Numerical analysis of a new volterra integro-differential equation involving fractal-fractional operators

Chaos, Solitons and Fractals 130 (2020) 109396

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

Numerical analysis of a new volterra integro-differential equation involving fractal-fractional operators ˘ Seda I˙ GRET ARAZ Siirt University, Department Mathematics, Faculty of Education, Turkey

a r t i c l e

i n f o

Article history: Received 4 July 2019 Revised 14 August 2019 Accepted 20 August 2019

Keywords: Fractal-fractional derivative Fractal-fractional integral Fractional integro-differential equation New numerical scheme

a b s t r a c t In this paper, we suggest a new integro-differential equation by utilizing from the concept of fractalfractional derivative and integral newly introduced by Atangana. We offer that under which conditions the solution of the suggested equation is exist and unique benefitting from Banach fixed-point theorem. Also, we construct a new numerical scheme for the numerical solution of our problem and we give numerical simulation and illustrations for different values of fractional order α and fractal order β . That’s why, this study will guide for researchers significantly in theory and applications.

1. Introduction Although origin of the fractional calculus dates back to almost a hundred years ago, in the recent times,it has been a wide range of applications in science and engineering. In addition, in the last years, new concept of differentiation and integration which combines the concept of fractional differentiation and the concept of fractal derivative has been studied as convolution of power law, exponential decay law, and generalized Mittag-Leffler law with fractal derivative and integral in [1,2]. In [3–18], applications in science and engineering are referred from fluid flow, rheology, dynamical processes, electrical networks, control theory of dynamical systems, mathematical biology, plasma physics and fusion, computational fluid mechanics, images processing, viscoelasticity, chemical physics and many others. The Volterra integral equations has been broadly employed in the various fields of engineering and science, for instance demography, viscoelastic materials, oscillation of a spring, financial mathematics, stochastic dynamical systems, electromagnetic waves, radiative energy transfer, the oscillation problems and mathematical biology. These equations have significant application areas in different branches in recent years. In the literature, until recently although differential equations integer order have been studied in [19–21]. Recently some important researches for fractional integro-differential equations have been handled by different authors. Tate et. al. investigate the existence and interval of existence of solutions, uniqueness,

E-mail address: [email protected] https://doi.org/10.1016/j.chaos.2019.109396 0960-0779/© 2019 Elsevier Ltd. All rights reserved.

© 2019 Elsevier Ltd. All rights reserved.

continuous dependence of solutions on initial conditions, estimates on solutions and continuous dependence on parameters and functions involved in the equations in [22]. Aghajani et. al. study the existence of solutions of a Cauchy type problem for a nonlinear fractional differential equation, via the techniques of measure of noncompactness in [23]. Baleanu et. al. offer a new method to investigate some fractional integro-differential equations involving the Caputo-Fabrizio derivation and they prove the existence of approximate solutions for these problems in [24]. Mahdy [25] present least squares method aid of Hermite polynomials for solving a linear system of fractional integro-differential equations with Caputo derivative. Diethelm et. al. [26] present existence, uniqueness, and structural stability of solutions of nonlinear differential equations of fractional order and relate our results to the selection of appropriate numerical schemes for the solution of fractional differential equations. Benchohra et. al. [27] investigate the existence of solutions on a compact interval for fractional integro-differential equations with state-dependent delay by using standard fixed point theorems. In this study, we deal with a new integro-differential equation which include mixed fractal-fractional operators. Briefly, this equation can be written such as; F F M α ,β Dt u 0

(x, t ) = f (x, t, u ) +F0F P Jt α,β [K (x, t, u )].

Here the derivative is Atangana-Baleanu fractal-fractional derivative and integral is Caputo fractal-fractional integral. Also, we can also consider the following equation F F P α ,β u 0 Dt

(x, t ) = f (x, t, u ) +F0F M Jt α,β [K (x, t, u )]

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

2

where derivative is Caputo fractal-fractional derivative and integral is Atangana-Baleanu fractal-fractional integral. Atangana offered a new definition which is fractal-fractional derivative of a given function with power law, exponential decay law and the generalized Mittag-Leffler function in [2]. In this study, we established a new integro-differential equation by using this new concept presented by Atangana. That’s why, this study will lead to new investigations for readers in modelling of physical and biological phenomenons. The plan of this study is as follows. In Section 2, we mention some necessary definitions about fractal-fractional calculus. In Section 3, we introduce a new fractional integro-differential equation with mixed fractal-fractional operators. We present necessary conditions for existence and uniqueness for solution of the suggested equation. In Section 4, we give a new numerical scheme for the considered equation and this numerical method is practical and useful in solving such equations. In Section 5, we give some illustrations for different values of fractional order α and fractal order β . In Section 6, we discuss the results about numerical solution of this equation.

3. Volterra equation with mixed fractal-fractional operators In this section, we handle new fractal-fractional integrodifferential equation F F M α ,β Dt u 0

(x, t ) = f (x, t, u ) +F0F P Jt α,β [K (x, t, u )]

where the differential is Atangana-Baleanu fractal-fractional derivative and integral is the Caputo fractal-fractional integral operator. We reformulate this equation such as; F F M α ,β Dt u 0

(x, t ) = f (x, t, u ) +

Definition 1. Suppose that f (t) be continuous and fractal differentiable on an open interval (a, b) with order β , then the fractalfractional derivative of f (t) with order α in the Riemann-Liouville sense having power law type kernel is given by; FFP

α ,β

D0,t ( f (t ) ) =

1 d (m − α ) dt β



t 0

u(x, t ) − u(x, 0 ) =

α ,β

AB(α ) d (1 − α ) dt β

 0

t



Eα −

α

1−α

 (t − s )α f (s )ds,

where 0 < α , β ≤ 1 and AB(α ) = 1 − α + (αα ) . Definition 3. Suppose that f (t) be continuous on an open interval (a, b), then the fractal-fractional integral of f (t) with order α having power law type kernel is given by; F F P α ,β J0,t

( f (t ) ) =

β (α )

 0

t

(t − s )α−1 sβ −1 f (s )ds.

Definition 4. Suppose that f (t) be continuous on an open interval (a, b), then the fractal-fractional integral of f (t) with order α having Mittag-Leffler type kernel is given by; F F M α ,β J0,t



t αβ sβ −1 f (s )(t − s )α −1 ds ( f (t ) ) = AB(α )(α ) 0 β (1 − α )t β −1 f (t ) + . AB(α )

Definition 5. Suppose that f (t) be continuous on opened interval I, the fractal-Laplace transform of order α is given by; F α Jp

( f (t ) ) =



∞ 0

sβ −1 (t − s )α −1 K (x, s, u )ds.





t αβ sβ −1 (t − s )α −1 f (x, s, u )ds AB(α )(α ) 0 β t β −1 (1 − α ) + f (x, t, u ) AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 K (x, s, u )ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 K (x, s, u )ds. (4) AB(α )(2α ) 0

u(x, t ) = u(x, 0 ) +

Now we shall define the mapping T.

Definition 2. Suppose that f (t) be continuous and fractal differentiable on an open interval (a, b) with order β , then the fractalfractional derivative of f (t) with order α in the Riemann-Liouville sense having generalized Mittag-Leffler type kernel is given by;

D0,t ( f (t ) ) =

0

t

t αβ sβ −1 (t − s )α −1 f (x, s, u )ds AB(α )(α ) 0 β t β −1 (1 − α ) + f (x, t, u ) AB(α )  t β α ,β + sβ −1 (t − s )α −1 F F0M Jl (α ) 0 (3) [K (x, l, u )]dlds

or equivalently

where m − 1 < α , β ≤ m ∈ N and

FFE



Now by taking Caputo fractal-fractional integral on both side, we can write Eq. (2) as follows;

(t − s )m−α−1 f (s )ds,

df (s ) f (t ) − f (s ) = lim . t→s dt β t β − sβ

β (α )

(2)

2. Preliminaries In this section, we refer from some important concepts on the fractal-fractional calculus which are used further in this paper.

(1)

exp (−pt )t α −1 f (t )dt,

α > 0.



t αβ sβ −1 (t − s )α −1 f (x, s, u )ds AB(α )(α ) 0 β t β −1 (1 − α ) + f (x, t, u ) AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 K (x, s, u )ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 K (x, s, u )ds. (5) AB(α )(2α ) 0

T u = u = u(x, 0 ) +

Taking norm on both side of Eq. (5), we have the following

   t β −1 u(x, 0 ) + AB(ααβ (t − s )α−1 f (x, s, u )ds )(α ) 0 s   β −1     + β t AB((α1−) α ) f (x, t, u )  T u  =   + β β t β −1 (1−α )  t sβ −1 (t − s )α−1 K (x, s, u )ds   (α ) AB(α ) 0     + αβ 2  t sβ −1 (t − s )2α−1 K (x, s, u )ds  AB(α )(2α ) 0 and then



(6)

t αβ sβ −1 (t − s )α −1  f (x, s, u )ds AB(α )(α ) 0 β t β −1 (1 − α ) +  f (x, t, u ) AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 K (x, s, u )ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 K (x, s, u )ds. (7) AB(α )(2α ) 0

T u ≤ u(x, 0 ) +

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

Assuming that K, f and u are continuous and bounded, we can write the following inequalities;

Thus, we can have the following

δn (x, t ) ≤

u(x, 0 ) ≤ M1  f (x, t, u ) ≤ M2 K (x, t, u ) ≤ M3 .

(8)

Let s = ty, then inequality (7) is rearranged as;

αβ β t β −1 (1 − α ) M2 t β +α −3 B(β , α ) + M2 AB(α )(α ) AB(α ) β β t β −1 (1 − α ) + M3 t β +α −3 B(β , α ) (α ) AB(α ) αβ 2 + M3 t β +2α −3 B(β , α ). (9) AB(α )(2α )

T u  ≤ M 1 +

Thus we have Tu < M < ∞. Also, we can write the following statement using the equality (6);    t β −1 αβ   s (t − s )α −1 [ f (x, s, u ) − f (x, s, v )]ds AB (α )(α ) 0     β −1 (1−α ) β t   + f x, t, u − f x, t, v [ ] ( ) ( ) AB (α )   T u − T v =  β β t β −1 1−α  t . α −1 ( ) β −1 +  s t − s K x, s, u − K x, s, v ds [ ] ( ) ( ) ( ) 0  (α ) AB(α )      + AB(ααβ)(2 2α ) 0t sβ −1 (t − s )2α−1 [K (x, s, u ) − K (x, s, v )]ds 



t αβ t β +α −3 B(β , α )H1 δn−1 (x, s )ds AB(α )(α ) 0  t β t β −1 (1 − α ) + H1 δn−1 (x, s )ds AB(α ) 0  t β β t β −1 (1 − α ) β +α−3 + H2 t B (β , α ) δn−1 (x, s )ds (α ) AB(α ) 0  t αβ 2 + H2 t β +2α −3 B(β , α ) δn−1 (x, s )ds. AB(α )(2α ) 0

 f (x, t, v ) − f (x, t, u ) < H1 u − v K (x, t, v ) − K (x, t, u ) < H2 u − v.

δn (x, t ) =

(11)

Thus we obtain the following inequality

β β +α−3 t B(β , α )l1 u − v (α ) β t β −1 (1 − α ) + l 1 u − v  AB(α ) β β t β −1 (1 − α ) β +α−3 + t B(β , α )l2 u − v (α ) AB(α ) αβ 2 + t β +2α −3 B(β , α )l2 u − v (12) AB(α )(2α )

and reorganize as follows;

⎛

T u − T v  ≤

(19)



t αβ  f (x, s, un−1 ) AB(α )(α ) 0 − f (x, s, un−2 )sβ −1 (t − s )α −1 ds β t β −1 (1 − α ) + [ f (x, t, un−1 ) − f (x, t, un−2 )] AB(α )  β β t β −1 (1−α ) t β −1 + s (t −s )α−1 K (x, s, un−1 ) (α ) AB(α ) 0 − K (x, s, un−2 )ds  t αβ 2 + sβ −1 (t − s )2α −1 [K (x, s, un−1 ) AB(α )(2α ) 0 − K (x, s, un−2 )]ds (20)

and we have

(14)

For L < 1, the contraction is obtained. Now, we calculate un (x, t ) − un−1 (x, t ). Thus we get; un (x, t ) − un−1 (x, t )  t αβ = sβ −1 (t − s )α −1 [ f (x, s, un−1 ) − f (x, s, un−2 )]ds AB(α )(α ) 0

β t β −1 (1 − α ) [ f (x, t, un−1 ) − f (x, t, un−2 )] AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α −1 [K (x, s, un−1 ) − K (x, s, un−2 )]ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 [K (x, s, un−1 ) − K (x, s, un−2 )]ds. (15) AB(α )(2α ) 0 +

We shall define the function δ n (x, t) such that

δn (x, t ) = un (x, t ) − un−1 (x, t ).

δn (x, t ) ≤

(13)

So we have

T u − T v ≤ Lu − v.



t αβ sβ −1 (t − s )α −1 AB(α )(α ) 0 [ f (x, s, un−1 ) − f (x, s, un−2 )]ds β t β −1 (1 − α ) + [ f (x, t, un−1 ) − f (x, t, un−2 )] AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 (α ) AB(α ) 0 [K (x, s, un−1 ) − K (x, s, un−2 )]ds  t αβ 2 + sβ −1 (t − s )2α −1 AB(α )(2α ) 0 [K (x, s, un−1 ) − K (x, s, un−2 )]ds

and providing that K and f are Lipschitz, then

⎞

β −1 β β +α −3 B(β , α )l1 + β t AB((α1−) α ) l1 (α ) t β −1 ⎜ ⎟ ⎝ + (βα ) β t AB((α1−) α ) t β +α−3 B(β , α )l2 ⎠u − v. 2 β +2α −3 B (β , α )l + AB(ααβ 2 )(2α ) t

(18)

Proof. By taking norm on both side of the equality (15),

Suppose that the functions f and K are Lipschitz, then we have

T u − T v  ≤

(17)

assuming that f and K holds the following inequality

(10)

 f (x, t, u ) − f (x, t, v ) ≤ l1 u − v K (x, t, u ) − K (x, t, v ) ≤ l2 u − v.

3

(16)

δn (x, t ) ≤



t αβ t β +α −3 B(β , α )H1 δn−1 (x, s )ds AB(α )(α ) 0  t β t β −1 (1 − α ) + H1 δn−1 (x, s )ds AB(α ) 0  t β β t β −1 (1 − α ) β +α−3 + H2 t B (β , α ) δn−1 (x, s )ds (α ) AB(α ) 0  t αβ 2 + H2 t β +2α −3 B(β , α ) δn−1 (x, s )ds. (21) AB(α )(2α ) 0

 Theorem 1. If K and f are continuous in 0 < τ < t < T < ∞ and −∞ < u(x, t ) < ∞. If in addition K and f are Lipschitz, then our equation has a solution. Proof. Suppose that the conditions of the theorem are hold, then we have that

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

4

δn (x, t ) ≤





t t αβ β t β −1 (1 − α ) t β +α −3 B(β , α )H1 H1 δn−1 (x, s )ds + δn−1 (x, s )ds AB(α )(α ) AB(α ) 0 0   t t β β t β −1 (1 − α ) β +α−3 αβ 2 + H2 t B (β , α ) H2 t β +2α −3 B(β , α ) δn−1 (x, s )ds + δn−1 (x, s )ds (α ) AB(α ) AB(α )(2α ) 0 0

and we get the following

δn (x, t ) ≤

+

 β −1 (β , α )H1 + β t AB((α1−) α ) H1 δ0 (x, t ) 2 β +2α −3 B (β , α ) (β , α ) + AB(ααβ )(2α ) H2 t

αβ t β +α −3 B AB(α )(α ) β −1 β β t ( 1 −α ) H2 t β +α −3 B (α ) AB(α )

<

(22)

n

β −1 αβ t β +α −3 B(β , α )H1 + β t AB((α1−) α ) H1 AB(α )(α ) β −1 2 β +2α −3 B (β , α ) + (βα ) β t AB((α1−) α ) H2 t β +α −3 B(β , α ) + AB(ααβ )(2α ) H2 t

 < max |u(x, 0 )| x∈[b,X ] t∈[a,T ]

max |u(x, 0 )|

x∈[b,X ] t∈[a,T ]

n β −1 (β , α )H1 + β t AB((α1−) α ) H1 . 2 β +2α −3 B (β , α ) (β , α ) + AB(ααβ )(2α ) H2 t

αβ t β +α −3 B AB(α )(α ) β −1 β β t ( 1 −α ) β + (α ) AB(α ) H2 t +α −3 B

(23)

With f and K being contraction H1 and H2 are less than 1, then let us consider

u(x, t ) =

n 

δi (x, t )

(24)

i=0

where the function u(x, t) exists and is continuous as finite summation of continuous functions. In addition, we now let

u(x, t ) = un (x, t ) − Rn (x, t )

(25)

where the function Rn (x, t) is the remainder of the difference approximate and exact solution

u(x, t ) − un (x, t ) =



t αβ β t β −1 (1 − α ) sβ −1 (t − s )α −1 f (x, s, u − un )ds + f (x, t, u − un ) AB(α )(α ) 0 AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 K (x, s, u − un )ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 K (x, s, u − un )ds. AB(α )(α )(α ) 0

But u − un = Rn , then



t αβ β t β −1 (1 − α ) sβ −1 (t − s )α −1 f (x, s, u − Rn )ds + f (x, t, u − Rn ) AB(α )(α ) 0 AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 K (x, s, u − Rn )ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 K (x, s, u − Rn )ds. AB(α )(α )(α ) 0

u(x, t ) − un (x, t ) = u(x, 0 ) +

It follows that

u(x, t ) − u(x, 0 ) −

(26)



(27)

t αβ β t β −1 (1 − α ) sβ −1 (t − s )α −1 f (x, s, u )ds − f (x, t, u ) AB(α )(α ) 0 AB(α )  β β t β −1 (1 − α ) t β −1 − s (t − s )α−1 K (x, s, u )ds (α ) AB(α ) 0  t  t αβ 2 αβ − sβ −1 (t − s )2α −1 K (x, s, u )ds = Rn (x, t ) + sβ −1 (t − s )α −1 [ f (x, s, u − Rn )− f (x, s, u )]ds AB(α )(2α ) 0 AB(α )(α ) 0  β t β −1 (1 − α ) β β t β −1 (1 − α ) t β −1 + s [ f (x, t, u − Rn ) − f (x, t, u )] + (t − s )α−1 [K (x, s, u − Rn ) − K (x, s, u )]ds AB(α ) (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 [K (x, s, u − Rn ) − K (x, s, u )]ds. (28) AB(α )(2α ) 0

Thus by taking the norm on both side equality (28), we have the following

  u(x, t ) − u(x, 0 ) − αβ 0t sβ −1 (t − s )α−1 f (x, s, u )ds − β t β −1 (1−α ) f (x, t, u ) AB(α )(α ) AB(α )   t β −1   − (βα ) β t AB((α1−) α ) 0 sβ −1 (t − s )α −1 K (x, s, u )ds      t β −1 αβ 2 2α −1   − s t − s K x, s, u ds ( ) ( ) AB(α )(2α )

0

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5



t αβ sβ −1 (t − s )α −1  f (x, s, u − Rn ) − f (x, s, u )ds AB(α )(α ) 0 β t β −1 (1 − α ) +  f (x, t, u − Rn ) − f (x, t, u ) AB(α )  β β t β −1 (1 − α ) t β −1 + s (t − s )α−1 K (x, s, u − Rn ) − K (x, s, u )ds (α ) AB(α ) 0  t αβ 2 + sβ −1 (t − s )2α −1 K (x, s, u − Rn ) − K (x, s, u )ds. AB(α )(2α ) 0

≤ Rn (x, t ) +

(29)

Now benefitting from the contraction of the function of f and K, then we have

  u(x, t ) − u(x, 0 ) − αβ  t sβ −1 (t − s )α−1 f (x, s, u )ds − β t β −1 (1−α ) f (x, t, u )   AB(α )(α ) 0 AB(α ) t β −1   − (βα ) β t AB((α1−) α ) 0 sβ −1 (t − s )α −1 K (x, s, u )ds      2 t αβ 2α −1 β −1   − AB(α )(2α ) 0 s K (x, s, u )ds (t − s ) αβ β t β −1 (1 − α ) t β +α −3 B(β , α )H1 Rn (x, t ) + H1 Rn (x, t ) AB(α )(α ) AB(α ) β β t β −1 (1 − α ) β +α−3 αβ 2 + H2 t B(β , α )Rn (x, t ) + H2 t β +2α −3 B(β , α )Rn (x, t ) (α ) AB(α ) AB(α )(2α ) ≤ Rn (x, t ) +

(30)

Since Rn (x, t) → 0, n → 0, thus we obtain the following equality

u(x, t ) − u(x, 0 ) −



t αβ β t β −1 (1 − α ) sβ −1 (t − s )α −1 f (x, s, u )ds − f (x, t, u ) AB(α )(α ) 0 AB(α )  β β t β −1 (1 − α ) t β −1 − s (t − s )α−1 K (x, s, u )ds (α ) AB(α ) 0  t αβ 2 − sβ −1 (t − s )2α −1 K (x, s, u )ds = 0. AB(α )(2α ) 0

(31)

This equality means that the function u (x, t) is the solution of our equation



t αβ β t β −1 (1 − α ) sβ −1 (t − s )α −1 f (x, s, u )ds + f (x, t, u ) AB(α )(α ) 0 AB(α )   t β β t β −1 (1 − α ) t β −1 αβ 2 + s sβ −1 (t − s )2α −1 K (x, s, u )ds (t − s )α−1 K (x, s, u )ds + (α ) AB(α ) AB(α )(2α ) 0 0

u(x, t ) = u(x, 0 ) +

(32)

and this function satisfies the following inequality;

⎛

⎜ u(x, t ) < ⎝

αβ

AB(α )(α )

⎞n β −1 t β +α −3 B(β , α )H1 + β t AB((α1−) α ) H1 β −1

+ (βα ) β t AB((α1−) α ) H2 t β +α −3 B(β , α ) 2 β +2α −3 B (β , α ) + AB(ααβ )(2α ) H2 t

⎟ |u(x, 0 )|. ⎠ xmax ∈ b,X [

(33)

]

t∈[a,T ]

Let us show that the function u (x, t) is unique in order to finalize our proof. Suppose that the functions u1 , u2 are two different solutions of our equation. Using the above inequality, we can write such as;

⎛

⎜ u1 (x, t ) − u2 (x, t ) ≤ ⎝

αβ

AB(α )(α )

⎞n β −1 t β +α −3 B(β , α )H1 + β t AB((α1−) α ) H1 β −1

+ (βα ) β t AB((α1−) α ) H2 t β +α −3 B(β , α ) 2 β +2α −3 B (β , α ) + AB(ααβ )(2α ) H2 t

⎟ |u(x, 0 )|. ⎠ xmax ∈ b,X [

]

(34)

t∈[a,T ]

Here H1 , H2 < 1, then

u1 (x, t ) = u2 (x, t ).

(35) 

4. Numerical scheme for volterra type In this section, we consider the general linear partial differential equation F F M α ,β Dt u 0

(x, t ) = R(x, t, u(x, t ))

(36)

where

u(x, 0 ) = f (x ).

(37)

The above equation can be converted such as; ABC α ,β u 0 Dt

(x, t ) = β t β −1 R(x, t, u(x, t )) = B(x, t, u(x, t )).

(38)

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6

Applying the integral, we have

1−α B(x, t, u(x, t ) ) AB(α )  t α + (t −τ )α−1 B(x, τ , u(x, τ ) )dτ . AB(α )(α ) 0

u(x, t ) − u(x, 0 ) =

Replacing F(xi , tj , ui ) by its value, we have

β (t )α (x, t, u ) = (α + 2 ) ⎧ ⎡ α ⎫ ⎤ ⎪ ⎨ ( j − k + 1) ⎪ ⎬   ⎢ t β −1 K x , t , uk ( j − k + 2 +α α ) ⎥ i k ⎢ ⎥ i k j−1 ⎢ ⎪ ⎪  ⎩ −( j − k ) ⎭ ⎥ ⎢ ⎥. j − k + 2 + 2 α ( ) × ⎢ % α +1 &⎥ ⎥ k=0 ⎢ j − k + 1 ) ⎣t β −1 K x , t , uk−1  ( ⎦ α − j − k ( ) i k −1 i k−1 ( j − k + 1 + α)

F F P α ,β K 0 Jt

(39) At t = tn , we have

u ( xi , tn ) = u ( xi , 0 ) + +

α

1−α B ( xi , tn , u ( xi , tn ) ) AB(α ) n−1  t j+1  (tn − τ )α−1 B(xi , τ , u(xi , τ ))dτ

AB(α )(α )

tj

j=0

Finally

(40)





R xi , t j , u xi , t j



or

uni = u0i + +

1−α B ( xi , tn , u ( xi , tn ) ) AB(α ) n−1  t j+1  α (tn − τ )α−1 B(xi , τ , u(xi , τ ))dτ . (41)

AB(α )(α )

j=0

tj

Thus we get the following 1−α B ( xi , tn , u ( xi , tn ) ) AB(α )

uni = u0i + +

  n−1    α (t )α γn B ( x 0 , t 0 , u ( x 0 , t 0 ) ) + βn− j B xi , t j , uij AB(α )(α + 2 ) j=0

where





β −1

B xi , t j , uij = β t j





R xi , t j , uij .

(43)

Thus





R xi , t j , u xi , t j





α ,β



α ,β 

(44)



K xi , t j , uij



.

Thus at xi , tj , we have F F P α ,β K 0 Jt



β (x, t, u )] = (α )

β (x, t, u ) = (α ) =

t

(t − s )α−1 sβ −1 K (x, s, u )ds.

0



tj

0

(46)

 α−1 sβ −1 t j − s K (xi , s, ui )ds

j−1  

β (α ) k=0

tk+1

s

tk

 β −1

Now we consider the general Volterra equation with power-law kernel derivative and the Atangana-Baleanu fractional integral F F P α ,β u 0 Dt

(x, t ) = f (x, t, u ) +F0F M Jt α,β [K (x, t, u )].

tj − s

α−1

RL α 0 Dt u

(x, t ) = β t β −1 R(x, t, u(x, t )) = B(x, t, u(x, t ))

R(x, t, u(x, t ) ) = f (x, t, u ) +F0F P Jt

α ,β

[K (x, t, u )].



u(x, 0 ) −α 1 t + (1 − α ) (1 − α )

t 0

(56)

Applying the integral on both sides, we obtain the following

u(x, t ) − u(x, 0 ) =



1

(α )

t

0

B(x, τ , u(x, τ ) )(t − τ )−α dτ

(47)

at the point (xi , tn+1 ), we have

(48)

u(xi , tn+1 ) − u(xi , 0 )=



1

(α )

tn+1 0

(57)

B(xi , τ , u(xi , τ ) )(tn+1 − τ )−α dτ (58)

then we have

(x, t, u ) =

(55)

d u(x, τ )(t − τ )−α dτ dτ

= B(x, t, u(x, t ) ).

K (xi , s, ui )ds

sβ −1 K (xi , s, ui ) = F (xi , s, ui )

(54)

Using the connection of Riemann-Liouville with Caputo, we obtain

We put

F F P α ,β K 0 Jt

(53)

where

(45)

We present the integral part F F P α ,β [K 0 Jt

⎧ ⎫ ⎤ ( j − k + 1 )α ⎪ ⎪ ⎨ ⎬ ⎢ t β −1 K x , t , uk  ( j − k + 2 +α α ) ⎥ i k ⎢ k ⎥ i j−1 ⎪ ⎪ ⎩ −( j − k ) ⎭ ⎥ β (t )α  ⎢ ⎢ j − k + 2 + 2α ) ⎥. ( + ⎢ % (α + 2 ) α +1 &⎥ ⎥ k=0 ⎢ j − k + 1) ⎣t β −1 F x , t , uk−1  ( ⎦ α −( j − k ) i k−1 i k−1 ( j − k + 1 + α) ⎡

Using the differentiability property of the integral the above is converted to

[K (x, t, u )].

= f xi , t j , uij +F0F P Jt j



(52) (42)

Here we shall take R(x, t, u(x, t)) as follows;

R(x, t, u(x, t ) ) = f (x, t, u ) +F0F P Jt



= f xi , t j , uij

(51)

β (α )

 j−1

k=0

tk+1

tk



tj − s

α−1

F (xi , s, ui )ds.

(49)

uni +1 = u0i +

Now applying the Atangana-Toufik method, we get

β (t )α (x, t, u ) = (α + 2 ) ⎡ ⎤    ( j − k + 1 )α ( j − k + 2 + α ) k F x , t , u α j−1 ⎢ i k i  − ( j − k ) ( j − k + 2 + 2α ) ⎥ ⎢ ⎥.  × ⎣   ⎦ ( j − k + 1 )α+1 k=0 F xi , tk−1 , uki −1 α −( j − k ) ( j − k + 1 + α )

and

F F P α ,β F 0 Jt

=

u0i

1

(α )

n   j=0

t j+1

tj

B(xi , τ , u(xi , τ ) )(tn+1 − τ )−α dτ

  n    (t )α j + γ B ( x , t , u ( x0 , t0 ) ) + βn− j B xi , t j , ui . (α + 2 ) n 0 0 j=0

(59)

Here

(50)

β −1

B ( x0 , t0 , u ( x0 , t0 ) ) = β t0 and

R ( x0 , t0 , u0 )

(60)

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

7

Fig. 1. Numerical simulation on xy plane for α =0.9 and β =0.7.

Fig. 2. Numerical simulation on xyz plane for α =0.9 and β =0.7.

R ( x0 , t0 , u0 ) = f ( x0 , t0 , u ( x0 , t0 ) ) +

1−α K (x0 , t0 , u(x0 , t0 ) ). AB(α )

So we have the following scheme

(61) We write β −1

B(xi , t j , u(xi , t j )) = β t j



R xi , t j , uij



R

xi , t j , uij



=f



xi , t j , uij



B ( xi , tk , u ( xi , tk ) ) = β tk

 1−α  + K xi , t j , uij AB(α )

⎧ ⎫ ⎤ ( j − k + 1 )α ⎪ ⎪ ⎨ ⎬ ⎢ t β −1 K x , t , uk  ( j − k + 2 +α α ) ⎥ i k ⎢ ⎥ i k −( j − k ) j ⎢ α  ⎪ ⎪ ⎥ ⎩ ⎭ β (t ) ⎢ ( j − k + 2 + α ) &⎥ + ⎢ ⎥. % (α + 2 ) α +1 ⎥ k=0 ⎢ j − k + 1) (   ⎣t β −1 K x , t , uk−1 ⎦ α −( j − k + 1 ) i k−1 i k−1 ( j − k + 2 + 2α )

' f



xi , tk , uki



(

 1−α  + K xi , tk , uki AB(α )

⎧ α ⎫ ⎤⎤ ⎪ ⎪ ⎨ ( j − k + 1) ⎬  ⎢ ⎢ t β −1 K x , t , uk ( j − k + 2 +α α ) ⎥⎥ i k ⎢ ⎢ k ⎥⎥ i −( j − k ) ⎪ ⎪ ⎢ ⎢ ⎥⎥ ⎩ ⎭ j α ⎢ ⎥⎥ ⎢ j − k + 2 + α) ( β −1 ⎢ β (t ) ⎢ ⎥⎥. +β tk ⎢ ⎢ ⎥ ⎧ ⎫⎥ ⎢ (α + 2 ) k=0 ⎢ α +1 ⎬⎥⎥ ⎨ ⎢ ⎢ ⎥⎥ ( j − k + 1)   α ⎣ ⎣t β −1 K x , t , uk−1 ⎦⎦ −( j − k + 1 ) i k−1 i k−1 ⎩( j − k + 2 + 2α )⎭ ⎡

(62)

and



β −1

⎡



(64)

⎡

(63)

5. Numerical illustrations and simulations Application, consider the following fractal-fractional differential equation with mixed operator F F M α ,β Dt u 0

where

(x, t ) = f (x, t ) +F0F P Jt α,β [K (x, t, u )]

8

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

Fig. 3. Numerical simulation on xz plane for α =0.9 and β =0.7.

Fig. 4. Numerical simulation on yz plane for α =0.9 and β =0.7.

Fig. 5. Numerical simulation on xy plane for α =0.9 and β =0.9.

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

Fig. 6. Numerical simulation on xyz plane for α =0.9 and β =0.9.

Fig. 7. Numerical simulation on xz plane for α =0.9 and β =0.9.

Fig. 8. Numerical simulation on yz plane for α =0.9 and β =0.9.

9

10

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

Fig. 9. Numerical simulation on xy plane for α =0.3 and β =0.3.

Fig. 10. Numerical simulation on xyz plane for α =0.3 and β =0.3.

Fig. 11. Numerical simulation on xz plane for α =0.3 and β =0.3.

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

Fig. 12. Numerical simulation on yz plane for α =0.3 and β =0.3.

Fig. 13. Numerical simulation on xy plane for α =0.03 and β =0.03.

Fig. 14. Numerical simulation on xyz plane for α =0.03 and β =0.03.

11

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

12

Fig. 15. Numerical simulation on xz plane for α =0.03 and β =0.03.

Fig. 16. Numerical simulation on yz plane for α =0.03 and β =0.03.

) f (x, t ) = sin xex

β t β −1 (1 − α )t 2α+2β +1 (α + 2β ) AB(α )(2α + 2β ) * αβ 2 3α+2β +1 (α + 2β ) + t . AB(α ) (3α + 2β )

We apply the suggested numerical scheme. The numerical solutions are depicted for different values of α , β in Fig. 1–16. 6. Conclusion In this paper, we offered newly introduced fractional integrodifferential equation where the derivative is Atangana-Baleanu fractal-fractional derivative and the integral is Caputo fractalfractional integral. The definition of fractal-fractional derivative and integral of a given function with power law, exponential decay law and the generalized Mittag-Leffler function was defined by Atangana in [2]. This concept opened new doors of investigation in real world problems, for instance it was used to examine chaotic behavior of some attractors from applied mathematics.

We combined with Atangana-Baleanu fractal-fractional derivative and Caputo fractal-fractional integral in a integro-differential equation in this study and especially this choice is the main contribution for our study. We hope that this mathematical equation will present new ideas and this study will shed light on theory and applications for our readers. We present general conditions about existence and uniqueness for solution of newly introduced equation. We establish the numerical algorithm for the suggested equation by using the newly introduced numerical scheme and we present numerical simulation for this equation for different values of fractional order α and fractal order β . The results show that the numerical scheme is highly efficient and useful in solving such equations and also the method converges rapidly to the exact solution.

Declaration of Competing Interest The author declare that there is no conflict of interests regarding the publication of this paper.

S.I.˙ ARAZ / Chaos, Solitons and Fractals 130 (2020) 109396

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