# Existence and Uniqueness Study of the Conformable Laplace Transform

Authors: Jihad Younis; Bouchenak Ahmed; Mazin AlJazzazi; Rasha Al Hejaj; Hassen Aydi
Journal: Journal of Mathematics
 (2022)

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## Abstract

This paper tackles the topic of conformable Laplace transform. The authors aim at discussing its existence by exploring and providing the kind of functions that possess a conformable Laplace transform. Furthermore, the comparison theorem of conformable improper integrals is presented to further explain and justify the existence of conformable Laplace transform for some functions. The uniqueness is also established in order to determine the inverse of conformable Laplace transform for functions. Moreover, we present a table of the conformable Laplace transform of the usual functions. Finally, as an application, we use the conformable Laplace transform for solving some fractional differential equations.


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## Body

## 1. Introduction

Many researchers tried to define a fractional derivative, like Riemann–Liouville fractional derivative and Caputo fractional derivative (see [1–7]). Most of the definitions give a numerical solution to the problems. However, the conformable derivative is a natural definition which gives us a simple and easy solution for the problems. For more different applications on conformable derivatives, we refer to [8–15].Many researchers use the Laplace transform because it is an operational tool for solving a lot of difficult problems such as non-homogeneous fractional differential equations with variable coefficients and fractional initial value problems. The solution can be made through the following main steps. The first step is simplifying the hard problem by transforming it to a simple equation. The second step is to obtain a solution for the transformed problem by using the theory of partial and ordinary differential equations. The final step is to insert the inverse fractional Laplace transform into the solution of the transformed problem to get the solution of the given hard problem.In 2015, Abdeljawad [9] introduced the conformable Laplace transform, which helps us to solve several strong problems and fractional differential equations. However, there is no study for the existence and uniqueness of the conformable Laplace transform. Therefore, in this paper, we will prove some results about the existence and uniqueness of the conformable Laplace transform as well as some new ideas concerning this topic. Then, we present a table containing the conformable Laplace transform for the usual functions. Finally, as an application, we solve a non-homogeneous fractional differential equation with variable coefficients using the conformable Laplace transform. For further results on conformable Laplace transform, see [16–20].


## 2. Basics of Conformable Laplace Transform

In 2015, the conformable Laplace transform was introduced by Abdeljawad [9].Definition 1.
Lety:0,∞⟶ℝ be a real-valued function and 0<γ≤1. Then, the conformable Laplace transform of y is defined as(1)Lγyt=Yγζ=∫0∞yte−ζtγ/γdγt=∫0∞yte−ζtγ/γtγ−1dt,provided the integral exists.
One of the nice results is the relation between the usual and the conformable Laplace transforms, which is given in the theorems below.Theorem 1 (see [9, 19]).
Lety:0,∞⟶ℝ be a function such that Lγyt=Yγζ exists. Then,(2)Lγytζ=Yγζ=Lyγt1/γζ,0<γ≤1,where L is the usual Laplace transform.Theorem 2 (see [9, 19]).
Lety:0,∞⟶ℝ, z:0,∞⟶ℝ and let λ, μ, a∈ℝ and 0<γ≤1. Then,(1)
Lγλyt+μzt=λYγζ+μZγζ, ζ>0.(2)
Lγe−atγ/γytζ=Yγζ+a, ζ>a.(3)
LγIγytζ=Yγζ/ζ, ζ>0.(4)
Lγtnγ/γnytζ=−1ndn/dζnYγζ, ζ>0.(5)
Lγy∗zt=YγζZγζ, ζ>0,where Yγ and Zγ are the conformable Laplace transforms of the functions y and z, respectively, y∗z is the convolution product of y and z, and Iγyt is the conformable integral.


## 3. Existence and Uniqueness of the Conformable Laplace Transform

The existence and uniqueness of the usual Laplace transform can be found in [21, 22]. When we go back to Definition 1, we notice that the integral defining a conformable Laplace transform represents an improper integral. The function yt is said to have a conformable Laplace transform if the improper integral converges. This leads us to wonder about the types of functions which have conformable Laplace transforms; in other words, what type of functions guarantees a convergent improper integral.Example 1.
We need to investigate the existence of conformable Laplace transform of the following functions:(3)ayt=eatγ/γ,byt=1,cyt=tγγ,dyt=etγ/γ2.
a First, using Definition 1, we see that(4)Lγeatγ/γ=∫0∞eatγ/γe−ζtγ/γdγt=∫0∞ea−ζtγ/γtγ−1dt=limT⟶∞∫0Tea−ζtγ/γtγ−1dt=limT⟶∞1a−ζea−ζtγ/γ0T=Tγγ,ifζ=a,1ζ−a+1ζ−ae−ζ−aTγ/γ,otherwise.
For the convergence of the improper integral,ζ>a is needed. In this case,(5)Lγeatγ/γ=1ζ−a,ζ>a.
b Similar to a, we obtain(6)Lγ1=∫0∞e−ζtγ/γdγt=limT⟶∞∫0Teζtγ/γtγ−1dt,=limT⟶∞−1ζe−ζtγ/γ0T=1ζ,ζ>0.
c We have(7)Lγtγγ=∫0∞tγγe−ζtγ/γdγt=∫0∞tγγtγ−1e−ζtγ/γdt.
By integration by part, we get(8)Lγtγγ=1ζ2,ζ>0.
d Again using Definition 1, we find(9)Lγetγ/γ2=∫0∞etγ/γ2e−ζtγ/γdγt=∫0∞etγ/γtγ/γ−ζdγt.
Ifζ≤0, then tγ/γtγ/γ−ζ≥0 so that etγ/γtγ/γ≥1 and this implies that ∫0∞etγ/γtγ/γ−ζdγt≥∫0∞dγt. The comparison theorem of improper integrals (see Theorem 9 below) revealed that the divergence of the right integral implies the divergence of the left integral. Now, if ζ>0, then ∫0∞etγ/γtγ/γ−ζdγt≥∫ζ∞dγt. By the same argumentation, the integral on the left is divergent. Consequently, the function yt=etγ/γ2 does not have a conformable fractional Laplace transform.
The previous example leads us to question which type or types of functions have a conformable Laplace transform. Through a closer examination of Example1 (a), we observe that for ζ>a, the integral ∫0∞e−ζ−adγt is convergent and a critical component for this convergence is the type of the function yt.
To be precise, ifyt is a continuous function subject to(10)yt≤Meatγ/γ,t≥C,where M≥0 and a and C are constants, then this condition yields(11)∫0∞yte−ζtγ/γdγt≤∫0Cyte−ζtγ/γdγt+M∫C∞e−ζ−atγ/γdγt.
Sinceyt is continuous in 0≤t≤C, by letting A=maxyt:0≤t≤C, we have(12)∫0Cyte−ζtγ/γdγt≤A∫0Ce−−ζtγ/γdγt=A1ζ−e−ζCγ/γζ<∞.
On the other hand, by Example1 (a), the integral ∫0∞e−ζ−adγt is convergent for ζ>a. By the comparison theorem of conformable improper integrals (see Theorem 9 below), the integral on the left is also convergent. That is, yt has a conformable Laplace transform.Definition 2.
A function that satisfies condition (10) is a function with a conformable exponential order at infinity. Graphically, this means that the graph of yt is contained in the region bounded by the graphs of gh=Meatγ/γ and gh=−Meatγ/γ for t≥C. Note also that this type of function controls the negative exponential in the transform integral to keep the integral from blowing up. If C=0, then we say that the function is conformable and exponentially bounded.Theorem 3.
Any bounded functionyt for t≥0 is conformable and exponentially bounded.Proof.
Sinceyt is bounded for t≥0, there is a positive constant M such that yt≤M for all t≥0. But this is the same as (1) with a=0 and C=0. Thus, yt is conformable and exponentially bounded.
The question that can be raised at this level is the following: is it possible to relax the condition of continuity on the functionyt looking at the following situation?Example 2.
We will show that the square wave function with its graph shown in Figure1 has a conformable Laplace transform.
Note that the function is periodic of period 2. Sinceyte−ζtγ/γ≤e−ζtγ/γ, we have ∫0∞yte−ζtγ/γdγt≤∫0∞e−ζtγ/γdγt. But, the integral on the right is convergent for ζ>0, so that the integral on the left is convergent as well. That is, Lγyt exists for ζ>0.
The function in the previous example belongs to a type of function that we define next.Figure 1
Square wave function.Definition 3.
(see [21, 22]). We say that a function is piecewise continuous on an interval if the interval can be broken into a finite number of subintervals on which the function is continuous on each open subinterval (i.e., the subinterval without its endpoints) and has a finite limit at the endpoints (jump discontinuities and no vertical asymptotes) of each subinterval.
Note that a piecewise continuous function is a function that has a finite number of breaks in it and does not blow up to infinity anywhere. A function defined fort≥0 is said to be piecewise continuous on the infinite interval if it is piecewise continuous on 0≤t≤T for all T>0.
Next, we will establish the existence of the conformable Laplace transform for all functions that are piecewise continuous and have a conformable exponential order at infinity. For that purpose, we need the following comparison conformable theorem.Theorem 4 (comparison theorem of conformable improper integrals).
Letyt and zt be both conformable integrable functions for all t≥t0 subject to yt≤zt for t≥t0. If ∫t0∞ztdγt is convergent, then ∫t0∞ytdγt is also convergent. On the other hand, if ∫t0∞ytdγt is divergent, ∫t0∞ztdγt is also divergent.Proof.
Since the conformable improper integrals are given by(13)∫t0∞ytdγt=∫t0∞yttγ−1dt,it is easier to use the same method of the proof of the comparison theorem of improper integrals.Theorem 5 (existence).
Letyt be piecewise continuous on t≥0 and have a conformable exponential order at infinity with yt≤Meatγ/γ for t≥C. Hence, the conformable Laplace transform(14)Lγyt=Yγζ=∫0∞e−ζtγ/γytdγt=∫0∞e−ζtγ/γyttγ−1dtexists as long as ζ>a.Proof.
The integral in the definition ofLγyt can be split into two integrals as follows:(15)∫0∞e−ζtγ/γytdγt=∫0Ce−ζtγ/γytdγt=∫C∞e−ζtγ/γytdγt.
Sinceyt is piecewise continuous in 0≤t≤C, it is bounded there. By letting A=maxyt:0≤t≤C, we have(16)∫0Cyte−ζtγ/γdγt≤A∫0Ce−ζtγ/γdγt=A1ζ−e−ζCγ/γζ<∞.
Now, by Example1 (a), the integral ∫0Cyte−ζtγ/γdγt is convergent for ζ>a. By Theorem 9, the integral on the left is also convergent. That is, yt has a conformable Laplace transform.Remark 1.
The two conditions above are sufficient but are not necessary forLγyt to exist.Example 3.
Let the functionyt=t−γ/2,0<γ≤1, be defined on any interval 0,T. The function yt is discontinuous since as t⟶0+ from right, the limit tends to ∞. However, the conformable Laplace transform of yt exists. Indeed, clearly yt is discontinuous at 0.
Now, to justify that the conformable Laplace transform ofyt exists, we must calculate it.
By applying Definition1, we have(17)Lγt−γ/2=∫0∞t−γ/2e−ζtγ/γdγt=∫0∞t−γ/2e−ζtγ/γtγ−1dt=∫0∞tγ/2−1e−ζtγ/γdt.
Put the variable change as(18)u=ζtγγ⇒t=γuζ1/γ,dt=γuζ1/γ−11ζdu.
Therefore,(19)Lγt−γ/2=∫0∞γuζ−1/2e−u1ζdu=1γζ∫0∞u−1/2e−udu=1γζΓ12=πγζ.
It is the result as required.
In what follows, we will denote the class of all piecewise continuous functions with conformable exponential order at infinity byPℰ. The next theorem proves that any linear combination of functions in Pℰ is also in Pℰ. Similarly, it is the case for the product of two functions in Pℰ.Theorem 6.
Suppose thatyt and zt are two elements of Pℰ with(20)yt≤M1ea1tγ/γ,t≥C1,zt≤M2ea2tγ/γ,t≥C2.(1)
For any constantsβ and μ, the function βyt+μzt is also a member of Pℰ. Moreover,(21)Lγβyt+μzt=βLγyt+μLγzt.(2)
The functionht=ytzt is an element of Pℰ.Proof.
(1)
It is easy to see thatβyt+μzt is a piecewise continuous function.Now, letC=C1+C2,a=maxa1,a2 and M=βM1+μM2. Then, for t≥C, we have(22)βyt+μzt≤βyt+|muzt≤βM1ea1tγ/γ+μM2ea2tγ/γ≤Meatγ/γ.This proves thatβyt+μzt is of conformable exponential order at infinity. On the other hand,(23)Lγβyt+μzt=βLγyt+μLγzt.(2)
It is clear thatht=ytzt is a piecewise continuous function. Now, letting C=C1+C2, M=M1M2, and a=a1+a2, then we see that for t≥C, we have(24)ht=ytzt≤M1M2ea1+a2tγ/γ=Meatγ/γ.Then,ht is of conformable exponential order at infinity. By Theorem 5, Lγht exists for ζ>a.
We next discuss the problem of how to determine the functionyt if Lγyt is given, that is, how to invert the transform. The following result on uniqueness provides a possible answer. This result establishes a one-to-one correspondence between the set Pℰ and its conformable Laplace transforms. Alternatively, the following theorem guarantees that the conformable Laplace transform of a member in Pℰ is unique.Theorem 7 (uniqueness).
Letyt and zt be two elements in Pℰ with conformable Laplace transforms Yγζ and Zγζ such that Yγζ=Zγζ for some ζ>a. Then, yt=zt for all t≥0 where both functions are continuous.
For a piecewise continuous functiony of conformable exponential order at infinity whose conformable Laplace transform is +, we call y the inverse conformable Laplace transform of Yγ and write y=Lγ−1Yζ. Symbolically,(25)yt=ℒγ−1Yγζ⇔Yγζ=ℒγyt.Proof.
The standard techniques used to prove this theorem (i.e., complex analysis, residue computations, and/or Fourier’s integral inversion theorem) are generally beyond the scope of an introductory differential equation course.
The previous theorem states that ifyt is continuous and has a conformable Laplace transform Yγζ, then there is no other function that possesses the same conformable Laplace transform. To find Lγ−1Yγζ, we can inspect tables of conformable Laplace transforms of known functions to find a particular yt that yields the given Yγζ.
When the functionyt is not continuous, the uniqueness of the inverse conformable Laplace transform is not assured. The following example explains the uniqueness issue.Example 4.
Consider the two functionsyt=1,if0≤t≤3,0,ift>3. and zt=1,if0≤t<3,0,ift≥3.(1)
The two functions are equal for allt≠3, and so they are not identical.(2)
We haveLγyt=Lγzt=∫03e−ζtγ/γdγt=−1/ζe−ζtγ/γ03=1−e−ζ3γ/γ/ζ,ζ>0.
Thus, both functionsyt and zt possess the same conformable Laplace transform even though they are not identical. However, they are equal on the interval where they are both continuous.
The inverse conformable Laplace transform satisfies a linear property as indicated in the following result.Theorem 8.
Given two conformable Laplace transformsYγζ and Zγζ, for any constants a and b, we have(26)ℒγ−1aYγζ+bZγζ=aℒγ−1Yγζ+bℒγ−1Zγζ.Proof.
Suppose thatYγζ=ℒγyt and Gγζ=ℒγzt.
Sinceℒγayt+bzt=aℒγyt+bℒγzt=aYγζ+bZγζ, we have(27)ℒγ−1ζ+baYγZγζ=ayt+bzt=aℒγ−1Yγζ+bℒγ−1Zγζ.
We conclude this work with the following table of conformable Laplace transform pairs (see Table1).Table 1
Conformable Laplace transform.
ytℒγyt=Yγζ,0<γ≤1.yt=1,ift≥0,0,otherwise.1/ζ,ζ>0.tp,p∈ℝγp/γΓ1+p/γ/ζ1+p/γ, ζ>0, p/γ>−1.eatγ/γ,a∈ℝ1/ζ−a, ζ>a.sinatγ/γ, a∈ℝa/ζ2+a2, ζ>0.cosatγ/γ, a∈ℝζ/ζ2+a2, ζ>0.sinhatγ/γ, a∈ℝa/ζ2−a2,   ζ>a.coshatγ/γ,   a∈ℝζ/ζ2−a2, ζ>a.ytℒγyt=Yγζ,0<γ≤1.e−atγ/γytYγζ+a, ζ>aIγyt, Iγis the conformable integralYγζ/ζ, ζ>ζ0.tnγ/γnyt,n∈ℕ−1ndn/dζnYγζ, ζ>ζ0.y∗zt,∗is the convolutionYγζZγζ,   ζ>ζ0.


## 4. Application

In order to study the solution of most difficult problems, like a non-homogeneous fractional differential equation with variables coefficients and fractional initial value problems, we are going to explore the following theorems.Theorem 9 (see [20]).
Lety:0,∞⟶ℝ be a continuous real-valued differentiable function and 0<γ≤1. Then,(28)LγDγyt=ζyγζ−y0,ζ>0.Theorem 10 (see [20]).
Lety:0,∞⟶ℝ be a continuous real-valued differentiable function and 0<γ≤1. Then,(29)LγD2γyt=ζ2yγζ−yγ0−ζy0,ζ>0.Example 5.
Consider the following conformable differential problem:(30)y2γt+tγγyγt−2yt=4,yγ0=y0=0.
Let us solve such a problem using a conformable Laplace transform. One writes(31)ℒγy2γt+ℒγtγγyγt−ℒγ2yt=ℒγ4.
Letℒγyt=yγζ. Then, equation (1) is reduced to(32)ζ2yγζ−ddζζyγζ−2yγζ=4s,which can be simplified to(33)yγ′ζ+3ζ−ζyγζ=−4ζ2.
Since(34)e∫3/ζ−ζdζ=ζ3e−ζ2/2,the solution of equation (33) is given by(35)yγζ=eζ2/2ζ3∫ζ3e−ζ2/2−4ζ2dζ+c,=eζ2/2ζ34e−ζ2/2+c.
Hence,(36)yγζ=4ζ3+cζ3eζ2/2,cis constant.
Since(37)limζ⟶∞yγζ=0,the constant c in the second term should be zero, i.e.,(38)yγζ=4ζ3.
Applying the inverse conformable Laplace transform to the both sides of equation (38), we obtain(39)yx=ℒγ−14ζ3=2γ2t2γ.
It is the result as required.


## 5. Conclusion

The authors in discrete and continuous dynamical systems introduced a more general form of Laplace transform to solve fractional dynamic equations with fractional operators on systems of functions with respect to another function. This method is more efficient especially in the discrete problems because the conformable Laplace transforms are not unique in this case. For more details, we refer the readers to [23–25]. However, the conformable Laplace transform gives perfect and easy and exact solution as we mentioned in Example 5.



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