Research ArticleNumerical Treatment of the Modified TimeFractional Fokker-Planck Equation

Yuxin Zhang

School of Mathematics and Statistics Tianshui Normal University Tianshui 741001 China

Correspondence should be addressed to Yuxin Zhang zhangyuxin2006163com

Received 26 December 2013 Revised 8 February 2014 Accepted 16 February 2014 Published 27 March 2014

Academic Editor Adem Kilicman

Copyright copy 2014 Yuxin ZhangThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

A numerical method for themodified time fractional Fokker-Planck equation is proposed Stability and convergence of themethodare rigorously discussed by means of the Fourier method We prove that the difference scheme is unconditionally stable andconvergence order is 119874(120591 + ℎ4) where 120591 and ℎ are the temporal and spatial step sizes respectively Finally numerical results aregiven to confirm the theoretical analysis

1 Introduction

Fractional differential equations have attracted considerableattention due to their frequent appearance in various appli-cations in fluid mechanics biology physics and engineering[1 2] Usually fractional differential equations do not haveanalytic solutions and can only be solved by some semian-alytical and numeric techniques Recently several semiana-lytical methods such as variational iteration method homo-topy perturbationmethod Adomian decompositionmethodhomotopy analysis method and collocation method havebeen used to solve fractional differential equations [3ndash7]Meanwhile some effective numerical techniques are devel-oped see [8ndash17]

In the present paper we aremotivated to study the follow-ing modified time fractional Fokker-Planck equation [18]

120597119906 (119909 119905)

120597119905= [120581120572

1205972

1205971199092minus ]120572

120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 119871 0 le 119905 le 119879

(1)

119906 (119909 0) = 120601 (119909) 0 lt 119909 lt 119871

119906 (0 119905) = 1205931(119905) 0 le 119905 le 119879

119906 (119871 119905) = 1205932(119905) 0 le 119905 le 119879

(2)

where 120581120572gt 0 is a fractional diffusion coefficient and ]

120572gt 0

is a fractional friction coefficient RL1198631minus120572

0119905119906(119909 119905) denotes the

temporal Riemann-Liouville derivative operator defined as[2]

RL1198631minus120572

0119905119906 (119909 119905) =

1

Γ (120572)

120597

120597119905int

119905

0

119906 (119909 119904)

(119905 minus 119904)1minus120572119889119904 (3)

where Γ(sdot) is the gamma functionThe outline of the paper is as follows In Section 2 an

effective numerical method for solving the modified timefractional Fokker-Planck equation is proposed The solvabil-ity stability and convergence of the numerical method arediscussed in Sections 3 and 4 respectively In Section 5 wegive some numerical results demonstrating the convergenceorders of the numerical method Also a conclusion is given inSection 6

2 The Construction of Numerical Method

Let 119909119895= 119895ℎ (119895 = 0 1 119872) and 119905

119896= 119896120591 (119896 = 0 1 119873)

where ℎ = 119871119872 and 120591 = 119879119873 are the uniform spatial andtemporal mesh sizes respectively and119872119873 are two positiveintegers

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 282190 10 pageshttpdxdoiorg1011552014282190

2 Abstract and Applied Analysis

Lemma 1 Suppose that

[120581120572

1205972

1205971199092minus ]120572

120597

120597119909] 119906 (119909

119895 119905) = 119892 (119909

119895 119905) (4)

then a fourth-order difference scheme for the above equation isgiven as follows

A119906 (119909119895 119905) =L119892 (119909

119895 119905) + O (ℎ

4) (5)

whereA andL are two difference operators and are defined by

A = 120581120572cosh(

radic6]120572ℎ

6120581120572

)1205752

119909minus ]120572120583119909120575119909

L = 119868 +ℎ2

12(1205752

119909minus]120572

120581120572

120583119909120575119909)

(6)

in which 119868 is an unit operator and 120583119909120575119909and 1205752119909are average cen-

tral and second central difference operators with respect to 119909and are defined by

120583119909120575119909119906 (119909119895 119905) =

119906 (119909119895+1 119905) minus 119906 (119909

119895minus1 119905)

2ℎ

1205752

119909119906 (119909119895 119905) =

119906 (119909119895+1 119905) minus 2119906 (119909

119895 119905) + 119906 (119909

119895minus1 119905)

ℎ2

(7)

Proof In view of Taylor expansion we can obtain

119864119895(119905) = A119906 (119909

119895 119905) minusL119892 (119909

119895 119905)

= 120581120572

infin

sum

119896=0

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

times (

1205972119906 (119909119895 119905)

1205971199092+ℎ2

12

1205974119906 (119909119895 119905)

1205971199094+ 119874 (ℎ

4))

minus ]120572(

120597119906 (119909119895 119905)

120597119909+ℎ2

6

1205973119906 (119909119895 119905)

1205971199093+ 119874 (ℎ

4))

minus (119892 (119909119895 119905) +

ℎ2

12

1205972119892 (119909119895 119905)

1205971199092+ 119874 (ℎ

4))

+]120572ℎ2

12120581120572

(

120597119892 (119909119895 119905)

120597119909+ 119874 (ℎ

2))

= [(120581120572

1205972119906 (119909119895 119905)

1205971199092minus ]120572

120597119906 (119909119895 119905)

120597119909) minus 119892 (119909

119895 119905)]

minus]120572ℎ2

12120581120572

[(120581120572

1205973119906 (119909119895 119905)

1205971199093minus ]120572

1205972119906 (119909119895 119905)

1205971199092)

minus

120597119892 (119909119895 119905)

120597119909]

+ℎ2

12[(120581120572

1205974119906 (119909119895 119905)

1205971199094minus ]120572

1205973119906 (119909119895 119905)

1205971199093)

minus

1205972119892 (119909119895 119905)

1205971199092] + 119874 (ℎ

4)

(8)

Noting (4) we easily obtain

119864119895(119905) = 119874 (ℎ

4) (9)

This completes the proof

Combing (1) (4) and (5) we obtain

L sdot

120597119906 (119909119895 119905119896)

120597119905

= A sdot RL1198631minus120572

0119905119906 (119909119895 119905119896) +L sdot 119891 (119909

119895 119905119896) + O (ℎ

4)

(10)

Using the relation of the Riemann-Liouville fractionalderivative RL119863

1minus120572

0119905119906(119909 119905) and Grunwald-Letnikov fractional

derivative GL1198631minus120574

0119905119906(119909 119905) we can approximate the Riemann-

Liouville fractional derivative RL1198631minus120572

0119905119906(119909 119905) by [2]

RL1198631minus120572

0119905119906 (119909119895 119905119896) =

1

1205911minus120572

119896

sum

119898=0

120603(1minus120572)

119898119906 (119909119895 119905119896minus119898

) + 119874 (120591)

(11)

where 120603(1minus120572)119898

= (minus1)119898( 1minus120572119898)

For first-order derivative 120597119906(119909119895 119905119896)120597119905 we apply the

following backward difference scheme

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896) minus 119906 (119909

119895 119905119896minus1)

120591+ 119874 (120591) (12)

Let 119906119896119894be the numerical approximation of 119906(119909

119894 119905119896) substi-

tuting (11) and (12) into (10) and omitting the error termO(120591+

ℎ4) we can obtain the following difference scheme for solving

(1)

L sdot (119906119896

119895minus 119906119896minus1

119895) = 120591120572A sdot

119896

sum

119898=0

120603(1minus120572)

119898119906119896minus119898

119895+ 120591L sdot 119891

119896

119895

(13)

Abstract and Applied Analysis 3

The discrete form of above system is

(1

12minus 1198881minus 1198882+ 1198883) 119906119896

119895+1+ (

5

6+ 21198882) 119906119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 119906119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 119906119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 119906119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 119906119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(119906119896minus119898

119895+1minus 119906119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(119906119896minus119898

119895+1minus 2119906119896minus119898

119895+ 119906119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119891119896

119895+1+5

6120591119891119896

119895+ 120591 (

1

12+ 1198881)119891119896

119895minus1

(14)

where 1198881= ]120572ℎ24120581

120572 1198882= 120591120572120581120572120579ℎ2 1198883= 120591120572]1205722ℎ and 120579 =

cosh(radic6]120572ℎ6120581120572)

The initial and boundary conditions can be discretized as

1199060

119895= 120601 (119895ℎ) 119895 = 0 1 119872

119906119896

0= 1205931(119896120591) 119906

119896

119872= 1205932(119896120591) 119896 = 0 1 119873

(15)

Obviously the local truncation error of difference scheme(14) is 119877119896

119895= 119874(120591 + ℎ

4)

3 The Solvability of the Difference Scheme

Firstly let us denote

U0 = (120601 (1199091) 120601 (119909

2) 120601 (119909

119872minus1))119879

U119896 = (1199061198961 119906119896

2 119906

119896

119872minus1)119879

F119896 = (1198911198961 119891119896

2 119891

119896

119872minus1)119879

119896 = 0 1 119873

(16)

Then we can give the compact form of difference scheme(14) as follows

AU119896 = BU119896minus1 +119896

sum

119898=2

C119898U119896minus119898 +DF119896 + G 119896 = 1 119873

(17)

where

A =(

11990421199043

119904111990421199043

d d d119904111990421199043

11990411199042

)

B =(

11990121199013

119901111990121199013

d d d119901111990121199013

11990111199012

)

C119898=(

11990221199023

119902111990221199023

d d d119902111990221199023

11990211199022

)

D =(

11990321199033

119903111990321199033

d d d119903111990321199033

11990311199032

)

119866 = (1198921 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119872minus3

119892119872minus1

)

119879

1199041=1

12+ 1198881minus 1198882minus 1198883 119904

2=5

6+ 21198882

1199043=1

12minus 1198881minus 1198882+ 1198883

1199011=1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1

1199012=5

6minus 21198882120603(1minus120572)

1

1199013=1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1

1199021= (1198882+ 1198883) 120603(1minus120572)

119898

1199022= minus2119888

2120603(1minus120572)

119898 119902

3= (1198882minus 1198883) 120603(1minus120572)

119898

1199031= 120591 (

1

12+ 1198881) 119903

2=5

6120591 119903

3= 120591 (

1

12minus 1198881)

(18)1198921= minus 119904

11205931(119905119896) + 11990111205931(119905119896minus1) + (1198882+ 1198883)

times

119896

sum

119898=2

120603(1minus120572)

1198981205931(119905119896minus119898

) + 1199031119891 (1199090 119905119896)

119892119872minus1

= minus 11990431205932(119905119896) + 11990131205932(119905119896minus1) + (1198882+ 1198883)

times

119896

sum

119898=2

120603(1minus120572)

1198981205932(119905119896minus119898

) + 1199033119891 (119909119872 119905119896)

(19)

4 Abstract and Applied Analysis

Theorem 2 Difference equation (17) is uniquely solvable

Proof It is well known that the eigenvalues of the matrix Aare

120582119896= 1199042+ 2radic11990411199043 cos(

119896120587

119872)

=5

6+2120591120572120581120572120579

ℎ2

+ 2radic(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

cos(119896120587119872)

(20)

where 119896 = 1 2 119872 minus 1Note that 120579 ge 1 and 120581

120572gt 0 if

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

lt 0 (21)

then we easily know that 120582119896= 0

If

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

ge 0 (22)

then

120582119896ge5

6+2120591120572120581120572120579

ℎ2minus 2

10038161003816100381610038161003816100381610038161003816

1

12minus120591120572120581120572120579

ℎ2

10038161003816100381610038161003816100381610038161003816

gt 0 (23)

At the moment we obtain det(A) = 0 that is to say thematrix A is invertible Hence difference equation (17) has aunique solution

4 Stability and Convergence Analysis

In this section we analyze the stability and convergence ofdifference scheme (17) by the Fourier method [8] Firstly wegive the stability analysis

Lemma 3 The coefficients 120603(1minus120572)119898

(119898 = 0 1 ) satisfy [8] asfollows

(1) 120603(1minus120572)0

= 1 120603(1minus120572)1

= 120572 minus 1 120603(1minus120574)119898

lt 0 119895 = 1 2

(2) suminfin119898=0

120603(1minus120574)

119898= 0

Let 119880119896119895be the approximate solution of (14) and define

120588119896

119895= 119906119896

119895minus 119880119896

119895 119895 = 1 2 119872 minus 1 119896 = 0 1 119873

120588119896= (120588119896

1 120588119896

2 120588

119896

119872minus1)119879

119896 = 0 1 119873

(24)

respectively

So we can easily obtain the following roundoff errorequation

(1

12minus 1198881minus 1198882+ 1198883) 120588119896

119895+1+ (

5

6+ 21198882) 120588119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 120588119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 120588119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 120588119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 120588119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 120588119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 2120588119896minus119898

119895+ 120588119896minus119898

119895minus1)

119895 = 1 2 119872 minus 1

120588119896

0= 120588119896

119872= 0 119896 = 0 1 119873

(25)

Now we define the grid functions

120588119896(119909) =

120588119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(26)

then 120588119896(119909) can be expanded in a Fourier series

120588119896(119909) =

infin

sum

119897=minusinfin

120585119896(119897) exp(2120587119897119909

119871119894) (27)

where

120585119896(119897) =

1

119871int

119871

0

120588119896(119909) exp(minus2120587119897119909

119871119894) 119889119909 (28)

We introduce the following norm

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816120588119896

119895

10038161003816100381610038161003816

2

)

12

= [int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909]

12

(29)

and according to the Parseval equality

int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909 =

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(30)

we obtain

1003817100381710038171003817100381712058811989610038171003817100381710038171003817

2

2=

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(31)

Abstract and Applied Analysis 5

Through the above analysis we can suppose that the solutionof (25) has the following form

120588119896

119895= 120585119896exp (119894120573119895ℎ) (32)

where 120573 = 2120587119897119871Substituting the above expression into (25) one gets

[1 minus (1

3minus 41198882) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1minus 1198883) sin (120573ℎ)] 120585

119896

= [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ] 120585

119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

119896 = 0 1 119873

(33)

Lemma 4 The following relation holds

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816le 1 (34)

where

119876 = [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ]

119878 = [1 minus (1

3minus 41198882) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) ]

(35)

Proof Because of

81198882minus 3211988811198883=8120591120572120581120572

ℎ2cosh(

radic6]120572ℎ

6120581120572

) minus2120591120572]2120572

3120581120572

=8120591120572120581120572

ℎ2[1 +

infin

sum

119896=2

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

] gt 0

(36)

we obtain

81205721198882[1 minus

1

3sin2 (1

2120573ℎ)] minus 32120572119888

11198883cos2 (1

2120573ℎ)

+ 161198882

2(1 minus 120603

(1minus120572)2

1) sin2 (1

2120573ℎ)

+ 161198882

3(1 minus 120603

(1minus120572)2

1) cos2 (1

2120573ℎ) gt 0

(37)

Furthermore we can rewrite the above inequality as

[1 minus1

3sin2 (1

2120573ℎ) minus 4119888

2120603(1minus120572)

1sin2 (1

2120573ℎ)]

2

+ 4(1198881+ 1198883120603(1minus120572)

1)2

sin2 (120573ℎ)

le [1 minus1

3sin2 (1

2120573ℎ) + 4119888

2sin2 (1

2120573ℎ)]

2

+ 4(1198881minus 1198883)2sin2 (120573ℎ)

(38)

that is

100381610038161003816100381610038161003816100381610038161003816

(1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1+ 1198883120603(1minus120572)

1)

times sin (120573ℎ)) times (1 minus (13minus 41198882) sin2 (1

2120573ℎ)

minus119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) )

minus1100381610038161003816100381610038161003816100381610038161003816

le 1

(39)

This completes the proof of Lemma 4

Lemma 5 Supposing that 120585119896(119896 = 1 119873) is the solution of

(33) then we have

10038161003816100381610038161205851198961003816100381610038161003816 le exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 119896 = 1 119873 (40)

Proof For 119896 = 0 from (33) we get

100381610038161003816100381612058511003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816

100381610038161003816100381612058501003816100381610038161003816

(41)

In the light of Lemma 4 it is clear that

100381610038161003816100381612058511003816100381610038161003816 le

100381610038161003816100381612058501003816100381610038161003816 = exp (119872 sdot 0120591)

100381610038161003816100381612058501003816100381610038161003816 (42)

Now we suppose that

1003816100381610038161003816120585ℓ1003816100381610038161003816 le exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 (ℓ = 1 2 119896 minus 1) (43)

For 119896 gt 0 from (33) with Lemmas 3 and 4 we have

10038161003816100381610038161205851198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816100381610038161003816

119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

1003816100381610038161003816100381610038161003816100381610038161003816

6 Abstract and Applied Analysis

=1

|119878|

1003816100381610038161003816100381610038161003816119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896minus1

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)] 120585

0

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120585119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120585119896minus1198981003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878| |119876| exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878||119876|minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

infin

sum

119898=1

120603(1minus120572)

119898

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

=1

|119878||119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le (1 +119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le exp (119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

= exp (119872 (119896 minus 1) 120591)100381610038161003816100381612058501003816100381610038161003816 = 119870

100381610038161003816100381612058501003816100381610038161003816

(44)

This finishes the proof of Lemma 5

Lemma 6 Difference scheme (14) is unconditionally stable

Proof According to Lemma 5 we obtain

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161003816120588119896

119894

10038161003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ1003816100381610038161003816120585119896 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205851198961003816100381610038161003816

2

)

12

le 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381612058501003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205850 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381610038161205880

119894

10038161003816100381610038161003816

2

)

12

=100381710038171003817100381710038171205880100381710038171003817100381710038172 119896 = 1 2 119873

(45)

which means that difference scheme (14) is unconditionallystable

Next we give the convergence analysis Suppose

E119896

119894= 119906 (119909

119894 119905119896) minus 119906119896

119894 119894 = 1 119872 minus 1 119896 = 1 119873

(46)

and denote

E119896= (E119896

1E119896

2 E

119896

119872minus1)119879

119877119896= (119877119896

1 119877119896

2 119877

119896

119872minus1)119879

119896 = 1 119873

(47)

From (14) we obtain

(1

12minus 1198881minus 1198882+ 1198883)E119896

119895+1+ (

5

6+ 21198882)E119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883)E119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1)E119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1)E119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1)E119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minusE119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minus 2E119896minus119898

119895+E119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119877119896

119895+1+5

6120591119877119896

119895

+ 120591 (1

12+ 1198881)119877119896

119895minus1

119895 = 1 2 119872 minus 1 119896 = 0 1 119873

(48)

Similar to the stability analysismethod we define the gridfunctions

E119896(119909) =

E119896119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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Function Spaces

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

Lemma 1 Suppose that

[120581120572

1205972

1205971199092minus ]120572

120597

120597119909] 119906 (119909

119895 119905) = 119892 (119909

119895 119905) (4)

then a fourth-order difference scheme for the above equation isgiven as follows

A119906 (119909119895 119905) =L119892 (119909

119895 119905) + O (ℎ

4) (5)

whereA andL are two difference operators and are defined by

A = 120581120572cosh(

radic6]120572ℎ

6120581120572

)1205752

119909minus ]120572120583119909120575119909

L = 119868 +ℎ2

12(1205752

119909minus]120572

120581120572

120583119909120575119909)

(6)

in which 119868 is an unit operator and 120583119909120575119909and 1205752119909are average cen-

tral and second central difference operators with respect to 119909and are defined by

120583119909120575119909119906 (119909119895 119905) =

119906 (119909119895+1 119905) minus 119906 (119909

119895minus1 119905)

2ℎ

1205752

119909119906 (119909119895 119905) =

119906 (119909119895+1 119905) minus 2119906 (119909

119895 119905) + 119906 (119909

119895minus1 119905)

ℎ2

(7)

Proof In view of Taylor expansion we can obtain

119864119895(119905) = A119906 (119909

119895 119905) minusL119892 (119909

119895 119905)

= 120581120572

infin

sum

119896=0

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

times (

1205972119906 (119909119895 119905)

1205971199092+ℎ2

12

1205974119906 (119909119895 119905)

1205971199094+ 119874 (ℎ

4))

minus ]120572(

120597119906 (119909119895 119905)

120597119909+ℎ2

6

1205973119906 (119909119895 119905)

1205971199093+ 119874 (ℎ

4))

minus (119892 (119909119895 119905) +

ℎ2

12

1205972119892 (119909119895 119905)

1205971199092+ 119874 (ℎ

4))

+]120572ℎ2

12120581120572

(

120597119892 (119909119895 119905)

120597119909+ 119874 (ℎ

2))

= [(120581120572

1205972119906 (119909119895 119905)

1205971199092minus ]120572

120597119906 (119909119895 119905)

120597119909) minus 119892 (119909

119895 119905)]

minus]120572ℎ2

12120581120572

[(120581120572

1205973119906 (119909119895 119905)

1205971199093minus ]120572

1205972119906 (119909119895 119905)

1205971199092)

minus

120597119892 (119909119895 119905)

120597119909]

+ℎ2

12[(120581120572

1205974119906 (119909119895 119905)

1205971199094minus ]120572

1205973119906 (119909119895 119905)

1205971199093)

minus

1205972119892 (119909119895 119905)

1205971199092] + 119874 (ℎ

4)

(8)

Noting (4) we easily obtain

119864119895(119905) = 119874 (ℎ

4) (9)

This completes the proof

Combing (1) (4) and (5) we obtain

L sdot

120597119906 (119909119895 119905119896)

120597119905

= A sdot RL1198631minus120572

0119905119906 (119909119895 119905119896) +L sdot 119891 (119909

119895 119905119896) + O (ℎ

4)

(10)

Using the relation of the Riemann-Liouville fractionalderivative RL119863

1minus120572

0119905119906(119909 119905) and Grunwald-Letnikov fractional

derivative GL1198631minus120574

0119905119906(119909 119905) we can approximate the Riemann-

Liouville fractional derivative RL1198631minus120572

0119905119906(119909 119905) by [2]

RL1198631minus120572

0119905119906 (119909119895 119905119896) =

1

1205911minus120572

119896

sum

119898=0

120603(1minus120572)

119898119906 (119909119895 119905119896minus119898

) + 119874 (120591)

(11)

where 120603(1minus120572)119898

= (minus1)119898( 1minus120572119898)

For first-order derivative 120597119906(119909119895 119905119896)120597119905 we apply the

following backward difference scheme

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896) minus 119906 (119909

119895 119905119896minus1)

120591+ 119874 (120591) (12)

Let 119906119896119894be the numerical approximation of 119906(119909

119894 119905119896) substi-

tuting (11) and (12) into (10) and omitting the error termO(120591+

ℎ4) we can obtain the following difference scheme for solving

(1)

L sdot (119906119896

119895minus 119906119896minus1

119895) = 120591120572A sdot

119896

sum

119898=0

120603(1minus120572)

119898119906119896minus119898

119895+ 120591L sdot 119891

119896

119895

(13)

Abstract and Applied Analysis 3

The discrete form of above system is

(1

12minus 1198881minus 1198882+ 1198883) 119906119896

119895+1+ (

5

6+ 21198882) 119906119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 119906119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 119906119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 119906119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 119906119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(119906119896minus119898

119895+1minus 119906119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(119906119896minus119898

119895+1minus 2119906119896minus119898

119895+ 119906119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119891119896

119895+1+5

6120591119891119896

119895+ 120591 (

1

12+ 1198881)119891119896

119895minus1

(14)

where 1198881= ]120572ℎ24120581

120572 1198882= 120591120572120581120572120579ℎ2 1198883= 120591120572]1205722ℎ and 120579 =

cosh(radic6]120572ℎ6120581120572)

The initial and boundary conditions can be discretized as

1199060

119895= 120601 (119895ℎ) 119895 = 0 1 119872

119906119896

0= 1205931(119896120591) 119906

119896

119872= 1205932(119896120591) 119896 = 0 1 119873

(15)

Obviously the local truncation error of difference scheme(14) is 119877119896

119895= 119874(120591 + ℎ

4)

3 The Solvability of the Difference Scheme

Firstly let us denote

U0 = (120601 (1199091) 120601 (119909

2) 120601 (119909

119872minus1))119879

U119896 = (1199061198961 119906119896

2 119906

119896

119872minus1)119879

F119896 = (1198911198961 119891119896

2 119891

119896

119872minus1)119879

119896 = 0 1 119873

(16)

Then we can give the compact form of difference scheme(14) as follows

AU119896 = BU119896minus1 +119896

sum

119898=2

C119898U119896minus119898 +DF119896 + G 119896 = 1 119873

(17)

where

A =(

11990421199043

119904111990421199043

d d d119904111990421199043

11990411199042

)

B =(

11990121199013

119901111990121199013

d d d119901111990121199013

11990111199012

)

C119898=(

11990221199023

119902111990221199023

d d d119902111990221199023

11990211199022

)

D =(

11990321199033

119903111990321199033

d d d119903111990321199033

11990311199032

)

119866 = (1198921 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119872minus3

119892119872minus1

)

119879

1199041=1

12+ 1198881minus 1198882minus 1198883 119904

2=5

6+ 21198882

1199043=1

12minus 1198881minus 1198882+ 1198883

1199011=1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1

1199012=5

6minus 21198882120603(1minus120572)

1

1199013=1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1

1199021= (1198882+ 1198883) 120603(1minus120572)

119898

1199022= minus2119888

2120603(1minus120572)

119898 119902

3= (1198882minus 1198883) 120603(1minus120572)

119898

1199031= 120591 (

1

12+ 1198881) 119903

2=5

6120591 119903

3= 120591 (

1

12minus 1198881)

(18)1198921= minus 119904

11205931(119905119896) + 11990111205931(119905119896minus1) + (1198882+ 1198883)

times

119896

sum

119898=2

120603(1minus120572)

1198981205931(119905119896minus119898

) + 1199031119891 (1199090 119905119896)

119892119872minus1

= minus 11990431205932(119905119896) + 11990131205932(119905119896minus1) + (1198882+ 1198883)

times

119896

sum

119898=2

120603(1minus120572)

1198981205932(119905119896minus119898

) + 1199033119891 (119909119872 119905119896)

(19)

4 Abstract and Applied Analysis

Theorem 2 Difference equation (17) is uniquely solvable

Proof It is well known that the eigenvalues of the matrix Aare

120582119896= 1199042+ 2radic11990411199043 cos(

119896120587

119872)

=5

6+2120591120572120581120572120579

ℎ2

+ 2radic(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

cos(119896120587119872)

(20)

where 119896 = 1 2 119872 minus 1Note that 120579 ge 1 and 120581

120572gt 0 if

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

lt 0 (21)

then we easily know that 120582119896= 0

If

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

ge 0 (22)

then

120582119896ge5

6+2120591120572120581120572120579

ℎ2minus 2

10038161003816100381610038161003816100381610038161003816

1

12minus120591120572120581120572120579

ℎ2

10038161003816100381610038161003816100381610038161003816

gt 0 (23)

At the moment we obtain det(A) = 0 that is to say thematrix A is invertible Hence difference equation (17) has aunique solution

4 Stability and Convergence Analysis

In this section we analyze the stability and convergence ofdifference scheme (17) by the Fourier method [8] Firstly wegive the stability analysis

Lemma 3 The coefficients 120603(1minus120572)119898

(119898 = 0 1 ) satisfy [8] asfollows

(1) 120603(1minus120572)0

= 1 120603(1minus120572)1

= 120572 minus 1 120603(1minus120574)119898

lt 0 119895 = 1 2

(2) suminfin119898=0

120603(1minus120574)

119898= 0

Let 119880119896119895be the approximate solution of (14) and define

120588119896

119895= 119906119896

119895minus 119880119896

119895 119895 = 1 2 119872 minus 1 119896 = 0 1 119873

120588119896= (120588119896

1 120588119896

2 120588

119896

119872minus1)119879

119896 = 0 1 119873

(24)

respectively

So we can easily obtain the following roundoff errorequation

(1

12minus 1198881minus 1198882+ 1198883) 120588119896

119895+1+ (

5

6+ 21198882) 120588119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 120588119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 120588119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 120588119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 120588119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 120588119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 2120588119896minus119898

119895+ 120588119896minus119898

119895minus1)

119895 = 1 2 119872 minus 1

120588119896

0= 120588119896

119872= 0 119896 = 0 1 119873

(25)

Now we define the grid functions

120588119896(119909) =

120588119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(26)

then 120588119896(119909) can be expanded in a Fourier series

120588119896(119909) =

infin

sum

119897=minusinfin

120585119896(119897) exp(2120587119897119909

119871119894) (27)

where

120585119896(119897) =

1

119871int

119871

0

120588119896(119909) exp(minus2120587119897119909

119871119894) 119889119909 (28)

We introduce the following norm

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816120588119896

119895

10038161003816100381610038161003816

2

)

12

= [int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909]

12

(29)

and according to the Parseval equality

int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909 =

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(30)

we obtain

1003817100381710038171003817100381712058811989610038171003817100381710038171003817

2

2=

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(31)

Abstract and Applied Analysis 5

Through the above analysis we can suppose that the solutionof (25) has the following form

120588119896

119895= 120585119896exp (119894120573119895ℎ) (32)

where 120573 = 2120587119897119871Substituting the above expression into (25) one gets

[1 minus (1

3minus 41198882) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1minus 1198883) sin (120573ℎ)] 120585

119896

= [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ] 120585

119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

119896 = 0 1 119873

(33)

Lemma 4 The following relation holds

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816le 1 (34)

where

119876 = [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ]

119878 = [1 minus (1

3minus 41198882) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) ]

(35)

Proof Because of

81198882minus 3211988811198883=8120591120572120581120572

ℎ2cosh(

radic6]120572ℎ

6120581120572

) minus2120591120572]2120572

3120581120572

=8120591120572120581120572

ℎ2[1 +

infin

sum

119896=2

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

] gt 0

(36)

we obtain

81205721198882[1 minus

1

3sin2 (1

2120573ℎ)] minus 32120572119888

11198883cos2 (1

2120573ℎ)

+ 161198882

2(1 minus 120603

(1minus120572)2

1) sin2 (1

2120573ℎ)

+ 161198882

3(1 minus 120603

(1minus120572)2

1) cos2 (1

2120573ℎ) gt 0

(37)

Furthermore we can rewrite the above inequality as

[1 minus1

3sin2 (1

2120573ℎ) minus 4119888

2120603(1minus120572)

1sin2 (1

2120573ℎ)]

2

+ 4(1198881+ 1198883120603(1minus120572)

1)2

sin2 (120573ℎ)

le [1 minus1

3sin2 (1

2120573ℎ) + 4119888

2sin2 (1

2120573ℎ)]

2

+ 4(1198881minus 1198883)2sin2 (120573ℎ)

(38)

that is

100381610038161003816100381610038161003816100381610038161003816

(1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1+ 1198883120603(1minus120572)

1)

times sin (120573ℎ)) times (1 minus (13minus 41198882) sin2 (1

2120573ℎ)

minus119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) )

minus1100381610038161003816100381610038161003816100381610038161003816

le 1

(39)

This completes the proof of Lemma 4

Lemma 5 Supposing that 120585119896(119896 = 1 119873) is the solution of

(33) then we have

10038161003816100381610038161205851198961003816100381610038161003816 le exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 119896 = 1 119873 (40)

Proof For 119896 = 0 from (33) we get

100381610038161003816100381612058511003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816

100381610038161003816100381612058501003816100381610038161003816

(41)

In the light of Lemma 4 it is clear that

100381610038161003816100381612058511003816100381610038161003816 le

100381610038161003816100381612058501003816100381610038161003816 = exp (119872 sdot 0120591)

100381610038161003816100381612058501003816100381610038161003816 (42)

Now we suppose that

1003816100381610038161003816120585ℓ1003816100381610038161003816 le exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 (ℓ = 1 2 119896 minus 1) (43)

For 119896 gt 0 from (33) with Lemmas 3 and 4 we have

10038161003816100381610038161205851198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816100381610038161003816

119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

1003816100381610038161003816100381610038161003816100381610038161003816

6 Abstract and Applied Analysis

=1

|119878|

1003816100381610038161003816100381610038161003816119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896minus1

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)] 120585

0

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120585119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120585119896minus1198981003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878| |119876| exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878||119876|minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

infin

sum

119898=1

120603(1minus120572)

119898

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

=1

|119878||119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le (1 +119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le exp (119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

= exp (119872 (119896 minus 1) 120591)100381610038161003816100381612058501003816100381610038161003816 = 119870

100381610038161003816100381612058501003816100381610038161003816

(44)

This finishes the proof of Lemma 5

Lemma 6 Difference scheme (14) is unconditionally stable

Proof According to Lemma 5 we obtain

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161003816120588119896

119894

10038161003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ1003816100381610038161003816120585119896 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205851198961003816100381610038161003816

2

)

12

le 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381612058501003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205850 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381610038161205880

119894

10038161003816100381610038161003816

2

)

12

=100381710038171003817100381710038171205880100381710038171003817100381710038172 119896 = 1 2 119873

(45)

which means that difference scheme (14) is unconditionallystable

Next we give the convergence analysis Suppose

E119896

119894= 119906 (119909

119894 119905119896) minus 119906119896

119894 119894 = 1 119872 minus 1 119896 = 1 119873

(46)

and denote

E119896= (E119896

1E119896

2 E

119896

119872minus1)119879

119877119896= (119877119896

1 119877119896

2 119877

119896

119872minus1)119879

119896 = 1 119873

(47)

From (14) we obtain

(1

12minus 1198881minus 1198882+ 1198883)E119896

119895+1+ (

5

6+ 21198882)E119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883)E119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1)E119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1)E119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1)E119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minusE119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minus 2E119896minus119898

119895+E119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119877119896

119895+1+5

6120591119877119896

119895

+ 120591 (1

12+ 1198881)119877119896

119895minus1

119895 = 1 2 119872 minus 1 119896 = 0 1 119873

(48)

Similar to the stability analysismethod we define the gridfunctions

E119896(119909) =

E119896119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

The discrete form of above system is

(1

12minus 1198881minus 1198882+ 1198883) 119906119896

119895+1+ (

5

6+ 21198882) 119906119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 119906119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 119906119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 119906119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 119906119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(119906119896minus119898

119895+1minus 119906119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(119906119896minus119898

119895+1minus 2119906119896minus119898

119895+ 119906119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119891119896

119895+1+5

6120591119891119896

119895+ 120591 (

1

12+ 1198881)119891119896

119895minus1

(14)

where 1198881= ]120572ℎ24120581

120572 1198882= 120591120572120581120572120579ℎ2 1198883= 120591120572]1205722ℎ and 120579 =

cosh(radic6]120572ℎ6120581120572)

The initial and boundary conditions can be discretized as

1199060

119895= 120601 (119895ℎ) 119895 = 0 1 119872

119906119896

0= 1205931(119896120591) 119906

119896

119872= 1205932(119896120591) 119896 = 0 1 119873

(15)

Obviously the local truncation error of difference scheme(14) is 119877119896

119895= 119874(120591 + ℎ

4)

3 The Solvability of the Difference Scheme

Firstly let us denote

U0 = (120601 (1199091) 120601 (119909

2) 120601 (119909

119872minus1))119879

U119896 = (1199061198961 119906119896

2 119906

119896

119872minus1)119879

F119896 = (1198911198961 119891119896

2 119891

119896

119872minus1)119879

119896 = 0 1 119873

(16)

Then we can give the compact form of difference scheme(14) as follows

AU119896 = BU119896minus1 +119896

sum

119898=2

C119898U119896minus119898 +DF119896 + G 119896 = 1 119873

(17)

where

A =(

11990421199043

119904111990421199043

d d d119904111990421199043

11990411199042

)

B =(

11990121199013

119901111990121199013

d d d119901111990121199013

11990111199012

)

C119898=(

11990221199023

119902111990221199023

d d d119902111990221199023

11990211199022

)

D =(

11990321199033

119903111990321199033

d d d119903111990321199033

11990311199032

)

119866 = (1198921 0 0⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟

119872minus3

119892119872minus1

)

119879

1199041=1

12+ 1198881minus 1198882minus 1198883 119904

2=5

6+ 21198882

1199043=1

12minus 1198881minus 1198882+ 1198883

1199011=1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1

1199012=5

6minus 21198882120603(1minus120572)

1

1199013=1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1

1199021= (1198882+ 1198883) 120603(1minus120572)

119898

1199022= minus2119888

2120603(1minus120572)

119898 119902

3= (1198882minus 1198883) 120603(1minus120572)

119898

1199031= 120591 (

1

12+ 1198881) 119903

2=5

6120591 119903

3= 120591 (

1

12minus 1198881)

(18)1198921= minus 119904

11205931(119905119896) + 11990111205931(119905119896minus1) + (1198882+ 1198883)

times

119896

sum

119898=2

120603(1minus120572)

1198981205931(119905119896minus119898

) + 1199031119891 (1199090 119905119896)

119892119872minus1

= minus 11990431205932(119905119896) + 11990131205932(119905119896minus1) + (1198882+ 1198883)

times

119896

sum

119898=2

120603(1minus120572)

1198981205932(119905119896minus119898

) + 1199033119891 (119909119872 119905119896)

(19)

4 Abstract and Applied Analysis

Theorem 2 Difference equation (17) is uniquely solvable

Proof It is well known that the eigenvalues of the matrix Aare

120582119896= 1199042+ 2radic11990411199043 cos(

119896120587

119872)

=5

6+2120591120572120581120572120579

ℎ2

+ 2radic(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

cos(119896120587119872)

(20)

where 119896 = 1 2 119872 minus 1Note that 120579 ge 1 and 120581

120572gt 0 if

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

lt 0 (21)

then we easily know that 120582119896= 0

If

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

ge 0 (22)

then

120582119896ge5

6+2120591120572120581120572120579

ℎ2minus 2

10038161003816100381610038161003816100381610038161003816

1

12minus120591120572120581120572120579

ℎ2

10038161003816100381610038161003816100381610038161003816

gt 0 (23)

At the moment we obtain det(A) = 0 that is to say thematrix A is invertible Hence difference equation (17) has aunique solution

4 Stability and Convergence Analysis

In this section we analyze the stability and convergence ofdifference scheme (17) by the Fourier method [8] Firstly wegive the stability analysis

Lemma 3 The coefficients 120603(1minus120572)119898

(119898 = 0 1 ) satisfy [8] asfollows

(1) 120603(1minus120572)0

= 1 120603(1minus120572)1

= 120572 minus 1 120603(1minus120574)119898

lt 0 119895 = 1 2

(2) suminfin119898=0

120603(1minus120574)

119898= 0

Let 119880119896119895be the approximate solution of (14) and define

120588119896

119895= 119906119896

119895minus 119880119896

119895 119895 = 1 2 119872 minus 1 119896 = 0 1 119873

120588119896= (120588119896

1 120588119896

2 120588

119896

119872minus1)119879

119896 = 0 1 119873

(24)

respectively

So we can easily obtain the following roundoff errorequation

(1

12minus 1198881minus 1198882+ 1198883) 120588119896

119895+1+ (

5

6+ 21198882) 120588119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 120588119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 120588119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 120588119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 120588119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 120588119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 2120588119896minus119898

119895+ 120588119896minus119898

119895minus1)

119895 = 1 2 119872 minus 1

120588119896

0= 120588119896

119872= 0 119896 = 0 1 119873

(25)

Now we define the grid functions

120588119896(119909) =

120588119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(26)

then 120588119896(119909) can be expanded in a Fourier series

120588119896(119909) =

infin

sum

119897=minusinfin

120585119896(119897) exp(2120587119897119909

119871119894) (27)

where

120585119896(119897) =

1

119871int

119871

0

120588119896(119909) exp(minus2120587119897119909

119871119894) 119889119909 (28)

We introduce the following norm

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816120588119896

119895

10038161003816100381610038161003816

2

)

12

= [int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909]

12

(29)

and according to the Parseval equality

int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909 =

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(30)

we obtain

1003817100381710038171003817100381712058811989610038171003817100381710038171003817

2

2=

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(31)

Abstract and Applied Analysis 5

Through the above analysis we can suppose that the solutionof (25) has the following form

120588119896

119895= 120585119896exp (119894120573119895ℎ) (32)

where 120573 = 2120587119897119871Substituting the above expression into (25) one gets

[1 minus (1

3minus 41198882) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1minus 1198883) sin (120573ℎ)] 120585

119896

= [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ] 120585

119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

119896 = 0 1 119873

(33)

Lemma 4 The following relation holds

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816le 1 (34)

where

119876 = [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ]

119878 = [1 minus (1

3minus 41198882) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) ]

(35)

Proof Because of

81198882minus 3211988811198883=8120591120572120581120572

ℎ2cosh(

radic6]120572ℎ

6120581120572

) minus2120591120572]2120572

3120581120572

=8120591120572120581120572

ℎ2[1 +

infin

sum

119896=2

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

] gt 0

(36)

we obtain

81205721198882[1 minus

1

3sin2 (1

2120573ℎ)] minus 32120572119888

11198883cos2 (1

2120573ℎ)

+ 161198882

2(1 minus 120603

(1minus120572)2

1) sin2 (1

2120573ℎ)

+ 161198882

3(1 minus 120603

(1minus120572)2

1) cos2 (1

2120573ℎ) gt 0

(37)

Furthermore we can rewrite the above inequality as

[1 minus1

3sin2 (1

2120573ℎ) minus 4119888

2120603(1minus120572)

1sin2 (1

2120573ℎ)]

2

+ 4(1198881+ 1198883120603(1minus120572)

1)2

sin2 (120573ℎ)

le [1 minus1

3sin2 (1

2120573ℎ) + 4119888

2sin2 (1

2120573ℎ)]

2

+ 4(1198881minus 1198883)2sin2 (120573ℎ)

(38)

that is

100381610038161003816100381610038161003816100381610038161003816

(1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1+ 1198883120603(1minus120572)

1)

times sin (120573ℎ)) times (1 minus (13minus 41198882) sin2 (1

2120573ℎ)

minus119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) )

minus1100381610038161003816100381610038161003816100381610038161003816

le 1

(39)

This completes the proof of Lemma 4

Lemma 5 Supposing that 120585119896(119896 = 1 119873) is the solution of

(33) then we have

10038161003816100381610038161205851198961003816100381610038161003816 le exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 119896 = 1 119873 (40)

Proof For 119896 = 0 from (33) we get

100381610038161003816100381612058511003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816

100381610038161003816100381612058501003816100381610038161003816

(41)

In the light of Lemma 4 it is clear that

100381610038161003816100381612058511003816100381610038161003816 le

100381610038161003816100381612058501003816100381610038161003816 = exp (119872 sdot 0120591)

100381610038161003816100381612058501003816100381610038161003816 (42)

Now we suppose that

1003816100381610038161003816120585ℓ1003816100381610038161003816 le exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 (ℓ = 1 2 119896 minus 1) (43)

For 119896 gt 0 from (33) with Lemmas 3 and 4 we have

10038161003816100381610038161205851198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816100381610038161003816

119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

1003816100381610038161003816100381610038161003816100381610038161003816

6 Abstract and Applied Analysis

=1

|119878|

1003816100381610038161003816100381610038161003816119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896minus1

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)] 120585

0

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120585119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120585119896minus1198981003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878| |119876| exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878||119876|minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

infin

sum

119898=1

120603(1minus120572)

119898

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

=1

|119878||119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le (1 +119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le exp (119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

= exp (119872 (119896 minus 1) 120591)100381610038161003816100381612058501003816100381610038161003816 = 119870

100381610038161003816100381612058501003816100381610038161003816

(44)

This finishes the proof of Lemma 5

Lemma 6 Difference scheme (14) is unconditionally stable

Proof According to Lemma 5 we obtain

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161003816120588119896

119894

10038161003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ1003816100381610038161003816120585119896 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205851198961003816100381610038161003816

2

)

12

le 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381612058501003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205850 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381610038161205880

119894

10038161003816100381610038161003816

2

)

12

=100381710038171003817100381710038171205880100381710038171003817100381710038172 119896 = 1 2 119873

(45)

which means that difference scheme (14) is unconditionallystable

Next we give the convergence analysis Suppose

E119896

119894= 119906 (119909

119894 119905119896) minus 119906119896

119894 119894 = 1 119872 minus 1 119896 = 1 119873

(46)

and denote

E119896= (E119896

1E119896

2 E

119896

119872minus1)119879

119877119896= (119877119896

1 119877119896

2 119877

119896

119872minus1)119879

119896 = 1 119873

(47)

From (14) we obtain

(1

12minus 1198881minus 1198882+ 1198883)E119896

119895+1+ (

5

6+ 21198882)E119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883)E119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1)E119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1)E119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1)E119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minusE119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minus 2E119896minus119898

119895+E119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119877119896

119895+1+5

6120591119877119896

119895

+ 120591 (1

12+ 1198881)119877119896

119895minus1

119895 = 1 2 119872 minus 1 119896 = 0 1 119873

(48)

Similar to the stability analysismethod we define the gridfunctions

E119896(119909) =

E119896119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Theorem 2 Difference equation (17) is uniquely solvable

Proof It is well known that the eigenvalues of the matrix Aare

120582119896= 1199042+ 2radic11990411199043 cos(

119896120587

119872)

=5

6+2120591120572120581120572120579

ℎ2

+ 2radic(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

cos(119896120587119872)

(20)

where 119896 = 1 2 119872 minus 1Note that 120579 ge 1 and 120581

120572gt 0 if

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

lt 0 (21)

then we easily know that 120582119896= 0

If

(1

12minus120591120572120581120572120579

ℎ2)

2

minus (]120572ℎ

24120581120572

minus120591120572]120572

2ℎ)

2

ge 0 (22)

then

120582119896ge5

6+2120591120572120581120572120579

ℎ2minus 2

10038161003816100381610038161003816100381610038161003816

1

12minus120591120572120581120572120579

ℎ2

10038161003816100381610038161003816100381610038161003816

gt 0 (23)

At the moment we obtain det(A) = 0 that is to say thematrix A is invertible Hence difference equation (17) has aunique solution

4 Stability and Convergence Analysis

In this section we analyze the stability and convergence ofdifference scheme (17) by the Fourier method [8] Firstly wegive the stability analysis

Lemma 3 The coefficients 120603(1minus120572)119898

(119898 = 0 1 ) satisfy [8] asfollows

(1) 120603(1minus120572)0

= 1 120603(1minus120572)1

= 120572 minus 1 120603(1minus120574)119898

lt 0 119895 = 1 2

(2) suminfin119898=0

120603(1minus120574)

119898= 0

Let 119880119896119895be the approximate solution of (14) and define

120588119896

119895= 119906119896

119895minus 119880119896

119895 119895 = 1 2 119872 minus 1 119896 = 0 1 119873

120588119896= (120588119896

1 120588119896

2 120588

119896

119872minus1)119879

119896 = 0 1 119873

(24)

respectively

So we can easily obtain the following roundoff errorequation

(1

12minus 1198881minus 1198882+ 1198883) 120588119896

119895+1+ (

5

6+ 21198882) 120588119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883) 120588119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1) 120588119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1) 120588119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1) 120588119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 120588119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(120588119896minus119898

119895+1minus 2120588119896minus119898

119895+ 120588119896minus119898

119895minus1)

119895 = 1 2 119872 minus 1

120588119896

0= 120588119896

119872= 0 119896 = 0 1 119873

(25)

Now we define the grid functions

120588119896(119909) =

120588119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(26)

then 120588119896(119909) can be expanded in a Fourier series

120588119896(119909) =

infin

sum

119897=minusinfin

120585119896(119897) exp(2120587119897119909

119871119894) (27)

where

120585119896(119897) =

1

119871int

119871

0

120588119896(119909) exp(minus2120587119897119909

119871119894) 119889119909 (28)

We introduce the following norm

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816120588119896

119895

10038161003816100381610038161003816

2

)

12

= [int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909]

12

(29)

and according to the Parseval equality

int

119871

0

10038161003816100381610038161003816120588119896(119909)10038161003816100381610038161003816

2

119889119909 =

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(30)

we obtain

1003817100381710038171003817100381712058811989610038171003817100381710038171003817

2

2=

infin

sum

119897=minusinfin

1003816100381610038161003816120585119896 (119897)1003816100381610038161003816

2

(31)

Abstract and Applied Analysis 5

Through the above analysis we can suppose that the solutionof (25) has the following form

120588119896

119895= 120585119896exp (119894120573119895ℎ) (32)

where 120573 = 2120587119897119871Substituting the above expression into (25) one gets

[1 minus (1

3minus 41198882) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1minus 1198883) sin (120573ℎ)] 120585

119896

= [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ] 120585

119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

119896 = 0 1 119873

(33)

Lemma 4 The following relation holds

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816le 1 (34)

where

119876 = [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ]

119878 = [1 minus (1

3minus 41198882) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) ]

(35)

Proof Because of

81198882minus 3211988811198883=8120591120572120581120572

ℎ2cosh(

radic6]120572ℎ

6120581120572

) minus2120591120572]2120572

3120581120572

=8120591120572120581120572

ℎ2[1 +

infin

sum

119896=2

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

] gt 0

(36)

we obtain

81205721198882[1 minus

1

3sin2 (1

2120573ℎ)] minus 32120572119888

11198883cos2 (1

2120573ℎ)

+ 161198882

2(1 minus 120603

(1minus120572)2

1) sin2 (1

2120573ℎ)

+ 161198882

3(1 minus 120603

(1minus120572)2

1) cos2 (1

2120573ℎ) gt 0

(37)

Furthermore we can rewrite the above inequality as

[1 minus1

3sin2 (1

2120573ℎ) minus 4119888

2120603(1minus120572)

1sin2 (1

2120573ℎ)]

2

+ 4(1198881+ 1198883120603(1minus120572)

1)2

sin2 (120573ℎ)

le [1 minus1

3sin2 (1

2120573ℎ) + 4119888

2sin2 (1

2120573ℎ)]

2

+ 4(1198881minus 1198883)2sin2 (120573ℎ)

(38)

that is

100381610038161003816100381610038161003816100381610038161003816

(1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1+ 1198883120603(1minus120572)

1)

times sin (120573ℎ)) times (1 minus (13minus 41198882) sin2 (1

2120573ℎ)

minus119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) )

minus1100381610038161003816100381610038161003816100381610038161003816

le 1

(39)

This completes the proof of Lemma 4

Lemma 5 Supposing that 120585119896(119896 = 1 119873) is the solution of

(33) then we have

10038161003816100381610038161205851198961003816100381610038161003816 le exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 119896 = 1 119873 (40)

Proof For 119896 = 0 from (33) we get

100381610038161003816100381612058511003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816

100381610038161003816100381612058501003816100381610038161003816

(41)

In the light of Lemma 4 it is clear that

100381610038161003816100381612058511003816100381610038161003816 le

100381610038161003816100381612058501003816100381610038161003816 = exp (119872 sdot 0120591)

100381610038161003816100381612058501003816100381610038161003816 (42)

Now we suppose that

1003816100381610038161003816120585ℓ1003816100381610038161003816 le exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 (ℓ = 1 2 119896 minus 1) (43)

For 119896 gt 0 from (33) with Lemmas 3 and 4 we have

10038161003816100381610038161205851198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816100381610038161003816

119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

1003816100381610038161003816100381610038161003816100381610038161003816

6 Abstract and Applied Analysis

=1

|119878|

1003816100381610038161003816100381610038161003816119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896minus1

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)] 120585

0

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120585119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120585119896minus1198981003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878| |119876| exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878||119876|minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

infin

sum

119898=1

120603(1minus120572)

119898

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

=1

|119878||119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le (1 +119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le exp (119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

= exp (119872 (119896 minus 1) 120591)100381610038161003816100381612058501003816100381610038161003816 = 119870

100381610038161003816100381612058501003816100381610038161003816

(44)

This finishes the proof of Lemma 5

Lemma 6 Difference scheme (14) is unconditionally stable

Proof According to Lemma 5 we obtain

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161003816120588119896

119894

10038161003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ1003816100381610038161003816120585119896 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205851198961003816100381610038161003816

2

)

12

le 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381612058501003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205850 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381610038161205880

119894

10038161003816100381610038161003816

2

)

12

=100381710038171003817100381710038171205880100381710038171003817100381710038172 119896 = 1 2 119873

(45)

which means that difference scheme (14) is unconditionallystable

Next we give the convergence analysis Suppose

E119896

119894= 119906 (119909

119894 119905119896) minus 119906119896

119894 119894 = 1 119872 minus 1 119896 = 1 119873

(46)

and denote

E119896= (E119896

1E119896

2 E

119896

119872minus1)119879

119877119896= (119877119896

1 119877119896

2 119877

119896

119872minus1)119879

119896 = 1 119873

(47)

From (14) we obtain

(1

12minus 1198881minus 1198882+ 1198883)E119896

119895+1+ (

5

6+ 21198882)E119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883)E119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1)E119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1)E119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1)E119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minusE119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minus 2E119896minus119898

119895+E119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119877119896

119895+1+5

6120591119877119896

119895

+ 120591 (1

12+ 1198881)119877119896

119895minus1

119895 = 1 2 119872 minus 1 119896 = 0 1 119873

(48)

Similar to the stability analysismethod we define the gridfunctions

E119896(119909) =

E119896119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Through the above analysis we can suppose that the solutionof (25) has the following form

120588119896

119895= 120585119896exp (119894120573119895ℎ) (32)

where 120573 = 2120587119897119871Substituting the above expression into (25) one gets

[1 minus (1

3minus 41198882) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1minus 1198883) sin (120573ℎ)] 120585

119896

= [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ] 120585

119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

119896 = 0 1 119873

(33)

Lemma 4 The following relation holds

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816le 1 (34)

where

119876 = [1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881+ 1198883120603(1minus120572)

1) sin (120573ℎ) ]

119878 = [1 minus (1

3minus 41198882) sin2 (1

2120573ℎ)

minus 119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) ]

(35)

Proof Because of

81198882minus 3211988811198883=8120591120572120581120572

ℎ2cosh(

radic6]120572ℎ

6120581120572

) minus2120591120572]2120572

3120581120572

=8120591120572120581120572

ℎ2[1 +

infin

sum

119896=2

1

(2119896)(radic6]120572ℎ

6120581120572

)

2119896

] gt 0

(36)

we obtain

81205721198882[1 minus

1

3sin2 (1

2120573ℎ)] minus 32120572119888

11198883cos2 (1

2120573ℎ)

+ 161198882

2(1 minus 120603

(1minus120572)2

1) sin2 (1

2120573ℎ)

+ 161198882

3(1 minus 120603

(1minus120572)2

1) cos2 (1

2120573ℎ) gt 0

(37)

Furthermore we can rewrite the above inequality as

[1 minus1

3sin2 (1

2120573ℎ) minus 4119888

2120603(1minus120572)

1sin2 (1

2120573ℎ)]

2

+ 4(1198881+ 1198883120603(1minus120572)

1)2

sin2 (120573ℎ)

le [1 minus1

3sin2 (1

2120573ℎ) + 4119888

2sin2 (1

2120573ℎ)]

2

+ 4(1198881minus 1198883)2sin2 (120573ℎ)

(38)

that is

100381610038161003816100381610038161003816100381610038161003816

(1 minus (1

3+ 41198882120603(1minus120572)

1) sin2 (1

2120573ℎ) minus 119894 sdot 2 (119888

1+ 1198883120603(1minus120572)

1)

times sin (120573ℎ)) times (1 minus (13minus 41198882) sin2 (1

2120573ℎ)

minus119894 sdot 2 (1198881minus 1198883) sin (120573ℎ) )

minus1100381610038161003816100381610038161003816100381610038161003816

le 1

(39)

This completes the proof of Lemma 4

Lemma 5 Supposing that 120585119896(119896 = 1 119873) is the solution of

(33) then we have

10038161003816100381610038161205851198961003816100381610038161003816 le exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 119896 = 1 119873 (40)

Proof For 119896 = 0 from (33) we get

100381610038161003816100381612058511003816100381610038161003816 =

1003816100381610038161003816100381610038161003816

119876

119878

1003816100381610038161003816100381610038161003816

100381610038161003816100381612058501003816100381610038161003816

(41)

In the light of Lemma 4 it is clear that

100381610038161003816100381612058511003816100381610038161003816 le

100381610038161003816100381612058501003816100381610038161003816 = exp (119872 sdot 0120591)

100381610038161003816100381612058501003816100381610038161003816 (42)

Now we suppose that

1003816100381610038161003816120585ℓ1003816100381610038161003816 le exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816 (ℓ = 1 2 119896 minus 1) (43)

For 119896 gt 0 from (33) with Lemmas 3 and 4 we have

10038161003816100381610038161205851198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816100381610038161003816

119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

1003816100381610038161003816100381610038161003816100381610038161003816

6 Abstract and Applied Analysis

=1

|119878|

1003816100381610038161003816100381610038161003816119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896minus1

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)] 120585

0

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120585119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120585119896minus1198981003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878| |119876| exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878||119876|minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

infin

sum

119898=1

120603(1minus120572)

119898

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

=1

|119878||119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le (1 +119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le exp (119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

= exp (119872 (119896 minus 1) 120591)100381610038161003816100381612058501003816100381610038161003816 = 119870

100381610038161003816100381612058501003816100381610038161003816

(44)

This finishes the proof of Lemma 5

Lemma 6 Difference scheme (14) is unconditionally stable

Proof According to Lemma 5 we obtain

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161003816120588119896

119894

10038161003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ1003816100381610038161003816120585119896 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205851198961003816100381610038161003816

2

)

12

le 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381612058501003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205850 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381610038161205880

119894

10038161003816100381610038161003816

2

)

12

=100381710038171003817100381710038171205880100381710038171003817100381710038172 119896 = 1 2 119873

(45)

which means that difference scheme (14) is unconditionallystable

Next we give the convergence analysis Suppose

E119896

119894= 119906 (119909

119894 119905119896) minus 119906119896

119894 119894 = 1 119872 minus 1 119896 = 1 119873

(46)

and denote

E119896= (E119896

1E119896

2 E

119896

119872minus1)119879

119877119896= (119877119896

1 119877119896

2 119877

119896

119872minus1)119879

119896 = 1 119873

(47)

From (14) we obtain

(1

12minus 1198881minus 1198882+ 1198883)E119896

119895+1+ (

5

6+ 21198882)E119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883)E119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1)E119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1)E119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1)E119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minusE119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minus 2E119896minus119898

119895+E119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119877119896

119895+1+5

6120591119877119896

119895

+ 120591 (1

12+ 1198881)119877119896

119895minus1

119895 = 1 2 119872 minus 1 119896 = 0 1 119873

(48)

Similar to the stability analysismethod we define the gridfunctions

E119896(119909) =

E119896119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

=1

|119878|

1003816100381610038161003816100381610038161003816119876120585119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896minus1

sum

119898=2

120603(1minus120572)

119898120585119896minus119898

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)] 120585

0

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120585119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120585119896minus1198981003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878| |119876| exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896minus1

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

100381610038161003816100381612058501003816100381610038161003816

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816120603(1minus120572)

119896

100381610038161003816100381612058501003816100381610038161003816

le1

|119878||119876|minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

infin

sum

119898=1

120603(1minus120572)

119898

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

=1

|119878||119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le (1 +119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

le exp (119872120591) exp (119872 (119896 minus 2) 120591)100381610038161003816100381612058501003816100381610038161003816

= exp (119872 (119896 minus 1) 120591)100381610038161003816100381612058501003816100381610038161003816 = 119870

100381610038161003816100381612058501003816100381610038161003816

(44)

This finishes the proof of Lemma 5

Lemma 6 Difference scheme (14) is unconditionally stable

Proof According to Lemma 5 we obtain

10038171003817100381710038171003817120588119896100381710038171003817100381710038172

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161003816120588119896

119894

10038161003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ1003816100381610038161003816120585119896 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= (

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205851198961003816100381610038161003816

2

)

12

le 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381612058501003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ10038161003816100381610038161205850 exp (119894120573119895ℎ)

1003816100381610038161003816

2

)

12

= 119870(

119872minus1

sum

119894=1

ℎ100381610038161003816100381610038161205880

119894

10038161003816100381610038161003816

2

)

12

=100381710038171003817100381710038171205880100381710038171003817100381710038172 119896 = 1 2 119873

(45)

which means that difference scheme (14) is unconditionallystable

Next we give the convergence analysis Suppose

E119896

119894= 119906 (119909

119894 119905119896) minus 119906119896

119894 119894 = 1 119872 minus 1 119896 = 1 119873

(46)

and denote

E119896= (E119896

1E119896

2 E

119896

119872minus1)119879

119877119896= (119877119896

1 119877119896

2 119877

119896

119872minus1)119879

119896 = 1 119873

(47)

From (14) we obtain

(1

12minus 1198881minus 1198882+ 1198883)E119896

119895+1+ (

5

6+ 21198882)E119896

119895

+ (1

12+ 1198881minus 1198882minus 1198883)E119896

119895minus1

= (1

12minus 1198881+ 1198882120603(1minus120572)

1minus 1198883120603(1minus120572)

1)E119896minus1

119895+1

+ (5

6minus 21198882120603(1minus120572)

1)E119896minus1

119895

+ (1

12+ 1198881+ 1198882120603(1minus120572)

1+ 1198883120603(1minus120572)

1)E119896minus1

119895minus1

minus 1198883

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minusE119896minus119898

119895minus1)

+ 1198882

119896

sum

119898=2

120603(1minus120572)

119898(E119896minus119898

119895+1minus 2E119896minus119898

119895+E119896minus119898

119895minus1)

+ 120591 (1

12minus 1198881)119877119896

119895+1+5

6120591119877119896

119895

+ 120591 (1

12+ 1198881)119877119896

119895minus1

119895 = 1 2 119872 minus 1 119896 = 0 1 119873

(48)

Similar to the stability analysismethod we define the gridfunctions

E119896(119909) =

E119896119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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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

Abstract and Applied Analysis 7

119877119896(119909) =

119877119896

119895 when 119909

119895minusℎ

2lt 119909 le 119909

119895+ℎ

2

119895 = 1 2 119872 minus 1

0 when 0 le 119909 le ℎ2or 119871 minus ℎ

2lt 119909 le 119871

(49)

then E119896(119909) and 119877119896(119909) can be expanded to the following

Fourier series respectively

E119896(119909) =

infin

sum

119897=minusinfin

120577119896(119897) exp(2120587119897119909

119871119894)

119877119896(119909) =

infin

sum

119897=minusinfin

120578119896(119897) exp(2120587119897119909

119871119894)

(50)

where

120577119896(119897) =

1

119871int

119871

0

E119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

120578119896(119897) =

1

119871int

119871

0

119877119896(119909) exp(minus2120587119897119909

119871119894) 119889119909

(51)

The same as before we also have

10038171003817100381710038171003817E119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816E119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120577119896 (119897)1003816100381610038161003816

2

)

12

(52)

10038171003817100381710038171003817119877119896100381710038171003817100381710038172

= (

119872minus1

sum

119895=1

ℎ10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816

2

)

12

= (

infin

sum

119897=minusinfin

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816

2

)

12

(53)

Based on the above analysis we can assume that E119896119894and 119877119896

119894

have the following forms

E119896

119895= 120577119896exp (119894120573119895ℎ)

119877119896

119895= 120578119896exp (119894120573119895ℎ)

(54)

respectively Substituting the above two expressions into (48)yields

119878120577119896= 119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

119896 = 1 2 119873

(55)

Lemma7 Let 120577119896(119896 = 0 1 119873) be the solution of (55) then

there exists a positive constant 1198623 so that

10038161003816100381610038161205771198961003816100381610038161003816 le 1198623119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 119896 = 0 1 119873

(56)

Proof FromE0 = 0 we have

1205770= 1205770(119897) = 0 (57)

In view of the convergence of the series of (53) there is apositive constant 119862

1 such that

10038161003816100381610038161205781198961003816100381610038161003816 =

1003816100381610038161003816120578119896 (119897)1003816100381610038161003816 le 1198621

100381610038161003816100381612057811003816100381610038161003816 = 1198621

10038161003816100381610038161205781 (119897)1003816100381610038161003816 (58)

For 119896 = 1 from (55) we have

1205771=119876

1198781205770+

120591 [1minus(13) sin2 ((12) 120573ℎ)]minus119894 sdot 21198881sin (120573ℎ)

1198781205781

=

120591 [1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 21198881sin (120573ℎ)

1198781205781

(59)

Noticing (58) then

100381610038161003816100381612057711003816100381610038161003816 le 120591

100381610038161003816100381612057811003816100381610038161003816 le 1198621120591 exp (119872 sdot 0120591)

100381610038161003816100381612057811003816100381610038161003816 (60)

Now we suppose that

1003816100381610038161003816120577ℓ1003816100381610038161003816 le 1198621ℓ120591 exp (119872 (ℓ minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816 ℓ = 1 119873 minus 1

(61)

Then when 119896 gt 1 we obtain

10038161003816100381610038161205771198961003816100381610038161003816 =

1

|119878|

1003816100381610038161003816100381610038161003816119876120577119896minus1

minus [41198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)]

times

119896

sum

119898=2

120603(1minus120572)

119898120577119896minus119898

+ 120591 [1 minus1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ) 120578

119896

1003816100381610038161003816100381610038161003816

le1

|119878||119876|

1003816100381610038161003816120577119896minus11003816100381610038161003816 minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times

119896

sum

119898=2

120603(1minus120572)

119898

1003816100381610038161003816120577119896minus1198981003816100381610038161003816

+ 120591

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

10038161003816100381610038161205781198961003816100381610038161003816

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

minus

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816(119896 minus 119898)

times

119896

sum

119898=2

120603(1minus120572)

119898exp (119872 (119896 minus 119898 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

8 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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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

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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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 Abstract and Applied Analysis

Table 1 The maximum error temporal and spatial convergence orders by difference scheme (14) for different 120572

120572 Themaximum error Temporal convergence order Spatial convergence order

01 ℎ = 14 120591 = 14 01650 mdash mdashℎ = 18 120591 = 164 00103 10004 40017

02 ℎ = 14 120591 = 14 01526 mdash mdashℎ = 18 120591 = 164 00095 10014 40057

03 ℎ = 14 120591 = 14 01400 mdash mdashℎ = 18 120591 = 164 00087 10021 40083

04 ℎ = 14 120591 = 14 01274 mdash mdashℎ = 18 120591 = 164 00080 09983 39932

05 ℎ = 14 120591 = 14 01150 mdash mdashℎ = 18 120591 = 164 00072 09994 39975

06 ℎ = 14 120591 = 14 01027 mdash mdashℎ = 18 120591 = 164 00065 09955 39819

07 ℎ = 14 120591 = 14 00907 mdash mdashℎ = 18 120591 = 164 00058 09917 39670

08 ℎ = 14 120591 = 14 00789 mdash mdashℎ = 18 120591 = 164 00051 09879 39515

09 ℎ = 14 120591 = 14 00676 mdash mdashℎ = 18 120591 = 164 00045 09773 39090

le1

|119878| |119876| (119896 minus 1) exp (119872 (119896 minus 2) 120591)

+ 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times (119896 minus 1) exp (119872 (119896 minus 1) 120591)

+

1003816100381610038161003816100381610038161003816[1minus

1

3sin2 (1

2120573ℎ)]minus119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816 1198621120591100381610038161003816100381612057811003816100381610038161003816

le1

|119878|(119896minus1) [|119876|+120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ)+119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

1003816100381610038161003816100381610038161003816[1 minus

1

3sin2 (1

2120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)

1003816100381610038161003816100381610038161003816

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le (119896 minus 1) [|119876| + 120572

100381610038161003816100381610038161003816100381641198882sin2 (1

2120573ℎ) + 119894 sdot 2119888

3sin (120573ℎ)

1003816100381610038161003816100381610038161003816]

+

10038161003816100381610038161003816[1 minus (13) sin2 ((12) 120573ℎ)] minus 119894 sdot 2119888

1sin (120573ℎ)10038161003816100381610038161003816

|119878|

times 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 119896 (1 +119872120591) 1198621120591 exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

le 1198621119896120591 exp (119872120591) exp (119872 (119896 minus 2) 120591)

100381610038161003816100381612057811003816100381610038161003816

= 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381610038161003816100381612057811003816100381610038161003816

(62)This completes the proof

Theorem 8 Difference scheme (14) is convergent and theconvergence order is 119874(120591 + ℎ4)

Proof Firstly we know that there are exist positive constants1198622 such that

10038161003816100381610038161003816119877119896

119895

10038161003816100381610038161003816le 1198622(120591 + ℎ

4) (63)

10038171003817100381710038171003817119877119896

119895

10038171003817100381710038171003817le 1198622radic(119872 minus 1) ℎ (120591 + ℎ

4) le 119862

2radic119871 (120591 + ℎ

4)

(64)

Using (52) and (64) with Lemma 7 we get10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 1198621119896120591 exp (119872 (119896 minus 1) 120591)

100381710038171003817100381710038171198771100381710038171003817100381710038172

le 11986211198622radic119871119896120591 exp (119872119879) (120591 + ℎ4) = 119862 (120591 + ℎ4)

(65)

Due to 119896 le 119873

119896120591 le 119879 (66)

so10038171003817100381710038171003817E119896100381710038171003817100381710038172

le 119862 (120591 + ℎ4) (67)

where 119862 = 11986211198622119879radic119871 exp(119872119879) This ends the proof

Remark 9 From above discussion we know that differencescheme is an implicit scheme and it is unconditionally stableand convergent If we take

120597119906 (119909119895 119905119896)

120597119905=

119906 (119909119895 119905119896+1) minus 119906 (119909

119895 119905119896)

120591+ 119874 (120591) (68)

in (1) then we can obtain an explicit scheme and it isconditionally stable and convergent

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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

Abstract and Applied Analysis 9

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 1 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1400 ℎ = 120)

5 Numerical Example

In this section a numerical example is presented to confirmour theoretical results

Example 10 Consider the following equation

120597119906 (119909 119905)

120597119905= [

1205972

1205971199092minus120597

120597119909] RL119863

1minus120572

0119905119906 (119909 119905) + 119891 (119909 119905)

0 lt 119909 lt 2 0 le 119905 le 1

(69)

with the initial and boundary conditions

119906 (119909 0) = 0 0 le 119909 le 2

119906 (0 119905) = 0 119906 (2 119905) = 0 0 lt 119905 le 1

(70)

where 119891(119909 119905) = 2119905 sin(119909) sin(2 minus 119909) + (21199051+120572Γ(2 +

120572))[2 cos 2(119909 minus 1) + sin 2(1 minus 119909)] The analytical solution ofthis equation is 119906(119909 119905) = 1199052 sin(119909) sin(2 minus 119909)

The maximum error temporal and spatial convergenceorders by difference scheme (14) for various 120572 are listed inTable 1 From the obtained results we can draw the followingconclusions the experimental convergence orders are appro-ximately 1 and 4 in temporal and spatial directions respec-tively Figures 1 and 2 show the comparison of the numericalsolution with the analytical solution at 119905 = 05 and 120572 = 08 fordifferent temporal and spatial mesh sizes

By Table 1 and Figures 1 and 2 it can be seen that thenumerical solution is in excellent agreement with the analyt-ical solution These results confirm our theoretical analysis

6 Conclusion

In this paper a computationally effective numericalmethod isproposed for simulating themodified time fractional Fokker-Planck equation It has proven the unconditional stability

0 05 1 15 2

0

002

004

006

008

01

012

014

016

018

x

Numerical solutionAnalytical solution

u(xt=05)

Figure 2 The comparison of the numerical solution with theanalytical solution at 119905 = 05 (120591 = 1200 ℎ = 150)

and solvability of proposed scheme Also we showed that themethod is convergent with order 119874(120591 + ℎ4) The numericalresults demonstrate the effectiveness of the proposed scheme

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The work was partially supported by Tianshui Normal Uni-versity ldquoQingLanrdquo Talent Engineering Funds

References

[1] K S Miller and B Ross An Introduction To the FractionalCalculus and Fractional Differential Equations JohnWiley NewYork NY USA 1993

[2] I Podlubny Fractional Differential Equations Academic Press1999

[3] A Barari M Omidvar A R Ghotbi and D D Ganji ldquoApplica-tion of homotopy perturbationmethod and variational iterationmethod to nonlinear oscillator differential equationsrdquo ActaApplicandae Mathematicae vol 104 no 2 pp 161ndash171 2008

[4] S Das ldquoAnalytical solution of a fractional diffusion equation byvariational iterationmethodrdquoComputers andMathematics withApplications vol 57 no 3 pp 483ndash487 2009

[5] D D Ganji and A Sadighi ldquoApplication of homotopy-perturb-ation and variational iterationmethods to nonlinear heat trans-fer and porous media equationsrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 24ndash34 2007

[6] E Hesameddini and F Fotros ldquoSolution for time-fractionalcoupled Klein-Gordon Schrodinger equation using decompo-sition methodrdquo International Mathematics Olympiad vol 7 pp1047ndash1056 2012

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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

10 Abstract and Applied Analysis

[7] G-C Wu and E W M Lee ldquoFractional variational iterationmethod and its applicationrdquo Physics Letters A vol 374 no 25pp 2506ndash2509 2010

[8] C-M Chen F Liu and K Burrage ldquoFinite difference methodsand a fourier analysis for the fractional reaction-subdiffusionequationrdquo Applied Mathematics and Computation vol 198 no2 pp 754ndash769 2008

[9] C Celik andM Duman ldquoCrank-Nicolson method for the frac-tional diffusion equation with the Riesz fractional derivativerdquoJournal of Computational Physics vol 231 no 4 pp 1743ndash17502012

[10] X Hu and L Zhang ldquoImplicit compact difference schemes forthe fractional cable equationrdquo Applied Mathematical Modellingvol 36 pp 4027ndash4043 2012

[11] M M Meerschaert and C Tadjeran ldquoFinite difference approxi-mations for two-sided space-fractional partial differential equa-tionsrdquoApplied Numerical Mathematics vol 56 no 1 pp 80ndash902006

[12] D K Salkuyeh ldquoOn the finite difference approximation to theconvection-diffusion equationrdquoAppliedMathematics and Com-putation vol 179 no 1 pp 79ndash86 2006

[13] E Sousa ldquoNumerical approximations for fractional diffusionequations via splinesrdquo Computers and Mathematics with Appli-cations vol 62 no 3 pp 938ndash944 2011

[14] E Sousa ldquoFinite difference approximations for a fractionaladvection diffusion problemrdquo Journal of Computational Physicsvol 228 no 11 pp 4038ndash4054 2009

[15] C Tadjeran and M M Meerschaert ldquoA second-order accuratenumerical method for the two-dimensional fractional diffusionequationrdquo Journal of Computational Physics vol 220 no 2 pp813ndash823 2007

[16] S B Yuste and L Acedo ldquoAn explicit finite difference methodand a new von Neumann-type stability analysis for fractionaldiffusion equationsrdquo SIAM Journal on Numerical Analysis vol42 no 5 pp 1862ndash1874 2005

[17] Q Yang F Liu and I Turner ldquoNumerical methods for frac-tional partial differential equations with Riesz space fractionalderivativesrdquo Applied Mathematical Modelling vol 34 no 1 pp200ndash218 2010

[18] B I Henry T A M Langlands and P Straka ldquoFractionalFokker-Planck equations for subdiffusion with space- andtime-dependent forcesrdquo Physical Review Letters vol 105 no17 Article ID 170602 2010

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