Some misinterpretations and lack of understanding in differential operators with no singular kernels

Abdon Atangana; Emile Franc Doungmo Goufo

doi:10.1515/phys-2020-0158

Article Open Access

Some misinterpretations and lack of understanding in differential operators with no singular kernels

Abdon Atangana and Emile Franc Doungmo Goufo

Published/Copyright: October 12, 2020

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From the journal Open Physics Volume 18 Issue 1

Abstract

Humans are part of nature, and as nature existed before mankind, mathematics was created by humans with the main aim to analyze, understand and predict behaviors observed in nature. However, besides this aspect, mathematicians have introduced some laws helping them to obtain some theoretical results that may not have physical meaning or even a representation in nature. This is also the case in the field of fractional calculus in which the main aim was to capture more complex processes observed in nature. Some laws were imposed and some operators were misused, such as, for example, the Riemann–Liouville and Caputo derivatives that are power-law-based derivatives and have been used to model problems with no power law process. To solve this problem, new differential operators depicting different processes were introduced. This article aims to clarify some misunderstandings about the use of fractional differential and integral operators with non-singular kernels. Additionally, we suggest some numerical discretizations for the new differential operators to be used when dealing with initial value problems. Applications of some nature processes are provided.

Keywords: fractional derivative with dimension; integration with dimensionalization; law-related exact solution; numerical approach; applications

1 Introduction

Differential and integral operators have attracted attention of several researchers in the last few decades due to their abilities to replicate processes found in real-world situations. Most of the real-world problems cannot be replicated using the concept of rate of change that gives birth to the classical differentiation initiated by Leibniz and applied by Newton in classical mechanics. This is due to the fact that many real-world problems follow a process different from the process of Markov, the initial conditions together with the generator function cannot really be used to predict the future behavior of the process. In addition to this, nature provides us with complex problems in which the observed facts follow more than one process. For example, the flow of subsurface water within a fracture aquifer, in which the flow within the matrix soil is different from that in a fracture and this situation becomes more complex to be replicated using the concept of rate of change. With all these observations in mind, researchers have embarked themselves in modeling real-world problems using non-local operators. Two non-local differential operators were in fashion in the literature including Riemann–Liouville and Caputo [1] fractional derivatives, and they are all based on the power law. Therefore, they are only able to replicate processes by following the power law. Regrettably enough, many real-world problems were modeled using these differential operators, even those with no power law process. However, it should be noted that the main aim of modeling real-world problems is to analyze, understand and predict. Hence, a mathematical model should be able to represent the observed facts. Therefore, modeling of real-world problems for which the process does not follow the power law with Caputo or Riemann–Liouville derivatives is misleading as the obtained results cannot be used to predict accurately the observed facts.

Nature therefore informs us that the power law cannot be used to replicate its complexities; therefore, new differential and integral operators are needed for better understanding, description and prediction of the concerned process. In 2015, Caputo and Fabrizio [2,3], as real-world problem solvers, unanimously agreed that power-law-based differential and integral operators cannot solve all the problems, in particular, problems following fading memory or as they called it fatigue effect. Indeed, they noted that this class of nature processes cannot be handled using Caputo or Riemann–Liouville derivatives. Thus, they suggested a new differential operator with an exponential decay kernel, a function able to replicate the fading memory or the fatigue effect. This differential operator was an advantage in the field of fractional calculus and applications. While the operator became popular, some researchers noted that the associated integral (or antiderivative) was just a superposition of the function and its classical integral. This observation led some researchers to doubt about the fractional character of such a new operator. The problem was solved later as Atangana and Baleanu [4] suggested a new differential operator with the generalized Mittag–Leffler function, and with this operator, the associated fractional integral was found to be a superposition of the function and its Riemann–Liouville integral, a great contribution as one was able to see a fractional character with memory [5,6,7,8,9,10,11,12]. While the new trends were welcome and acclaimed by the research community, new theoretical results were obtained, new models were constructed, some real-world problems were modeled accurately and some papers with no real contributions were published with the aim of imposing the use of power-law-based fractional operators. In particular, recently, we noted that a paper published in arxiv with the title “A mistake to define fractional derivative with non-singular kernel,” like other already published papers, lacks physical and theoretical arguments underpinning their claims. They claimed that initial value problems with Atangana–Baleanu and Caputo–Fabrizio derivatives do not have exact solutions that satisfy the initial conditions. However, they failed to include the effect of non-localities that could lead to the wrong conclusion. In a comment written by Baleanu [13], he stressed on the fact that modelers always forget the dimension of real-world problems; indeed, this concept is not new but rather, its use has been forgotten. In this article, we present some interesting results to clarify the mistake and lack of understanding for those writing against derivatives with non-singular kernels. We start Section 2 by answering some open questions related to the equality to a constant (using different fractional operators) and those related to fixed point in Section 3. Then, in Section 4 a new local differential operator is introduced and used to construct new fractional differential and integral operators that take into account the effect of dimensionalization. A novel operator is used to address the solvability of some related equations in Section 6, where exact solutions are studied. In Section 7, we propose a discretization scheme applicable to the novel operator before providing some useful applications in Section 8. Let us now start with some critical issues and answers.

2 Critical questions and answers: the constant

In already published papers, it was asked to find the solution of the following equations:

(1) D t α 0 CF ( u ( t ) ) = a , where a ≠ 0 , t ≥ 0 ,

(2) D t α 0 A B C ( u ( t ) ) = a , where a ≠ 0 , t ≥ 0 .

Here, we are going to look at that question and bring some views using simple high school type of mathematics. For (1), we have

(3) D t α 0 A B C ( u ( t ) ) = W ( α ) ( 1 − α ) ∫ 0 t d u ( τ ) d τ E α − α 1 − α ( t − τ ) α d τ = a .

Can this equation have a solution! If yes, can the solution satisfy the initial condition? The answer is no and why is it so? The kernel used here is continuous. Hence, this leads to the following:

∫ 0 t f ( τ ) d τ = a = constant ,

where f is assumed to be continuous. The aforementioned equation makes no sense due to the fact that the operator integral is a non-local operator as it accumulates previous information. Therefore, in a practical point of view, the sum of information over a variable upper boundary cannot be a constant except may be zero, which means adding values with opposite signs.

An example usually given as high school exercise is

∫ 0 t a d τ = a t ≠ constant , t ≥ 0 .

Thus, the general Caputo–Fabrizio equation:

(4) D t α 0 CF ( u ( t ) ) = M ( α ) ( 1 − α ) ∫ 0 t d u ( τ ) d τ exp − α 1 − α ( t − τ ) d τ = a .

except where f ( t ) = 0 and a = 0 . The same conclusion applies to the general Atangana–Baleanu equation:

(5) D t α 0 A B C ( u ( t ) ) = W ( α ) ( 1 − α ) ∫ 0 t d u ( τ ) d τ E α − α 1 − α ( t − τ ) α d τ = a .

2.1 Specific Caputo–Fabrizio’s case

The later equations can also be solved using the Laplace transform. Hence, expressed with the convolution operator *, equation (4) yields

d u ( t ) d t × M ( α ) ( 1 − α ) exp − α 1 − α ( t − τ ) d τ = a .

Taking the Laplace transform on both sides of that equation gives

[ s u ˜ ( s ) − u ( 0 ) ] M ( α ) ( 1 − α ) 1 s + α 1 − α = a s ,

where u ˜ is the Laplace transform of u . More development leads to

s u ˜ ( s ) − u ( 0 ) = a s ( 1 − α ) M ( α ) s + α 1 − α ,

u ˜ ( s ) = u ( 0 ) s + a s 2 ( 1 − α ) M ( α ) s + α 1 − α ,

u ˜ ( s ) = u ( 0 ) s + a M ( α ) ( 1 − α ) s + a s 2 1 − α M ( α ) α ( 1 − α ) .

Taking the inverse Laplace transform yields

(6) u ( t ) = u ( 0 ) + a ( 1 − α ) M ( α ) + a α M ( α ) t .

For this solution to satisfy the initial condition, t = 0 we shall have a ( 1 − α ) M ( α ) = 0 leading to α = 1 , which reduces to the classical case. Can we confirm at this stage that function (6) is the exact solution of the problem (4)? To verify it we pose

D t α 0 CF u ( 0 ) + a ( 1 − α ) M ( α ) + a α M ( α ) t = M ( α ) ( 1 − α ) ∫ 0 t a α M ( α ) exp − α 1 − α ( t − τ ) d τ = a α ( 1 − α ) − 1 − α α exp − α 1 − α y 0 t = a α ( 1 − α ) 1 − α α − 1 − α α exp − α t 1 − α = a 1 − exp − α t 1 − α .

Here when α → 1 we obtain the constant a, and (6) is the exact solution. Hence, problem (4) has an exact solution that satisfies the initial condition u ( 0 ) = 0 only when α = 1 .

2.2 Specific Atangana–Baleanu’s case

We proceed in a similar way for problem (5).

D t α 0 A B C ( u ( t ) ) = a , where a ≠ 0 , t ≥ 0

d u ( t ) d t × W ( α ) ( 1 − α ) E α − α 1 − α ( t ) α = a .

Taking the Laplace transform on both sides of that equation gives

W ( α ) ( 1 − α ) ( s u ˜ ( s ) − u ( 0 ) ) s α − 1 s α + α 1 − α = a s ,

s u ˜ ( s ) − u ( 0 ) = ( 1 − α ) W ( α ) a s s α + α 1 − α s α − 1 ,

u ˜ ( s ) = u ( 0 ) s + ( 1 − α ) W ( α ) a s 2 s α + α 1 − α s α − 1 .

The inverse Laplace transform yields

u ( t ) = u ( 0 ) + ( 1 − α ) a W ( α ) + a α ( 1 − α ) ( 1 − α ) W ( α ) t α Γ ( α + 1 ) ,

giving

(7) u ( t ) = u ( 0 ) + ( 1 − α ) a W ( α ) + a α W ( α ) t α Γ ( α + 1 ) .

For this solution to satisfy the initial condition, t = 0 we shall have ( 1 − α ) a W ( α ) = 0 leading to α = 1 , which reduces to the classical case. Can we again confirm at this stage that function (7) is the exact solution of problem (5)? To verify it we pose

D t α 0 A B C u ( 0 ) + ( 1 − α ) a W ( α ) + a α W ( α ) t α Γ ( α + 1 ) = a α 2 W ( α ) Γ ( α + 1 ) W ( α ) 1 − α ∫ 0 t τ α − 1 E α − α 1 − α ( t − τ ) α d τ = a α 2 Γ ( α + 1 ) 1 1 − α ∫ 0 t ∑ n = 0 ∞ − α 1 − α n ( t − τ ) n α τ α − 1 Γ ( n α + 1 ) d τ = a α 2 Γ ( α + 1 ) 1 1 − α ∑ n = 0 ∞ − α 1 − α n Γ ( n α + 1 ) ∫ 0 t ( t − τ ) n α τ α − 1 d τ .

Now letting τ = y t , we obtain

(8) D t α 0 A B C u ( 0 ) + ( 1 − α ) a W ( α ) + a α W ( α ) t α Γ ( α + 1 ) = a α 2 Γ ( α + 1 ) 1 1 − α ∑ n = 0 ∞ − α 1 − α n Γ ( n α + 1 ) × ∫ 0 1 t n α + α y α − 1 ( 1 − y ) n α d y = a α 2 t α Γ ( α + 1 ) 1 1 − α × ∑ n = 0 ∞ − α t α 1 − α n Γ ( n α + 1 ) B ( α , n α + 1 ) = a α 2 t α Γ ( α + 1 ) 1 1 − α × ∑ n = 0 ∞ − α t α 1 − α n Γ ( n α + 1 ) Γ ( α ) Γ ( n α + 1 ) Γ ( n α + α + 1 ) = a α t α 1 − α E α , α + 1 − α t α 1 − α .

Now putting Z = − α t α 1 − α we have Z → − ∞ as α → 1 and using E 1 , 2 ( Z ) = e Z − 1 Z we finally get a α t α 1 − α E α , α + 1 − α t α 1 − α = − a Z E α , α + 1 ( Z ) → − a Z e Z − 1 Z as α → 1 .

Here when α → 1 we obtain the constant a , and (7) is the exact solution. Hence, problem (5) has an exact solution that satisfies the initial condition u ( 0 ) = 0 only when α = 1 .

2.3 Specific Caputo’s case (power law)

Now we deal with the power law case by considering the classical Caputo derivative

(9) D t α 0 C ( u ( t ) ) = a , where a ≠ 0 , t ≥ 0 ,

which yields the solution

(10) u ( t ) = u ( 0 ) + 1 Γ ( α ) ∫ 0 t a ( t − τ ) α − 1 d τ = u ( 0 ) + a t α Γ ( α + 1 ) ,

which satisfies the initial condition t = 0 . Let us now check the following

(11) D t α 0 C u ( 0 ) + a t α Γ ( α + 1 ) = 1 Γ ( 1 − α ) ∫ 0 t a α τ α − 1 Γ ( α + 1 ) ( t − τ ) − α d τ = a α ( Γ ( 1 − α ) ) Γ ( α + 1 ) ∫ 0 t τ α − 1 ( t − τ ) − α d τ = a α ( Γ ( 1 − α ) ) Γ ( α + 1 ) ∫ 0 t τ α − 1 ( t − τ ) − α d τ = a α ( Γ ( 1 − α ) ) Γ ( α + 1 ) B ( α , 1 − α ) = a α ( Γ ( 1 − α ) ) Γ ( α + 1 ) Γ ( α ) Γ ( 1 − α ) Γ ( 1 ) = a .

Thus, (10) is the exact solution of problem (9) satisfying the initial condition u ( 0 ) = 0 .

We can now ask the real question as follows: How can a non-local operator of a non-zero function give zero, especially when the function is positive? In other words, why do we have for instance

(12) D t α 0 C ( u ( t ) ) = 1 Γ ( 1 − α ) ∫ 0 t d u ( τ ) d τ ( t − τ ) − α d τ = a ? , where a ≠ 0 , t ≥ 0 .

The answer to this question relies on the work done by [14], where the authors provided a detailed analysis for each of the kernels involved. Indeed, they considered the following kernels:

K α ( t ) = 1 Γ ( 1 − α ) t − α ,

K 1 , α ( t ) = M ( α ) 1 − α exp − α 1 − α t ,

K 2 , α ( t ) = W ( α ) 1 − α E α − α 1 − α t α .

The power law kernel K α ( t ) leads to a non-crossover from waiting time distribution to probability distribution. This means that the power law kernel is unable to describe processes taking place in different states. This power law kernel will therefore cause loss of memory in the process. Assuming that the process displays two different types of behaviors, for instance, the random walk in an earlier stage and the power law in a later stage. The power law kernel will only be able to replicate the power law process and will ignore the random walk process. This will result in a loss of memory for the whole process as the effect of the random walk process will be lost.

On the other hand, in their analysis, the authors in [14] discovered that the exponential decay law kernel can capture processes taking place in states, as it can depict processes with distribution that follow the Gaussian and the non-Gaussian laws with a steady state. In their work, they wondered why the process following the steady state can take place without external force. Well, the answer comes from the properties of the exponential function which is known as the queen of classical calculus as many other functions can be derived from the exponential function. Finally, the generalized Mittag–Leffler function that is greater than the exponential function has the same property as the power law for late time.

3 Critical questions and answers: the fixed point

The second question that arises in many researcher’s minds is the following: Can we obtain a fixed-point with a non-singular kernel differential operator? In other terms, can we have, for instance,

(13) D t α 0 CF ( u ( t ) ) = u ( t ) , where t ≥ 0 ?

(14) D t α 0 A B C ( u ( t ) ) = u ( t ) , where t ≥ 0 ?

3.1 Specific Caputo–Fabrizio’s case

If we assume that the aforementioned equations are well posed, then by the Laplace transform, we have from equation (13)

s u ˜ ( s ) − u ( 0 ) s + α 1 − α = u ˜ ( s ) 1 − α M ( α ) ,

u ˜ ( s ) = u ( 0 ) s + u ˜ ( s ) s + α 1 − α s 1 − α M ( α ) ,

u ˜ ( s ) 1 − s + α 1 − α s 1 − α M ( α ) = u ( 0 ) s ,

u ˜ ( s ) = u ( 0 ) s 1 1 − 1 − α M ( α ) − α s ,

u ˜ ( s ) = u ( 0 ) 1 − 1 − α M ( α ) 1 s − α M ( α ) M ( α ) − 1 + α .

By the inverse Laplace transform, we have

u ( t ) = c u ( 0 ) M ( α ) M ( α ) − 1 + α exp α M ( α ) M ( α ) − 1 + α t .

To satisfy the initial condition, we put c = M ( α ) − 1 + α M ( α ) so that

(15) u ( t ) = u ( 0 ) exp α M ( α ) M ( α ) − 1 + α t .

Now we can ask the question: Is (15) the exact solution for (13)?

D t α 0 CF u ( t ) = u ( 0 ) M ( α ) ( 1 − α ) ∫ 0 t α M ( α ) M ( α ) − 1 + α × exp α M ( α ) M ( α ) − 1 + α τ exp − α ( t − τ ) 1 − α d τ = u ( 0 ) 1 ( 1 − α ) α ( M ( α ) ) 2 M ( α ) − 1 + α × ∫ 0 t exp α M ( α ) M ( α ) − 1 + α τ exp − α ( t − τ ) 1 − α d τ = A ( α ) ∫ 0 t exp [ B ( α ) τ ] exp [ − λ ( α ) ( t − τ ) ] d τ = A ( α ) ∫ 0 t exp [ ( B ( α ) + λ ( α ) ) τ − λ ( α ) t ] d τ = A ( α ) ∫ 0 t exp [ C ( α ) τ − λ ( α ) t ] d τ , ( C ( α ) = B ( α ) + λ ( α ) ) ,

where A ( α ) = u ( 0 ) 1 ( 1 − α ) α ( M ( α ) ) 2 M ( α ) − 1 + α , B ( α ) = α M ( α ) M ( α ) − 1 + α and λ ( α ) = α 1 − α .

Now changing the variable as y = C ( α ) τ − λ ( α ) t we have

A ( α ) ∫ 0 t exp [ C ( α ) τ − λ ( α ) t ] d τ = A ( α ) ∫ − λ ( α ) t C ( α ) τ − λ ( α ) t exp [ y ] 1 C ( α ) d y = A ( α ) C ( α ) exp [ ( C ( α ) − λ ( α ) ) t ] − exp [ − λ ( α ) t ] .

Now if α = 1 , then we recover the function u ( t ) = e t .

3.2 Specific Atangana–Baleanu’s case

For the case with the Atangana–Baleanu derivative, can we also have

(16) D t α 0 A B C ( u ( t ) ) = u ( t ) , where t ≥ 0 ?

The obvious solution is u ( t ) = 0 meaning that there is no dynamical process. Is it possible to have a non-trivial solution u ( t ) ≠ 0 to (16)? By using Sumudu transform S we have

[ S ( u ( t ) ) − u ( 0 ) ] W ( α ) 1 − α + α s α = S ( u ( t ) ) ,

S ( u ( t ) ) = u ( 0 ) α W ( α ) B ( α ) 1 − α s α B ( α ) .

Taking the inverse Sumudu transform S − 1 gives directly

(17) u ( t ) = u ( 0 ) c W ( α ) B ( α ) E α α B ( α ) t α ,

where B ( α ) = W ( α ) − 1 + α . Now if α = 1 , then we recover the function u ( t ) = e t , u ( t ) = u ( 0 ) e t , for c = 1 .

Now the question is to know whether solution (17) is the exact solution of (16). To check it, we pose

D t α 0 A B C u ( 0 ) c W ( α ) B ( α ) E α α B ( α ) t α = u ( 0 ) c ( W ( α ) ) 2 ( 1 − α ) B ( α ) ∫ 0 t d d τ E α α B ( α ) τ α × E α − α 1 − α ( t − τ ) − α d τ = u ( 0 ) c ( W ( α ) ) 2 ( 1 − α ) B ( α ) ∫ 0 t ∑ j = 0 ∞ d d τ α B ( α ) τ α j Γ ( j α + 1 ) × ∑ k = 0 ∞ − α 1 − α ( t − τ ) − α k Γ ( k α + 1 ) d τ = u ( 0 ) c ( W ( α ) ) 2 ( 1 − α ) B ( α ) ∑ j = 0 ∞ α B ( α ) j Γ ( j α + 1 ) × ∑ k = 0 ∞ j α − α 1 − α k Γ ( k α + 1 ) ∫ 0 t τ j α − 1 ( t − τ ) k α d τ = u ( 0 ) c ( W ( α ) ) 2 ( 1 − α ) B ( α ) ∑ j = 0 ∞ α B ( α ) j Γ ( j α + 1 ) × ∑ k = 0 ∞ j α − α 1 − α k Γ ( k α + 1 ) t k α + j α B ( j α , k α + 1 ) = u ( 0 ) c ( W ( α ) ) 2 ( 1 − α ) B ( α ) ∑ j = 0 ∞ α B ( α ) j Γ ( j α + 1 ) × ∑ k = 0 ∞ j α − α 1 − α k Γ ( k α + 1 ) t k α + j α Γ ( j α ) Γ ( k α + 1 ) Γ ( j α + k α + 1 ) = u ( 0 ) c ( W ( α ) ) 2 ( 1 − α ) B ( α ) ∑ k = 0 ∞ − α t α 1 − α k E α , k α + 1 α t α B ( α ) ,

where we have integrated following the same technique used in (8). Hence, we have

D t α 0 A B C u ( 0 ) c W ( α ) B ( α ) E α α B ( α ) t α ≠ u ( 0 ) c W ( α ) B ( α ) E α α B ( α ) t α .

3.3 Specific Caputo’s case (power law)

We know that for the power law case, using the Laplace transform on the equation

(18) D t α 0 C ( u ( t ) ) = u ( t ) , where t ≥ 0 ,

leads to the solution

(19) u ( t ) = u ( 0 ) E α [ t α ] .

Indeed, we use the fact that

s α u ˜ ( s ) − s α − 1 u ( 0 ) = u ˜ ( s ) ,

then,

u ˜ ( s ) = s α − 1 u ( 0 ) s α − 1 ,

where the inverse Laplace transform yields solution (19). At this stage, can we affirm that solution (19) is the exact solution to model (18)? In other words, do we have D t α 0 C ( u ( 0 ) E α [ t α ] ) = u ( 0 ) E α [ t α ] ? To check it we pose

D t α 0 C ( E α [ t α ] ) = 1 Γ ( 1 − α ) ∫ 0 t d d τ E α [ τ α ] ( t − τ ) − α d τ = 1 Γ ( 1 − α ) ∫ 0 t d d τ ∑ k = 0 ∞ τ k α Γ ( k α + 1 ) ( t − τ ) − α d τ = 1 Γ ( 1 − α ) ∫ 0 t ∑ k = 0 ∞ k α t k α − 1 Γ ( k α + 1 ) ( t − τ ) − α d τ = 1 Γ ( 1 − α ) ∑ k = 0 ∞ k α Γ ( k α + 1 ) ∫ 0 t t k α − 1 ( t − τ ) − α d τ = 1 Γ ( 1 − α ) ∑ k = 0 ∞ k α Γ ( k α + 1 ) t k α − α B ( k α , 1 − α ) = 1 Γ ( 1 − α ) ∑ k = 0 ∞ t k α − α Γ ( k α ) Γ ( k α ) Γ ( 1 − α ) Γ ( k α + 1 − α ) = t − α E α , 1 − α ( t α ) .

Hence,

D t α 0 C ( u ( 0 ) E α [ t α ] ) = u ( 0 ) t − α E α , 1 − α ( t α ) ,

which result is different from u ( 0 ) E α [ t α ] .

4 Dimensionalization

Very recently, a comment by Dumitru Baleanu [13] suggested that some physical problems modeled by fractional derivatives violated the concept of dimension. Can we perhaps use the dimensionalization to obtain the fixed point? Or even obtain the exact solution that satisfies the initial condition.

4.1 Dimensionalization: specific Caputo–Fabrizio’s case

In this case, we consider the following differential equation:

(20) D t α 0 CF ( u ( t ) ) = G t 1 − α a , where t ≥ 0

and where a is a constant and G t 1 − α is the dimensional function. Hence, applying the Laplace transform ℒ on both sides of (20) yields

[ s u ˜ ( s ) − u ( 0 ) ] M ( α ) ( 1 − α ) 1 s + α 1 − α = ℒ G t 1 − α a = G ˜ s 1 − α a .

More development leads to

s u ˜ ( s ) − u ( 0 ) = G ˜ s 1 − α a ( 1 − α ) M ( α ) s + α 1 − α ,

u ˜ ( s ) = u ( 0 ) s + G ˜ s 1 − α a s ( 1 − α ) M ( α ) s + α 1 − α ,

u ˜ ( s ) = u ( 0 ) s + G ˜ s 1 − α a M ( α ) ( 1 − α ) + α G ˜ s 1 − α a M ( α ) s .

Taking the inverse Laplace transform yields

(21) u ( t ) = u ( 0 ) + ( 1 − α ) G t 1 − α a M ( α ) + α M ( α ) ∫ 0 t G τ 1 − α a d τ .

Choosing for instance G t 1 − α = t 1 − α , hence

u ( t ) = u ( 0 ) + ( 1 − α ) t 1 − α a M ( α ) + a α ( 2 − α ) M ( α ) t 2 − α .

At t = 0 , we obviously have the initial condition u ( 0 ) that is satisfied. Taking the Caputo–Fabrizio derivative of that solution, will it give

D t α 0 CF u ( 0 ) + ( 1 − α ) t 1 − α a M ( α ) + a α ( 2 − α ) M ( α ) t 2 − α = at 1 − α ?

To check it we have

(22) D t α 0 CF u ( 0 ) + ( 1 − α ) t 1 − α a M ( α ) + a α ( 2 − α ) M ( α ) t 2 − α = M ( α ) 1 − α ∫ 0 t ( 1 − α ) 2 a M ( α ) τ − α + a α M ( α ) τ 1 − α × exp − α 1 − α ( t − τ ) d τ .

The main task here is to evaluate the following integrals:

I − α = ∫ 0 t τ − α exp − α 1 − α ( t − τ ) d τ ,

I 1 − α = ∫ 0 t τ 1 − α exp − α 1 − α ( t − τ ) d τ .

Using a simple integration by parts yields a relation between I 1 − α and I − α as follows:

I 1 − α = ( 1 − α ) τ 1 − α α exp − α 1 − α ( t − τ ) 0 t − ∫ 0 t ( 1 − α ) 2 τ − α α exp − α 1 − α ( t − τ ) d τ = ( 1 − α ) t 1 − α α − ( 1 − α ) 2 α × ∫ 0 t τ − α exp − α 1 − α ( t − τ ) d τ = ( 1 − α ) t 1 − α α − ( 1 − α ) 2 α I − α .

The substitution of the later relation into (22) leads to

D t α 0 CF u ( 0 ) + ( 1 − α ) t 1 − α a M ( α ) + a α ( 2 − α ) M ( α ) t 2 − α = ( 1 − α ) a I − α + a α 1 − α I 1 − α = ( 1 − α ) a I − α + a α 1 − α ( 1 − α ) t 1 − α α − ( 1 − α ) 2 α I − α = a α 1 − α ( 1 − α ) t 1 − α α = a t 1 − α ,

which proves that the dimensionalization statement clearly works with the Caputo–Fabrizio fractional derivative for the dimensional function G t 1 − α = t 1 − α . Hence, with that chosen dimensional function, it is clear that solution (21) satisfies not only the initial condition but also verifies equation (20).

4.2 Dimensionalization: specific Caputo’s (power law) case

We consider the following differential equation with the Caputo derivative as:

(23) D t α 0 C ( u ( t ) ) = G t 1 − α a , where t ≥ 0 ,

integrating using the Riemann–Liouville integral yields the solution

(24) u ( t ) = u ( 0 ) + a Γ ( α ) ∫ 0 t G τ 1 − α ( t − τ ) α − 1 d τ .

Choosing again G t 1 − α = t 1 − α , hence

(25) u ( t ) = u ( 0 ) + a Γ ( α ) ∫ 0 t τ 1 − α ( t − τ ) α − 1 d τ = u ( 0 ) + a t Γ ( 2 − α ) .

Again at t = 0 , we obviously have the initial condition u ( 0 ) that is satisfied.

Taking the Caputo derivative of that solution, will it give

D t α 0 C [ u ( 0 ) + a t Γ ( 2 − α ) ] = at 1 − α ?

To check it we have

D t α 0 C [ u ( 0 ) + a t Γ ( 2 − α ) ] = 1 Γ ( 1 − α ) ∫ 0 t [ a Γ ( 2 − α ) ( t − τ ) − α ] d τ = a Γ ( 2 − α ) Γ ( 1 − α ) ∫ 0 t ( t − τ ) − α d τ = a Γ ( 2 − α ) Γ ( 1 − α ) Γ ( 1 − α ) Γ ( 2 − α ) t 1 − α = a t 1 − α ,

which is the expected result and we obtain a similar result to the one obtained in Caputo–Fabrizio’s case.

4.3 Dimensionalization: specific Atangana–Baleanu’s case

Finally, we consider the following differential equation with the Atangana–Baleanu–Caputo derivative as:

(26) D t α 0 A B C ( u ( t ) ) = G t 1 − α a , where t ≥ 0 ,

integrating using the Atangana–Baleanu integral yields the solution

(27) u ( t ) = u ( 0 ) + 1 − α W ( α ) G t 1 − α a + α Γ ( α ) W ( α ) ∫ 0 t G τ 1 − α a ( t − τ ) α − 1 d τ .

Choosing again G t 1 − α = t 1 − α , hence

(28) u ( t ) = u ( 0 ) + 1 − α W ( α ) t 1 − α a + a t α Γ ( 2 − α ) W ( α ) .

Again at t = 0 , we have the initial condition u (0) that is satisfied. This means that the idea of dimension applied in a number of real-world problems in the last few years has not realized that such an idea plays an important role in the analysis of problems with initial conditions.

5 New differentiation and integration with dimension

While the concept of dimension has been used and applied with great success in recent years, no clear differential and integral operators have been suggested and that derive from the classical concepts. Therefore, we propose in this section some useful operators that will help in that regard. We start by the investigation using the classical differentiation.

5.1 New differentiation and integration with dimension and power law

Let us set

(29) d t α 0 AG ( f ( t ) ) = d d t α f ( t ) = lim t → t 1 f ( t ) − f ( t 1 ) G t 2 − α − G t 1 2 − α ( 2 − α ) , where t > 0 , 0 < α ≤ 1 .

Since we are trying to stay in the aim of change of rate and taking into account the modeling aspect, we restrict ourselves to the case G t 2 − α = t 2 − α .

Proposition 5.1

If the function f is differentiable, then,

d t α 0 AG ( f ( t ) ) = f ′ ( t ) t 1 − α f o r a n y t > 0 , 0 < α ≤ 1 .

Proof

Let t 1 > 0 then,□

d t α 0 AG f ( t 1 ) = lim t → t 1 f ( t ) − f ( t 1 ) G t 2 − α − G t 2 − α ( 2 − α ) = lim t → t 1 f ( t ) − f ( t 1 ) ( t − t 1 ) ( t − t 1 ) ( t 2 − α − t 1 2 − α ) ( 2 − α ) = f ′ ( t ) 2 − α ( 2 − α ) t 1 1 − α = f ′ ( t 1 ) t 1 1 − α .

We call operation (29) the AG-differentiation and its equivalent integration process leads to the following lemma.

Lemma 5.2

If the function f is differentiable, then the integral operator associated with d t α 0 AG reads as

(30) I t α 0 AG f ( t ) = ∫ 0 t τ 1 − α f ( τ ) d τ .

Note that this integral is not our main aim here. We are interested in the fractional version. Let f be a continuous function non-necessarily differentiable. We define the following differential operator:

(31) D t α 0 AG p ( f ( t ) ) = d t α 0 AG 1 Γ ( 1 − α ) ∫ 0 t ( t − τ ) − α f ( τ ) d τ .

This operator takes into account the dimensionalization aspect since, according to Proposition (5.1), we have

(32) D t α 0 AG p ( f ( t ) = 1 t 1 − α d d t 1 Γ ( 1 − α ) ∫ 0 t ( t − τ ) − α f ( τ ) d τ

D t α 0 AG p ( f ( t ) ) = d d t 1 Γ ( 1 − α ) ∫ 0 t ( t − τ ) − α f ( τ ) d ( t , τ ) ,

where d ( t , τ ) = d τ t 1 − α .

The integral operator associated with D t α 0 AG p reads as

(33) I t α 0 AG p f ( t ) = 1 Γ ( α ) ∫ 0 t τ 1 − α f ( τ ) ( t − τ ) α − 1 d τ .

5.2 New differentiation and integration with dimension and exponential law

A similar version to definition (31) with the exponential kernel reads as

(34) D t α 0 AG e ( f ( t ) ) = d t α 0 AG f ( t ) × M ( α ) 1 − α exp α 1 − α t = M ( α ) 1 − α d d t ∫ 0 t f ( τ ) exp α 1 − α ( t − τ ) d τ t 1 − α = M ( α ) 1 − α d d t ∫ 0 t f ( τ ) exp α 1 − α ( t − τ ) d ( t , τ ) .

The integral operator associated with D t α 0 AG e reads as

(35) I t α 0 AG e f ( t ) = 1 − α M ( α ) t 1 − α f ( t ) + α M ( α ) ∫ 0 t τ 1 − α f ( τ ) d τ .

5.3 New differentiation and integration with dimension and Mittag–Leffler law

A similar version to definition (31) with the Mittag–Leffler kernel reads as

(36) D t α 0 AG m ( f ( t ) ) = d t α 0 AG f ( t ) × W ( α ) 1 − α E α − α 1 − α t α = W ( α ) 1 − α d d t ∫ 0 t f ( τ ) E α − α 1 − α ( t − τ ) α d τ t 1 − α = W ( α ) 1 − α d d t ∫ 0 t f ( τ ) E α − α 1 − α ( t − τ ) α d ( t , τ ) .

The integral operator associated with D t α 0 AG m reads as

(37) I t α 0 AG m f ( t ) = 1 − α W ( α ) t 1 − α f ( t ) + α W ( α ) Γ ( α ) ∫ 0 t τ 1 − α f ( τ ) ( t − τ ) α − 1 d τ .

6 Solvability of some related equations: exact solutions

In this section, we present some exact solutions of related given equations. Let us start by considering the following equations:

(Exact solution related to the exponential function)
(38) D t α 0 AG e ( f ( t ) ) = t λ .
1. We applied the associated integral operator (35) on both sides of this equation that yields
  f ( t ) = 1 − α M ( α ) t λ ⋅ t 1 − α + α M ( α ) ∫ 0 t τ 1 − α τ λ d τ
2. or
  f ( t ) = 1 − α M ( α ) ⋅ t 1 − α + λ + α ( 2 − α + λ ) M ( α ) t 2 − α + λ .
3. At t = 0 , we have f ( 0 ) = 0 .
(Exact solution related to the gamma function)
(39) D t α 0 AG p ( f ( t ) ) = t λ .
1. We applied the associated integral operator (33) on both sides of this equation that yields
  f ( t ) = 1 Γ ( α ) ∫ 0 t τ 1 − α τ λ ( t − τ ) α − 1 d τ
2. or
f ( t ) = t 1 + λ Γ ( λ + 2 − α ) Γ ( λ + 2 ) .
(Exact solution related to the gamma function)
(40) D t α 0 AG m ( f ( t ) ) = t λ .
1. We applied the associated integral operator (37) on both sides of this equation that yields
  f ( t ) = 1 − α W ( α ) t 1 − α t λ + α W ( α ) Γ ( α ) ∫ 0 t τ 1 − α τ λ ( t − τ ) α − 1 d τ
2. or
f ( t ) = 1 − α W ( α ) t 1 − α + λ + α W ( α ) t 1 + λ Γ ( λ + 2 − α ) Γ ( λ + 2 ) .
(Exact solution related to the incomplete beta function)
(41) D t α 0 AG e ( f ( t ) ) = ( 1 − t θ ) α − 1 .
1. We applied the associated integral operator (35) on both sides of this equation that yields
  f ( t ) = 1 − α M ( α ) ( 1 − t θ ) α − 1 ⋅ t 1 − α + α M ( α ) ∫ 0 t τ 1 − α ( 1 − τ θ ) α − 1 d τ .
2. Then, using the transformation y = τ θ yields
  f ( t ) = 1 − α M ( α ) ( 1 − t θ ) α − 1 ⋅ t 1 − α + α M ( α ) ∫ 0 t θ y 1 − α θ ( 1 − y ) α − 1 y 1 − θ θ θ d y . = 1 − α M ( α ) ( 1 − t θ ) α − 1 ⋅ t 1 − α + α θ M ( α ) ∫ 0 t θ y 2 − α θ − 1 ( 1 − y ) α − 1 d y . = 1 − α M ( α ) ( 1 − t θ ) α − 1 ⋅ t 1 − α + α θ M ( α ) B t θ ; 2 − α θ , α ,
3. where B ( x ; a , b ) is the incomplete beta function, a generalization of the beta function, reading as
B ( t ; a , b ) = ∫ 0 t τ a − 1 ( 1 − τ ) b − 1 d τ .
(Exact solution related to the (lower) incomplete gamma function)
(42) D t α 0 AG e ( f ( t ) ) = t 2 α − 2 e − t .
1. We applied the associated integral operator (35) on both sides of this equation that yields
  f ( t ) = 1 − α M ( α ) t 2 α − 2 e − t ⋅ t 1 − α + α M ( α ) ∫ 0 t τ 1 − α τ 2 α − 2 e − τ d τ .
2. Then,
  f ( t ) = 1 − α M ( α ) t α − 1 e − t + α M ( α ) ∫ 0 t τ α − 1 e − τ d τ = 1 − α M ( α ) t α − 1 e − t + α M ( α ) γ ( α , t ) ,
3. where γ ( α , t ) is the lower incomplete gamma function given by
γ ( α , t ) = ∫ 0 t τ α − 1 e − τ d τ .
(Exact solution related to the)
(43) D t α 0 AG m ( f ( t ) ) = t 2 α − 2 e − t .
1. We applied the associated integral operator (37) on both sides of this equation that yields
f ( t ) = 1 − α W ( α ) t 1 − α t 2 α − 2 e − t + α W ( α ) Γ ( α ) ∫ 0 t τ 1 − α τ 2 α − 2 e − τ ( t − τ ) α − 1 d τ = 1 − α W ( α ) t α − 1 e − t + α W ( α ) Γ ( α ) ∫ 0 t τ α − 1 e − τ ( t − τ ) α − 1 d τ .
(44) D t α 0 AG m ( f ( t ) ) = sin t .
1. We applied the associated integral operator (37) on both sides of this equation that yields
  f ( t ) = 1 − α W ( α ) t 1 − α sin t + α W ( α ) Γ ( α ) ∫ 0 t τ 1 − α sin τ ( t − τ ) α − 1 d τ .
2. Then,
  ∫ 0 t τ 1 − α sin τ ( t − τ ) α − 1 d τ = ∫ 0 t τ 1 − α ( t − τ ) α − 1 ∑ j = 0 ∞ ( − 1 ) j ( 2 j + 1 ) ! τ 2 j + 1 d τ = ∑ j = 0 ∞ ( − 1 ) j ( 2 j + 1 ) ! ∫ 0 t τ 1 − α ( t − τ ) α − 1 τ 2 j + 1 d τ = ∑ j = 0 ∞ ( − 1 ) j ( 2 j + 1 ) ! t 2 + 2 j Γ ( α ) Γ ( 3 − α + 2 j ) Γ ( 3 − 2 j ) .
3. This finally leads to the solution
f ( t ) = 1 − α W ( α ) t 1 − α sin t + α t 2 W ( α ) ∑ j = 0 ∞ ( − t 2 ) j ( 2 j + 1 ) ! Γ ( 3 − α + 2 j ) Γ ( 3 − 2 j ) .
(45) D t α 0 AG e ( f ( t ) ) = sin t .
1. We applied the associated integral operator (35) on both sides of this equation that yields
  f ( t ) = 1 − α M ( α ) sin t ⋅ t 1 − α + α M ( α ) ∫ 0 t τ 1 − α sin τ d τ .
2. Then,

f ( t ) = 1 − α M ( α ) sin t ⋅ t 1 − α + α t 3 − α M ( α ) ∑ j = 0 ∞ ( − t 2 ) j ( 2 j + 1 ) ! ( 2 j + 3 − α ) .

7 Discretization

We consider the infinite norm

∥ f ∥ = sup t ∈ D f ∣ f ( t ) ∣

and the discretization at the point t n = n Δ t , x m = m Δ x reads as

d AG d t β f ( t ) t = t n ≃ f ( t n + 1 ) − f ( t n ) ( Δ t ) 2 − β n 1 − β ,

d AG d x β f ( t ) x = x i ≃ f ( x i + 1 ) − f ( x i − 1 ) 2 ( Δ x ) 2 − β i 1 − β .

Discretization with power law kernel:
d AG d t β f ( t ) × t − α Γ ( 1 − α ) t = t n ≃ t β − 1 Γ ( 1 − α ) d d t ∫ 0 t n ( t n − τ ) − α f ( τ ) d τ ≃ t n β − 1 Γ ( 1 − α ) ∫ 0 t n + 1 ( t n + 1 − τ ) − α f ( τ ) d τ − ∫ 0 t n ( t n − τ ) − α f ( τ ) d τ Δ t ,
where
∫ 0 t n + 1 ( t n + 1 − τ ) − α f ( τ ) d τ ≃ ∑ j = 0 n f ( t j + 1 ) + f ( t j ) 2 ∫ t j + 1 t j ( t n + 1 − τ ) − α d τ ≃ ∑ j = 0 n f j + 1 + f j 2 ( 1 − α ) ( t n + 1 − t j ) 1 − α − ( t n + 1 − t j + 1 ) 1 − α = ( Δ t ) 1 − α 2 ( 1 − α ) ∑ j = 0 n { f j + 1 + f j } { ( n + 1 − j ) 1 − α − ( n − j + 1 ) 1 − α } = ( Δ t ) 1 − α 2 ( 1 − α ) ∑ j = 0 n { f j + 1 + f j } δ n , j α

∫ 0 t n ( t n − τ ) − α f ( τ ) d τ ≃ ∑ j = 0 n − 1 f ( t j + 1 ) + f ( t j ) 2 ∫ t j + 1 t j ( t n − τ ) − α d τ ≃ ∑ j = 0 n − 1 f j + 1 + f j 2 ( 1 − α ) ( Δ t ) 1 − α { ( n − j ) 1 − α − ( n − j − 1 ) 1 − α } = ∑ j = 0 n − 1 f j + 1 − f j 2 ( 1 − α ) Δ t 1 − α δ n − 1 , j α .
Discretization with exponential law kernel:
d AG d t β f ( t ) × M ( α ) 1 − α exp − α 1 − α t | t = t n + 1 ≃ M ( α ) 1 − α d d t ∫ 0 t n + 1 exp − α 1 − α ( t n + 1 − τ ) t n + 1 β − 1 f ( τ ) d τ ≃ M ( α ) 1 − α ( Δ t ( n + 1 ) ) β − 1 d d t F ( t n + 1 ) ≃ M ( α ) 1 − α ( Δ t ( n + 1 ) ) β − 1 F ( t n + 1 ) − F ( t n ) Δ t ,
where
F ( t n + 1 ) = ∫ 0 t n + 1 f ( τ ) exp − α 1 − α ( t n + 1 − τ ) d τ ,

F ( t n ) = ∫ 0 t n f ( τ ) exp − α 1 − α ( t n − τ ) d τ .
Without loss of generality we evaluate
F ( t n + 1 ) = ∑ j = 0 n ∫ t j + 1 t j f ( t j + 1 ) + f ( t j ) 2 × exp − α 1 − α ( t n + 1 − τ ) d τ = ∫ t j + 1 t j ∑ j = 0 n f ( t j + 1 ) + f ( t j ) 2 × exp − α 1 − α ( t n + 1 − τ ) d τ = ∑ j = 0 n f j + 1 + f j 2 δ n , j α
with
δ n , j α = 1 − α α exp − α 1 − α Δ t ( n − j ) − exp − α 1 − α Δ t ( n + 1 − j )
and
F ( t n ) = ∑ j = 0 n f j + 1 + f j 2 δ n − 1 , j α
with
δ n − 1 , j α = 1 − α α exp − α 1 − α Δ t ( n − 1 − j ) − exp − α 1 − α Δ t ( n − j ) .

Thus,
d AG d t β f ( t ) × M ( α ) 1 − α exp − α 1 − α t t = t n + 1 ≃ M ( α ) Δ t α ( Δ t ( n + 1 ) ) β − 1 × ∑ j = 0 n f j + 1 + f j 2 exp − α 1 − α Δ t ( n − j ) − exp − α 1 − α Δ t ( n − j + 1 ) − ∑ j = 0 n f j + 1 + f j 2 exp − α 1 − α Δ t ( n − j − 1 ) − exp − α 1 − α Δ t ( n − j ) .
Discretization with the Mittag–Leffler law kernel:

d AG d t β f ( t ) ∗ W ( α ) 1 − α E α − α 1 − α t α | t = t n + 1 ≃ W ( α ) 1 − α d d t ∫ 0 t n + 1 f ( τ ) E α − α 1 − α ( t n + 1 − τ ) α × ( Δ t ( n + 1 ) ) β − 1 d τ ≃ W ( α ) 1 − α ( Δ t ( n + 1 ) ) β − 1 F ( t n + 1 ) − F ( t n ) Δ t

and

F ( t n + 1 ) = ∫ 0 t n + 1 f ( τ ) E α − α 1 − α ( t n + 1 − τ ) α d τ ,

F ( t n ) = ∫ 0 t n f ( τ ) E α − α 1 − α ( t n − τ ) α d τ ,

where

F ( t n + 1 ) = ∑ j = 0 n f j + 1 + f j 2 δ n , j α , F ( t n ) = ∑ j = 0 n − 1 f j + 1 + f j 2 δ n − 1 , j α ,

with

δ n , j α = ( n − j ) Δ t E α , 2 − α 1 − α ( ( n − j ) Δ t ) α − ( ( n − j + 1 ) Δ t ) E α , 2 − α 1 − α ( ( n − j + 1 ) Δ t ) α

δ n − 1 , j α = Δ t ( n − j − 1 ) E α , 2 − α 1 − α ( Δ t ( n − j − 1 ) ) α − ( n − j ) Δ t E α , 2 − α 1 − α ( ( n − j ) Δ t ) α

8 Applications

We apply in this section the new operators and use the aforementioned numerical discretization to some known systems.

8.1 Attractors

The major aim of this subsection is to see what happens when the new differential operators defined here are applied to some systems. We start with the Lü-chen system [15,16] well known to be a special attractor with many wings. We use the aforementioned scheme to evaluate the effects on such a model and see whether there is preservation of the multi-wing state. Hence, the original model is given by

(46) x ̇ ( t ) = a y − x , y ̇ ( t ) = ν + x [ 1 − z ] + c y , z ̇ ( t ) = x y − b z ,

where a , b , c , u are real numbers parameterizing the system, with − 15 ≤ ν ≤ 15 . The time-dependent system states are represented by x = x ( t ) , y = y ( t ) and z = z ( t ) . Numerical representations of the Lü-chen system (46) plotted for the parameters ( a , b , c ) = ( 34 , 2 , 20 ) and ν = − 12 , ν = 1 and ν = 12 are depicted in Figure 1(a–c), respectively [15]. The initial conditions considered are ( x ( 0 ) , y ( 0 ) , z ( 0 ) ) = ( 1 , 1 , 12 ) , and the figures show the presence of multi-scroll chaotic attractors that change together with the parameter ν .

$Figure 1 Multi-scroll chaotic attractors generated by the original Lü-chen model (46), plotted for the parameters ( a , b , c ) = ( 34 , 2 , 20 ) (a,b,c)\hspace{-0.2em}=\hspace{-0.2em}(34,2,20) and ν = − 12 , ν = 1 \nu =-12,\hspace{.5em}\nu =1 and ν = 12 \nu =12 .$

Figure 1

Multi-scroll chaotic attractors generated by the original Lü-chen model (46), plotted for the parameters ( a , b , c ) = ( 34 , 2 , 20 ) and ν = − 12 , ν = 1 and ν = 12 .

Now, with the derivative D t α 0 AG p the fractional derivative with dimension and power law given by (32), the model takes the form:

(47) D t α 0 AG p x ( t ) = a y − x , D t α 0 AG p y ( t ) = ν + x [ 1 − z ] + c y , D t α 0 AG p z ( t ) = x y − b z .

Numerical representations of the Lü-chen system (47) plotted for α = 0.9 and the parameters ( a , b , c ) = ( 34 , 2 , 20 ) and ν = − 12 , ν = 1 and ν = 12 are depicted in Figure 2(a–c), respectively. It shows that the model is still chaotic and there is preservation of the multi-wing state.

$Figure 2 Multi-scroll chaotic attractors generated by the application of power law operator D t α 0 AG p {}_{0}{}^{\text{AG}p}D_{t}^{\alpha } on the Lü-chen model (47), plotted for α = 0.9 \alpha =0.9 and the parameters ( a , b , c ) = ( 34 , 2 , 20 ) (a,b,c)=(34,2,20) and ν = − 12 , ν = 1 \nu =-12,\hspace{.5em}\nu =1 and ν = 12 \nu =12 . As the parameter ν \nu varies, the system is still chaotic and there is a preservation of the multi-wing state for the attractor.$

Figure 2

Multi-scroll chaotic attractors generated by the application of power law operator D t α 0 AG p on the Lü-chen model (47), plotted for α = 0.9 and the parameters ( a , b , c ) = ( 34 , 2 , 20 ) and ν = − 12 , ν = 1 and ν = 12 . As the parameter ν varies, the system is still chaotic and there is a preservation of the multi-wing state for the attractor.

With the derivative D t α 0 AG e , the fractional derivative with dimension and exponential law given by (34), the model takes the form:

(48) D t α 0 AG e x ( t ) = a y − x , D t α 0 AG e y ( t ) = ν + x [ 1 − z ] + c y , D t α 0 AG e z ( t ) = x y − b z .

Numerical representations of (48) are depicted in Figure 3 with α = 0.9 . It shows similar dynamics to Figures 1 and 2, still characterized by chaos and preservation of the multi-wing state that changed with parameters ν and α .

$Figure 3 Multi-scroll chaotic attractors generated by the application of exponential law operator D t α 0 AG e {}_{0}{}^{\text{AG}e}D_{t}^{\alpha } on the Lü-chen model (48), plotted for α = 0.9 \alpha =0.9 , and the parameters ( a , b , c ) = ( 34 , 2 , 20 ) (a,b,c)=(34,2,20) and ν = − 12 , ν = 1 \nu =-12,\hspace{.5em}\nu =1 and ν = 12 \nu =12 . As the parameter ν \nu varies, the system is still chaotic and there is a preservation of the multi-wing state for the attractor.$

Figure 3

Multi-scroll chaotic attractors generated by the application of exponential law operator D t α 0 AG e on the Lü-chen model (48), plotted for α = 0.9 , and the parameters ( a , b , c ) = ( 34 , 2 , 20 ) and ν = − 12 , ν = 1 and ν = 12 . As the parameter ν varies, the system is still chaotic and there is a preservation of the multi-wing state for the attractor.

With the derivative D t α 0 AG m , the fractional derivative with dimension and the Mittag–Leffler law given by (36), the model takes the form:

(49) D t α 0 AG m x ( t ) = a y − x , D t α 0 AG m y ( t ) = ν + x [ 1 − z ] + c y , D t α 0 AG m z ( t ) = x y − b z .

Numerical representations of (49) are depicted in Figure 4 with for α = 0.9 . It shows similar dynamics to Figures 1–3, still characterized by chaos and preservation of the multi-wing state that changed with parameters ν and α .

$Figure 4 Multi-scroll chaotic attractors generated by the application of Mittag–Leffler law operator D t α 0 AG m {}_{0}{}^{\text{AG}m}D_{t}^{\alpha } on the Lü-chen model (49), plotted for the parameters ( a , b , c ) = ( 34 , 2 , 20 ) (a,b,c)=(34,2,20) and ν = − 12 , ν = 1 \nu =-12,\hspace{.5em}\nu =1 and ν = 12 \nu =12 . As the parameter ν \nu varies, the system is still chaotic and there is a preservation of the multi-wing state for the attractor.$

Figure 4

Multi-scroll chaotic attractors generated by the application of Mittag–Leffler law operator D t α 0 AG m on the Lü-chen model (49), plotted for the parameters ( a , b , c ) = ( 34 , 2 , 20 ) and ν = − 12 , ν = 1 and ν = 12 . As the parameter ν varies, the system is still chaotic and there is a preservation of the multi-wing state for the attractor.

8.2 Triple covers of the proto-Lorenz system

Applying D t α 0 AG m the fractional derivative with dimension and the Mittag–Leffler law given by (36) to the equations for the triple covers of the proto-Lorenz system [18,17] leads to

(50) D t α 0 AG m p ( t ) = 1 3 [ − ( r 1 + 1 ) p + ( r 1 − r 2 + z ) q ] + 1 3 ( p 2 + q 2 ) 1 / 2 [ ( 1 − r 1 ) ( p 2 − q 2 ) + 2 ( r 2 + r 1 − z ) p q ] , D t α 0 AG m q ( t ) = 1 3 [ ( r 2 − r 1 − z ) p − ( r 1 + 1 ) q ] + 1 3 ( p 2 + q 2 ) 1 / 2 [ 2 ( r 1 − 1 ) p q + ( r 2 + r 1 − z ) ( p 2 − q 2 ) ] , D t α 0 AG m z ( t ) = 1 2 ( 3 p 2 q − q 3 ) − r 3 z ,

where ( p , q , z ) represent the coordinates that belong to a copy space containing the real coordinates [18,17]. Numerical representations of (50) are depicted in Figure 5 for parameters r 1 = 10 , r 2 = 28 and r 3 = 8 / 3 and show that such a 3-cover system maintains the covers in a chaotic attractor with three scrolls and taking the form of a jet fighter.

$Figure 5 Graphical simulations generated by the application of Mittag–Leffler law operator D t α 0 AG m {}_{0}{}^{\text{AG}m}D_{t}^{\alpha } on the triple cover model (50). It is a chaotic attractor with three scrolls, taking the form of a jet fighter. The parameters used are r 1 = 10 , r 2 = 28 and r 3 = 8 / 3 {r}_{1}=10,\hspace{.5em}{r}_{2}=28\hspace{.5em}\text{and}\hspace{.5em}{r}_{3}=8/3 .$

Figure 5

Graphical simulations generated by the application of Mittag–Leffler law operator D t α 0 AG m on the triple cover model (50). It is a chaotic attractor with three scrolls, taking the form of a jet fighter. The parameters used are r 1 = 10 , r 2 = 28 and r 3 = 8 / 3 .

Applying D t α 0 AG e the fractional derivative with dimension and the Mittag–Leffler law given by (34) to the equations for the triple covers of the proto-Lorenz system leads to

(51) D t α 0 AG e p ( t ) = 1 3 [ − ( r 1 + 1 ) p + ( r 1 − r 2 + z ) q ] + 1 3 ( p 2 + q 2 ) 1 / 2 [ ( 1 − r 1 ) ( p 2 − q 2 ) + 2 ( r 2 + r 1 − z ) p q ] , D t α 0 AG e q ( t ) = 1 3 [ ( r 2 − r 1 − z ) p − ( r 1 + 1 ) q ] + 1 3 ( p 2 + q 2 ) 1 / 2 [ 2 ( r 1 − 1 ) p q + ( r 2 + r 1 − z ) ( p 2 − q 2 ) ] , D t α 0 AG e z ( t ) = 1 2 ( 3 p 2 q − q 3 ) − r 3 z .

Numerical representations of (51) are depicted in Figure 6 for parameters r 1 = 10 , r 2 = 28 and r 3 = 8 / 3 and show that such a 3-cover system maintains the covers in a chaotic attractor with three scrolls and taking the form of a jet fighter.

$Figure 6 Graphical simulations generated by the application of exponential law operator D t α 0 AG e {}_{0}{}^{\text{AG}e}D_{t}^{\alpha } on the triple cover model (51). It is a chaotic attractor with three scrolls, taking the form of a jet fighter. The parameters used are r 1 = 10 , r 2 = 28 and r 3 = 8 / 3 {r}_{1}=10,\hspace{.5em}{r}_{2}=28\hspace{.5em}\text{and}\hspace{.5em}{r}_{3}=8/3 .$

Figure 6

Graphical simulations generated by the application of exponential law operator D t α 0 AG e on the triple cover model (51). It is a chaotic attractor with three scrolls, taking the form of a jet fighter. The parameters used are r 1 = 10 , r 2 = 28 and r 3 = 8 / 3 .

8.3 Application in wave motion

Applying D t α 0 AG p , D t α 0 AG e and D t α 0 AG m to the Sawada–Kotera equation, a nonlinear differential equation that falls into the class of Korteweg–de Vries (KdV)-type equations [19,20] leads to

(52) D t α 0 AG p g ( t , x ) = − g x x x x x ( t , x ) − 15 g ( t , x ) g x x x ( t , x ) − 15 g x ( t , x ) g x x ( t , x ) − 45 g 2 ( t , x ) g x ( t , x ) ,

(53) D t α 0 AG e g ( t , x ) = − g x x x x x ( t , x ) − 15 g ( t , x ) g x x x ( t , x ) − 15 g x ( t , x ) g x x ( t , x ) − 45 g 2 ( t , x ) g x ( t , x )

and

(54) D t α 0 AG m g ( t , x ) = − g x x x x x ( t , x ) − 15 g ( t , x ) g x x x ( t , x ) − 15 g x ( t , x ) g x x ( t , x ) − 45 g 2 ( t , x ) g x ( t , x ) .

Numerical representation are depicted in Figures 7 and 8, when α = 0.9 with the initial condition h ( x ) = sec h 2 ( π x ) . It shows a dynamics similar to the classical Sawada–Kotera equation well known in the wave motion theory.

$Figure 7 (a–c) Graphical simulations of solution g ( t , x ) g(t,x) to models (52)–(54), when α = 0.9 \alpha =0.9 with the initial condition h ( x ) = sec h 2 ( π x ) h(x)=\text{sec}{h}^{2}(\pi x) there is a dynamics similar to the classical Sawada–Kotera equation well known in the wave motion theory.$

Figure 7

(a–c) Graphical simulations of solution g ( t , x ) to models (52)–(54), when α = 0.9 with the initial condition h ( x ) = sec h 2 ( π x ) there is a dynamics similar to the classical Sawada–Kotera equation well known in the wave motion theory.

$Figure 8 (a–c) Graphical simulations of solution g ( t , x ) g(t,x) to models (52)–(54), when α = 0.6 \alpha =0.6 with the initial condition h ( x ) = sec h 2 ( π x ) h(x)=\text{sec}{h}^{2}(\pi x) there is a dynamics similar to the classical Sawada–Kotera equation well known in the wave motion theory.$

Figure 8

(a–c) Graphical simulations of solution g ( t , x ) to models (52)–(54), when α = 0.6 with the initial condition h ( x ) = sec h 2 ( π x ) there is a dynamics similar to the classical Sawada–Kotera equation well known in the wave motion theory.

8.4 Conclusion

We have introduced new differential operators depicting different processes observed in nature sciences in order to clarify some misunderstandings about the use of fractional differential and integral operators with non-singular kernels. In fact, many recent works and publications have proved that some operators related to certain laws are being misused, such as, for example, the Riemann–Liouville and Caputo derivatives that are power-law-related derivatives but are being used to model and describe problems not associated with power law process. We have also suggested some numerical discretizations for the newly defined differential operators which will be important to address the solvability of some initial values problems. Applications to the chaos theory and wave motion have been provided to show the validity of the differential operators. Doors are now open for more investigations and use of those new operators in the domain of mathematical modeling.

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Received: 2020-05-06

Revised: 2020-07-02

Accepted: 2020-07-03

Published Online: 2020-10-12

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/phys-2020-0158

Keywords for this article

fractional derivative with dimension; integration with dimensionalization; law-related exact solution; numerical approach; applications

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