250023)

119876 = 625119895

49678 70433 19759 49636 66110120579119895

(03022 10115) (06036 50003) (04966 100061) (07025 150035) (02078 250015)

119876 = 900119895

49803 70981 20104 50513 66630120579119895

(03030 10103) (06052 49991) (05019 100072) (07047 150059) (02065 250024)


120598120574

00909 01818 02727 03636 04545 05455 06364 07273 08182 09091001 00231 00121 00388 00266 00137 00250 00184 00233 00822 005890001 00083 00045 00088 00080 00089 00114 00099 00091 00096 00070


120598120574

00909 01818 02727 03636 04545

001(02986 02844)(05017 07031)(07023 02998)

(02983 02987)(05013 06993)(07018 03014)

(03053 03013)(05022 06979)(07030 03061)

(02951 02875)(05005 07029)(07036 03016)

(03035 03049)(05009 06985)(07022 03030)

0001(03002 02937)(05010 07010)(07024 03018)

(03031 02956)(05002 07002)(07016 03016)

(03019 02951)(05006 07003)(07020 03017)

(03021 02950)(05006 07003)(07021 03019)

(02996 02947)(05008 07008)(07024 03018)


0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(a)

0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(b)



120598120574

05455 06364 07273 08182 09091

001(03029 02955)(05021 07004)(07029 03036)

(03003 03007)(05007 06983)(07010 03029)

(03083 03026)(04974 06976)(06995 03008)

(02957 02992)(05050 06980)(07018 03088)

(02830 02881)(04980 07038)(07056 03000)

0001(03011 02940)(05010 07006)(07025 03022)

(03029 02946)(05006 07002)(07021 03021)

(03017 02943)(05009 07006)(07021 03017)

(02992 02954)(05004 07008)(07026 03019)

(03023 02972)(05005 06996)(07021 03024)


Exact 120582119895

1 3 5120579119895

(03 03) (05 07) (07 03)

119879 = 01119895

10108 30096 50584120579119895

(02886 02917) (05007 07032) (07043 03034)

119879 = 1119895

10039 30035 48805120579119895

(03265 03068) (04966 06984) (06993 03001)

119879 = 2119895

08916 29666 50708120579119895

(02959 02993) (05046 06980) (07004 03093)

119879 = 5119895

08560 29201 51635120579119895

(02805 03130) (05134 06938) (07026 03100)









5 Conclusion


Acknowledgments


References
















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Exact 120582119895

11 1 21 minus5 minus7120579119895

(05 1) (05 5) (05 10) (05 15) (05 25)

119876 = 25119895

116978 10539 220974 minus51806 minus73625120579119895

(05196 09876) (05349 49513) (05173 100002) (05059 149989) (05129 250005)

119876 = 100119895

117866 09881 221760 minus51460 minus73026120579119895

(05226 09925) (05196 49688) (05188 100002) (05059 150016) (05102 250001)

119876 = 256119895

114531 10006 221702 minus51226 minus71781120579119895

(05136 09929) (05219 49689) (05190 100005) (05027 150023) (05046 249995)

119876 = 400119895

116968 10962 221453 minus50899 minus71799120579119895

(05213 09977) (05521 49829) (05186 100005) (05009 150027) (05047 250008)

119876 = 625119895

119726 10230 220421 minus51289 minus71438120579119895

(05274 09951) (05309 50000) (05171 100004) (05033 150033) (05028 249993)

119876 = 900119895

115229 10794 223220 minus49636 minus71709120579119895

(05163 09975) (05477 49907) (05209 100004) (04919 150005) (05035 250000)


Exact 120582119895

5 7 2 5 7120579119895

(03 1) (06 5) (05 10) (07 15) (02 25)

119876 = 25119895

49670 73015 18876 51820 69783120579119895

(03016 10113) (06133 50007) (04816 100010) (07097 149972) (01985 249974)

119876 = 100119895

49553 71001 19929 50736 69830120579119895

(03009 10117) (06057 49999) (04993 100079) (07061 150017) (01982 249966)

119876 = 256119895

49891 70950 20070 49055 67970120579119895

(03041 10112) (06054 49992) (05018 100075) (07003 150016) (02031 250043)

119876 = 400119895

49690 70538 19133 49015 67483120579119895

(03020 10114) (06041 49998) (04859 100036) (07002 150014) (02032 250023)

119876 = 625119895

49678 70433 19759 49636 66110120579119895

(03022 10115) (06036 50003) (04966 100061) (07025 150035) (02078 250015)

119876 = 900119895

49803 70981 20104 50513 66630120579119895

(03030 10103) (06052 49991) (05019 100072) (07047 150059) (02065 250024)


120598120574

00909 01818 02727 03636 04545 05455 06364 07273 08182 09091001 00231 00121 00388 00266 00137 00250 00184 00233 00822 005890001 00083 00045 00088 00080 00089 00114 00099 00091 00096 00070


120598120574

00909 01818 02727 03636 04545

001(02986 02844)(05017 07031)(07023 02998)

(02983 02987)(05013 06993)(07018 03014)

(03053 03013)(05022 06979)(07030 03061)

(02951 02875)(05005 07029)(07036 03016)

(03035 03049)(05009 06985)(07022 03030)

0001(03002 02937)(05010 07010)(07024 03018)

(03031 02956)(05002 07002)(07016 03016)

(03019 02951)(05006 07003)(07020 03017)

(03021 02950)(05006 07003)(07021 03019)

(02996 02947)(05008 07008)(07024 03018)


0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(a)

0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(b)



120598120574

05455 06364 07273 08182 09091

001(03029 02955)(05021 07004)(07029 03036)

(03003 03007)(05007 06983)(07010 03029)

(03083 03026)(04974 06976)(06995 03008)

(02957 02992)(05050 06980)(07018 03088)

(02830 02881)(04980 07038)(07056 03000)

0001(03011 02940)(05010 07006)(07025 03022)

(03029 02946)(05006 07002)(07021 03021)

(03017 02943)(05009 07006)(07021 03017)

(02992 02954)(05004 07008)(07026 03019)

(03023 02972)(05005 06996)(07021 03024)


Exact 120582119895

1 3 5120579119895

(03 03) (05 07) (07 03)

119879 = 01119895

10108 30096 50584120579119895

(02886 02917) (05007 07032) (07043 03034)

119879 = 1119895

10039 30035 48805120579119895

(03265 03068) (04966 06984) (06993 03001)

119879 = 2119895

08916 29666 50708120579119895

(02959 02993) (05046 06980) (07004 03093)

119879 = 5119895

08560 29201 51635120579119895

(02805 03130) (05134 06938) (07026 03100)









5 Conclusion


Acknowledgments


References
















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(a)

0 02 04 06 08 10

01

02

03

04

05

06

07

08

09

1


x

y

(b)



120598120574

05455 06364 07273 08182 09091

001(03029 02955)(05021 07004)(07029 03036)

(03003 03007)(05007 06983)(07010 03029)

(03083 03026)(04974 06976)(06995 03008)

(02957 02992)(05050 06980)(07018 03088)

(02830 02881)(04980 07038)(07056 03000)

0001(03011 02940)(05010 07006)(07025 03022)

(03029 02946)(05006 07002)(07021 03021)

(03017 02943)(05009 07006)(07021 03017)

(02992 02954)(05004 07008)(07026 03019)

(03023 02972)(05005 06996)(07021 03024)


Exact 120582119895

1 3 5120579119895

(03 03) (05 07) (07 03)

119879 = 01119895

10108 30096 50584120579119895

(02886 02917) (05007 07032) (07043 03034)

119879 = 1119895

10039 30035 48805120579119895

(03265 03068) (04966 06984) (06993 03001)

119879 = 2119895

08916 29666 50708120579119895

(02959 02993) (05046 06980) (07004 03093)

119879 = 5119895

08560 29201 51635120579119895

(02805 03130) (05134 06938) (07026 03100)









5 Conclusion


Acknowledgments


References
















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











5 Conclusion


Acknowledgments


References
















































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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