Hindawi Publishing CorporationAdvances in Mathematical PhysicsVolume 2013 Article ID 476154 8 pageshttpdxdoiorg1011552013476154

Research ArticleOn the Inverse Problem of the Fractional Heat-Like PartialDifferential Equations Determination of the Source Function

Guumllcan Oumlzkum1 Ali Demir1 Sertaccedil Erman1 Esra Korkmaz2 and Berrak Oumlzguumlr1

1 Department of Mathematics Science and Letter Faculty Kocaeli University Umuttepe Campus 41380 Kocaeli Turkey2 Ardahan University 75000 Ardahan Turkey

Correspondence should be addressed to Gulcan Ozkum gulcankocaeliedutr

Received 22 May 2013 Revised 12 September 2013 Accepted 12 September 2013

Academic Editor H Srivastava

Copyright copy 2013 Gulcan Ozkum 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

The study in this paper mainly concerns the inverse problem of determining an unknown source function in the linear fractionaldifferential equation with variable coefficient using Adomian decomposition method (ADM) We apply ADM to determine thecontinuous right hand side functions 119891(119909) and 119891(119905) in the heat-like diffusion equations 119863120572

119905119906(119909 119905) = ℎ(119909)119906

119909119909(119909 119905) + 119891(119909) and

119863120572119905119906(119909 119905) = ℎ(119909)119906

119909119909(119909 119905) + 119891(119905) respectively The results reveal that ADM is very effective and simple for the inverse problem of

determining the source function

1 Introduction

Fractional differential equations (FDEs) are obtained by gen-eralizing differential equations to an arbitrary orderThey areused to model physical systems with memory Since FDEshave memory nonlocal relations in space and time complexphenomena can be modeled by using these equations Dueto this fact materials with memory and hereditary effectsfluid flow rheology diffusive transport electrical networkselectromagnetic theory and probability signal processingand many other physical processes are diverse applicationsof FDEs Since FDEs are used to model complex phenomenathey play a crucial role in engineering physics and appliedmathematics Therefore they are generating an increasinginterest from engineers and scientist in the recent years Asa result FDEs are quite frequently encountered in differentresearch areas and engineering applications [1]

The book written by Oldham and Spanier [2] played anoutstanding role in the development of the fractional calcu-lus Also it was the first book that was entirely devoted to asystematic presentation of the ideas methods and applica-tions of the fractional calculus Afterwards several funda-mental works on various aspects of the fractional calculusinclude extensive survey on fractional differential equationsby Miller and Ross [3] Podlubny [4] and others Further

several references to the books by Oldham and Spanier [2]Miller and Ross [3] and Podlubny [4] show that appliedscientists need first of all an easy introduction to the theoryof fractional derivatives and fractional differential equationswhich could help them in their initial steps in adopting thefractional calculus as a method of research [5]

In general FDEs do not have exact analytical solutionshence the approximate and numerical solutions of theseequations are studied [6ndash8] Analytical approximations forlinear and nonlinear FDEs are obtained by variational iter-ation method Adomian decomposition method homotopyperturbationmethod Lagrangemultipliermethod BPs oper-ational matrices method and so forth An effective and easy-to-use method for solving such equations is needed Largeclasses of linear and nonlinear differential equations bothordinary and partial can be solved by the Adomian decom-position method [9ndash12]

Solving an equationwith certain data in a specified regionis called direct problem On the other hand determiningan unknown input by using output is called an inverseproblem This unknown input could be some coefficientsor it could be a source function in equation Based on thisunknown input the inverse problem is called inverse problemof coefficient identification or inverse problem of sourceidentification respectively Generally inverse problems are

2 Advances in Mathematical Physics

ill-posed problems that is they are very sensitive to errorsin measured input In order to deal with this ill-posednessregularization methods have been developed Inverse prob-lems have many practical applications such as geophysicsoptics quantum mechanics astronomy medical imagingandmaterials testing X-ray tomography and photoelasticityTheoretical and applied aspects of inverse problems havebeen under intense study lately especially for the fractionalequation [13ndash16]

In this paper we investigate inverse problems of thelinear heat-like differential equations of fractional orders119863120572119905119906(119909 119905) = ℎ(119909)119906

119909119909(119909 119905) + 119891(119909) and 119863120572

119905119906(119909 119905) =

ℎ(119909)119906119909119909(119909 119905) + 119891(119905) where the function 119906(119909 119905) is assumed

to be a causal function of time and space Time fractionalderivative operator 119863120572

119905is considered as in Caputo sense [17]

We use the Adomian decompositionmethod [9 10] to obtainsource functions 119891(119909) and 119891(119905) under the initial and mixedboundary conditions By this method we determine thesource functions 119891(119909) and 119891(119905) in a rapidly converging seriesform when they exist Compared with previous researches[10 12 18] the method we use in this paper is more effectiveand accurate

The structure of this paper is given as follows First wegive some basic definitions of fractional calculus Inverseproblem of finding the source function in one-dimensionalfractional heat-like equations with mixed boundary condi-tions is given in Section 2 After that we give some illustrativeexamples of this method for all cases in Section 3 Finally theconclusion is given in Section 4

11 Fractional Calculus In this section we give basic defini-tions and properties of the fractional calculus [17 18]

Definition 1 A real function 119891(119909) 119909 gt 0 is said to be in thespace 119862

120583 120583 isin R if there exists a real number 119901 gt 120583 such that

119891(119909) = 1199091199011198911(119909) where 119891

1(119909) isin 119862[0infin) and it is said to be

in the space 119862119898120583if 119891(119898) isin 119862

120583119898 isin N

Definition 2 TheRiemann-Liouville fractional integral oper-ator of order 120572 ge 0 of a function 119891 isin 119862

120583 120583 ge minus1 is defined

as

119869120572119891 (119909) =1

Γ (120572)int119909

0

(119909 minus 119905)120572minus1119891 (119905) 119889119905 120572 gt 0 119909 gt 0

1198690119891 (119909) = 119891 (119909)

(1)

Some of the basic properties of this operator are given asfollows

For 119891 isin 119862120583 120583 ge minus1 120572 120573 ge 0 and 120574 gt minus1

(1) 119869120572119869120573119891 (119909) = 119869120572+120573119891 (119909)

(2) 119869120572119869120573119891 (119909) = 119869120573119869120572119891 (119909)

(3) 119869120572119909120574 =Γ (120574 + 1)

Γ (120572 + 120574 + 1)119909120572+120574

(2)

The other properties can be found in [17]

Definition 3 The fractional derivative of 119891(119909) in the Caputosense is defined as

119863120572119891 (119909) = 119869119898minus120572119863119898119891 (119909)

=1

Γ (119898 minus 120572)int119909

0

(119909 minus 119905)119898minus120572minus1119891(119898) (119905) 119889119905

(3)

where119898 minus 1 lt 120572 le 119898119898 isin N 119909 gt 0 119891 isin 119862119898minus1

Useful properties of119863120572 are given as follows

Lemma 4 If 119898 minus 1 lt 120572 le 119898 119898 isin N and 119891 isin 119862119898120583 120583 ge minus1

then

119863120572119869120572119891 (119909) = 119891 (119909)

119869120572119863120572119891 (119909) = 119891 (119909) minus119898minus1

sum119896=0

119891(119896) (0+)119909119896

119896 119909 gt 0

(4)

Since traditional initial and boundary conditions are allowedin problems including Caputo fractional derivatives it isconsidered here In this paper we deal with the fractional heat-like equations where the unknown function 119906 = 119906(119909 119905) is anarbitrary function of time and space

Definition 5 The Caputo time fractional derivative operatorof order 120572 gt 0 is defined as follows where 119898 is the smallestinteger that exceeds 120572

119863120572119905119906 (119909 119905)

=120597120572119906 (119909 119905)

120597119905120572

=

1

Γ (119898 minus 120572)

timesint119905

0

(119905 minus 120591)119898minus120572minus1

120597119898119906 (119909 120591)

120597119905119898119889120591 for 119898 minus 1 lt 120572 lt 119898

120597119898119906 (119909 119905)

120597119905119898 for 120572 = 119898 isin IN

(5)

For more details about Caputo fractional differential opera-tor we refer to [17]

Definition 6 The Mittag-Leffler function with two-parameters is defined by the series expansion as shownbelow where the real part of 120572 is strictly positive [19]

119864120572120573

(119911) =infin

sum119899=0

119911119899

Γ (120572119899 + 120573) (6)

2 Inverse Problem ofDetermining Source Function

In this section we deal with inverse problem of findingthe source function in one-dimensional fractional heat-likeequationswithmixed boundary conditions To determine theunknown source function we have developed new methodsthrough ADM as in the following subsections

Advances in Mathematical Physics 3

21 Determination of Unknown Source Functions Dependingon 119909 We consider the following inverse problem of deter-mining the source function 119891(119909)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119909)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(7)

where the functions ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) and ℎ(119909) 119891(119909)

1198911(119909) isin 119862infin[0infin) In order to determine the source function

for this kind of inverse problems we apply ADM Firstwe apply the time-dependent Riemann-Liouville fractionalintegral operator 119869120572

119905to both sides of (7) to get rid of fractional

derivative119863120572119905as shown below

119869120572119905119863120572119905119906 (119909 119905) = 119869120572

119905(ℎ (119909) 119906

119909119909(119909 119905)) + 119869120572

119905119891 (119909) (8)

Then we get

119906 (119909 119905) = 119906 (119909 0) + 119869120572119905119891 (119909) + 119869120572

119905(ℎ (119909) 119906

119909119909(119909 119905)) (9)

In ADM the solution 119906(119909 119905) is written in the following seriesform [9]

119906 (119909 119905) =infin

sum119899=0

119906119899(119909 119905) (10)

where 119906 and 119906119899 119899 isin N are defined in 119862infin[0infin) times 1198621

120583[0infin)

After substituting the decomposition (10) into (9) and settingthe recurrence scheme as follows

1199060(119909 119905) = 119906 (119909 0) + 119869120572

119905119891 (119909)

119906119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (119906

119899)119909119909

(119909 119905)) 119899 = 0 1 (11)

we get ADM polynomials below

1199060(119909 119905) = 119891

1(119909) + 119891 (119909)

119905120572

Γ (120572 + 1)

1199061(119909 119905) = 119869120572

119905(ℎ (119909) (119906

0)119909119909

(119909))

= ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ (119909) 11989110158401015840

(119909)1199052120572

Γ (2120572 + 1)

1199062(119909 119905)

= 119869120572119905(ℎ (119909) (119906

1)119909119909

(119909))

= ℎ (119909) [ℎ10158401015840

(119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1)

(12)

After writing these polynomials in (10) the solution 119906(119909 119905) isgiven by

119906 (119909 119905)

= 1198911(119909) + 119891 (119909)

119905120572

Γ (120572 + 1)+ ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ (119909) 11989110158401015840

(119909)1199052120572

Γ (2120572 + 1)

+ ℎ (119909) [ℎ10158401015840

(119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(13)

If we arrange it with respect to like powers of 119905 then we get

119906 (119909 119905)

= 1198911(119909) + [119891 (119909) + ℎ (119909) 119891

10158401015840

1(119909)]

119905120572

Γ (120572 + 1)

+ ℎ (119909) [11989110158401015840

(119909) + ℎ10158401015840 (119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

4 Advances in Mathematical Physics

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(14)

To determine the unknown source function first we expandthe boundary conditions 119906(0 119905) = ℎ

1(119905) and 119906

119909(0 119905) =

ℎ2(119905) into the following series for the space whose bases are

119905119899120572Γ(119899120572 + 1)infin

119899=0 0 lt 120572 le 1

ℎ1(119905) = ℎ

1(0) + ℎ1015840

1(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158401(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(15)

ℎ2(119905) = ℎ

2(0) + ℎ1015840

2(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158402(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(16)

On the other hand if we rewrite the boundary conditions119906(0 119905) and 119906

119909(0 119905) from (14) then we have

ℎ1(119905) = 119891

1(0) + [119891 (0) + ℎ (0) 119891

10158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0)+ ℎ (0) 119891

(119894V)1

(0)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(17)

ℎ2(119905) = 1198911015840

1(0) + [1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ (0) 119891101584010158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ1015840 (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0)

+ 2ℎ10158401015840 (0) 11989110158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+ 2ℎ1015840 (0) 119891(119894V)1

(0)

+ ℎ (0) 119891(V)1

(0)]1199052120572

Γ (2120572 + 1)

+ ℎ1015840 (0) [ℎ10158401015840

(0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0) ]

+ ℎ (0) [ℎ101584010158401015840

(0) 11989110158401015840

(0) + 3ℎ10158401015840 (0) 119891101584010158401015840

(0)

+ 3ℎ1015840 (0) 119891(119894V)

(0)

+ ℎ (0) 119891(V)

(0) ]1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(18)

Equating (15) and (17) yields the following

ℎ1(0) = 119891

1(0)

ℎ10158401(0) = 119891 (0) + ℎ (0) 119891

10158401015840

1(0)

ℎ101584010158401(0) = ℎ (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

(19)

and equating (16) and (18) yields the following

ℎ2(0) = 1198911015840

1(0)

ℎ10158402(0) = 1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0) + ℎ (0) 119891

101584010158401015840

1(0)

(20)

ℎ101584010158402(0) = ℎ1015840 (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0) + 2ℎ10158401015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+2ℎ1015840 (0) 119891(119894V)1

(0) + ℎ (0) 119891(V)1

(0)]

(21)

Using the above data in the following Taylor series expansionof unknown function 119891(119909) we get

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2+ 119891101584010158401015840 (0)

1199093

3+ sdot sdot sdot

(22)

Consequently we determine 119891(119909) as follows

119891 (119909) = [ℎ10158401(0) minus ℎ (0) 119891

10158401015840

1(0)]

+ [ℎ10158402(0) minus ℎ1015840 (0) 119891

10158401015840

1(0) minus ℎ (0) 119891

101584010158401015840

1(0)] 119909

Advances in Mathematical Physics 5

+ [ℎ101584010158401(0)

ℎ (0)minus ℎ10158401015840 (0) 119891

1015840

1(0) minus ℎ1015840 (0) 119891

10158401015840

1(0)

minus ℎ1015840 (0) 119891101584010158401015840

1(0) minus ℎ (0) 119891

(119894V)1

(0)]1199092

2+ sdot sdot sdot

(23)

where ℎ(0) = 0

22 Determination of Unknown Source Functions Dependingon 119905 Weconsider the following inverse problemof determin-ing the source function 119891(119905)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119905)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(24)

where ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) ℎ(119909) 119891

1(119909) isin 119862infin[0infin) and

119891(119905) isin 1198621120583[0infin) 120583 ge minus1 As in the previous case we apply

ADM to determine the unknown function 119891(119905)First to reduce the problem we define new functions in

the following form

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (25)

Then the reduced problem is given as follows

119863120572119905V (119909 119905) = ℎ (119909) V

119909119909(119909 119905) (26)

with the following initial and mixed boundary conditions

V (119909 0) = 1198911(119909) minus 119908 (0)

V (0 119905) = ℎ1(119905) minus 119908 (119905)

V119909(0 119905) = ℎ

2(119905)

(27)

By using ADM as in the previous section we determine thefunction 119908(119905) which leads to the source function 119891(119905) Let usapply 119869120572

119905to both sides of (26) as shown below

119869120572119905119863120572119905V (119909 119905) = 119869120572

119905(ℎ (119909) V

119909119909(119909 119905)) (28)

Then we get

V (119909 119905) = V (119909 0) + 119869120572119905(ℎ (119909) V

119909119909(119909 119905)) (29)

Now we define the solution V(119909 119905) by the following decom-position series according to ADM

V (119909 119905) =infin

sum119899=0

V119899(119909 119905) (30)

Substituting (30) into (29) we obtain

V0(119909 119905) = V (119909 0)

V119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (V

119899)119909119909

(119909 119905)) 119899 = 0 1 (31)

Hence the recurrence scheme is obtained as followsV0(119909 119905) = V (119909 0) = 119891

1(119909) minus 119908 (0)

V1(119909 119905) = 119869120572

119905(ℎ (119909) (V

0)119909119909

(119909)) = ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

V2(119909 119905) = 119869120572

119905(ℎ (119909) (V

1)119909119909

(119909)) = ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)

(32)

Consequently from (30) the solution V(119909 119905) is given as shownbelow

V (119909 119905) = V0(119909 119905) + V

1(119909 119905) + V

2(119909 119905) + sdot sdot sdot

= 1198911(119909) minus 119908 (0) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(33)

By using the boundary condition V(0 119905) = ℎ1(119905) + 119908(119905) and

119908(0) = 0 we have

119908 (119905) = 1198911(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(34)

which implies the following

119869120572119905119891 (119905) = 119891

1(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(35)

Since 119863120572119905119908(119909 119905) = 119863120572

119905119869120572119905119891(119905) = 119891(119905) we obtain the source

function 119891(119905) as follows

119891 (119905) = 119863120572119905[1198911(119909) minus ℎ

1(119905) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot ]

(36)

3 Examples

Example 1 We consider the inverse problem of determiningsource function 119891(119909) in the following one-dimensional frac-tional heat-like PDE

119863120572119905119906 (119909 119905) = 2119906

119909119909(119909 119905) + 119891 (119909) 119909 gt 0 0 lt 120572 le 1 119905 gt 0

(37)

6 Advances in Mathematical Physics

subject to the following initial and nonhomogeneous mixedboundary conditions

119906 (119909 0) = 119890119909 + sin119909

119906 (0 119905) = 1198902119905

119906119909(0 119905) = 1198902119905+1

(38)

Now let us apply the time-dependent Riemann Liouvillefractional integral operator 119869120572

119905to both sides of (37)

119869120572119905119863120572119905119906 (119909 119905) = 2119869120572

119905119906119909119909

(119909 119905) + 119869120572119905119891 (119909) (39)

which implies

119906 (119909 119905) minus 119906 (119909 0) = 2119869120572119905119906119909119909

(119909 119905) + 119891 (119909) 119869120572

119905(1) (40)

Then from the initial condition we get

119906 (119909 119905) = 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119869120572119905119906119909119909

(119909 119905) (41)

Now we apply ADM to the problem In (41) the sum of thefirst three terms is identified as 119906

0 So

1199060= 119890119909 + sin119909 + 119891 (119909)

119905120572

Γ (120572 + 1)

119906119896+1

= 2119869120572119905(119906119896)119909119909

(119909 119905) 119896 ge 0

(42)

For 119896 = 0 we have

1199061= 2119869120572119905(1199060)119909119909

(119909 119905)

= 2119890119909119905120572

Γ (120572 + 1)minus 2 sin119909 119905120572

Γ (120572 + 1)

+ 211989110158401015840 (119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(43)

similarly for 119896 = 1 we have

1199062= 2119869120572119905(1199061)119909119909

(119909 119905)

= 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(44)

and for 119896 = 2 we have

1199063=2119869120572119905(1199062)119909119909

(119909 119905)

= 81198901199091199053120572

Γ (3120572 + 1)minus 8 sin119909 1199053120572

Γ (3120572 + 1)

+ 8119891(V119894) (119909)1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(45)

Then using ADM polynomials we get the solution 119906(119909 119905) asfollows119906 (119909 119905) = 119906

0+ 1199061+ 1199062+ 1199063+ sdot sdot sdot

= 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119890119909

119905120572

Γ (120572 + 1)

minus 2 sin119909 119905120572

Γ (120572 + 1)+ 211989110158401015840 (119909)

1199052120572

Γ (2120572 + 1)

+ 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ 8119890119909

1199053120572

Γ (3120572 + 1)

minus 8 sin119909 1199053120572

Γ (3120572 + 1)+ 8119891(V119894) (119909)

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(46)

After arranging it according to like powers of 119905 we have

119906 (119909 119905) = 119890119909 + sin119909 +119905120572

Γ (120572 + 1)[119891 (119909) + 2119890119909 minus 2 sin119909]

+1199052120572

Γ (2120572 + 1)[211989110158401015840 (119909) + 4119890119909 + 4 sin119909]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (119909) + 8119890119909 minus 8 sin119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (119909) + 16119890119909 + 16 sin119909] + sdot sdot sdot

(47)

Now by applying the boundary condition given in (38) weobtain

119906 (0 119905) = 1 +119905120572

Γ (120572 + 1)[119891 (0) + 2] +

1199052120572

Γ (2120572 + 1)[211989110158401015840 (0) + 4]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (0) + 8]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (0) + 16] + sdot sdot sdot

(48)

From (15) it must be equal to the following Taylorseries expansion of 1198902119905 in the space whose bases are119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 = 1 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(49)

Hence from the equality of the coefficients of correspondingterms we get

119891 (0) = 11989110158401015840 (0) = 119891(119894V) (0) = 119891(V119894) (0) = sdot sdot sdot = 0 (50)

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Advances in Mathematical Physics

ill-posed problems that is they are very sensitive to errorsin measured input In order to deal with this ill-posednessregularization methods have been developed Inverse prob-lems have many practical applications such as geophysicsoptics quantum mechanics astronomy medical imagingandmaterials testing X-ray tomography and photoelasticityTheoretical and applied aspects of inverse problems havebeen under intense study lately especially for the fractionalequation [13ndash16]

In this paper we investigate inverse problems of thelinear heat-like differential equations of fractional orders119863120572119905119906(119909 119905) = ℎ(119909)119906

119909119909(119909 119905) + 119891(119909) and 119863120572

119905119906(119909 119905) =

ℎ(119909)119906119909119909(119909 119905) + 119891(119905) where the function 119906(119909 119905) is assumed

to be a causal function of time and space Time fractionalderivative operator 119863120572

119905is considered as in Caputo sense [17]

We use the Adomian decompositionmethod [9 10] to obtainsource functions 119891(119909) and 119891(119905) under the initial and mixedboundary conditions By this method we determine thesource functions 119891(119909) and 119891(119905) in a rapidly converging seriesform when they exist Compared with previous researches[10 12 18] the method we use in this paper is more effectiveand accurate

The structure of this paper is given as follows First wegive some basic definitions of fractional calculus Inverseproblem of finding the source function in one-dimensionalfractional heat-like equations with mixed boundary condi-tions is given in Section 2 After that we give some illustrativeexamples of this method for all cases in Section 3 Finally theconclusion is given in Section 4

11 Fractional Calculus In this section we give basic defini-tions and properties of the fractional calculus [17 18]

Definition 1 A real function 119891(119909) 119909 gt 0 is said to be in thespace 119862

120583 120583 isin R if there exists a real number 119901 gt 120583 such that

119891(119909) = 1199091199011198911(119909) where 119891

1(119909) isin 119862[0infin) and it is said to be

in the space 119862119898120583if 119891(119898) isin 119862

120583119898 isin N

Definition 2 TheRiemann-Liouville fractional integral oper-ator of order 120572 ge 0 of a function 119891 isin 119862

120583 120583 ge minus1 is defined

as

119869120572119891 (119909) =1

Γ (120572)int119909

0

(119909 minus 119905)120572minus1119891 (119905) 119889119905 120572 gt 0 119909 gt 0

1198690119891 (119909) = 119891 (119909)

(1)

Some of the basic properties of this operator are given asfollows

For 119891 isin 119862120583 120583 ge minus1 120572 120573 ge 0 and 120574 gt minus1

(1) 119869120572119869120573119891 (119909) = 119869120572+120573119891 (119909)

(2) 119869120572119869120573119891 (119909) = 119869120573119869120572119891 (119909)

(3) 119869120572119909120574 =Γ (120574 + 1)

Γ (120572 + 120574 + 1)119909120572+120574

(2)

The other properties can be found in [17]

Definition 3 The fractional derivative of 119891(119909) in the Caputosense is defined as

119863120572119891 (119909) = 119869119898minus120572119863119898119891 (119909)

=1

Γ (119898 minus 120572)int119909

0

(119909 minus 119905)119898minus120572minus1119891(119898) (119905) 119889119905

(3)

where119898 minus 1 lt 120572 le 119898119898 isin N 119909 gt 0 119891 isin 119862119898minus1

Useful properties of119863120572 are given as follows

Lemma 4 If 119898 minus 1 lt 120572 le 119898 119898 isin N and 119891 isin 119862119898120583 120583 ge minus1

then

119863120572119869120572119891 (119909) = 119891 (119909)

119869120572119863120572119891 (119909) = 119891 (119909) minus119898minus1

sum119896=0

119891(119896) (0+)119909119896

119896 119909 gt 0

(4)

Since traditional initial and boundary conditions are allowedin problems including Caputo fractional derivatives it isconsidered here In this paper we deal with the fractional heat-like equations where the unknown function 119906 = 119906(119909 119905) is anarbitrary function of time and space

Definition 5 The Caputo time fractional derivative operatorof order 120572 gt 0 is defined as follows where 119898 is the smallestinteger that exceeds 120572

119863120572119905119906 (119909 119905)

=120597120572119906 (119909 119905)

120597119905120572

=

1

Γ (119898 minus 120572)

timesint119905

0

(119905 minus 120591)119898minus120572minus1

120597119898119906 (119909 120591)

120597119905119898119889120591 for 119898 minus 1 lt 120572 lt 119898

120597119898119906 (119909 119905)

120597119905119898 for 120572 = 119898 isin IN

(5)

For more details about Caputo fractional differential opera-tor we refer to [17]

Definition 6 The Mittag-Leffler function with two-parameters is defined by the series expansion as shownbelow where the real part of 120572 is strictly positive [19]

119864120572120573

(119911) =infin

sum119899=0

119911119899

Γ (120572119899 + 120573) (6)

2 Inverse Problem ofDetermining Source Function

In this section we deal with inverse problem of findingthe source function in one-dimensional fractional heat-likeequationswithmixed boundary conditions To determine theunknown source function we have developed new methodsthrough ADM as in the following subsections

Advances in Mathematical Physics 3

21 Determination of Unknown Source Functions Dependingon 119909 We consider the following inverse problem of deter-mining the source function 119891(119909)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119909)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(7)

where the functions ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) and ℎ(119909) 119891(119909)

1198911(119909) isin 119862infin[0infin) In order to determine the source function

for this kind of inverse problems we apply ADM Firstwe apply the time-dependent Riemann-Liouville fractionalintegral operator 119869120572

119905to both sides of (7) to get rid of fractional

derivative119863120572119905as shown below

119869120572119905119863120572119905119906 (119909 119905) = 119869120572

119905(ℎ (119909) 119906

119909119909(119909 119905)) + 119869120572

119905119891 (119909) (8)

Then we get

119906 (119909 119905) = 119906 (119909 0) + 119869120572119905119891 (119909) + 119869120572

119905(ℎ (119909) 119906

119909119909(119909 119905)) (9)

In ADM the solution 119906(119909 119905) is written in the following seriesform [9]

119906 (119909 119905) =infin

sum119899=0

119906119899(119909 119905) (10)

where 119906 and 119906119899 119899 isin N are defined in 119862infin[0infin) times 1198621

120583[0infin)

After substituting the decomposition (10) into (9) and settingthe recurrence scheme as follows

1199060(119909 119905) = 119906 (119909 0) + 119869120572

119905119891 (119909)

119906119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (119906

119899)119909119909

(119909 119905)) 119899 = 0 1 (11)

we get ADM polynomials below

1199060(119909 119905) = 119891

1(119909) + 119891 (119909)

119905120572

Γ (120572 + 1)

1199061(119909 119905) = 119869120572

119905(ℎ (119909) (119906

0)119909119909

(119909))

= ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ (119909) 11989110158401015840

(119909)1199052120572

Γ (2120572 + 1)

1199062(119909 119905)

= 119869120572119905(ℎ (119909) (119906

1)119909119909

(119909))

= ℎ (119909) [ℎ10158401015840

(119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1)

(12)

After writing these polynomials in (10) the solution 119906(119909 119905) isgiven by

119906 (119909 119905)

= 1198911(119909) + 119891 (119909)

119905120572

Γ (120572 + 1)+ ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ (119909) 11989110158401015840

(119909)1199052120572

Γ (2120572 + 1)

+ ℎ (119909) [ℎ10158401015840

(119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(13)

If we arrange it with respect to like powers of 119905 then we get

119906 (119909 119905)

= 1198911(119909) + [119891 (119909) + ℎ (119909) 119891

10158401015840

1(119909)]

119905120572

Γ (120572 + 1)

+ ℎ (119909) [11989110158401015840

(119909) + ℎ10158401015840 (119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

4 Advances in Mathematical Physics

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(14)

To determine the unknown source function first we expandthe boundary conditions 119906(0 119905) = ℎ

1(119905) and 119906

119909(0 119905) =

ℎ2(119905) into the following series for the space whose bases are

119905119899120572Γ(119899120572 + 1)infin

119899=0 0 lt 120572 le 1

ℎ1(119905) = ℎ

1(0) + ℎ1015840

1(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158401(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(15)

ℎ2(119905) = ℎ

2(0) + ℎ1015840

2(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158402(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(16)

On the other hand if we rewrite the boundary conditions119906(0 119905) and 119906

119909(0 119905) from (14) then we have

ℎ1(119905) = 119891

1(0) + [119891 (0) + ℎ (0) 119891

10158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0)+ ℎ (0) 119891

(119894V)1

(0)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(17)

ℎ2(119905) = 1198911015840

1(0) + [1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ (0) 119891101584010158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ1015840 (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0)

+ 2ℎ10158401015840 (0) 11989110158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+ 2ℎ1015840 (0) 119891(119894V)1

(0)

+ ℎ (0) 119891(V)1

(0)]1199052120572

Γ (2120572 + 1)

+ ℎ1015840 (0) [ℎ10158401015840

(0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0) ]

+ ℎ (0) [ℎ101584010158401015840

(0) 11989110158401015840

(0) + 3ℎ10158401015840 (0) 119891101584010158401015840

(0)

+ 3ℎ1015840 (0) 119891(119894V)

(0)

+ ℎ (0) 119891(V)

(0) ]1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(18)

Equating (15) and (17) yields the following

ℎ1(0) = 119891

1(0)

ℎ10158401(0) = 119891 (0) + ℎ (0) 119891

10158401015840

1(0)

ℎ101584010158401(0) = ℎ (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

(19)

and equating (16) and (18) yields the following

ℎ2(0) = 1198911015840

1(0)

ℎ10158402(0) = 1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0) + ℎ (0) 119891

101584010158401015840

1(0)

(20)

ℎ101584010158402(0) = ℎ1015840 (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0) + 2ℎ10158401015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+2ℎ1015840 (0) 119891(119894V)1

(0) + ℎ (0) 119891(V)1

(0)]

(21)

Using the above data in the following Taylor series expansionof unknown function 119891(119909) we get

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2+ 119891101584010158401015840 (0)

1199093

3+ sdot sdot sdot

(22)

Consequently we determine 119891(119909) as follows

119891 (119909) = [ℎ10158401(0) minus ℎ (0) 119891

10158401015840

1(0)]

+ [ℎ10158402(0) minus ℎ1015840 (0) 119891

10158401015840

1(0) minus ℎ (0) 119891

101584010158401015840

1(0)] 119909

Advances in Mathematical Physics 5

+ [ℎ101584010158401(0)

ℎ (0)minus ℎ10158401015840 (0) 119891

1015840

1(0) minus ℎ1015840 (0) 119891

10158401015840

1(0)

minus ℎ1015840 (0) 119891101584010158401015840

1(0) minus ℎ (0) 119891

(119894V)1

(0)]1199092

2+ sdot sdot sdot

(23)

where ℎ(0) = 0

22 Determination of Unknown Source Functions Dependingon 119905 Weconsider the following inverse problemof determin-ing the source function 119891(119905)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119905)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(24)

where ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) ℎ(119909) 119891

1(119909) isin 119862infin[0infin) and

119891(119905) isin 1198621120583[0infin) 120583 ge minus1 As in the previous case we apply

ADM to determine the unknown function 119891(119905)First to reduce the problem we define new functions in

the following form

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (25)

Then the reduced problem is given as follows

119863120572119905V (119909 119905) = ℎ (119909) V

119909119909(119909 119905) (26)

with the following initial and mixed boundary conditions

V (119909 0) = 1198911(119909) minus 119908 (0)

V (0 119905) = ℎ1(119905) minus 119908 (119905)

V119909(0 119905) = ℎ

2(119905)

(27)

By using ADM as in the previous section we determine thefunction 119908(119905) which leads to the source function 119891(119905) Let usapply 119869120572

119905to both sides of (26) as shown below

119869120572119905119863120572119905V (119909 119905) = 119869120572

119905(ℎ (119909) V

119909119909(119909 119905)) (28)

Then we get

V (119909 119905) = V (119909 0) + 119869120572119905(ℎ (119909) V

119909119909(119909 119905)) (29)

Now we define the solution V(119909 119905) by the following decom-position series according to ADM

V (119909 119905) =infin

sum119899=0

V119899(119909 119905) (30)

Substituting (30) into (29) we obtain

V0(119909 119905) = V (119909 0)

V119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (V

119899)119909119909

(119909 119905)) 119899 = 0 1 (31)

Hence the recurrence scheme is obtained as followsV0(119909 119905) = V (119909 0) = 119891

1(119909) minus 119908 (0)

V1(119909 119905) = 119869120572

119905(ℎ (119909) (V

0)119909119909

(119909)) = ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

V2(119909 119905) = 119869120572

119905(ℎ (119909) (V

1)119909119909

(119909)) = ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)

(32)

Consequently from (30) the solution V(119909 119905) is given as shownbelow

V (119909 119905) = V0(119909 119905) + V

1(119909 119905) + V

2(119909 119905) + sdot sdot sdot

= 1198911(119909) minus 119908 (0) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(33)

By using the boundary condition V(0 119905) = ℎ1(119905) + 119908(119905) and

119908(0) = 0 we have

119908 (119905) = 1198911(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(34)

which implies the following

119869120572119905119891 (119905) = 119891

1(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(35)

Since 119863120572119905119908(119909 119905) = 119863120572

119905119869120572119905119891(119905) = 119891(119905) we obtain the source

function 119891(119905) as follows

119891 (119905) = 119863120572119905[1198911(119909) minus ℎ

1(119905) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot ]

(36)

3 Examples

Example 1 We consider the inverse problem of determiningsource function 119891(119909) in the following one-dimensional frac-tional heat-like PDE

119863120572119905119906 (119909 119905) = 2119906

119909119909(119909 119905) + 119891 (119909) 119909 gt 0 0 lt 120572 le 1 119905 gt 0

(37)

6 Advances in Mathematical Physics

subject to the following initial and nonhomogeneous mixedboundary conditions

119906 (119909 0) = 119890119909 + sin119909

119906 (0 119905) = 1198902119905

119906119909(0 119905) = 1198902119905+1

(38)

Now let us apply the time-dependent Riemann Liouvillefractional integral operator 119869120572

119905to both sides of (37)

119869120572119905119863120572119905119906 (119909 119905) = 2119869120572

119905119906119909119909

(119909 119905) + 119869120572119905119891 (119909) (39)

which implies

119906 (119909 119905) minus 119906 (119909 0) = 2119869120572119905119906119909119909

(119909 119905) + 119891 (119909) 119869120572

119905(1) (40)

Then from the initial condition we get

119906 (119909 119905) = 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119869120572119905119906119909119909

(119909 119905) (41)

Now we apply ADM to the problem In (41) the sum of thefirst three terms is identified as 119906

0 So

1199060= 119890119909 + sin119909 + 119891 (119909)

119905120572

Γ (120572 + 1)

119906119896+1

= 2119869120572119905(119906119896)119909119909

(119909 119905) 119896 ge 0

(42)

For 119896 = 0 we have

1199061= 2119869120572119905(1199060)119909119909

(119909 119905)

= 2119890119909119905120572

Γ (120572 + 1)minus 2 sin119909 119905120572

Γ (120572 + 1)

+ 211989110158401015840 (119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(43)

similarly for 119896 = 1 we have

1199062= 2119869120572119905(1199061)119909119909

(119909 119905)

= 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(44)

and for 119896 = 2 we have

1199063=2119869120572119905(1199062)119909119909

(119909 119905)

= 81198901199091199053120572

Γ (3120572 + 1)minus 8 sin119909 1199053120572

Γ (3120572 + 1)

+ 8119891(V119894) (119909)1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(45)

Then using ADM polynomials we get the solution 119906(119909 119905) asfollows119906 (119909 119905) = 119906

0+ 1199061+ 1199062+ 1199063+ sdot sdot sdot

= 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119890119909

119905120572

Γ (120572 + 1)

minus 2 sin119909 119905120572

Γ (120572 + 1)+ 211989110158401015840 (119909)

1199052120572

Γ (2120572 + 1)

+ 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ 8119890119909

1199053120572

Γ (3120572 + 1)

minus 8 sin119909 1199053120572

Γ (3120572 + 1)+ 8119891(V119894) (119909)

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(46)

After arranging it according to like powers of 119905 we have

119906 (119909 119905) = 119890119909 + sin119909 +119905120572

Γ (120572 + 1)[119891 (119909) + 2119890119909 minus 2 sin119909]

+1199052120572

Γ (2120572 + 1)[211989110158401015840 (119909) + 4119890119909 + 4 sin119909]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (119909) + 8119890119909 minus 8 sin119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (119909) + 16119890119909 + 16 sin119909] + sdot sdot sdot

(47)

Now by applying the boundary condition given in (38) weobtain

119906 (0 119905) = 1 +119905120572

Γ (120572 + 1)[119891 (0) + 2] +

1199052120572

Γ (2120572 + 1)[211989110158401015840 (0) + 4]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (0) + 8]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (0) + 16] + sdot sdot sdot

(48)

From (15) it must be equal to the following Taylorseries expansion of 1198902119905 in the space whose bases are119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 = 1 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(49)

Hence from the equality of the coefficients of correspondingterms we get

119891 (0) = 11989110158401015840 (0) = 119891(119894V) (0) = 119891(V119894) (0) = sdot sdot sdot = 0 (50)

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Advances in Mathematical Physics 3

21 Determination of Unknown Source Functions Dependingon 119909 We consider the following inverse problem of deter-mining the source function 119891(119909)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119909)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(7)

where the functions ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) and ℎ(119909) 119891(119909)

1198911(119909) isin 119862infin[0infin) In order to determine the source function

for this kind of inverse problems we apply ADM Firstwe apply the time-dependent Riemann-Liouville fractionalintegral operator 119869120572

119905to both sides of (7) to get rid of fractional

derivative119863120572119905as shown below

119869120572119905119863120572119905119906 (119909 119905) = 119869120572

119905(ℎ (119909) 119906

119909119909(119909 119905)) + 119869120572

119905119891 (119909) (8)

Then we get

119906 (119909 119905) = 119906 (119909 0) + 119869120572119905119891 (119909) + 119869120572

119905(ℎ (119909) 119906

119909119909(119909 119905)) (9)

In ADM the solution 119906(119909 119905) is written in the following seriesform [9]

119906 (119909 119905) =infin

sum119899=0

119906119899(119909 119905) (10)

where 119906 and 119906119899 119899 isin N are defined in 119862infin[0infin) times 1198621

120583[0infin)

After substituting the decomposition (10) into (9) and settingthe recurrence scheme as follows

1199060(119909 119905) = 119906 (119909 0) + 119869120572

119905119891 (119909)

119906119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (119906

119899)119909119909

(119909 119905)) 119899 = 0 1 (11)

we get ADM polynomials below

1199060(119909 119905) = 119891

1(119909) + 119891 (119909)

119905120572

Γ (120572 + 1)

1199061(119909 119905) = 119869120572

119905(ℎ (119909) (119906

0)119909119909

(119909))

= ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ (119909) 11989110158401015840

(119909)1199052120572

Γ (2120572 + 1)

1199062(119909 119905)

= 119869120572119905(ℎ (119909) (119906

1)119909119909

(119909))

= ℎ (119909) [ℎ10158401015840

(119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1)

(12)

After writing these polynomials in (10) the solution 119906(119909 119905) isgiven by

119906 (119909 119905)

= 1198911(119909) + 119891 (119909)

119905120572

Γ (120572 + 1)+ ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ (119909) 11989110158401015840

(119909)1199052120572

Γ (2120572 + 1)

+ ℎ (119909) [ℎ10158401015840

(119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(13)

If we arrange it with respect to like powers of 119905 then we get

119906 (119909 119905)

= 1198911(119909) + [119891 (119909) + ℎ (119909) 119891

10158401015840

1(119909)]

119905120572

Γ (120572 + 1)

+ ℎ (119909) [11989110158401015840

(119909) + ℎ10158401015840 (119909) 1198911015840

1(119909) + ℎ1015840 (119909) 119891

10158401015840

1(119909)

+ ℎ1015840 (119909) 119891101584010158401015840

1(119909) + ℎ (119909) 119891

(119894V)1

(119909)]1199052120572

Γ (2120572 + 1)

4 Advances in Mathematical Physics

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(14)

To determine the unknown source function first we expandthe boundary conditions 119906(0 119905) = ℎ

1(119905) and 119906

119909(0 119905) =

ℎ2(119905) into the following series for the space whose bases are

119905119899120572Γ(119899120572 + 1)infin

119899=0 0 lt 120572 le 1

ℎ1(119905) = ℎ

1(0) + ℎ1015840

1(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158401(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(15)

ℎ2(119905) = ℎ

2(0) + ℎ1015840

2(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158402(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(16)

On the other hand if we rewrite the boundary conditions119906(0 119905) and 119906

119909(0 119905) from (14) then we have

ℎ1(119905) = 119891

1(0) + [119891 (0) + ℎ (0) 119891

10158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0)+ ℎ (0) 119891

(119894V)1

(0)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(17)

ℎ2(119905) = 1198911015840

1(0) + [1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ (0) 119891101584010158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ1015840 (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0)

+ 2ℎ10158401015840 (0) 11989110158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+ 2ℎ1015840 (0) 119891(119894V)1

(0)

+ ℎ (0) 119891(V)1

(0)]1199052120572

Γ (2120572 + 1)

+ ℎ1015840 (0) [ℎ10158401015840

(0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0) ]

+ ℎ (0) [ℎ101584010158401015840

(0) 11989110158401015840

(0) + 3ℎ10158401015840 (0) 119891101584010158401015840

(0)

+ 3ℎ1015840 (0) 119891(119894V)

(0)

+ ℎ (0) 119891(V)

(0) ]1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(18)

Equating (15) and (17) yields the following

ℎ1(0) = 119891

1(0)

ℎ10158401(0) = 119891 (0) + ℎ (0) 119891

10158401015840

1(0)

ℎ101584010158401(0) = ℎ (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

(19)

and equating (16) and (18) yields the following

ℎ2(0) = 1198911015840

1(0)

ℎ10158402(0) = 1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0) + ℎ (0) 119891

101584010158401015840

1(0)

(20)

ℎ101584010158402(0) = ℎ1015840 (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0) + 2ℎ10158401015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+2ℎ1015840 (0) 119891(119894V)1

(0) + ℎ (0) 119891(V)1

(0)]

(21)

Using the above data in the following Taylor series expansionof unknown function 119891(119909) we get

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2+ 119891101584010158401015840 (0)

1199093

3+ sdot sdot sdot

(22)

Consequently we determine 119891(119909) as follows

119891 (119909) = [ℎ10158401(0) minus ℎ (0) 119891

10158401015840

1(0)]

+ [ℎ10158402(0) minus ℎ1015840 (0) 119891

10158401015840

1(0) minus ℎ (0) 119891

101584010158401015840

1(0)] 119909

Advances in Mathematical Physics 5

+ [ℎ101584010158401(0)

ℎ (0)minus ℎ10158401015840 (0) 119891

1015840

1(0) minus ℎ1015840 (0) 119891

10158401015840

1(0)

minus ℎ1015840 (0) 119891101584010158401015840

1(0) minus ℎ (0) 119891

(119894V)1

(0)]1199092

2+ sdot sdot sdot

(23)

where ℎ(0) = 0

22 Determination of Unknown Source Functions Dependingon 119905 Weconsider the following inverse problemof determin-ing the source function 119891(119905)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119905)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(24)

where ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) ℎ(119909) 119891

1(119909) isin 119862infin[0infin) and

119891(119905) isin 1198621120583[0infin) 120583 ge minus1 As in the previous case we apply

ADM to determine the unknown function 119891(119905)First to reduce the problem we define new functions in

the following form

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (25)

Then the reduced problem is given as follows

119863120572119905V (119909 119905) = ℎ (119909) V

119909119909(119909 119905) (26)

with the following initial and mixed boundary conditions

V (119909 0) = 1198911(119909) minus 119908 (0)

V (0 119905) = ℎ1(119905) minus 119908 (119905)

V119909(0 119905) = ℎ

2(119905)

(27)

By using ADM as in the previous section we determine thefunction 119908(119905) which leads to the source function 119891(119905) Let usapply 119869120572

119905to both sides of (26) as shown below

119869120572119905119863120572119905V (119909 119905) = 119869120572

119905(ℎ (119909) V

119909119909(119909 119905)) (28)

Then we get

V (119909 119905) = V (119909 0) + 119869120572119905(ℎ (119909) V

119909119909(119909 119905)) (29)

Now we define the solution V(119909 119905) by the following decom-position series according to ADM

V (119909 119905) =infin

sum119899=0

V119899(119909 119905) (30)

Substituting (30) into (29) we obtain

V0(119909 119905) = V (119909 0)

V119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (V

119899)119909119909

(119909 119905)) 119899 = 0 1 (31)

Hence the recurrence scheme is obtained as followsV0(119909 119905) = V (119909 0) = 119891

1(119909) minus 119908 (0)

V1(119909 119905) = 119869120572

119905(ℎ (119909) (V

0)119909119909

(119909)) = ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

V2(119909 119905) = 119869120572

119905(ℎ (119909) (V

1)119909119909

(119909)) = ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)

(32)

Consequently from (30) the solution V(119909 119905) is given as shownbelow

V (119909 119905) = V0(119909 119905) + V

1(119909 119905) + V

2(119909 119905) + sdot sdot sdot

= 1198911(119909) minus 119908 (0) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(33)

By using the boundary condition V(0 119905) = ℎ1(119905) + 119908(119905) and

119908(0) = 0 we have

119908 (119905) = 1198911(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(34)

which implies the following

119869120572119905119891 (119905) = 119891

1(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(35)

Since 119863120572119905119908(119909 119905) = 119863120572

119905119869120572119905119891(119905) = 119891(119905) we obtain the source

function 119891(119905) as follows

119891 (119905) = 119863120572119905[1198911(119909) minus ℎ

1(119905) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot ]

(36)

3 Examples

Example 1 We consider the inverse problem of determiningsource function 119891(119909) in the following one-dimensional frac-tional heat-like PDE

119863120572119905119906 (119909 119905) = 2119906

119909119909(119909 119905) + 119891 (119909) 119909 gt 0 0 lt 120572 le 1 119905 gt 0

(37)

6 Advances in Mathematical Physics

subject to the following initial and nonhomogeneous mixedboundary conditions

119906 (119909 0) = 119890119909 + sin119909

119906 (0 119905) = 1198902119905

119906119909(0 119905) = 1198902119905+1

(38)

Now let us apply the time-dependent Riemann Liouvillefractional integral operator 119869120572

119905to both sides of (37)

119869120572119905119863120572119905119906 (119909 119905) = 2119869120572

119905119906119909119909

(119909 119905) + 119869120572119905119891 (119909) (39)

which implies

119906 (119909 119905) minus 119906 (119909 0) = 2119869120572119905119906119909119909

(119909 119905) + 119891 (119909) 119869120572

119905(1) (40)

Then from the initial condition we get

119906 (119909 119905) = 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119869120572119905119906119909119909

(119909 119905) (41)

Now we apply ADM to the problem In (41) the sum of thefirst three terms is identified as 119906

0 So

1199060= 119890119909 + sin119909 + 119891 (119909)

119905120572

Γ (120572 + 1)

119906119896+1

= 2119869120572119905(119906119896)119909119909

(119909 119905) 119896 ge 0

(42)

For 119896 = 0 we have

1199061= 2119869120572119905(1199060)119909119909

(119909 119905)

= 2119890119909119905120572

Γ (120572 + 1)minus 2 sin119909 119905120572

Γ (120572 + 1)

+ 211989110158401015840 (119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(43)

similarly for 119896 = 1 we have

1199062= 2119869120572119905(1199061)119909119909

(119909 119905)

= 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(44)

and for 119896 = 2 we have

1199063=2119869120572119905(1199062)119909119909

(119909 119905)

= 81198901199091199053120572

Γ (3120572 + 1)minus 8 sin119909 1199053120572

Γ (3120572 + 1)

+ 8119891(V119894) (119909)1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(45)

Then using ADM polynomials we get the solution 119906(119909 119905) asfollows119906 (119909 119905) = 119906

0+ 1199061+ 1199062+ 1199063+ sdot sdot sdot

= 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119890119909

119905120572

Γ (120572 + 1)

minus 2 sin119909 119905120572

Γ (120572 + 1)+ 211989110158401015840 (119909)

1199052120572

Γ (2120572 + 1)

+ 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ 8119890119909

1199053120572

Γ (3120572 + 1)

minus 8 sin119909 1199053120572

Γ (3120572 + 1)+ 8119891(V119894) (119909)

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(46)

After arranging it according to like powers of 119905 we have

119906 (119909 119905) = 119890119909 + sin119909 +119905120572

Γ (120572 + 1)[119891 (119909) + 2119890119909 minus 2 sin119909]

+1199052120572

Γ (2120572 + 1)[211989110158401015840 (119909) + 4119890119909 + 4 sin119909]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (119909) + 8119890119909 minus 8 sin119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (119909) + 16119890119909 + 16 sin119909] + sdot sdot sdot

(47)

Now by applying the boundary condition given in (38) weobtain

119906 (0 119905) = 1 +119905120572

Γ (120572 + 1)[119891 (0) + 2] +

1199052120572

Γ (2120572 + 1)[211989110158401015840 (0) + 4]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (0) + 8]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (0) + 16] + sdot sdot sdot

(48)

From (15) it must be equal to the following Taylorseries expansion of 1198902119905 in the space whose bases are119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 = 1 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(49)

Hence from the equality of the coefficients of correspondingterms we get

119891 (0) = 11989110158401015840 (0) = 119891(119894V) (0) = 119891(V119894) (0) = sdot sdot sdot = 0 (50)

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Advances in Mathematical Physics

+ [ℎ10158401015840 (119909) 11989110158401015840

(119909) + 2ℎ1015840 (119909) 119891101584010158401015840

(119909)

+ ℎ (119909) 119891(119894V)

(119909)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(14)

To determine the unknown source function first we expandthe boundary conditions 119906(0 119905) = ℎ

1(119905) and 119906

119909(0 119905) =

ℎ2(119905) into the following series for the space whose bases are

119905119899120572Γ(119899120572 + 1)infin

119899=0 0 lt 120572 le 1

ℎ1(119905) = ℎ

1(0) + ℎ1015840

1(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158401(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(15)

ℎ2(119905) = ℎ

2(0) + ℎ1015840

2(0)

119905120572

Γ (120572 + 1)+ ℎ101584010158402(0)

1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(16)

On the other hand if we rewrite the boundary conditions119906(0 119905) and 119906

119909(0 119905) from (14) then we have

ℎ1(119905) = 119891

1(0) + [119891 (0) + ℎ (0) 119891

10158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0)+ ℎ (0) 119891

(119894V)1

(0)]1199052120572

Γ (2120572 + 1)

+ [ℎ10158401015840 (0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0)]1199053120572

Γ (3120572 + 1) + sdot sdot sdot

(17)

ℎ2(119905) = 1198911015840

1(0) + [1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ (0) 119891101584010158401015840

1(0)]

119905120572

Γ (120572 + 1)

+ ℎ1015840 (0) [11989110158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0)

+ 2ℎ10158401015840 (0) 11989110158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+ 2ℎ1015840 (0) 119891(119894V)1

(0)

+ ℎ (0) 119891(V)1

(0)]1199052120572

Γ (2120572 + 1)

+ ℎ1015840 (0) [ℎ10158401015840

(0) 11989110158401015840

(0) + 2ℎ1015840 (0) 119891101584010158401015840

(0)

+ ℎ (0) 119891(119894V)

(0) ]

+ ℎ (0) [ℎ101584010158401015840

(0) 11989110158401015840

(0) + 3ℎ10158401015840 (0) 119891101584010158401015840

(0)

+ 3ℎ1015840 (0) 119891(119894V)

(0)

+ ℎ (0) 119891(V)

(0) ]1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(18)

Equating (15) and (17) yields the following

ℎ1(0) = 119891

1(0)

ℎ10158401(0) = 119891 (0) + ℎ (0) 119891

10158401015840

1(0)

ℎ101584010158401(0) = ℎ (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

(19)

and equating (16) and (18) yields the following

ℎ2(0) = 1198911015840

1(0)

ℎ10158402(0) = 1198911015840 (0) + ℎ1015840 (0) 119891

10158401015840

1(0) + ℎ (0) 119891

101584010158401015840

1(0)

(20)

ℎ101584010158402(0) = ℎ1015840 (0) [119891

10158401015840

(0) + ℎ10158401015840 (0) 1198911015840

1(0) + ℎ1015840 (0) 119891

10158401015840

1(0)

+ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ (0) 119891

(119894V)1

(0)]

+ ℎ (0) [119891101584010158401015840

(0) + ℎ101584010158401015840 (0) 1198911015840

1(0) + 2ℎ10158401015840 (0) 119891

10158401015840

1(0)

+ ℎ1015840 (0) 119891101584010158401015840

1(0) + ℎ10158401015840 (0) 119891

101584010158401015840

1(0)

+2ℎ1015840 (0) 119891(119894V)1

(0) + ℎ (0) 119891(V)1

(0)]

(21)

Using the above data in the following Taylor series expansionof unknown function 119891(119909) we get

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2+ 119891101584010158401015840 (0)

1199093

3+ sdot sdot sdot

(22)

Consequently we determine 119891(119909) as follows

119891 (119909) = [ℎ10158401(0) minus ℎ (0) 119891

10158401015840

1(0)]

+ [ℎ10158402(0) minus ℎ1015840 (0) 119891

10158401015840

1(0) minus ℎ (0) 119891

101584010158401015840

1(0)] 119909

Advances in Mathematical Physics 5

+ [ℎ101584010158401(0)

ℎ (0)minus ℎ10158401015840 (0) 119891

1015840

1(0) minus ℎ1015840 (0) 119891

10158401015840

1(0)

minus ℎ1015840 (0) 119891101584010158401015840

1(0) minus ℎ (0) 119891

(119894V)1

(0)]1199092

2+ sdot sdot sdot

(23)

where ℎ(0) = 0

22 Determination of Unknown Source Functions Dependingon 119905 Weconsider the following inverse problemof determin-ing the source function 119891(119905)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119905)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(24)

where ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) ℎ(119909) 119891

1(119909) isin 119862infin[0infin) and

119891(119905) isin 1198621120583[0infin) 120583 ge minus1 As in the previous case we apply

ADM to determine the unknown function 119891(119905)First to reduce the problem we define new functions in

the following form

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (25)

Then the reduced problem is given as follows

119863120572119905V (119909 119905) = ℎ (119909) V

119909119909(119909 119905) (26)

with the following initial and mixed boundary conditions

V (119909 0) = 1198911(119909) minus 119908 (0)

V (0 119905) = ℎ1(119905) minus 119908 (119905)

V119909(0 119905) = ℎ

2(119905)

(27)

By using ADM as in the previous section we determine thefunction 119908(119905) which leads to the source function 119891(119905) Let usapply 119869120572

119905to both sides of (26) as shown below

119869120572119905119863120572119905V (119909 119905) = 119869120572

119905(ℎ (119909) V

119909119909(119909 119905)) (28)

Then we get

V (119909 119905) = V (119909 0) + 119869120572119905(ℎ (119909) V

119909119909(119909 119905)) (29)

Now we define the solution V(119909 119905) by the following decom-position series according to ADM

V (119909 119905) =infin

sum119899=0

V119899(119909 119905) (30)

Substituting (30) into (29) we obtain

V0(119909 119905) = V (119909 0)

V119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (V

119899)119909119909

(119909 119905)) 119899 = 0 1 (31)

Hence the recurrence scheme is obtained as followsV0(119909 119905) = V (119909 0) = 119891

1(119909) minus 119908 (0)

V1(119909 119905) = 119869120572

119905(ℎ (119909) (V

0)119909119909

(119909)) = ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

V2(119909 119905) = 119869120572

119905(ℎ (119909) (V

1)119909119909

(119909)) = ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)

(32)

Consequently from (30) the solution V(119909 119905) is given as shownbelow

V (119909 119905) = V0(119909 119905) + V

1(119909 119905) + V

2(119909 119905) + sdot sdot sdot

= 1198911(119909) minus 119908 (0) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(33)

By using the boundary condition V(0 119905) = ℎ1(119905) + 119908(119905) and

119908(0) = 0 we have

119908 (119905) = 1198911(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(34)

which implies the following

119869120572119905119891 (119905) = 119891

1(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(35)

Since 119863120572119905119908(119909 119905) = 119863120572

119905119869120572119905119891(119905) = 119891(119905) we obtain the source

function 119891(119905) as follows

119891 (119905) = 119863120572119905[1198911(119909) minus ℎ

1(119905) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot ]

(36)

3 Examples

Example 1 We consider the inverse problem of determiningsource function 119891(119909) in the following one-dimensional frac-tional heat-like PDE

119863120572119905119906 (119909 119905) = 2119906

119909119909(119909 119905) + 119891 (119909) 119909 gt 0 0 lt 120572 le 1 119905 gt 0

(37)

6 Advances in Mathematical Physics

subject to the following initial and nonhomogeneous mixedboundary conditions

119906 (119909 0) = 119890119909 + sin119909

119906 (0 119905) = 1198902119905

119906119909(0 119905) = 1198902119905+1

(38)

Now let us apply the time-dependent Riemann Liouvillefractional integral operator 119869120572

119905to both sides of (37)

119869120572119905119863120572119905119906 (119909 119905) = 2119869120572

119905119906119909119909

(119909 119905) + 119869120572119905119891 (119909) (39)

which implies

119906 (119909 119905) minus 119906 (119909 0) = 2119869120572119905119906119909119909

(119909 119905) + 119891 (119909) 119869120572

119905(1) (40)

Then from the initial condition we get

119906 (119909 119905) = 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119869120572119905119906119909119909

(119909 119905) (41)

Now we apply ADM to the problem In (41) the sum of thefirst three terms is identified as 119906

0 So

1199060= 119890119909 + sin119909 + 119891 (119909)

119905120572

Γ (120572 + 1)

119906119896+1

= 2119869120572119905(119906119896)119909119909

(119909 119905) 119896 ge 0

(42)

For 119896 = 0 we have

1199061= 2119869120572119905(1199060)119909119909

(119909 119905)

= 2119890119909119905120572

Γ (120572 + 1)minus 2 sin119909 119905120572

Γ (120572 + 1)

+ 211989110158401015840 (119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(43)

similarly for 119896 = 1 we have

1199062= 2119869120572119905(1199061)119909119909

(119909 119905)

= 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(44)

and for 119896 = 2 we have

1199063=2119869120572119905(1199062)119909119909

(119909 119905)

= 81198901199091199053120572

Γ (3120572 + 1)minus 8 sin119909 1199053120572

Γ (3120572 + 1)

+ 8119891(V119894) (119909)1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(45)

Then using ADM polynomials we get the solution 119906(119909 119905) asfollows119906 (119909 119905) = 119906

0+ 1199061+ 1199062+ 1199063+ sdot sdot sdot

= 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119890119909

119905120572

Γ (120572 + 1)

minus 2 sin119909 119905120572

Γ (120572 + 1)+ 211989110158401015840 (119909)

1199052120572

Γ (2120572 + 1)

+ 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ 8119890119909

1199053120572

Γ (3120572 + 1)

minus 8 sin119909 1199053120572

Γ (3120572 + 1)+ 8119891(V119894) (119909)

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(46)

After arranging it according to like powers of 119905 we have

119906 (119909 119905) = 119890119909 + sin119909 +119905120572

Γ (120572 + 1)[119891 (119909) + 2119890119909 minus 2 sin119909]

+1199052120572

Γ (2120572 + 1)[211989110158401015840 (119909) + 4119890119909 + 4 sin119909]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (119909) + 8119890119909 minus 8 sin119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (119909) + 16119890119909 + 16 sin119909] + sdot sdot sdot

(47)

Now by applying the boundary condition given in (38) weobtain

119906 (0 119905) = 1 +119905120572

Γ (120572 + 1)[119891 (0) + 2] +

1199052120572

Γ (2120572 + 1)[211989110158401015840 (0) + 4]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (0) + 8]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (0) + 16] + sdot sdot sdot

(48)

From (15) it must be equal to the following Taylorseries expansion of 1198902119905 in the space whose bases are119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 = 1 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(49)

Hence from the equality of the coefficients of correspondingterms we get

119891 (0) = 11989110158401015840 (0) = 119891(119894V) (0) = 119891(V119894) (0) = sdot sdot sdot = 0 (50)

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

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Stochastic AnalysisInternational Journal of

Advances in Mathematical Physics 5

+ [ℎ101584010158401(0)

ℎ (0)minus ℎ10158401015840 (0) 119891

1015840

1(0) minus ℎ1015840 (0) 119891

10158401015840

1(0)

minus ℎ1015840 (0) 119891101584010158401015840

1(0) minus ℎ (0) 119891

(119894V)1

(0)]1199092

2+ sdot sdot sdot

(23)

where ℎ(0) = 0

22 Determination of Unknown Source Functions Dependingon 119905 Weconsider the following inverse problemof determin-ing the source function 119891(119905)

119863120572119905119906 (119909 119905) = ℎ (119909) 119906

119909119909(119909 119905) + 119891 (119905)

119909 gt 0 119905 gt 0 0 lt 120572 le 1

119906 (119909 0) = 1198911(119909)

119906 (0 119905) = ℎ1(119905)

119906119909(0 119905) = ℎ

2(119905)

(24)

where ℎ1(119905) ℎ2(119905) isin 119862infin[0infin) ℎ(119909) 119891

1(119909) isin 119862infin[0infin) and

119891(119905) isin 1198621120583[0infin) 120583 ge minus1 As in the previous case we apply

ADM to determine the unknown function 119891(119905)First to reduce the problem we define new functions in

the following form

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (25)

Then the reduced problem is given as follows

119863120572119905V (119909 119905) = ℎ (119909) V

119909119909(119909 119905) (26)

with the following initial and mixed boundary conditions

V (119909 0) = 1198911(119909) minus 119908 (0)

V (0 119905) = ℎ1(119905) minus 119908 (119905)

V119909(0 119905) = ℎ

2(119905)

(27)

By using ADM as in the previous section we determine thefunction 119908(119905) which leads to the source function 119891(119905) Let usapply 119869120572

119905to both sides of (26) as shown below

119869120572119905119863120572119905V (119909 119905) = 119869120572

119905(ℎ (119909) V

119909119909(119909 119905)) (28)

Then we get

V (119909 119905) = V (119909 0) + 119869120572119905(ℎ (119909) V

119909119909(119909 119905)) (29)

Now we define the solution V(119909 119905) by the following decom-position series according to ADM

V (119909 119905) =infin

sum119899=0

V119899(119909 119905) (30)

Substituting (30) into (29) we obtain

V0(119909 119905) = V (119909 0)

V119899+1

(119909 119905) = 119869120572119905(ℎ (119909) (V

119899)119909119909

(119909 119905)) 119899 = 0 1 (31)

Hence the recurrence scheme is obtained as followsV0(119909 119905) = V (119909 0) = 119891

1(119909) minus 119908 (0)

V1(119909 119905) = 119869120572

119905(ℎ (119909) (V

0)119909119909

(119909)) = ℎ (119909) 11989110158401015840

1(119909)

119905120572

Γ (120572 + 1)

V2(119909 119905) = 119869120572

119905(ℎ (119909) (V

1)119909119909

(119909)) = ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)

(32)

Consequently from (30) the solution V(119909 119905) is given as shownbelow

V (119909 119905) = V0(119909 119905) + V

1(119909 119905) + V

2(119909 119905) + sdot sdot sdot

= 1198911(119909) minus 119908 (0) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(33)

By using the boundary condition V(0 119905) = ℎ1(119905) + 119908(119905) and

119908(0) = 0 we have

119908 (119905) = 1198911(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(34)

which implies the following

119869120572119905119891 (119905) = 119891

1(0) minus ℎ

1(119905) + ℎ (0) 119891

10158401015840

1(0)

119905120572

Γ (120572 + 1)

+ ℎ2 (0) 119891(119894V)1

(0)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(35)

Since 119863120572119905119908(119909 119905) = 119863120572

119905119869120572119905119891(119905) = 119891(119905) we obtain the source

function 119891(119905) as follows

119891 (119905) = 119863120572119905[1198911(119909) minus ℎ

1(119905) + ℎ (119909) 119891

10158401015840

1(119909)

119905120572

Γ (120572 + 1)

+ ℎ2 (119909) 119891(119894V)1

(119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot ]

(36)

3 Examples

Example 1 We consider the inverse problem of determiningsource function 119891(119909) in the following one-dimensional frac-tional heat-like PDE

119863120572119905119906 (119909 119905) = 2119906

119909119909(119909 119905) + 119891 (119909) 119909 gt 0 0 lt 120572 le 1 119905 gt 0

(37)

6 Advances in Mathematical Physics

subject to the following initial and nonhomogeneous mixedboundary conditions

119906 (119909 0) = 119890119909 + sin119909

119906 (0 119905) = 1198902119905

119906119909(0 119905) = 1198902119905+1

(38)

Now let us apply the time-dependent Riemann Liouvillefractional integral operator 119869120572

119905to both sides of (37)

119869120572119905119863120572119905119906 (119909 119905) = 2119869120572

119905119906119909119909

(119909 119905) + 119869120572119905119891 (119909) (39)

which implies

119906 (119909 119905) minus 119906 (119909 0) = 2119869120572119905119906119909119909

(119909 119905) + 119891 (119909) 119869120572

119905(1) (40)

Then from the initial condition we get

119906 (119909 119905) = 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119869120572119905119906119909119909

(119909 119905) (41)

Now we apply ADM to the problem In (41) the sum of thefirst three terms is identified as 119906

0 So

1199060= 119890119909 + sin119909 + 119891 (119909)

119905120572

Γ (120572 + 1)

119906119896+1

= 2119869120572119905(119906119896)119909119909

(119909 119905) 119896 ge 0

(42)

For 119896 = 0 we have

1199061= 2119869120572119905(1199060)119909119909

(119909 119905)

= 2119890119909119905120572

Γ (120572 + 1)minus 2 sin119909 119905120572

Γ (120572 + 1)

+ 211989110158401015840 (119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(43)

similarly for 119896 = 1 we have

1199062= 2119869120572119905(1199061)119909119909

(119909 119905)

= 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(44)

and for 119896 = 2 we have

1199063=2119869120572119905(1199062)119909119909

(119909 119905)

= 81198901199091199053120572

Γ (3120572 + 1)minus 8 sin119909 1199053120572

Γ (3120572 + 1)

+ 8119891(V119894) (119909)1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(45)

Then using ADM polynomials we get the solution 119906(119909 119905) asfollows119906 (119909 119905) = 119906

0+ 1199061+ 1199062+ 1199063+ sdot sdot sdot

= 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119890119909

119905120572

Γ (120572 + 1)

minus 2 sin119909 119905120572

Γ (120572 + 1)+ 211989110158401015840 (119909)

1199052120572

Γ (2120572 + 1)

+ 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ 8119890119909

1199053120572

Γ (3120572 + 1)

minus 8 sin119909 1199053120572

Γ (3120572 + 1)+ 8119891(V119894) (119909)

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(46)

After arranging it according to like powers of 119905 we have

119906 (119909 119905) = 119890119909 + sin119909 +119905120572

Γ (120572 + 1)[119891 (119909) + 2119890119909 minus 2 sin119909]

+1199052120572

Γ (2120572 + 1)[211989110158401015840 (119909) + 4119890119909 + 4 sin119909]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (119909) + 8119890119909 minus 8 sin119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (119909) + 16119890119909 + 16 sin119909] + sdot sdot sdot

(47)

Now by applying the boundary condition given in (38) weobtain

119906 (0 119905) = 1 +119905120572

Γ (120572 + 1)[119891 (0) + 2] +

1199052120572

Γ (2120572 + 1)[211989110158401015840 (0) + 4]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (0) + 8]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (0) + 16] + sdot sdot sdot

(48)

From (15) it must be equal to the following Taylorseries expansion of 1198902119905 in the space whose bases are119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 = 1 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(49)

Hence from the equality of the coefficients of correspondingterms we get

119891 (0) = 11989110158401015840 (0) = 119891(119894V) (0) = 119891(V119894) (0) = sdot sdot sdot = 0 (50)

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

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Stochastic AnalysisInternational Journal of

6 Advances in Mathematical Physics

subject to the following initial and nonhomogeneous mixedboundary conditions

119906 (119909 0) = 119890119909 + sin119909

119906 (0 119905) = 1198902119905

119906119909(0 119905) = 1198902119905+1

(38)

Now let us apply the time-dependent Riemann Liouvillefractional integral operator 119869120572

119905to both sides of (37)

119869120572119905119863120572119905119906 (119909 119905) = 2119869120572

119905119906119909119909

(119909 119905) + 119869120572119905119891 (119909) (39)

which implies

119906 (119909 119905) minus 119906 (119909 0) = 2119869120572119905119906119909119909

(119909 119905) + 119891 (119909) 119869120572

119905(1) (40)

Then from the initial condition we get

119906 (119909 119905) = 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119869120572119905119906119909119909

(119909 119905) (41)

Now we apply ADM to the problem In (41) the sum of thefirst three terms is identified as 119906

0 So

1199060= 119890119909 + sin119909 + 119891 (119909)

119905120572

Γ (120572 + 1)

119906119896+1

= 2119869120572119905(119906119896)119909119909

(119909 119905) 119896 ge 0

(42)

For 119896 = 0 we have

1199061= 2119869120572119905(1199060)119909119909

(119909 119905)

= 2119890119909119905120572

Γ (120572 + 1)minus 2 sin119909 119905120572

Γ (120572 + 1)

+ 211989110158401015840 (119909)1199052120572

Γ (2120572 + 1)+ sdot sdot sdot

(43)

similarly for 119896 = 1 we have

1199062= 2119869120572119905(1199061)119909119909

(119909 119905)

= 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

(44)

and for 119896 = 2 we have

1199063=2119869120572119905(1199062)119909119909

(119909 119905)

= 81198901199091199053120572

Γ (3120572 + 1)minus 8 sin119909 1199053120572

Γ (3120572 + 1)

+ 8119891(V119894) (119909)1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(45)

Then using ADM polynomials we get the solution 119906(119909 119905) asfollows119906 (119909 119905) = 119906

0+ 1199061+ 1199062+ 1199063+ sdot sdot sdot

= 119890119909 + sin119909 + 119891 (119909)119905120572

Γ (120572 + 1)+ 2119890119909

119905120572

Γ (120572 + 1)

minus 2 sin119909 119905120572

Γ (120572 + 1)+ 211989110158401015840 (119909)

1199052120572

Γ (2120572 + 1)

+ 41198901199091199052120572

Γ (2120572 + 1)+ 4 sin119909 1199052120572

Γ (2120572 + 1)

+ 4119891(119894V) (119909)1199053120572

Γ (3120572 + 1)+ 8119890119909

1199053120572

Γ (3120572 + 1)

minus 8 sin119909 1199053120572

Γ (3120572 + 1)+ 8119891(V119894) (119909)

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(46)

After arranging it according to like powers of 119905 we have

119906 (119909 119905) = 119890119909 + sin119909 +119905120572

Γ (120572 + 1)[119891 (119909) + 2119890119909 minus 2 sin119909]

+1199052120572

Γ (2120572 + 1)[211989110158401015840 (119909) + 4119890119909 + 4 sin119909]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (119909) + 8119890119909 minus 8 sin119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (119909) + 16119890119909 + 16 sin119909] + sdot sdot sdot

(47)

Now by applying the boundary condition given in (38) weobtain

119906 (0 119905) = 1 +119905120572

Γ (120572 + 1)[119891 (0) + 2] +

1199052120572

Γ (2120572 + 1)[211989110158401015840 (0) + 4]

+1199053120572

Γ (3120572 + 1)[4119891(119894V) (0) + 8]

+1199054120572

Γ (4120572 + 1)[8119891(V119894) (0) + 16] + sdot sdot sdot

(48)

From (15) it must be equal to the following Taylorseries expansion of 1198902119905 in the space whose bases are119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 = 1 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(49)

Hence from the equality of the coefficients of correspondingterms we get

119891 (0) = 11989110158401015840 (0) = 119891(119894V) (0) = 119891(V119894) (0) = sdot sdot sdot = 0 (50)

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Advances in Mathematical Physics 7

From (47) we have

119906119909(119909 119905) = 119890119909 + cos119909 +

119905120572

Γ (120572 + 1)[1198911015840 (119909) + 2119890119909 minus 2 cos119909]

+1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (119909) + 4119890119909 + 4 cos119909]

+1199053120572

Γ (3120572 + 1)[4119891(V) (119909) + 8119890119909 minus 8 cos119909]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (119909) + 16119890119909 + 16 cos119909] + sdot sdot sdot

(51)

So

119906119909(0 119905) = 2 +

119905120572

Γ (120572 + 1)1198911015840 (0) +

1199052120572

Γ (2120572 + 1)[2119891101584010158401015840 (0) + 8]

+1199053120572

Γ (3120572 + 1)[4119891(V) (0)]

+1199054120572

Γ (4120572 + 1)[8119891(V119894119894) (0) + 32] + sdot sdot sdot

(52)

From the derivative boundary condition given in (38) it mustbe equal to the following series expansion of 1198902119905+1 in the spacewhose bases are 119905119899120572Γ(119899120572 + 1)

infin

119899=0 0 lt 120572 le 1

1198902119905 + 1 = 2 + 2119905120572

Γ (120572 + 1)+ 4

1199052120572

Γ (2120572 + 1)

+ 81199053120572

Γ (3120572 + 1)+ 16

1199054120572

Γ (4120572 + 1)+ sdot sdot sdot

(53)

Then we find the following data

1198911015840 (0) = 2 119891101584010158401015840 (0) = minus2 119891(V) (0) = 2

119891(V119894119894) (0) = minus2 (54)

Next using (50) and (54) we have the Taylor series expansionof 119891(119909) as follows

119891 (119909) = 119891 (0) + 1198911015840 (0) 119909 + 11989110158401015840 (0)1199092

2

+ 119891101584010158401015840 (0)1199093

3+ 119891(119894V) (0)

1199094

4+ sdot sdot sdot

= 2119909 minus 21199093

3+ 2

1199095

5+ 2

1199097

7+ sdot sdot sdot

(55)

That is

119891 (119909) = 2 [119909 minus1199093

3+1199095

5+1199097

7+ sdot sdot sdot ] (56)

which is the series expansion of the function 2 sin119909 Conse-quently we determine the source function 119891(119909) as

119891 (119909) = 2 sin119909 (57)

Example 2 We consider the inverse problem of determiningsource function 119891(119905) in the following one-dimensional frac-tional heat-like diffusion equation

119863120572119905119906 (119909 119905) =

1

21199092119906119909119909

(119909 119905) + 119891 (119905)

119909 gt 0 0 lt 120572 le 1 119905 gt 0

(58)

subject to following initial and mixed boundary conditions

119906 (119909 0) = 1199092 +1

2 119906 (0 119905) =

1198902119905

2 119906

119909(0 119905) = 0

(59)

Now let us determine the source function 119891(119905) To reduce theproblem we define new functions as follows

119908 (119905) = 119869120572119905119891 (119905)

119906 (119909 119905) = V (119909 119905) + 119908 (119905) (60)

Then our reduced problem is given as follows

119863120572119905V (119909 119905) =

1

21199092V119909119909

(119909 119905) 0 lt 120572 le 1 119905 gt 0

V (119909 0) = 119906 (119909 0) minus 119908 (0) = 1199092 +1

2

V (0 119905) = 119906 (0 119905) minus 119908 (119905) =1198902119905

2minus 119908 (119905)

V119909(0 119905) = 0

(61)

Applying 119869120572119905to both sides of (61) then we get

V (119909 119905) minus V (119909 0) =1

21199092119869120572119905V119909119909

(119909 119905) (62)

which implies

V (119909 119905) = 1199092 +1

2+1

21199092119869120572119905V119909119909

(119909 119905) (63)

By using ADM for (63) we obtain

V0= 1199092 +

1

2

V119896+1

=1

21199092119869120572119905(V119896)119909119909

(119909 119905) 119896 ge 0

(64)

Then for 119896 = 0 we get

V1=

1

21199092119869120572119905(V0)119909119909

(119909 119905)

= 1199092119905120572

Γ (120572 + 1)

(65)

similarly for 119896 = 1 we get

V2=

1

21199092119869120572119905(V1)119909119909

(119909 119905)

= 11990921199052120572

Γ (2120572 + 1)

(66)

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Advances in Mathematical Physics

and for 119896 = 2 we get

V3=

1

21199092119869120572119905(V2)119909119909

(119909 119905)

= 11990921199053120572

Γ (3120572 + 1)

(67)

As a result we get the solution V as follows

V (119909 119905) = V0+ V1+ V2+ V3+ sdot sdot sdot

= 1199092 +1

2+ 1199092

119905120572

Γ (120572 + 1)

+ 11990921199052120572

Γ (2120572 + 1)+ 1199092

1199053120572

Γ (3120572 + 1)+ sdot sdot sdot

=1

2+ 1199092 [1 +

119905120572

Γ (120572 + 1)+

1199052120572

Γ (2120572 + 1)

+1199053120572

Γ (3120572 + 1)+ sdot sdot sdot ]

(68)

Therefore from the boundary condition we have

119908 (119905) =1198902119905

2minus1

2 (69)

Using (69) in the definition 119863120572119905119908(119905) = 119863120572

119905119869120572119905119891(119905) =

119891(119905) finally we obtain the source function 119891(119905) as 119891(119905) =

119863120572119905((11989021199052) minus (12)) that is

119891 (119905) =1

2119863120572119905(1198902119905) (70)

Here

119863120572119905(1198902119905) = 119905minus120572119864

11minus120572(2119905) (71)

where 11986411minus120572

is Mittag-Leffler function with two parametersgiven as (6)

4 Conclusion

Thebest part of this method is that one can easily apply ADMto the fractional partial differential equations like applyingADM to ordinary differential equations

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] K B Oldham and J SpanierThe Fractional Calculus AcademicPress New York NY USA 1974

[3] K SMiller and B RossAn Introduction to the Fractional Calcu-lus and Fractional Differential Equations A Wiley-IntersciencePublication John Wiley amp Sons New York NY USA 1993

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

[5] S S Ray K S Chaudhuri and R K Bera ldquoAnalytical approx-imate solution of nonlinear dynamic system containing frac-tional derivative by modified decomposition methodrdquo AppliedMathematics and Computation vol 182 no 1 pp 544ndash5522006

[6] K Diethelm ldquoAn algorithm for the numerical solution of differ-ential equations of fractional orderrdquo Electronic Transactions onNumerical Analysis vol 5 no Mar pp 1ndash6 1997

[7] D Baleanu K Diethelm E Scalas and J J Trujillo FractionalCalculus Models and Numerical Methods vol 3 of Series onComplexity Nonlinearity and Chaos World Scientific Publish-ing Hackensack NJ USA 2012

[8] M Alipour and D Baleanu ldquoApproximate analytical solutionfor nonlinear system of fractional differential equations by BPsoperational matricesrdquo Advances in Mathematical Physics vol2013 Article ID 954015 9 pages 2013

[9] G Adomian Solving Frontier Problems of Physics The Decom-position Method vol 60 of Fundamental Theories of PhysicsKluwer Academic Publishers Dordrecht The Netherlands1994

[10] G Adomian ldquoSolutions of nonlinear P D Erdquo Applied Mathe-matics Letters vol 11 no 3 pp 121ndash123 1998

[11] K Abbaoui and Y Cherruault ldquoThe decomposition methodapplied to the Cauchy problemrdquo Kybernetes vol 28 no 1 pp68ndash74 1999

[12] D Kaya and A Yokus ldquoA numerical comparison of partialsolutions in the decompositionmethod for linear and nonlinearpartial differential equationsrdquo Mathematics and Computers inSimulation vol 60 no 6 pp 507ndash512 2002

[13] D A Murio ldquoTime fractional IHCP with Caputo fractionalderivativesrdquo Computers amp Mathematics with Applications vol56 no 9 pp 2371ndash2381 2008

[14] A N Bondarenko and D S Ivaschenko ldquoNumerical methodsfor solving inverse problems for time fractional diffusion equa-tion with variable coefficientrdquo Journal of Inverse and Ill-PosedProblems vol 17 no 5 pp 419ndash440 2009

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

[16] M Kirane and S A Malik ldquoDetermination of an unknownsource term and the temperature distribution for the linear heatequation involving fractional derivative in timerdquoApplied Math-ematics and Computation vol 218 no 1 pp 163ndash170 2011

[17] M Caputo ldquoLinear models of dissipation whose Q is almostfrequency independentmdashIIrdquo Geophysical Journal Internationalvol 13 no 5 pp 529ndash539 1967

[18] S T Mohyud-Din A Yıldırım and M Usman ldquoHomotopyanalysis method for fractional partial differential equationsrdquoInternational Journal of Physical Sciences vol 6 no 1 pp 136ndash145 2011

[19] G Mittag-Leffler ldquoSur la representation analytique drsquounebranche uniforme drsquoune fonction monogenerdquo Acta Mathemat-ica vol 29 no 1 pp 101ndash181 1905

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of