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  • 7/30/2019 AN EFFICIENT CLASS OF FDM BASED ON HERMITE FORMULA FOR SOLVING FRACTIONAL REACTION-SUBDIFFUSION

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    International Journal of Mathematics and Computer

    Applications Research (IJMCAR)

    ISSN 2249-6955Vol. 2 Issue 4 Dec 2012 61-75

    TJPRC Pvt. Ltd.,

    AN EFFICIENT CLASS OF FDM BASED ON HERMITE FORMULA FORSOLVING

    FRACTIONALREACTION-SUBDIFFUSIONEQUATIONS

    N. H. SWEILAM1, M. M. KHADER

    2& M. ADEL

    1

    1Department of Mathematics, Faculty of Science, Cairo University, Giza, Egypt

    2Department of Mathematics, Faculty of Science, Benha University, Benha, Egypt

    ABSTRACT

    In this article, a numerical study is introduced for the fractional reaction-subdiffusion equations by using an

    efficient class of finite difference methods (FDM). The proposed scheme is based on Hermite formula. The stability

    analysis and the convergence of the proposed methods are given by a recently proposed procedure similar to the standard

    John von Neumann stability analysis. Simple and accurate stability criterion valid for different discretization schemes of

    the fractional derivative, arbitrary weight factor, and arbitrary order of the fractional derivative, are given and checked

    numerically. Finally, numerical examples are carried out to confirm the theoretical results.

    KEYWORDS: Finite Difference Methods, Hermite Formula, Fractional Reaction-Subdiffusion Equation, Stability and

    Convergence Analysis

    INTRODUCTION

    In the last few years, there are many studies for the fractional differential equations (FDEs), because of their

    important applications in many areas like physics, medicine and engineering, and this field allows us to study fractal

    phenomena which can not be studied by the ordinary case. There are many applications of the FDEs see (Chang Chen et al.

    (2007) - Hilfer (2000), Liu (2004), Metzler (2000)), the studied models have received a great deal of attention like in the

    fields of viscoelastic materials (Bagley (1999)), control theory (Podlubny (1999)), advection and dispersion of solutes in

    natural porous or fractured media (Benson et al. (2000)), anomalous diffusion. Most FDEs do not have exact solutions, so

    approximate and numerical techniques must be used. Recently, several numerical methods to solve FDEs have been given

    such as, variational iteration method (Sweilam et al.(2007)), homotopy perturbation method (Khader (2012)), Adomian

    decomposition method (Yu (2008)), homotopy analysis method (Sweilam and Khader (2011)), collocation method (Khader

    (2007), Sweilam and Khader (2010)) and finite difference method (Sweilam et al. (2011), Sweilam et al. (2012)).

    In this section, we employed the Caputo definitions of fractional derivative operatorD and its properties (Liu et

    al. (2007a), Metzler (2000), Podlubny (1999)) .

    DEFINITION 1

    The Riemann-Liouville derivative of order of the function )(xy is defined by

    0,>0,>,)(

    )(

    )(

    1=)(

    10

    xd

    x

    y

    dx

    d

    nxyD

    n

    x

    n

    n

    x +

    where n is the smallest integer exceeding and (.) is the Gamma function. If Nm= , then the above definition

    coincides with the classicalthm derivative ).(

    )(xy m

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    62 N. H. Sweilam, M. M. Khader & M. Adel

    FRACTIONAL REACTION-SUBDIFFUSION EQUATION

    The standard meanfield model for the evolution of the concentrations ),( txa and ),( txb ofA and B particles

    is given by the reaction-diffusion equations

    ),,(),(),(=),( 2

    2

    txbtxatxaxDtxat

    (1)

    ),,(),(),(=),(2

    2

    txbtxatxbx

    Dtxbt

    (2)

    where D is the diffusion coefficient assumed in this paper to equal for species and is the rate constant for the

    bimolecular reaction.

    In order to generalize the reaction-diffusion problem to a reaction-subdiffusion problem, we must deal with the

    subdiffusive motion of the particles. Seki et al. (2003) and Yuste et al. (2004) replaced Eqs. (1) and (2) with the set of

    reaction-subdiffusion equations in which both the motion and the reaction terms are affected by the subdiffusive character

    of the process

    )],,(),(),([=),(2

    21 txbtxatxa

    xkDtxa

    tt

    (3)

    )],,(),(),([=),(2

    21 txbtxatxb

    xkDtxb

    tt

    (4)

    where k is the generalized diffusion coefficient andtD is the Riemann-Liouville fractional partial derivative of order

    with respect to t. The fractional reaction-subdiffusion equations (3) and (4) are decoupled, which are equivalent to

    solve the following fractional reaction-subdiffusion equation

    ,

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    An Efficient Class of FDM Based on Hermite Formula forSolving 63FractionalReaction-SubdiffusionEquations

    FINITE DIFFERENCE SCHEME OF THE FRACTIONAL REACTION-SUBDIFFUSION

    EQUATION

    In this section, we introduce an efficient class of FDM and use it to obtain the discretization finite difference

    formula of the reaction-subdiffusion equation (6). For some positive constant numbers Nand M , we use the notations

    x and t , at space-step length and time-step length, respectively. The coordinates of the mesh points are xjxj =

    )0,1,...,=( Nj , and tmtm = , ( = 0,1,..., )m M and the values of the solution ),( txu on these grid points are

    ;UmJ

    m

    jmj utxu ),( whereN

    Lhx == ,

    M

    Tt == .

    For more details about discretization in fractional calculus see (Lubich (1986), Richtmyer et al. (1967), Yuste and

    Acedo (2005)).

    In the first step, the ordinary differential operators are discretized as follows (Zhang (2006)).

    ),(

    )2

    1(1

    1=| ,

    O

    u

    t

    u

    t

    mjt

    mt

    jx +

    +

    +

    (9)

    where+

    t denotes the backward difference operator with respect to t. Now, using the formula (9), we can derive an

    efficient approximate formula of the fractional derivative for ),( txut

    , 1)

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    64 N. H. Sweilam, M. M. Khader & M. Adel

    The above formula can be rewritten as

    1

    ,

    =1

    | = ( ),1

    (1 )2

    m rmt j

    x t rj m

    rt

    uuA O

    t

    + +

    +

    +

    (10)

    where)(2

    =

    A and 11 1)(= rrr .

    We must note thatr

    satisfy the following facts

    (1) 0>r , 1,2,...=r .

    (2) 1> +rr , 1,2,...=r .

    Now, we are going to obtain the finite difference scheme of the fractional reaction-subdiffusion equation (6). In our study

    we take 1== k .

    To achieve this aim we evaluate this equation at the points of the grid ),(mj

    tx ,

    ).,(),(),(

    =),(2

    2

    mjmj

    mj

    mj txgtxux

    txutxu

    t+

    (11)

    Using Eqs.(10) and (11), we have

    ).(),(),(

    )2

    1(1

    =),(

    1

    1=2

    2

    Otxgtxu

    uA

    x

    txumjmj

    t

    rm

    jt

    r

    m

    r

    mj++

    +

    ++

    (12)

    In order to get two additional equations, replace j by 1j and 1+j , respectively, in the above equation, so we have

    ),(),(),(

    )2

    1(1

    =),(

    11

    1

    1

    1=2

    1

    2

    Otxgtxu

    uA

    x

    txumjmj

    t

    rm

    jt

    r

    m

    r

    mj++

    +

    +

    +

    (13)

    and

    ).(),(),(

    )2

    1(1

    =),(

    11

    1

    1

    1=2

    1

    2

    Otxgtxu

    uA

    x

    txumjmj

    t

    rm

    jt

    r

    m

    r

    mj++

    ++

    +

    +

    +

    +

    + (14)

    The Hermite formula with two-order derivatives at the grid point ),( mj tx is

    ).(=)),(),(2),((12),(),(

    10),(

    4

    1122

    1

    2

    2

    2

    2

    1

    2

    hOtxutxutxuhx

    txu

    x

    txu

    x

    txumjmjmj

    mjmjmj

    +

    ++

    +

    +

    (15)

    Substitute from Eqs.(12)-(14) into Eq.(15), and denote ),( mj txu bym

    ju , ),( mj txg bym

    jg , then with neglect the high

    order terms, we have

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    An Efficient Class of FDM Based on Hermite Formula forSolving 65FractionalReaction-SubdiffusionEquations

    ,)(12)52(12)(12=]

    )21(1

    [

    ]

    )2

    1(1

    [10]

    )2

    1(1

    [

    1

    22

    1

    2

    1

    1

    1

    1=

    2

    1

    1=

    2

    1

    1

    1

    1=

    2

    m

    j

    m

    j

    m

    j

    m

    j

    t

    rm

    jt

    r

    m

    r

    m

    j

    t

    rm

    jt

    r

    m

    r

    m

    j

    t

    rm

    jt

    r

    m

    r

    uhuhuhgu

    Ah

    gu

    Ahgu

    Ah

    +++

    +

    +

    +

    +

    ++

    +

    +

    +

    ++

    +

    +

    (16)

    under some simplifications, we can obtain the following form

    .)12

    1(1)

    6

    5(2)

    12

    1(1

    )12

    1

    (1)6

    5

    (2)12

    1

    (1=)(12

    1

    )(6

    5)(

    12

    1)(

    6

    1

    )(3

    5)(

    6

    1

    )12

    1

    6

    5

    12

    1(2)

    12

    1

    6

    5

    12

    1(2

    1

    1

    2121

    1

    2

    1

    22

    1

    21

    11

    2

    121

    11

    2

    1

    1

    12=

    2

    1

    2=

    2

    1

    1

    12=

    2

    1

    1

    11

    1

    2

    11

    2

    +

    +

    ++

    +

    +

    +

    +

    +

    +

    +

    ++

    ++++

    ++

    ++

    +++++

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    rm

    j

    rm

    jr

    m

    r

    rm

    j

    rm

    jr

    m

    r

    rm

    j

    rm

    jr

    m

    r

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    uhuhuh

    uhuhuhggh

    gghgghuuAh

    uuAhuuAh

    uuuAhuuuAh

    (17)

    Letm

    jU be the approximate solution, and letm

    j

    m

    j

    m

    j UuT = , Nj 1,2,...,= , Mm 1,2,...,= be the error, so

    we have the error formula

    .)12

    1(1)

    6

    5(2)

    12

    1(1

    )12

    1

    (1)6

    5

    (2)12

    1

    (1=)(6

    1

    )(3

    5)(

    6

    1

    )12

    1

    6

    5

    12

    1(2)

    12

    1

    6

    5

    12

    1(2

    1

    1

    2121

    1

    2

    1

    22

    1

    2

    1

    1

    12=

    2

    1

    2=

    2

    1

    1

    12=

    2

    1

    1

    11

    1

    2

    11

    2

    +

    +

    +

    +

    +

    +

    +

    +

    +

    ++

    +++

    ++

    +++++

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    rm

    j

    rm

    jr

    m

    r

    rm

    j

    rm

    jr

    m

    r

    rm

    j

    rm

    jr

    m

    r

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    m

    j

    ThThTh

    ThThThTTAh

    TTAhTTAh

    TTTAhTTTAh

    (18)

    with

    .1,2,...,=0,==0 MmTTm

    N

    m(19)

    PROPOSITION 1

    Assuming that the solution of (18) has the formjh

    m

    m

    j eT i= , then

    2 2 2

    12 2 2

    2

    12 2 2 =2

    5 1 1 5( ( )) (1 ) ( ) (1 )6 6 12 12=5 1 5 1

    ( ( )) (1 ) (1 ) ( )6 6 12 12

    5 1( ( ))6 6 ( ),

    5 1 5 1( ( )) (1 ) (1 ) ( )6 6 12 12

    m m

    m

    r m r m r

    r

    cos h h A h cos h h

    cos h h A h h cos h

    h A cos h

    cos h h A h h cos h

    +

    + + +

    + + +

    +

    + + +

    (20)

    where m 2= .

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    66 N. H. Sweilam, M. M. Khader & M. Adel

    PROOF

    Substitute in (18) byjh

    m

    m

    j eT i= and divide by jhe i , we get

    0.=)(6

    1)(

    3

    5

    )(6

    1])

    12

    1(1)

    6

    5(2

    )12

    1[(1])

    12

    1(1)

    6

    5(2)

    12

    1(1[

    )212

    1

    6

    5

    12

    1()2

    12

    1

    6

    5

    12

    1(

    i

    12=

    2

    12=

    2

    i

    12=

    2

    1

    i22

    i2i22i2

    1

    2ii2ii

    h

    rmrmr

    m

    rrmrmr

    m

    r

    h

    rmrmr

    m

    rm

    h

    h

    m

    hh

    m

    hh

    m

    hh

    eAhAh

    eAhehh

    ehehheh

    AheeAhee

    ++

    +

    ++

    +++

    ++++

    (21)

    Using some trigonometric formulae and some simplifications we can obtain

    0.=)())(6

    1

    6

    5(

    )12

    5(1)()

    12

    1(1))(

    6

    1

    6

    5(

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5

    12=

    2

    1

    222

    222

    rmrmr

    m

    r

    m

    m

    hcosAh

    hhcoshAhhcos

    hcoshhAhhcos

    +

    +

    +++

    +++

    (22)

    from which we can obtain the required formula and this completes the proof.

    STABILITY ANALYSIS

    In this section, we use the John von Neumann method to study the stability analysis of the finite difference scheme (17).

    PROPOSITION 2

    Assume that

    ,)(2)(6

    22

    2

    +

    h

    h(23)

    then

    1.

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )12

    5(1)()

    12

    1(1))(

    6

    1

    6

    5(

    0222

    222

    +++

    +++

    hcoshhAhhcos

    hhcoshAhhcos

    (24)

    PROOF

    Since )(2=

    A , then )(2

    1

    =

    A , and by our assumption that )(2)(6

    22

    2

    + h

    h

    , so we can

    obtain

    ,6

    212

    2

    h

    h

    A +

    so,

    ,3

    1

    6

    1

    2

    12

    2222

    AhAhhh +

    and since 1)(2

    hsin , so we have

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    An Efficient Class of FDM Based on Hermite Formula forSolving 67FractionalReaction-SubdiffusionEquations

    ,2

    1)()

    3

    1

    6

    1(2 22222 hAhhsinAhh +

    i.e.,

    0,)()

    3

    1)

    12

    1(2(1

    2

    1 22222+ hsinAhhhAh

    0,)(3

    1)()

    12

    12(1)

    12

    5(1)

    12

    1(1 2222222 ++ hsinAhhsinhhhAh

    so,

    0,))(2(12

    1)

    12

    12(1)

    12

    5(1)

    12

    1(1))](2(1

    2

    1

    3

    1[1 2222 ++ hcoshhhhcosAh

    i.e.,

    0,)]12

    5(1))(2)

    12

    1[(1))(2

    6

    1

    6

    5( 222 +++ hhcoshhcosAh

    and since

    ),12

    5(1))(2)

    12

    1(1))(2)

    12

    1(1)

    12

    5(1 2222 hhcoshhcoshh ++

    we have that

    0,)])(2)12

    1(1)

    12

    5[(1))(2

    6

    1

    6

    5( 222 +++ hcoshhhcosAh

    which completes the proof.

    PROPOSITION 3

    Assume that m , be the solution of (20), with the condition (23), then |||| 0 m , )1,2,...,=( Mm .PROOF

    If we take 1=m in Eq.(20), we obtain

    .||

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )12

    5(1)()

    12

    1(1))(

    6

    1

    6

    5(

    |=| 0222

    222

    1

    hcoshhAhhcos

    hhcoshAhhcos

    +++

    +++

    (25)

    From Proposition 2, we have that |||| 01 .

    Suppose that |||| 0 s , 11,2,...,= ms . For 1>m

    .|

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )())(6

    1

    6

    5(

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )12

    5(1)()

    12

    1(1))(

    6

    1

    6

    5(

    |=||

    222

    1

    2=

    2

    1222

    222

    hcoshhAhhcos

    hcosAh

    hcoshhAhhcos

    hhcoshAhhcos

    rmrmr

    m

    r

    mm

    +++

    +

    +++

    +++

    +

    (26)

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    68 N. H. Sweilam, M. M. Khader & M. Adel

    Now, since

    ,)(=)( 011

    2=121

    2=

    mrmrrm

    rmrmrmr

    m

    r

    +

    +(27)

    we have

    .|

    )()121(1)

    125(1))(

    61

    65(

    ))(6

    1

    6

    5(

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )())(6

    1

    6

    5(

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )12

    5(1)()

    12

    1(1)(1))(

    6

    1

    6

    5(

    |=||

    222

    0

    2

    222

    1

    1

    2=

    2

    1222

    22

    2

    2

    hcoshhAhhcos

    hcosAh

    hcoshhAhhcos

    hcosAh

    hcoshhAhhcos

    hhcoshAhhcos

    m

    rmrr

    m

    r

    mm

    +++

    +

    +

    +++

    +

    +

    +++

    +++

    +

    (28)

    Using triangle inequality

    .|||

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    ))(6

    1

    6

    5(

    |

    |||)()12

    1(1)12

    5(1))(6

    1

    6

    5(

    )())(6

    1

    6

    5(

    |

    |||

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )12

    5(1)()

    12

    1(1)(12))(

    6

    1

    6

    5(

    |||

    0222

    2

    222

    1

    1

    2=

    2

    1222

    22

    2

    2

    hcoshhAhhcos

    hcosAh

    hcoshhAhhcos

    hcosAh

    hcoshhAhhcos

    hhcoshAhhcos

    m

    rm

    rr

    m

    r

    mm

    +++

    +

    +

    +++

    +

    +

    +++

    +++

    +

    (29)

    Now, since |||| 01 m , |||| 0 rm , and also since mrrmr +

    2112= =)( , so the above inequality takes the

    form

    |,||

    )()12

    1(1)

    12

    5(1))(

    6

    1

    6

    5(

    )125(1)()

    121(1))(

    61

    65(

    ||| 0222

    222

    hcoshhAhhcos

    hhcoshAhhcos

    m

    +++

    +++

    (30)

    and using Proposition 2 directly we complete the proof.

    We know that

    .|)(|= 2

    =

    2

    2NT m

    N

    (31)

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    An Efficient Class of FDM Based on Hermite Formula forSolving 69FractionalReaction-SubdiffusionEquations

    THEOREM 1

    The difference scheme (17) is stable under the condition (23).

    PROOF

    From Proposition 3, (31) and the formula2

    0

    2TTm , Mm 1,2,..,= , which means that the difference scheme is

    stable.

    By the Lax equivalence theorem (Richtmyer et al. (1967)) we can show that the numerical solution converges to the exact

    solution as 0, h .

    NUMERICAL RESULTS

    In this section, we will test the proposed method by considering the following numerical examples.

    EXAMPLE 1

    Consider the initial-boundary value problem of fractional reaction-subdiffusion equation

    ,01,

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    70 N. H. Sweilam, M. M. Khader & M. Adel

    Figure 2: The Behavior of the Exact Solution and the Numerical Solution of (32) by Means of the Proposed Method

    at 0.3= ,60

    1=h ,

    70

    1= and 1=T .

    Figure 3: The Behavior of the Exact Solution and the Numerical Solution of (32) by Means of the Proposed Method

    at 0.5= ,60

    1=h ,

    20

    1= , and 2=T .

    Figure 4: The Behavior of the Numerical Solution of (32) by Means of the Proposed Method at

    20

    1=h ,

    20

    1= , and 0.5=T , with different values of

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    An Efficient Class of FDM Based on Hermite Formula forSolving 71FractionalReaction-SubdiffusionEquations

    Figure 5: The Behavior of the Numerical Solution of (32) By Means of the Proposed Method at

    40

    1=h ,

    40

    1= , 0.7= , with different values of T

    EXAMPLE 2

    Consider the following initial-boundary problem of the fractional reaction-subdiffusion equation

    1,

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    72 N. H. Sweilam, M. M. Khader & M. Adel

    Figure 7: The Behavior of the Exact Solution and the Numerical Solution of (33) by Means of the Proposed Method

    at 0.2= ,60

    1=h ,

    20

    1= , and 0.2=T

    Figure 8: The Behavior of the Exact Solution and the Numerical Solution of (33) by Means Of The Proposed

    Method at 0.9= ,50

    1=h ,

    150

    1= and 0.01=T

    Figure 9: The Behavior of the Numerical Solution of the Proposed Problem (33) by Means of the Proposed Metho

    at 0.4= ,40

    1=h ,

    100

    1= , and 1=T

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    An Efficient Class of FDM Based on Hermite Formula forSolving 73FractionalReaction-SubdiffusionEquations

    From the previous figure 9, we can see that the numerical solution is unstable, since the stability condition (32) is not

    satisfied.

    The following two tables 1 and 2 show the magnitude of the maximum error between the numerical solution and the exact

    solution obtained by using the proposed finite difference scheme with different values ofh

    ,

    and the final time

    T.

    Table 1: The Maximum Error with Different Values of h , at 0.5= , and 0.2=T

    h MaximumError

    50

    1

    50

    1

    0.01602

    70

    1

    80

    1

    0.01303

    120

    1

    100

    1

    0.01216

    140

    1

    150

    1

    0.01118

    160

    1

    180

    1

    0.01091

    Table 2: The Maximum Error with Different Values of h , at 0.2= , and 0.1=T

    h MaximumError

    10

    1

    10

    1

    0.01013

    10

    1

    20

    1

    0.00290

    20

    1

    20

    1

    0.00283

    20

    1

    20

    1

    0.00281

    30

    1

    30

    1

    0.00023

    CONCLUSIONS AND REMARKS

    This paper presented a class of numerical methods for solving the fractional reaction-subdiffusion

    equations. This class of methods depends on the finite difference method based on Hermite formula. Special attention is

    given to study the stability and convergence of the fractional finite difference scheme. To execute this aim we have

    resorted to the kind of fractional John von Neumann stability analysis. From the theoretical study we can conclude that,

    this procedure is suitable for the proposed fractional finite difference scheme and leads to very good predictions for the

    stability bounds. Numerical solution and exact solution of the proposed problem are compared and the derived stability

    condition is checked numerically. From this comparison, we can conclude that, the numerical solutions are in excellent

    agreement with the exact solution. All computations in this paper are running using the Matlab programming.

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