A Novel Integral Operator Transform and Its Application to Some FODE and FPDE with Some Kind of Singularities

Abstract

We introduced a novel integral transform operator. We proved the existence and the uniqueness of the relatively new operator. We presented some useful properties of the new operator. We presented the application of this operator for solving some kind of fractional ordinary and partial differential equation containing some kind of singularity.

1. Introduction

Mathematical notation aside, the motivation behind integral transforms is easy to understand. There are many classes of problems that are difficult to solve or at least quite unwieldy algebraically in their original representations. An integral transform “maps” an equation from its original “domain” into another domain [1–3]. Manipulating and solving the equation in the target domain can be much easier than manipulation and solution in the original domain. The solution is then mapped back to the original domain with the inverse of the integral transform. There exist few integral transform operators in the literature [1–3], which are commonly used to solve partial fractional and fractional ordinary differential equations.

The Fourier transform, named after Joseph Fourier, is a mathematical transform with many applications in physics and engineering [4–11]. Very commonly it transforms a mathematical function of time, 𝑓(𝑡), into a new function, sometimes denoted by 𝐹, whose argument is frequency with units of cycles per second or (hertz) or radians per second. The new function is then known as the Fourier transform and/or the frequency spectrum of the function 𝑓. The Fourier transform is also a reversible operation. Thus, given the function 𝑓 one can determine the original function 𝑓; see in [8].

The Laplace transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems [12–17]. The Laplace transform is particularly useful in solving linear ordinary differential equations or partial fractional differential equations such as those arising in the analysis of groundwater pollution model [13] and electronic circuits [14].

In mathematics, the Mellin transform [15] is an integral transform that may be regarded as the multiplicative version of the two-sided Laplace transform. This integral transform is closely connected to the theory of Dirichlet series and is often used in number theory and the theory of asymptotic expansions; it is closely related to the Laplace transform and the Fourier transform and the theory of the gamma function and allied special functions.

The Mellin transform is widely used in computer science because of its scale invariance property [18]. The magnitude of the Mellin transform of a scaled function is identical to the magnitude of the original function [18]. This scale invariance property is analogous to the Fourier transform’s shift invariance property. The magnitude of a Fourier transform of a time-shifted function is identical to the original function. This property is useful in image recognition. An image of an object is easily scaled when the object is moved towards or away from the camera [19].

In mathematics, the Sumudu transform is an integral transform similar to the Laplace transform, introduced in the early 1990s by Watugala to solve differential equations and control engineering problems [20–27]. It is equivalent to the Laplace-Carson transform with the substitution 𝑝=1/𝑢.

However, there exists some kind of fractional ordinary and partial differential equations with some kind of singularities that cannot be solved directly via the above integral transform operators. In particular, the following kind of fractional ordinary and partial differential equations ₀𝐷^𝛼_𝑥𝑦(𝑥)+1𝑥^𝑛𝑦(𝑥)=𝑓(𝑥) (1) or ₀𝐷^𝛼_𝑥𝑢()+1𝑥,𝑡𝑥^𝑛𝑢(𝑥,𝑡)=ℎ(𝑥,𝑡), (2) where 𝛼>0, ₀𝐷^𝛼_𝑥, is the fractional derivative (Riemann-Liouville or Caputo) and 𝑛≥1.

To solve the above equations, some scholars make use of the Frobenius method, to obtain the solutions in series form. The Laplace transform of the product of two functions is different from the product of the Laplace transform of the two functions. The Fourier transform of the product of two functions is equivalent to the convolution of the Fourier transform of the two functions. This renders it very difficult to apply directly either the Laplace transform or the Fourier transform operators to solve this type of equation. Therefore some scholars multiply 𝑥^𝑛 on both sides of the above equations and then apply the Fourier or the Laplace transform. It is therefore worth to define an integral transform similar to Laplace or Laplace-Carson transform to transform such equation to an ordinary or partial differential equation without any additional transformation.

The aim of this work is to further introduce an integral transform operator that can be used to solve some kind of ordinary, partial and fractional ordinary, partial differential equation with some kind of singularities. We will start with the definition and present some theorems.

2. Definitions and Theorems

Definition 1. Let 𝑓(𝑥) be a continuous function over an open interval (0,∞) such that its Laplace transform is 𝑛 time differentiable; then, the new integral transform of order 𝑛 of 𝑓 is defined as follows: 𝑀_𝑛(𝑠)=𝑀_𝑛𝑓(𝑥)(𝑠)=^∞₀𝑥^𝑛𝑒^−𝑥𝑠𝑓(𝑥)𝑑𝑥 (3) and the inverse of the new integral transform of order 𝑛 is defined as 𝑓(𝑥)=𝑀_𝑛⁻¹𝑀_𝑛𝑓(𝑥=())−1^𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥(−1)^𝑛1Γ()×𝑛−1_𝑠0(𝑠−𝑡)^𝑛−1𝑀_𝑛(𝑡)+𝑑𝑡_{𝑛−1𝑘=0}𝑠^𝑘𝑦𝑘!^𝑘𝑦𝑑𝑠,_𝑘=𝜕^𝑘𝐹(0)𝜕𝑠^𝑘, (4) where 𝐹(𝑠) is the Laplace transform of 𝑓(𝑥). Before we continue, we will prove that the above definition is indeed the inverse operator transform of order 𝑛. In fact from the definition of new transform of order 𝑛 of a function 𝑓(𝑥), we have that 𝑀_𝑛(𝑠)=𝑀_𝑛𝑓(𝑥𝑠)=)(_∞0𝑥^𝑛𝑒^−𝑥𝑠𝑓(𝑥)𝑑𝑥=(−1)^𝑛𝑑^𝑛𝐹(𝑠)𝑑𝑠^𝑛, (5) thus 1Γ(𝑛−1)^𝑠₀(𝑠−𝑡)^𝑛−1𝑀_𝑛(𝑡)=()𝑑𝑡−1^𝑛𝐹(𝑠)−^𝑛−1_𝑘=0𝑠^𝑘𝑦𝑘!^𝑘. (6) It follows that (−1)^𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥(−1)^𝑛1Γ(𝑛−1)×^𝑠₀(𝑠−𝑡)^𝑛−1𝑀_𝑛(𝑡)+𝑑𝑡_{𝑛−1𝑘=0}𝑠^𝑘𝑦𝑘!^𝑘=()𝑑𝑠−1^𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥(−1)^𝑛[𝐹(𝑠𝑀)]𝑑𝑠_𝑛−1𝑀_𝑛𝑓(𝑥=())−1^2𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥[[𝐹(𝑠)]]𝑑𝑠=𝑓(𝑥). (7) Therefore the inverse of the new integral transform is well defined. Our next concern is to prove the uniqueness and the existence of the new integral transform.

Theorem 2. Let 𝑓(𝑥) and 𝑔(𝑥) be continuous functions defined for 𝑥≥0and having new transforms of order 𝑛, 𝐹(𝑝), and 𝐺(𝑝), respectively. If 𝐹(𝑝)=𝐺(𝑝), then 𝑓(𝑥)=𝑔(𝑥).

Proof. From the definition of the inverse of the new transform of order 𝑛, if 𝛼is sufficiently large, then the integral expression, by 𝑓(𝑥)=(−1)^𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥(−1)^𝑛𝐹𝑝𝑑𝑝 (8) for the inverse of the new integral transform of order 𝑛, can be used to obtain 𝑓(𝑥)=(−1)^2𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥𝐹𝑝𝑑𝑝. (9) By hypothesis we have that 𝐹(𝑝)=𝐺(𝑝); then replacing this in the above expression we have the following: 𝑓(𝑥)=(−1)^𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥(−1)^𝑛𝐺𝑝𝑑𝑝 (10) which boils down to 𝑓(𝑥)=(−1)^𝑛2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥(−1)^𝑛𝐺𝑝𝑑𝑝=𝑔(𝑥) (11) and this proves the uniqueness of the new integral transform of order 𝑛.

Theorem 3. If 𝑓(𝑡) is a piecewise continuous on every finite interval in [0,𝑡₀) and satisfies: 𝑡^𝑛𝑓(𝑡)≤𝑀𝑒^𝛼𝑡 (12) for all 𝑡∈[𝑡₀,∞), then 𝑀_𝑛[𝑓(𝑥)](𝑠) exists for all 𝑠>𝛼.

Proof. To prove the theorem we must show that the improper integral converges for 𝑠>𝑎. Splitting the improper integral into two parts, we have ^∞₀𝑡^𝑛𝑒^−𝑠𝑡𝑓(𝑡)=𝑑𝑡_𝑡00𝑡^𝑛𝑒^−𝑠𝑡𝑓(𝑡)𝑑𝑡+^∞_𝑡0𝑡^𝑛𝑒^−𝑠𝑡𝑓(𝑡)𝑑𝑡. (13) The first integral on the right side exists by hypothesis1; hence the existence of the new integral transform of order 𝑛, 𝑀_𝑛(𝑠), depends on the convergence of the second integral. By hypothesis2, we have 𝑡^𝑛𝑒^−𝑠𝑡𝑓(𝑡)≤𝑀𝑒^𝛼𝑡𝑒^−𝑠𝑡=𝑀𝑒^{(𝛼−𝑠)𝑡}. (14) Now ^∞_𝑡0𝑀𝑒^{(𝛼−𝑠)𝑡}𝑒𝑑𝑡=𝑀^{(𝛼−𝑠)𝑡0},𝛼−𝑠 (15) this converges for 𝛼<𝑠. Then, by the comparison test for improper integrals theorem, 𝑀_𝑛(𝑠), exists for 𝛼<𝑠.

Remark 4. There is a relationship between the Laplace transform and the new integral transform of order 𝑛 as follows: 𝐿𝑓(𝑥)(𝑠)=𝑀_𝑛1𝑥^𝑛𝑓(𝑥𝑠),𝐿𝑓(𝑥)𝑠))((=𝑀₀𝑓(𝑥)𝑠),𝑀_𝑛(𝑓(𝑥)(𝑠)=(−1)^𝑛𝑑^𝑛[𝐹(𝑠)]𝑑𝑠^𝑛, (16) where 𝐹(𝑠) is the Laplace transform of 𝑓(𝑥).

Remark 5. There is a relationship between the Laplace-Carson transform and then new integral transform of order 𝑛  as follows: 𝐿_𝑐𝑓(𝑥)(𝑠)=𝑀₁𝑓(𝑥)(𝑠). (17)

Theorem 6. A function 𝑓(𝑥) which is continuous on [0,∞) and satisfies the growth condition 𝑓(𝑥) can be recovered from the Laplace transform 𝐹(𝑝) as follows: 𝑓(𝑥)=lim_𝑛→∞(−1)^𝑛𝑛𝑛!𝑥^𝑛+1𝑀_𝑛𝑛𝑠. (18) Evidently, the main difficulty in using Theorem 6 for computing the inverse Laplace transform is the repeated symbolic differentiation of 𝐹(𝑝).

3. Some Properties of the New Integral Transform

In this section, we consider some of the properties of the new integral transform that will enable us to find further transform pairs {𝑓(𝑥),𝑀_𝑛(𝑠)} without having to compute consider the following: (I)𝑀_𝑛[𝑠+𝑐]=𝑀_𝑛𝑒^−𝑐𝑥𝑓(𝑥,()𝑀)II_𝑛𝑓(𝑠)=1𝑎𝑥)(𝑎𝑀_𝑛𝑠𝑎,(III)^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥𝑀_𝑛(𝑠)𝑑𝑠=𝑥^𝑛𝑓(𝑥),(IV)𝑀_𝑛(𝑥)(𝑥𝑠)=𝑎𝑓+𝑏𝑔)(𝑎𝑀_𝑛𝑓(𝑥)+𝑏𝑀_𝑛𝑔(𝑥)𝑠)(V)𝑀_𝑛(𝑓(𝑥)𝑥^𝑛𝑠)𝑓(𝑥𝑠),()𝑀(=𝐿)(VI_𝑛𝑓(𝑥)(𝑥𝑠)=()∗ℎ)(−1^𝑛𝑛_𝑘=0𝐶^𝑘_𝑛𝑑^𝑘(𝐺(𝑠))𝑑𝑠^𝑘×𝑑^𝑛−𝑘(𝐹(𝑠))𝑑𝑠^𝑛−𝑘(VII)𝑀_𝑛𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛𝑠)=()(−1^𝑛𝑛_𝑘=0𝐶^𝑘_𝑛𝑑^𝑘𝑠^𝑛𝑑𝑠^𝑘×𝑑^𝑛−𝑘(𝐹(𝑠))𝑑𝑠^𝑛−𝑘. (19) Let us verify the above properties. We will start with I, by definition we have the following: 𝑀_𝑛𝑒^−𝑐𝑥𝑓(𝑥=_∞0)𝑥^𝑛𝑒^−𝑐𝑥𝑒^−𝑠𝑥𝑓(𝑥=)𝑑𝑥_∞0𝑥^𝑛𝑒^{−(𝑐+𝑠)𝑥}𝑓(𝑥)𝑑𝑥=𝑀_𝑛[𝑠+𝑐] (20) and then the first property is verified.

For II we have the following by definition: 𝑀_𝑛𝑓(𝑠)=𝑎𝑥)(_∞0𝑥^𝑛𝑒^−𝑥𝑠𝑓(𝑎𝑥)𝑑𝑥=(−1)^𝑛𝑑^𝑛𝑑𝑠^𝑛𝐿𝑓(𝑎𝑥)(𝑠). (21) Now using the property of the Laplace transform 𝐿[𝑓(𝑎𝑥)](𝑠)=(1/𝑎)𝐹(𝑠/𝑎), from this we can further obtain 𝑀_𝑛𝑓(𝑠)=()𝑎𝑥)(−1^𝑛𝑑^𝑛𝑑𝑠^𝑛1𝑎𝐹𝑠𝑎=1𝑎(−1)^𝑛𝑑^𝑛𝑑𝑠^𝑛𝐹𝑠𝑎=1𝑎𝑀_𝑛𝑠𝑎 (22) and then, the property number II is verified.

For number III we have the following: Let 𝑔(𝑥)=𝑥^𝑛𝑓(𝑥), then ^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥𝑀_𝑛(𝑠)=𝑑𝑠_{𝛼+𝑖∞𝛼−𝑖∞}𝑒^𝑠𝑥∞₀𝑒^−𝑥𝑠𝑥^𝑛𝑓(𝑥)=𝑑𝑥𝑑𝑠_{𝛼+𝑖∞𝛼−𝑖∞}𝑒^𝑠𝑥∞₀𝑒^−𝑥𝑠𝑔(𝑥)𝑑𝑥𝑑𝑠. (23) By the theorem of inverse Laplace transform we obtain ^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑠𝑥𝑀_𝑛(𝑠)𝑑𝑠=𝑔(𝑥)=𝑥^𝑛𝑓(𝑥), (24) numbers IV and V are obvious to be verified. For number VI we have the following by definition: 𝑀_𝑛𝑓(𝑥)(𝑥𝑠)=∗ℎ)(_∞0𝑥^𝑛𝑒^−𝑠𝑥𝑓(𝑥)(𝑥=()∗ℎ)−1^𝑛𝑑^𝑛𝑑𝑠^𝑛𝐿𝑓(𝑥)∗ℎ(𝑥)(𝑠), (25) now using the property of Laplace transform of the convolution, we obtain the following: 𝐿𝑓(𝑥)∗ℎ(𝑥)(𝑠)=𝐹(𝑠)⋅𝐺(𝑠) (26) and then, using the property of the derivative of order 𝑛 for the product of two functions, we obtain 𝑀_𝑛𝑓(𝑥)(𝑥𝑠)=()∗ℎ)(−1^𝑛𝑑^𝑛𝑑𝑠^𝑛[𝐹(𝑠)(𝑠=()⋅𝐺)]−1^𝑛𝑛_𝑘=0𝐶^𝑘_𝑛𝑑^𝑘(𝐺(𝑠))𝑑𝑠^𝑘×𝑑^𝑛−𝑘(𝐹(𝑠))𝑑𝑠^𝑛−𝑘 (27) and then, the property number VI is verified.

For number VII, by definition, we have the following: 𝑀_𝑛𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛𝑠)=_∞0(𝑥^𝑛𝑒^−𝑠𝑥𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛=()𝑑𝑥−1^𝑛𝑑^𝑛𝑑𝑠^𝑛𝐿𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛(𝑠), (28) now using the property of the Laplace transform, 𝐿𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛(𝑠)=𝑠^𝑛𝐹(𝑠)−^𝑛−1_𝑘=0𝑠^{𝑛−𝑘−1}𝑑^𝑘𝑓(0)𝑑𝑥^𝑘 (29) now deriving the above expression 𝑛 times, we obtain the following expression: (−1)^𝑛𝑑^𝑛𝑑𝑠^𝑛𝐿𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛(𝑠=())−1^𝑛𝑛_𝑘=0𝐶^𝑘_𝑛𝑑^𝑘𝑠^𝑛𝑑𝑠^𝑘×𝑑^𝑛−𝑘(𝐹(𝑠))𝑑𝑠^𝑛−𝑘 (30) that is: 𝑀_𝑛𝑑^𝑛𝑓(𝑥)𝑑𝑥^𝑛(𝑠)=(−1)^𝑛𝑛_𝑘=0𝐶^𝑘_𝑛𝑑^𝑘𝑠^𝑛𝑑𝑠^𝑘×𝑑^𝑛−𝑘(𝐹(𝑠))𝑑𝑠^𝑛−𝑘. (31) This completes the proof of number VI.

4. Application to FODE and FPDE

Recently, the differential equations of fractional order derivative with singularities have been the focus of many studies due to their frequent appearance in various applications in fluid mechanics, viscoelasticity, biology physics, engineering, and groundwater models, in particular the monitoring of the flow through the geological formation and the pollution migration. Consequently, considerable attention has been given to the solutions of fractional differential equations and integral equations with singularity of physical interest. There exists in the literature some integral transform method that can be used to derive exact and approximate solutions for such equations; see, for instance, Laplace transform method [4–11], the Fourier transform method [12–17], the Mellin transform method [18, 19], the Sumudu transform method [20–27], the Adomian decomposition method [28, 29], and the homotopy decomposition method [30–33]. In this section we present the application of the proposed integral operator to the Cauchy-type of fractional ordinary differential and partial differential equations. We will start with the fractional ordinary differential equation. Here we consider the Cauchy-type equation of the following form: 𝐷^𝛼_𝑟𝑟Φ(𝑟)+1𝑟^𝑛Φ(𝑟)=0,𝑙−1<𝛼≤𝑙. (32) To solve the above equation, we apply on both sides the new integral transform of order 𝑛 to obtain the following: (−1)^𝑛𝑑^𝑛𝑑𝑠^𝑛𝐷^𝛼_𝑟𝑟Φ(𝑠)+Φ(𝑠)=0. (33) The new integral transform has gotten rid of the singularity; the new equation is just an ordinary fractional differential equation, which can be solved with, for instance, the homotopy decomposition method. Let us find the exact solution of the above equation for 𝑛=1 given below as 𝐷^𝛼_𝑟𝑟Φ(𝑟)+1𝑟Φ(𝑟)=0,𝑙−1<𝛼≤𝑙. (34) We will make use of the new integral transform to derive analytical solution of (34). Applying the new transform of order 1 on both sides of the above equation,  we obtain the following expression: 𝑑[𝐿(Φ)(𝑠)]+𝛼𝑑𝑠𝑠+1𝑠^𝛼(𝐿(Φ𝑠=)())_𝑙𝑚=2𝑑_𝑚(𝑚−1)𝑠^{𝑚−2−𝛼}, (35) where 𝑑_𝑚=𝐷₀^𝛼−𝑚+Φ(0⁺) (𝑚=2,…,𝑙). Now, one can derive the solution of the ordinary order differential equation with respect to the Laplace transform of Θ(s)=𝐿(Φ(𝑟)): Θ(𝑠)=𝑠^−𝛼−𝑠exp^1−𝛼×𝑎1−𝛼₁+^𝑙_𝑚=2𝑑_𝑚(𝑚−1)𝑠^𝑚−2−𝑠exp^1−𝛼,1−𝛼𝑑𝑠 (36) with 𝑎₁ an arbitrary real constant that will be obtained via the initial condition. We next expand the exponential function in the integrand in a series, and using term-by-term integration, we arrive at the following expression: Θ(𝑠)=𝑐Θ₁(𝑠)+^𝑙_𝑚=2𝑑_𝑚(𝑚−1)Θ^∗_𝑚(𝑠) (37) with of course Θ₁(𝑠)=𝑠^−𝛼−𝑠exp^1−𝛼,Θ1−𝛼_∗𝑚(𝑠)=𝑠^−𝛼𝑠exp^1−𝛼×𝛼−1_∞𝑗=011−𝛼^𝑗𝑠^{(1−𝛼)𝑗+𝑚−1}).(1−𝛼𝑗+𝑚−1𝑗! (38) Now, applying the inverse Laplace transform on Θ₁(𝑠) and using the fact that 𝑠^{−[𝛼+(𝛼−1)𝑗]}𝑟=𝐿^{𝛼+(𝛼−1)𝑗−1}Γ𝛼+(𝛼−1)𝑗 (39) we obtain Φ₁(𝑟)=𝑟^𝛼−1oΨ₁)∣𝑥(𝛼,𝛼−1^𝛼−1𝛼−1 (40) with oΨ₁[] the generalized Wright function for 𝑝=1 and 𝑞=2 [34–37]. We next expand the exponential function exp[−𝑠^1−𝛼/(1−𝛼)] in power series, multiplying the resulting two series; in addition to this if we consider the number 𝑏_𝑘(𝛼,𝑚) defined for 𝛼>0,𝑚=2,…𝑙(𝛼≠(𝑝+𝑚−1)/𝑝,𝑝∉ℕ), and 𝑘∈ℕ₀, 𝑏_𝑘(𝛼,𝑚)=^𝑙_{𝑝,𝑗=0,…,𝑘,𝑝+𝑗=𝑘}(−1)^𝑞().𝑝!𝑗!1−𝛼𝑞+𝑚−1 (41) The above family of number possesses satisfies the following recursive formula: 𝑏_𝑘(𝛼,𝑚)𝑏_𝑘+1()=𝛼,𝑚𝛼−𝑚𝛼−1+𝑘, (42) which produces the explicit expression for 𝑏_𝑘(𝛼,𝑚) in the form of 𝑏_𝑘()=Γ)/(𝛼,𝑚[(𝛼−𝑚𝛼−1)](𝑚−1)Γ[((𝛼−𝑚)/(𝛼−1))+𝑘],𝑘∈ℕ₀. (43) Now having the above expression on hand, we can derive that Θ^∗_𝑚(𝑠)=𝑠^{𝑚−𝛼−1∞}_𝑗=011−𝛼^𝑗𝑠^{(1−𝛼)𝑝}×𝑝!_∞𝑝=011−𝛼^𝑝(−1)^𝑝)𝑠(1−𝛼𝑝+𝑚−1^{(1−𝛼)𝑗}=𝑝!_∞𝑘=0𝑏_𝑘()1𝛼,𝑚1−𝛼^𝑘×𝑠^{(1−𝛼)𝑘+𝑚−𝛼−1}(𝑚=2,…,𝑙). (44) However, remembering (40) with 𝛽=(𝛼−1)𝑘+𝛼+1−𝑚, we can further derive the following expression for Φ^∗_𝑚(𝑟) as Φ^∗_𝑚(𝑟)=^∞_𝑘=0𝑏_𝑘(1𝛼,𝑚)1−𝛼^𝑘×Γ(𝑘+1)Γ[()𝑘]𝑥𝛼+1−𝑚+𝛼−1𝑘!^{(𝛼−1)𝑘+𝛼−𝑚} (45) or in the simplified version we have Φ^∗_𝑚(𝑟)=Γ[(𝛼−𝑚)/(𝛼−1)]()Φ𝑚−1^𝑚(𝑟), (46) where Φ_𝑚(𝑟)=𝑟₁^𝛼−𝑚Ψ₂×()(),1,1𝛼+1−𝑚,𝛼−1𝛼−𝑚∣𝑟𝛼−1,1^𝛼−1.𝛼−1 (47) It follows that the solution of the Cauchy-type equation is in the form of Φ(𝑟)=𝑎₁𝑟^𝛼−1oΨ₁)∣𝑥(𝛼,𝛼−1^𝛼−1𝛼−1+𝑎₂^𝑙_𝑚=2𝑏_𝑚()Γ)/(𝑚−1[(𝛼−𝑚𝛼−1)]()𝑟𝑚−1^1𝛼−𝑚Ψ₂×()(),1,1𝛼+1−𝑚,𝛼−1𝛼−𝑚∣𝑟𝛼−1,1^𝛼−1.𝛼−1 (48) We will examine the solution of the following fractional partial differential equation of the following form: ^𝐶₀𝐷^𝛼_𝑡𝑢()=1𝑥,𝑡𝑥𝜕²𝑢(𝑥,𝑡)𝜕𝑥²,0<𝛼≤1, (49) with initial and boundary conditions of the form 𝑢(𝑥,0)=0,𝑢𝑥₀(𝑡),𝜕,𝑡=ℎ_𝑥𝑢(0,𝑡)=𝑢(0,𝑡)=0(𝑡≥0). (50) To solve the above problem, the first step consists of applying the new integral transform on both sides of (49) to obtain 𝜕_𝑠^𝐶₀𝐷^𝛼_𝑡𝑈(𝑠,𝑡)=−𝑠²𝑈(𝑠,𝑡), (51) where 𝑠 is the Laplace variable. The next step in this derivation is to apply the Fourier transform in time to obtain 𝑖𝑝^𝛼𝜕_𝑠𝑈₁𝑠,𝑝=−𝑠²𝑈₁𝑠,𝑝, (52) where𝑝 is the Fourier variable. It follows that the solution of the above equation is simply given as 𝑈₁𝑝−𝑠𝑠,𝑝=𝑐exp³3𝑖𝑝^−𝛼. (53) The next step is to put exponential function in series form as follows −𝑠exp³3𝑖𝑝^−𝛼=^∞_𝑘=0−𝑠³/3𝑖𝑝^−𝛼𝑘=𝑘!^∞_𝑘=0−𝑠³/3^𝑘𝑖𝑝^−𝑘𝛼.𝑘! (54) Then, we first apply the inverse Laplace in both sides of the above equation to obtain 𝑈₁𝑥,𝑝=𝐿⁻¹𝑐𝑝^∞_𝑘=0−𝑠³/3^𝑘𝑖𝑝^−𝑘𝛼.𝑘! (55) Making use of the linearity to the inverse Laplace transform, we obtain 𝑈₁𝑥,𝑝=^∞_𝑘=0𝐿⁻¹−𝑠³/3^𝑘𝑐(𝑝)𝑖𝑝^−𝑘𝛼.𝑘! (56) And finally making use of the inverse Fourier transform and its linearity, we obtain 𝑢(𝑥,𝑡)=^∞_𝑘=0𝐿⁻¹−𝑠³/3^𝑘𝐹⁻¹𝑐(𝑝)𝑖𝑝^−𝑘𝛼.𝑘! (57) This, produces the solution of (49).

5. Conclusion

We introduced a new integral operator transform. We presented its existence and uniqueness. We presented some properties and its application for solving some kind of ordinary and partial fractional differential equations that arise in many fields of sciences.

Conflict of Interests

The authors declare that they have no conflict of interests.

Authors’ Contribution

A. Atangana wrote the first draft and A. Kilicman corrected the final version. All authors read and approved the final draft.

Acknowledgments

The authors would like to thank the referee for some valuable comments and helpful suggestions. Special thanks go to the editor for his valuable time spent to evaluate this paper.