A Numerical approach of fractional advection-diffusion equation with Atangana–Baleanu derivative

Chaos, Solitons and Fractals 130 (2020) 109527

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

A Numerical approach of fractional advection-diffusion equation with Atangana–Baleanu derivative H. Tajadodi Department of Mathematics, University of Sistan and Baluchestan, Zahedan, Iran

a r t i c l e

i n f o

Article history: Received 24 August 2019 Revised 31 October 2019 Accepted 15 November 2019

2010 MSC: 34A08 65Mxx

a b s t r a c t In the current paper, a new approach is applied to solve time fractional advection-diffusion equation. The utilized fractional derivative operator is the Atangana–Baleanu (AB) derivative in Caputo sense. The mentioned fractional derivative involves the Mittag–Leffler function as the kernel that is both non-singular and non-local. A new operational matrix of AB fractional integration is obtained for the Bernstein polynomials (Bps). By applying the aforesaid matrix, the considered problems are reduced to a system of equations. The approximate solution is derived by solving the yielded system. Also, the error bound is studied. The obtained results show that the applied scheme is simple and powerful tool in finding numerical solutions of fractional equations.

Keywords: Advection-Diffusion Atangana-Baleanu derivative Operational matrix Bernstein polynomials

1. Introduction The advection-diffusion equation is a partial differential equation which describes many quantities such as mass, heat, energy, etc [1,2]. This equation has been used to describe heat transfer in a draining film [3], water transfer in soils [4], pollutant transport in rivers and streams [5], the dispersion of dissolved materials in estuaries and coastal seas [6], long-range transport of pollutants in the atmosphere [7], dispersion of dissolved salts in groundwater [8] and flow in porous media [9]. There have been many special attempts for solving aforesaid equation. In [10], a highorder accurate method suggested for solving the one-dimensional heat and advection-diffusion equations. Dehghan (2004) [11] studied weighted finite difference techniques for solution of this problem. He also studied numerical solution of the three-dimensional advection-diffusion equation in [12]. In [13], collocation method was applied for solving of this equation by radial basis function. The meshless methods is employed for solving this kind of partial differential equations [14]. In the current study, we investigate the one-dimensional advection-diffusion equation of time fractional derivative type. Fractional calculus (FC) has a long history. It has been used in diverse fields such as mathematics, physics, chemistry, engineering, economics, finance, biology, etc [15,16]. There are various kinds of fractional integral and differential operators.

E-mail address: [email protected] https://doi.org/10.1016/j.chaos.2019.109527 0960-0779/© 2019 Published by Elsevier Ltd.

© 2019 Published by Elsevier Ltd.

The most important of these operators are Riemann–Liouville and Caputo. Kernel of these definitions are singular and local. Since fractional differential equations with singular kernel can not better describ nonlocal dynamics systems. In 2015, a new definition of fractional derivation is introduced by Caputo and Fabrizio [17]. Losada and Nieto considered properties of the newly definiton [18]. Kernel of this fractional derivative is non-singular and exponential [19–21]. After that, another type of fractional derivatives with nonsingular kernel is proposed by Atangana and Baleanu which the kernel of this derivative contains the Mittag-Leffler function [22]. kernel of this operators are nonlocal and non-singular. This new definition is called Atangana–Baleanu (AB) derivative and gives the better description of the dynamics systems with memory effect [23–27] In the current study, we consider the time fractional onedimensional advection-diffusion equation as: ABC

γ

Dt z(x, t ) + λ

∂ z(x, t ) ∂ 2 z(x, t ) =κ , ∂x ∂ x2

(1)

with initial condition

z(x, 0 ) =  (x ),

(2)

and boundary conditions

z(0, t ) = ρ0 (t ),

z(1, t ) = ρ1 (t ),

(3)

where λ is an arbitrary constant which denote the advection coefficient, κ > 0 the diffusion coefficient, ϖ(x), ρ 0 (t) and ρ 1 (t) are

2

H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

γ

known functions. In Eq. (1), ABC Dt represents Atangana–Baleanu (AB) derivative of Caputo type. In the current research work, a numerical method based on Bernstein polynomials is utilized to solve the advection-diffusion equation involving Atangana–Baleanu derivative. In this process, we compute operational matrix of Atangana–Baleanu fractional integration. In consequence, by using this operational matrix, the considered problem converts to systems of equations. Then, the approximate solution is derived by solving the yielded system. The organization of this paper is as: In Section2, definition and properties of AB fractional derivatives are presented. In Section 3, Bernstein polynomials and the error bound are given. After that, we compute the operational matrix of Atangana–Baleanu fractional integration. Section 4 is dedicated the numerical technique for solving the time fractional advectiondiffusion equation. In Section 5, the aforesaid technique is applied to solve some examples. Conclusion is drawn in the last Section.

In this part, we introduce some definitions and results connected to the Atangana–Baleanu fractional derivative. Definition 1. The AB fractional derivative of order γ in Caputo sense is exhibited as [22,24,25,28]:

M (γ ) (1 − γ )

γ

Dt z(t ) =

ABC

M (γ ) 1−γ

γ

Dt z(t ) =

∞ M (γ )  = 1−γ



n=0

∞ M (γ )  = 1−γ



n=0

∞  n=0



t

0

tn

(nγ + 1 )

−γ 1−γ

z  ( ζ )Eγ ( −

γ

(t − ζ )γ )dζ ,

1−γ

(4)

AB γ ABC γ It Dt z

(

AB γ ABC γ It Dt z

(

(t )) = + = =

γ + (γ ) .

+



t 0

(t − ζ )γ −1 z(ζ )dζ , (5)

=

we can write: AB γ It z

(1 − γ ) γ RL γ z(t ) + I z(t ), M (γ ) M (γ ) t

(t ) =

−

(6)

=

γ

where RL It is the Riemann–Liouville fractional integral of order γ as the following:

1

RL γ It z(t ) =



derive the classical integral.

AB γ η It t

=

C M (γ )

[1 − γ +

tγ

(γ )

],

C∈R

(t )] = AB Itη [AB Itγ z(t )].

(8)

γ

It z(t ) =

M (γ ) 1−γ

(t − ζ )nγ z (ζ )dζ

(13)

(t ).

(14)

(15)

(1 − γ ) ABC γ Dt z(t ) M (γ )  t γ (t − ζ )γ −1 ABC Dtγ z(ζ )dζ . M (γ ) (γ ) 0  (1 − γ ) ABC γ γ RL γ ABC γ Dt z(t ) + I Dt z(t ) M (γ ) M (γ ) t  n ∞  −γ RL γ n+1  It z (t ) 1−γ n=0   n ∞ γ RL γ  −γ RL γ n+1  It It z (t ) 1−γ 1−γ n=0  n ∞  −γ RL γ n+1  It z (t ) 1−γ n=0  n+1 ∞  −γ RL γ +γ n+1  It z (t ) 1−γ n=0 RL 1  It z

(t ) = z(t ) − z(0 ),

3. Bernstein polynomials

∞  n=0



−γ 1−γ

n

Bi,m (t ) =

m

0 ≤ i ≤ m.

(11)

  m−i

  Bi,m (t ) =

=

m i

m−i  k=0

(t ).

t i (1 − t )m−i ,

i

(16)

The another form of (16) can be obtained as:

(10)

RL nγ +1  It z

 

(9)

Theorem 3. The AB derivative is defined in the following explicit form: ABC

0

The Bernstein polynomials of mth degree are stated [29]:

tη γ (η + 1 ) γ [1 − γ + t ], M (γ ) (γ + η + 1 )

AB γ AB η It [ It z

t



We have the following results and theorems of AB integral [28]:

=

RL nγ +1  It z



t

(t − ζ )γ −1 z(ζ )dζ . (7) (γ ) 0 In Eq. (5), if γ = 0 we get the initial function, and if γ = 1 we

AB γ It C

1 (γ n + 1 )

(12)

Proof. By using AB integral and Eq. (11) we have

,

(1 − γ ) γ z(t ) + M (γ ) M (γ ) (γ )

(t ) =

(−γ )n (t − ζ )nγ dζ (1 − γ )n (γ n + 1 )

n=0

n

n

∞ 

(t )) = z(t ) − z(0 ).

Definition 2. Let 0 < γ < 1, the fractional integral associated to the AB derivative is stated below: AB γ It z

z ( ζ )

Theorem 4. Let 0 < γ < 1, then for every z(t) we have:

M ( 0 ) = M ( 1 ) = 1. and M (γ ) = 1 − γ

t 0

−γ 1−γ

in Eq. (4), Eγ (.) and M(γ ) are called the Mittag–Leffler and normalization function respectively, where

Eγ (t ) =





2. Preliminary tools

ABC

Proof. In view of the fact that the power series defining Eγ (t) is convergent, we can rewritten AB derivative by substituting the series expansion of Mittag–Leffler function into Eq. (4) as:

m−i

t (1 − t ) i

=

  (−1 )k

m

t

i



m

m−i

i

k

i



 (−1 )

k=0

k



m−i k

t i+k , i = 0, 1, . . . , m.

We can exhibit that (t ) = ATm (t ), where

tk

(17)

H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

T (t ) = B0,m (t ), B1,m (t ), · · · , Bm,m (t ) ,

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

m

(−1 )0 ..

(−1 )1

0

.

m

m−0

0

1−0



0

(−1 )0

. . .

..

m

...

 ...

i

..

.

...

0

.

0



Tm (t ) = [1, t, . . . , t m ]T

and A is an upper triangular matrix given by      ⎡

 

(−1 )m−0

m

=⎝

≤ =

m

m 

zk Bi,m (t ) = Z T (t ),

(18)

i=0

1 0

⎛

m−0





1



0

1 0



1



m+1 r =0

0

M (m + 1 )!

0

1 0

dxdt

i

r m+1−r ∂ t r ∂ xm+1−r t x

(m+1 )!

M (x+t )m+1 (m+1 )!

 1 

12

2

m ∂ m+1 z(ξ ,η )

0

 1  1  0

(x ∂∂x + t ∂∂t )m+1 z(ξ , η ) (m + 1 )!

2

⎞ 12

dxdt ⎠

 12

2 dxdt

(x + t )2m+2 dxdt

 12

√ M 2 2m+1 ≤ ,  (m + 1 )! (2m + 3 )(m + 2 )

If z(t) ∈ L2 [0, 1], then any function z(t) can be expanded in terms of Bps:

z(t ) ≈



=

⎤ m−0

⎥ ⎥ ⎥ . ⎥ . ⎥ .   ⎥ ⎥ m m − i ⎥ (−1 )m−i ⎥. ⎥ i m−i ⎥ ⎥ . . ⎥ . ⎥ ⎥   ⎦ m 0 (−1 ) 0

3

where ξ ∈ [0, x], η ∈ [0, t], we have the above bound.

(26) 

3.2. Operational matrix of AB fractional integration

where

Z = [z0 , . . . , zm ]T ,

(19)

(t ) = [B0,m (t ), B1,m (t ), . . . , Bm,m (t )]T .

(20)



1

Z =

0



z(t ) (t )dt Q −1 . T

where

Q = (t ), T (t ) =



1 0

(21)

(t )T (t )dt.

m  m 



AB γ

L2 ([0,

1] × [0, 1]), may be

zi, j Bi,m (x )B j,m (t ) = T (x )Z (t ),

 (m+1 )×(m+1 )

(23)

and zi, j = Bi,m (x ), z(x, t ), B j,m (t ).

RL γ It

Theorem 5. Suppose that z(x, t ) ∈ C m+1 [0, 1] × [0, 1], Y = span{B0,m (t ), . . . , Bm,m (t )} ⊂ L2 [0, 1] and   Y ∈ span{B0,m (x ), · · · , Bm,m (x )} ⊂ L2 [0, 1]. If zm (x, t ) = Y × Y is the best approximation of z(x, t) by means of Bps, then the error bound is presented as:

√ M 2 2m+1

.  (m + 1 )! (2m + 3 )(m + 2 )

(24)

m+1



t (1 − γ ) γ (t ) + (τ )(t − τ )γ −1 dτ . M (γ ) M (γ ) (γ ) 0 (1 − γ ) γ RL γ = (t ) + I (t ). (28) M (γ ) M (γ ) t

RL I γ (t ) t

using Bernstein basis:

(t ) Fγ (t ),

(29)

where Fγ is operational matrix of the R-L fractional integration. For more details, you can see [30,31]. By substituting Fγ in Eq. (28), we get AB γ

(1 − γ ) γ RL γ (t ) + I (t ) M (γ ) M (γ ) t   (1 − γ ) γ γ = I+ F (t ), M (γ ) 1−γ

I (t ) =

3.1. Error analysis

z(x, t ) − zm (x, t ) 2 ≤

(27)

I (t ) =

(22)

i=0 j=0

where Z = zi, j

(t ) Iγ (t ),

Now, we expand

An arbitrary function z(x, t) defined over expanded in Bernstein basis as:

z(x, t ) ≈

AB γ It

where Iγ is called the operational matrix of fractional integration for Bps. We obtain the matrix Iγ as follows:

Then, vector Z can be obtained by: T

The AB fractional integral of (t) can be approximated in terms of Bps as:

(30)

where I(m+1 )×(m+1 ) is an identity matrix. In consequence, the operational matrix of AB fractional integration is equaled:

Iγ =

  (1 − γ ) γ γ I+ F . M (γ ) 1−γ

(31)

4. Description of the proposed method

x,t ) where M = max(x,t )∈[0,1]×[0,1] | ∂∂t r ∂ xmz(+1 −r |, and r = 0, · · · , m + 1.

In this section, we consider Eq. (1) with the initial-boundary conditions (2) and (3). For solving the considered equation, we ap-

Proof. z(x, t) can be expanded into Taylor series

2 ) proximate ∂ ∂z(xx,t using the Bernstein Polynomials as: 2

∂ ∂ 1 ∂ ∂ + t )z ( 0, 0 ) + ( x + t )2 z ( 0, 0 ) ∂x ∂t 2! ∂ x ∂t 1 ∂ ∂ + ··· + ( x + t )m z ( 0, 0 ), (25) m! ∂ x ∂t

z1 (x, t ) = z(0, 0 ) + (x

Since zm (x, t) is the best approximation z(x, t) out of Y × Y , z1 (x, t) ∈ Y × Y , we have

z(x, t ) − zm (x, t ) 2 ≤ z(x, t ) − z1 (x, t ) 2    m+1   1 ∂ ∂   = x +t z ( ξ , η ) ∂t  (m + 1 )! ∂ x 

2

m  m ∂ 2 z(x, t )  = zi, j Bi,m (t )B j,m (t ) = T (x )Z (t ). 2 ∂x i=0 j=0

(32)

Integrating with respect x once yields

∂ z(x, t ) ∂ z(0, t ) = + T (x )FT Z (t ). ∂x ∂x

(33)

Again, integrating Eq. (33) with respect x, we get

z(x, t ) = ρ0 (t ) + x

 T ∂ z ( 0, t ) + T (x ) F2 Z (t ). ∂x

(34)

4

H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

Putting x = 1, in Eq. (34) we get

 T ∂ z ( 0, t ) = ρ1 (t ) − ρ0 (t ) − T (1 ) F2 Z (t ), ∂x

(35)

by substituting Eq. (35) in Eq. (33), we have

 T ∂ z(x, t ) = T (x )FT Z (t ) − T (1 ) F2 Z (t ) − ρ0 (t ) + ρ1 (t ) ∂x = T (x )FT Z (t )−T (x )dT (1 )(F2 )T Z (t )−T (x )dP0T (t ) + T (x )dP1T (t )

 = T (x ) FT Z −dT (1 )(F2 )T Z −dP0T +dP1T (t ) = T (x )M1 (t ),

(36)

where d = [1, 1, · · · , 1]T and P0 and P1 can be expanded by Bps as:

ρ0 (t ) = P0T (t ),

(37)

ρ1 (t ) = P1T (t ),

(38)

Fig. 1. Numerical solution with γ = 0.99 and m = 3 for Ex. 1.

where P0 = [ρ00 , ρ01 , · · · , ρ0m ]T , P1 = [ρ10 , ρ11 , · · · , ρ1m ]T . Now, by substituting Eq. (35) in Eq. (34), we have

 T

z (x, t ) = T (x ) F2



Table 1 The absolute errors for various values γ and m = 4 at t = 0.8 for Ex. 1.

Z (t ) + ρ0 (t )

  T + x ρ1 (t ) − ρ0 (t ) − T (1 ) F2 Z (t )  T = T (x ) F2 Z (t ) + T (x )dP0T (t ) + T (x )CP1T (t )  T − T (x )CP0T (t ) − T (x )C T (1 ) F2 Z (t )  T  T  = T (x ) F2 Z + dP0T + CP1T − CP0T − C T (1 ) F2 Z (t ) = T (x )M2 (t ).

(39)

x

γ = 0.99

γ = 0.97

γ = 0.9

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

2.7498 × 10−4 6.3345 × 10−4 6.02346 × 10−4 2.09987 × 10−5 8.25677 × 10−4 1.61134 × 10−3 2.19108 × 10−3 2.60144 × 10−3 2.83754 × 10−3 2.4489 × 10−3

2.93242 × 10−4 1.49675 × 10−3 1.42231 × 10−3 1.2294 × 10−4 1.87481 × 10−3 3.97505 × 10−3 5.73539 × 10−3 6.86698 × 10−3 7.01159 × 10−3 5.3375 × 10−3

9.10476 × 10−5 2.85031 × 10−3 2.11944 × 10−3 1.40019 × 10−3 6.50327 × 10−3 1.18847 × 10−2 1.63624 × 10−2 1.88775 × 10−2 1.82711 × 10−2 1.28807 × 10−2

where C = [c0 , c0 , · · · , cm ]T is obtanied as the following: x = C T (x ). Now applying

(40) AB γ It

In this section we provide the computational results of the mentioned method on some examples.

to both sides of Eq. (1), we get

 ∂ z(x, t ) ∂ 2 z(x, t ) − λ κ ∂x ∂ x2  2    ∂ z(x, t ) γ ∂ z (x, t ) AB γ =  (x ) + κ AB It − λ I , t ∂x ∂ x2

z(x, t ) =  (x ) +

AB γ It



Example 1. We firstly consider the following advection-diffusion equation as [10]:

(41)

by expanding ϖ(x) using Bernstein polynomials and substituting Eqs. (32), (36) and Eq. (39) in Eq. (41), we get

  AB γ I T (x )Z (t )

T (x )M2 (t ) = T (x )W dT (t ) + κ  γ − λ AB It T (x )M1 (t )



T

T

γ



= T (x ) W dT + κ Z Iγ − λM1 Iγ φ (t ),

T

γ



  (x ) M2 − W d − κ Z I + λM1 I φ (t ) = 0. T

γ

γ

(43)

Finally M2 − W dT − κ Z Iγ + λM1 Iγ = 0

γ

Dt z(x, t ) + λ

∂ z(x, t ) ∂ 2 z(x, t ) =κ , ∂x ∂ x2

(45)

with the initial condition

z(x, 0 ) = exp{−

( x − λ )2 }, 4κ

(46)

1 ( ( 1 + t )λ )2 z ( 0, t ) = √ exp{− }, 4κ ( 1 + t ) 1+t

(42)

where  (x ) = W T (x ), W = [0 , 1 , · · · , m ]T . In consequence, we have T

ABC

and the boundary conditions

t

=  (x )W d (t )+κ  (x )Z I (t )−λ (x )M1 I (t ) T

5. Illustrative examples

(44)

By solving Eq. (44), we can get Z. Then, the approximation z(x, t) can be obtained using Eq. (39).

z ( 1, t ) = √

1 1+t

exp{−

( 1 − ( 1 + t )λ )2 }. 4κ ( 1 + t )

(47)

(48)

The exact solution of this example when γ = 1 is z(x, t ) =

√1 1+t

t )λ ) exp{− (x−4κ(1+ (1+t ) }. For this problem, we put λ = 2

0.25. By using the aforementioned method, Figs. 1 and 2 show the numerical solutions for m = 3, 6 with γ = 0.99 and κ = 0.1. In Figs. 3 and 4, we can see the exact solution and numerical solutions with different values of γ when κ = 0.1, 0.025 respectively. In Table 1, the absolute errors is presented for m = 4, κ = 0.1 and various values of γ . Also, Table 2 demonstrates the absolute errors of z(x, t) for γ = 0.98, κ = 0.025 and various values of m.

H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

5

Table 2 The absolute errors for various values m and γ = 0.98, x = 0.01 for Ex. 1. t

m=4

m=6

m=8

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1.08611 × 10−1 6.67039 × 10−2 4.33799 × 10−2 3.12633 × 10−2 2.48122 × 10−2 2.02145 × 10−2 1.5319 × 10−2 9.58834 × 10−3 4.06707 × 10−3 1.3578 × 10−3

8.08534 × 10−4 3.41577 × 10−3 3.5976 × 10−3 3.1522 × 10−3 2.68129 × 10−3 2.25369 × 10−3 1.81808 × 10−3 1.41291 × 10−3 1.16105 × 10−3 1.06073 × 10−3

4.42185 × 10−3 1.37383 × 10−3 8.81477 × 10−4 5.22709 × 10−4 2.21051 × 10−4 6.20983 × 10−5 6.98557 × 10−5 2.41276 × 10−4 3.38745 × 10−4 3.22167 × 10−4

and the boundary conditions

z(0, t ) = a exp(bt ),

(51) √

z(1, t ) = a exp(bt − c¯ )

Fig. 2. Numerical solution with γ = 0.99 and m = 6 for Ex. 1.

Example 2. We investigate another problem as the following form [14]: ABC

γ

Dt z(x, t ) + λ

∂ z(x, t ) ∂ 2 z(x, t ) =κ , ∂x ∂ x2

(49)

(52)

The exact solution for γ = 1 is z(x, t ) = a exp(bt − c¯x ). For this problem, we put λ = 1, κ = 0.1, a = 1, b = 0.2. The approximate solutions of Eq. (49) for m = 4, 6 and γ = 0.99 are demonstrated in Fig. 5 and Fig. 6. The absolute errors are reported for different values of γ and m in Table 3 and 4. Example 3. We consider the following advection-diffusion equation with λ = 1, κ = 0.1 [32]

with the initial condition

z(x, 0 ) = a exp(−c¯x ),

λ ± λ2 + 4κλ > 0. c¯ = 2κ

(50)

ABC

γ

Dt z(x, t ) + λ

∂ z(x, t ) ∂ 2 z(x, t ) =κ , ∂x ∂ x2

Fig. 3. The numerical solutions for κ = 0.1 with different values of γ at t = 0.3, 0.6, 0.9 for Ex. 1.

(53)

6

H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

Fig. 4. The numerical solutions for κ = 0.025, γ = 0.98 with different values of m and t for Ex. 1.

Fig. 5. Numerical solution with γ = 0.99 and m = 4 for Ex. 2.

and the initial condition



z(x, 0 ) = exp(5x ) 0.25 sin

πx 2

+ cos

 π x  2

,

Fig. 6. Numerical solution with γ = 0.99 and m = 6 for Ex. 2.

(54)

5t π 2t ) exp(− ), 2 40

z(1, t ) = 0.25 exp(5(1 −

t π 2t )) exp(− ). 2 40

(55) (56)











z(x, t ) = exp(5(x − 2t )) exp(− π40t ) 0.25 sin π2x + cos π2x . Fig. 7 and Fig. 8 are shown the approximate solutions for γ = 0.99 and m = 6, 12. Also, comparison between the exact solution and the numerical solutions for various values of γ and m = 8 at t = 0.25, 0.5 are demonstrated in Fig. 9. The absolute errors for γ = 0.97, 0.99 and m = 12 at x = 0.01 are reported in Table 5. 2

and the boundary conditions

z(0, t ) = exp(−

For γ = 1, the exact solution of Eq. (53) is

H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

Table 3 The absolute errors for various values γ and x = 0.75, m = 4 for Ex. 2. t

γ = 0.95

γ = 0.99

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1.71859 × 10−3 1.75931 × 10−3 1.71957 × 10−5 1.62395 × 10−4 1.49224 × 10−3 1.33947 × 10−3 1.17588 × 10−3 1.00691 × 10−3 8.33239 × 10−4 6.50761 × 10−4 4.50582 × 10−4

3.79438 × 10−4 3.49969 × 10−4 3.18707 × 10−4 2.85508 × 10−4 2.50304 × 10−4 2.13106 × 10−4 1.73998 × 10−4 1.33143 × 10−4 9.07809 × 10−5 4.72258 × 10−5 2.86994 × 10−6

Table 4 The absolute errors for various values m and x = 0.75, γ = 0.99 for Ex. 2. t 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

m=3

m=4 −4

7.75301 × 10 1.13874 × 10−3 1.41664 × 10−3 1.62234 × 10−3 1.76917 × 10−3 1.87045 × 10−3 1.93953 × 10−3 1.98971 × 10−3 2.03434 × 10−3 2.08671 × 10−3 2.16016 × 10−3

2.72343 × 10−4 3.6244 × 10−4 3.55417 × 10−4 3.30913 × 10−4 3.10176 × 10−4 2.89645 × 10−4 2.62649 × 10−4 2.2919 × 10−4 1.93803 × 10−4 1.5151 × 10−4 6.18581 × 10−5

Fig. 7. Numerical solution with γ = 0.99 and m = 6 for Ex. 3.

Table 5 The absolute errors for various values γ and x = 0.01, m = 12 for Ex. 3. t 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

γ = 0.97

γ = 0.99 −2

1.42113 × 10 3.73734 × 10−3 9.66706 × 10−4 3.54766 × 10−4 9.71242 × 10−4 1.3628 × 10−3 1.43421 × 10−3 1.50647 × 10−3 1.43038 × 10−3 1.40509 × 10−3 1.48575 × 10−3

7.52415 × 10−3 1.14029 × 10−3 1.96678 × 10−4 1.5248 × 10−4 3.83808 × 10−4 4.99935 × 10−4 4.77636 × 10−4 5.5642 × 10−4 4.48081 × 10−4 5.33854 × 10−4 7.70512 × 10−4

Fig. 8. Numerical solution with γ = 0.99 and m = 12 for Ex. 3.

Fig. 9. The numerical solutions with different values of γ at t = 0.25, 0.5 for Ex. 3.

7

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H. Tajadodi / Chaos, Solitons and Fractals 130 (2020) 109527

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