Restricted fractional differential transform for solving irrational order fractional differential equations

Chaos, Solitons and Fractals 101 (2017) 81–85

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Chaos, Solitons and Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos

Restricted fractional differential transform for solving irrational order fractional differential equations Ayad R. Khudair∗, S.A.M. Haddad, Sanaa L. khalaf Department of Mathematics, Faculty of Science, University of Basrah, Basrah, Iraq

a r t i c l e

i n f o

Article history: Received 16 March 2017 Revised 18 April 2017 Accepted 17 May 2017

Keywords: Fractional differential equations Two dimensional differential transformation Fractional differential transform Restricted fractional differential transform

a b s t r a c t Arikoglu and Ozkol developed a new semi-analytical numerical technique, fractional differential transform method (FDTM), for solving fractional differential equations (FDEs). FDTM was not achieved for solving irrational order fractional differential equations. Here we develop a new method to be applicable for solving rational or irrational order FDEs. This method is called the restricted fractional differential transform method (RFDTM). In fact, RFDTM is based on the restriction of the classical two dimensional differential transform methods. A useful theorem is provided, and Several FDEs are solved by using RFDTM. Moreover, several illustrative examples are presented to demonstrate the accuracy and effectiveness of the proposed method. © 2017 Elsevier Ltd. All rights reserved.

1. Introduction Fractional calculus and FDEs are widely explored subject due to the great importance of many applications in fluid mechanics, biology, control theory of dynamical systems, probability and statistics, viscoelasticity, polymer modeling, finance, physics and engineering, see e.g. Schneider and Wyss [1], Mainardi [2], Magin et al. [3], Magin [4], Metzler and Klafter [5], Beyer and Kempfle [6], Lederman et al. [7], Bagley and Torvik [8], Riewe [9], Kulish and Lage [10], Wyss [11], Song and Wang [12], and the works by Diethelm and Freed cf. Keil et al. [13]. Comprehensive reviews of literature concerning the application of fractional differential equations may be found in the books by Oldham and Spanier [14], Diethelm [15], and Podlubny [16]. There are very few of FDEs can be solved analytically. Thus, accurate and efficient numerical techniques are needed. Various semi-numerical techniques have been proposed for approximate solutions of the fractional order differential and fractional order integral equations, For example, the Adomian decomposition method [17–21], variational iteration method [18,22–26], homotopy perturbation method [27–30], Wavelet Method [31,32], fractional difference method [17,33], and extrapolation method [34]. Another efficient and accurate semi-numerical method, such as FDTM, was introduced by Arikoglu and Ozkol [35] to solve linear and nonlinear initial value problems of fractional order, which utilize the form of fractional power series as the approximation to the exact solution.

∗

Corresponding author. E-mail address: [email protected] (A.R. Khudair).

http://dx.doi.org/10.1016/j.chaos.2017.05.026 0960-0779/© 2017 Elsevier Ltd. All rights reserved.

It is appropriate to note that FDTM is currently of considerable interest for solving FDEs, see, e.g. ErtÄurk and Momani [36], Oturanc et al. [37], Arikoglu and Ozkol [38], and later work followed by Nazari and Shahmorad [39]. The motivation for the present article is the work of Arikoglu and Ozkol [35,38]. Arikoglu and Ozkol [35] developed the differential transform method (DTM) to introduce a new analytical technique for solving FDEs that is named as FDTM. They argued that one should expand the analytic solution function as the following fractional power series [40],

y (x ) =

∞ 

k

F ( k ) ( x − x0 ) γ

(1)

k=0

where γ is the order of the fraction to be selected and F(k) is the kth fractional differential transform of y(x)given by

F (k ) =

⎧ ⎨ ⎩



1 dk/γ f (x )  (k/γ )! dxk/γ x=x 0

k/γ ∈ Z + 0

k/γ ∈ /Z

(2)

+

Now, if the considering FDEs include a fractional derivative, sayDα , where α is irrational number, then one cannot find γ in Eqs. (1) and (2). So that, it is impossible to apply FDTM for solving FDEs. The object of this work is to present a new technique, RFDTM, to be applicable to solve FDEs, which including rational or irrational fractional derivative. In the following sections RFDTM will be presented.

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A.R. Khudair et al. / Chaos, Solitons and Fractals 101 (2017) 81–85



2. Restricted fractional differential transform method

× There are many definitions of fractional order derivative, e.g. Riemann–Liouville derivative, Grünwald–Letnikov derivative, Caputo derivative, Sonin–Letnikov derivative, Miller–Ross derivative, Hadamard derivative, Weyl derivative, Marchaud derivative, Riesz–Miller derivative, Erdélyi–Kober derivative [15,16,41]. Caputo derivative is always used in FDEs to express of many real world physical problems since it has the advantage of defining integer order initial conditions. The Caputo derivative for any analytic function u(x) is defined by [40] C

Dαx0 u(x ) =

1 (n − α )



x0

(x − t )α−n+1

n − 1 < α < n, n ∈ Z +

dt,

 =

i=0 j=0

By setting

1 F (i, j ) = i! j!



∂ i+ j f (x, y ) ∂ xi ∂ y j

∂ i+ j f (x, y ) ∂ xi ∂ y j



=

∞ 

(4)

( x 0 ,y0 )

 xα j

∞ 

i=0 U (i,

∞ 

∞  ∞ 



=

(6)



∂ i+ j f (x, y ) ∂ xi ∂ y j



(7) ( x 0 ,y0 )

U (i, j )(x − x0 )i+α j

(8)

i=0 j=0

Eq. (7) is called the restricted fractional differential transform (RFDTM), while Eq. (8) is called inverse of RFDTM. Now, let u(x), v(x) and w(x) can be express as u(x ) =

∞ ∞

∞ ∞ i+ α j , v ( x ) = i+ α j and w (x ) = i=0 j=0 U (i, j )x i=0 j=0 V (i, j )x

∞ ∞ i + α j respectively, then the fundamental mathei=0 j=0 W (i, j )x matical operations performed by RFDTM are introduced in the following theorems.



j ∞  i  

j 

Proof ((1)). It hold directly.

j ∞  i  

∞  ∞ 



U (i, k )xi

i=0

βk γ j−k xα j

∞ 

V (i, j − k )xi

i=0

U (i, k )V (i − r, j − k )xi U (i, k )V (i − r, j − k )xi

W (i, j )xi+α j =

=

∞ 

ω j xα j

j=0

j ∞  ∞  i  

U (i, k )V (i − r, j − k )xi+α j ,

By comparing the coefficients of x, one can have

W (i, j ) =

j i  

V (i − r, k ) for i ≥ 0, j ≥ 0.

j

i

k=0 r=0

U (i, k )V (i − r, j − k )

k=0 r=0

Theorem (3). If v(x ) = xm+α n u(x ), where m and n are an integer number then

V (i, j ) = 0 for i < m or j < n V (i, j ) = U (i − m, j − n ) for i ≥ m and j ≥ n Proof (3).

v ( x ) = xm+α n u ( x ), xm+α n u ( x ) =

∞  ∞ 

U (i, j ) xi+m+α j+α n

i=0 j=0

v (x ) =

m −1  n−1 

V (i, j ) xi+α j +

i=0 j=0

+

m −1  n−1 

V (i, j ) xi+α j +

+

V (i, j ) xi+α j

∞  ∞ 

V (i, j ) xi+α j

i=m j=n

V (i, j ) xi+α j +

i=0 j=0

U (r, j − k )

m −1  ∞  i=0 j=n

∞  n−1  i=m j=0

=

Theorem (2). If w(x ) = u(x )v(x ) then W (i, j ) =

m −1  ∞ 

V (i, j ) xi+α j

i=0 j=n

∞  n−1 

V (i, j ) xi+α j

i=m j=0

Proof ((2)).

j=0 i=0

j ∞  

j )xi

βk γ j−k

k=0

j ∞  

i=0 V (i,

j=0 k=0

ω j xα j , ω j =

Theorem (1). If w(x ) = u(x ) + v(x ) then W (i, j ) = U (i, j ) + V (i, j ), for i ≥ 0 , j ≥ 0.

∞  ∞ 



∞

j=0 i=0 k=0 r=0

Clearly, F(i, j) in Eq. (5) is the two dimensions differential transform of the functionf(x, y), while Eq. (6) represent the differential inverse transform of F(i, j). Now, If u(x ) = f (x, y )|y=(x−x )α +y , where α > 0, that is, the two 0 0 dimensional function f(x, y) is restricted to one dimensional function u(x), then Eq. (5) and Eq. (6) respectively, become

w ( x ) = u ( x )v ( x )

j )xi , β j =

i=0 k=0 r=0

(5) ( x 0 ,y0 )

i=0 j=0

∞  ∞ 

γj

xα j

k=0 i=0 r=0

F (i, j )(x − x0 )i (y − y0 ) j

1 U (i, j ) = F (i, j ) = i! j!



γ j xα j =

j=0 i=0

f (x, y ) =

∞ 

j=0

k=0

=

i+α j

j=0

∞

β j xα j

ωj =



( x − x0 )i ( y − y0 ) j

βj



j=0

Then

u (x ) =

∞ 

where γ j =

Let f(x, y): R2 → R be analytical function then it can be express as multi Taylor series about (x0 , y0 )as follows:



V (i, j )x

j=0

(3)

∞  ∞  1 f (x, y ) = i! j!



j=0 i=0

j=0

u(n ) (t )

x

∞  ∞ 

+



W (i, j )xi+α j =

∞  ∞  j=0 i=0

U (i, j )xi+α j

∞  ∞ 

V (i + m, j + n ) xi+m+α j+α n

i=0 j=0

=

∞  ∞  i=0 j=0

U (i, j ) xi+m+α j+α n

A.R. Khudair et al. / Chaos, Solitons and Fractals 101 (2017) 81–85

By comparing the coefficients of x, one can have

Proof (5).

V (i, j ) = 0 for i < m or j < n V (i, j ) = U (i − m, j − n ) Theorem (4). If v(x ) =

u (x ) =

for i ≥ m and j ≥ n

d m u (x ) d xm

V (i, j ) = U (i + m, j )

+

C

Proof (4).

u (x ) =

U (i, j ) x

=

i=0 j=0

+

∞  ∞ 

U (i, 0 )x

∞  ∞ 

(i + α j + 1 ) i+α j−α x , (i + α j − α + 1 )

U (i, j + 1 )

V (i, j ) = U (i, j + 1 )

U (i, j ) (i + α j )(i + α j − 1 )

(i + α j + α + 1 ) i+α j x , (i + α j + 1 )

(i + α j + α + 1 ) for i ≥ 0, j ≥ 0, (i + α j + 1 )

3. Illustrated example This section is devoted to demonstrating the applicability of RFDTM to solve several initial value problems of FDEs, including the fractional Bagley–Torvik differential equations and fractional Ricatti differential equations.

i=0

m −1  ∞ 

U (i, j ) (i + α j )(i + α j − 1 )

Example 1. Consider the fractional differential equation

i=0 j=1

C

· · · (i + α j − m + 1 )xi+α j−m ∞  ∞  + U (i, j ) (i + α j )(i + α j − 1 )

U (i, 0 ) = 0, i = 2, 3, . . . i+α j+1 ) U (i, j + 1 ) = − (( U (i, j ), i+α j+α +1 )

i=0 j=1

· · · (i + α j − m + 1 )xi+α j−m ∞  ∞  + U (i + m, j ) (i + m + α j )(i + m + α j − 1 ) i=0 j=1

j = 0, 1, 2, . . .

u ( 0 ) = U ( 0, 0 ) = A u ( 0 ) = U ( 1, 0 ) = B Now, we solve the above linear forward system by using Maple software to have

u(x ) = A + Bx −

i+α j

· · · ( i + α j + 1 )x ∞  ∞ ∞   v (x ) = V (i, j ) xi+α j = V (i, 0 ) xi

+

i=0

−

V (i, j ) xi+α j

i=0 j=1

−

U (i, j ) = 0,

, j≥0

+

Theorem (5). If v(x ) = C Dα u(x ), [α ] < α < [α ] + 1 then

+

i≥0

i = 0, 1, 2, . . . ,

Also, we apply RFDTM to the initial conditions Eq. (10), to find

U (i, j ) (i + α j )(i + α j − 1 )


V (i, j ) = U (i + m, j ) (i + α j + k ),

(10)

By applying RFDTM to Eq. (9), one get

i=0

i=0 j=0

(9)

u(0 ) = A, u (0 ) = B.

· · · (i + α j − m + 1 )xi+α j−m ∞  d m u (x ) ( i + m )! i = U (i + m, 0 ) x d xm i! m −1  ∞ 

Dα u ( x ) + u ( x ) = 0 , 1 < α < 2 with initial conditions

i=m j=1

+

U (i, j )

i=0 j=0

· · · (i + α j − m + 1)xi+α j−m ∞  d m u (x ) ( i + m )! i = U (i + m, 0 ) x m dx i!

∞  ∞ 

∞  ∞ 

i=0 j=0

i=0 j=1

+

i=0 j=1

[α ] < α < [α ] + 1 ∞  ∞  v (x ) = V (i, j ) xi+α j

i+α j

i=0

+

U (i, j ) xi+α j

[α ] < α < [α ] + 1 ∞  (i + 1 ) i−α U (i, 0 ) x (i − α + 1 )

+

∞  d m u (x ) ( i + m )! i = U (i + m, 0 ) x d xm i!

+

∞  ∞ 

i=[α ]+1 i

i=0 j=1

∞  ∞ 

Dα u ( x ) =

i=0

U (i, j ) x

U (i, 0 )xi +

i=0

i=0 j=1

(i + α j + k ) for i ≥ 0 , j ≥ 0

∞ 

∞ 

(i + 1 ) i−α C α D u (x ) = U (i, 0 ) x ( i − α + 1) i=[α ]+1

then

i+α j

U (i, j ) xi+α j =

∞ 

k=1

∞  ∞ 

∞  ∞  i=0 j=0

U (i, j ) = 0 for i < m , j ≥ 1 m

83

k=1

U (i, 0 ) = 0, i = [α ] + 1, [α ] + 2, [α ] + 3, . . . (i + α j + α + 1 ) V (i, j ) = U (i, j + 1 ) for i ≥ 0, j ≥ 0 (i + α j + 1 )

−

Bx1+α

(2 + α )

Ax3α Bx1+2α − (2 + 2α ) (1 + 3α )

+

Ax2α

(1 + 2α )

Bx1+3α Ax5α Ax4α Bx1+4α + + − (2 + 3α ) (1 + 4α ) (2 + 4α ) (1 + 5α ) Bx1+5α Ax6α + (2 + 5α ) (1 + 6α )

Ax7α Bx7α +1 Bx6α +1 Ax8α − − + (2 + 6α ) (1 + 7α ) (2 + 7α ) (1 + 8α ) Bx8α +1

(2 + 8α )

and hence

u (x ) =

Axα

(1 + α )

∞  k=0

(−1 )k



− ···

A x kα Bx1+kα + (1 + kα ) (2 + kα )

 (11)

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A.R. Khudair et al. / Chaos, Solitons and Fractals 101 (2017) 81–85

Indeed Eq. (11) represents the general exact solution of Eq. (9) for any 1 < α < 2 which is conceding with the exact solution [41, page 232]. From Eq. (11) when u(0 ) = 1 ,and u (0 ) = 0the solution is given by

u (x ) =

∞  k=0

(−xα )k (α k + 1 )

On the other hand, when 1 < α < 2 is not specific as a given √ real number. Also when α = π that is, when we have the follow√ C π ing FDE D u(x ) + u(x ) = 0, u(0 ) = A, u (0 ) = B. The FDTM will be not applicable since it is impossible to apply Eqs. (2) and (4) in [35]. Example 2. Consider the fractional Bagley–Torvik differential equation:

u (x ) + Dα u(x ) + u(x ) = 1 + x,

x ∈ [0, 1],

1<α<2

(12)

k=1

Indeed Eq. (14) represents the general solution of Eq. (12) for any 1 < α < 2, u(0 ) = A, and u (0 ) = B. when u(0 ) = 1, and u (0 ) = 1the solution is u(x ) = 1 + x for any 1 < α < 2 which is conceding with the exact solution [35] in case α = 32 . Example 3. Consider the fractional differential equation:

Dα u ( x ) = u 2 ( x ) + 1 ,

with initial conditions

u(0 ) = A, u (0 ) = B.

(1 − B )x1+2α (1 − A )x3α + (2 + 2α ) (1 + 3α ) (1 − B )x1+3α (1 − A )x4α (1 − B )x1+4α (1 − A )x5α + − − + (2 + 3α ) (1 + 4α ) (2 + 4α ) (1 + 5α ) (1 − B )x1+5α + − ··· (2 + 5α )  ∞  (1 − A )xkα (1 − B )x1+kα u(x ) = A + Bx + (−1 )k+1 + (14) (1 + kα ) (2 + kα ) −

By applying RFDTM to Eqs. (12) and (13), one get

U (1, 0 ) = u (0 ) = B, U (i, 0 ) = 0, i = 2, 3, 4, . . . U ( 0, j ) = 0, j = 1, 2, 3, . . . U ( 1, j ) = 0, j = 1, 2, 3, . . . U (i + 2, j )(i + α + 2 )(i + α + 1 ) (i + α j + α + 1 ) +U (i, j + 1 ) (i + α j + 1 ) +U (i, j ) = 1, i = 0, 1, j = 0 U (i + 2, j )(i + α + 2 )(i + α + 1 ) (i + α j + α + 1 ) +U (i, j + 1 ) (i + α j + 1 ) +U (i, j ) = 0, i = 2, . . . j = 0 U (i + 2, j )(i + α + 2 )(i + α + 1 ) (i + α j + α + 1 ) +U (i, j + 1 ) (i + α j + 1 ) +U (i, j ) = 0, i = 0, 1, 2, . . . , j = 1, 2, . . . and hence

U (0, 0 ) = u(0 ) = A,

U (1, 0 ) = u (0 ) = B, U (i, 0 ) = 0, i = 2, 3, 4, . . . U ( 0, j ) = 0, j = 1, 2, 3, . . . U ( 1, j ) = 0, j = 1, 2, 3, . . . 1 ( 1 − A ), (α + 1 ) 1 U ( 1, 1 ) = ( 1 − B ), (α + 2 ) U (i, 1 ) = 0, i = 2, 3, 4, . . . (i + α j + 1 ) U (i, j + 1 ) = − (U (i, j ) (i + α j + α + 1 ) +U (i + 2, j )(i + α + 2 )(i + α + 1 )), U ( 0, 1 ) =

u

(k )

( 0 ) = 0, k = 0, 1, 2, . . . , m − 1

(1 − A )xα (1 − B )x1+α (1 − A )x2α u(x ) = A + Bx + + − (1 + α ) (2 + α ) (1 + 2α )

(16)

By applying RFDTM to Eqs. (15) and (16), one get

U (i, 0 ) = 0, i = 0, 1, 2, . . . U ( 0, 1 ) = U 2 ( 0, 0 ) + 1 = 1 , U (i, 1 ) =

i 

U (r, 0 )U (i − r, 0 ) = 0 ,

i = 1, 2, 3, . . . ,

r=0 i  (i + α j + 1 )  U (i − r, k )U (r, j − k ), (i + α j + α + 1 ) r=0 j

U (i, j + 1 ) =

k=0

i = 0, 1, 2, . . . , j = 1, 2, . . . Now, we solve the above linear forward system by using Maple software to have

(1 + 2α )x3α (1 + 4α )(1 + 2α )x5α u ( x ) = xα + + (1 + 3α ) (1 + 5α )(1 + 3α )



+ +

4(1 + 4α )(1 + 2α )(1 + 6α ) (1 + 7α )(1 + 5α )(1 + 3α )

(1 + 2α )2 (1 + 6α ) 7α x + ··· (1 + 7α )(1 + 3α )2

(17)

Eq.(17) is consistent with the result in [42, page 525]. Example 4. Consider the following Riccati fractional differential equation:

Dα u ( x ) = 2 u ( x ) − u 2 ( x ) + 1 ,

u ( 0 ) = 0, 0 < α < 1

where the exact solution for α = 1 is



u (x ) = 1 +

√ √ 1 2 tanh 2t + log 2

√

2−1 √ 2−1

(18)

.

(19)

By applying RFDTM to Eq. (18), one get

U (0, 0 ) = A, U (i, 0 ) = 0, i = 1, 2, 3, . . . U (0, 1 ) = 2U (0, 0 ) − U 2 (0, 0 ) + 1, U (i, 1 ) = 2U (0, 0 ) −

i = 0, 1, 2, . . . , j = 1, 2, . . . Now, solve the above linear forward system by using Maple software to have

(15)

with the initial conditions

(13)

U (0, 0 ) = u(0 ) = A,

m−1<α
i 

U (r, 0 )U (i − r, 0 )

i = 1, 2, 3, . . . ,

r=0

U (i, j + 1 ) =

(i + α j + 1 ) (i + α j + α + 1 )  × 2U (i, j ) −

j i   k=0 r=0

U (i, k )U (i − r, j − k ) ,

A.R. Khudair et al. / Chaos, Solitons and Fractals 101 (2017) 81–85

i = 0, 1, 2, . . . , j = 1, 2, . . . We now solve the foregoing linear forward system by using Maple software to obtain u (x ) =

xα

(α + 1 )

++

2 x2α

(1 + 2α )

−

  (1 + 2α ) − 4(α + 1 )2 x3α

(α + 1 )2 (1 + 3α )   2 π (1 + 4α ) (1 + 2α ) −4(α + 1)2 (1 + 2α ) + 2(1 + 3α )(α + 1 ) 16−a x4α −  3  2 + 2α (α + 1 )2 (1 + 2α )2 √

+···

(20) If we now pick α = 1 then the solution is

u ( x ) = x + x2 +

7 5 7 6 1 3 1 4 53 7 71 8 x − x − x − x + x + x − ··· 3 3 15 45 315 315 (21)

Eq. (21) is consistent with the exact solution Eq. (19). Also If 1 2 then the solution is

α=

√ 2 x 16 (π − 1 )x3/2 ( π − 4 )x 2 u ( x ) = √ + 2x + + 3 / 2 3 π π π





2 5/2 32 3π + 44π − 32 x − 45 π 5/2   3 2 1 333π + 284π − 512 x − + ··· 36 π2 √ = 1.128379167 x + 2.x + 2.051213127x3/2

−0.2732395433x2 − 5.521879267x5/2 −10.32009894x3 − · · ·

(22)

Eq. (22) is agreement with the result in, Example 6 [35] and Example 3.2 [17]. 4. Conclusion The differential transforms method is a powerful tool for solving differential equations. Arikoglu and Ozkol was modified the differential transform method to be suitable for solving fractional differential equation. This modification is called FDTM. In fact, FDTM is excellent when the order of fractional derivative in the considering FDEs is a specific rational value. There are two cases FDTM cannot be applied. The first is, when the order of fractional derivative is unspecific value in a given interval, say, 0 < α < 1. The second is, when the order √ of fractional derivative is a specific irrational number, like α = 2. To overcome these difficulties we introduce RFDTM to solve the fractional differential equation even if the order of fractional derivative is real (irrational or rational) or unspecific value. This method is not entailed any integration or any complex manipulations even if the FRDEs is content high nonlinearity terms as in the examples (3–4). Also, it is explicit and easy to compute by using Maple software or by setting a computer code. The accuracy and effectiveness of RFDTM depend on the number of terms as shown in illustrated examples. References [1] Schneider W, Wyss W. Fractional diffusion and wave equations. J Math Phys 1989;30(1):134–44. [2] Gorenflo R, Mainardi F. Fractional calculus. Springer; 1997. [3] Magin RL, Ingo C, Colon-Perez L, Triplett W, Mareci TH. Characterization of anomalous diffusion in porous biological tissues using fractional order derivatives and entropy. Microporous Mesoporous Mater 2013;178:39–43. [4] Magin RL. Fractional calculus models of complex dynamics in biological tissues. Comput Math Appl 2010;59(5):1586–93. [5] Metzler R, Klafter J. Boundary value problems for fractional diffusion equations. Physica A 20 0 0;278(1):107–25. [6] H. Beyer, S. Kempfle, Definition of physically consistent damping laws with fractional derivatives, ZAMM-J Appl Math Mech/Zeitschrift fÄur Angewandte Mathematik und Mechanik 75 (8) (1995) 623–635.

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