EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6340 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Conformable Double Laplace-Shehu Transform Monther Al-Momani1, Baha’ Abughazaleh2,∗ 1 Department of Basic Sciences, Al-Ahliyya Amman University, Amman, Jordan 2 Department of Mathematics, Isra University, Amman, Jordan Abstract. We present a new transform called the conformable double Laplace-Shehu transform. This tool helps in solving fractional partial differential equations. These equations come up often in science and engineering. The transform is built using the idea of the conformable derivative. We explain the basic rules of the transform and show how it can be used. To show its use, we solve two equations that are well known. These are the wave and heat equations. 2020 Mathematics Subject Classifications: 44A05 Key Words and Phrases: Laplace transform, Shehu transform, double transform, conformable double Laplace-Shehu transform 1. Introduction Fractional partial differential equations are used to model many problems in physics, electric circuits, fluid flow, optics, and biology. The conformable derivative, as introduced in [1], keeps most of the main ideas of classical derivatives while working for fractional cases. Several methods have been developed for solving conformable fractional equations. The conformable double Laplace transform was discussed in [2, 3], while the conformable double Sumudu transform appeared in [4]. More work on these types of transforms can be found in [5–7]. Later, a new method called the double Laplace-Shehu transform was proposed in [8]. It was applied successfully to different types of partial differential equations. Further studies related to integral transforms are available in [9–16]. In this paper, we presented a new method called the conformable double Laplace-Shehu transform (CL-SH) for solving a specific type of partial differential equations. We first outline the basic properties of this transform such as the conditions that make it valid for use and how it interacts with mathematical derivatives then we demonstrate through practical examples how it can be applied to solve these equations. This method offers a different perspective for studying mathematical problems and may pave the way for developing new ideas in applied mathematics. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6340 Email addresses: montheralmomani72@gmail.com (M. Al-Momani), baha.abughazaleh@iu.edu.jo (B. Abughazaleh) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 2 of 12 2. Conformable Fractional Derivative We give the main definitions and results for conformable fractional derivatives in this section. Definition 1. [1] Let 0 < γ ≤ 1 and r : [0,∞)→ R. The conformable fractional derivative of order γ is defined as: dγ dλγ r(λ) = lim υ→0 r(λ+ υλ1−γ)− r(λ) υ where λ ≥ 0, and ∂γ ∂λγ is referred to as the fractional derivative of order γ. Definition 2. [17] Let 0 < γ1, γ2 ≤ 1 and r(λ, µ) : [0,∞)×[0,∞)→ R. The conformable partial derivatives of orders γ1 and γ2 of the function r(λ, µ) are defined as: ∂γ1 ∂λγ1 r(λ, µ) = lim υ→0 r(λ+ υλ1−γ1 , µ)− r(λ, µ) υ ∂γ2 ∂µγ2 r(λ, µ) = lim υ→0 r(λ, µ+ υµ1−γ2)− r(λ, µ) υ where λ, µ ≥ 0, ∂γ1 ∂λγ1 and ∂γ2 ∂µγ2 are referred to as fractional derivatives of orders γ1 and γ2, respectively. Theorem 1. [18]Suppose that r(λ, µ) is differentiable at a point λ, µ ≥ 0, 0 < γ1, γ2 ≤ 1, then: ∂γ1r ∂λγ1 = λ1−γ1 ∂r ∂λ , ∂γ2r ∂µγ2 = µ1−γ2 ∂r ∂µ . 3. The Conformable Double Laplace-Shehu transform This section defines the CL-SH and explains its basic properties. We start by introduc- ing the individual conformable Laplace and Shehu transforms, then define the combined form. We prove that the CL-SH is a linear operator. We also give the conditions under which the transform exists. Several examples for basic functions are included. Finally, we show how the CL-SH interacts with conformable partial derivatives through a set of key identities. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 3 of 12 3.1. Definition and basic properties of the conformable double Laplace- Shehu transform Definition 3. Let r(λ, µ) be a continuous function on [0,∞)×[0,∞). Then 1- The conformable Laplace transformation (CL) of r(λ, µ), denoted by Lγ λ[r(λ, µ)], is defined as: A (ς) = Lγ λ(r(λ, µ)) = ∞∫ 0 e −ς λγ γ r(λ, µ)λγ−1dλ, ς ∈ C 2- The conformable Shehu transformation (CSH) of r(λ, µ), denoted by Lγ µ[r(λ, µ)], is defined as: B (υ) = Hγ µ(r(λ, µ)) = ∞∫ 0 e −τ µγ υγ r(λ, µ)µγ−1dµ, υ ∈ C 3- The CL-SH of r(λ, µ), denoted by Lγ1 λ Hγ2 µ [r(λ, µ)], is defined as: R(ς, υ) = Lγ1 λ Hγ2 µ [r(λ, µ)] = ∫ ∞ 0 ∫ ∞ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) r(λ, µ)λγ1−1µγ2−1dλdµ. Theorem 2. Assume that r : [0,∞)×[0,∞)→ R such that R(ς, υ) = Lγ1 λ Hγ2 µ [r(λ γ1 γ1 , µ γ2 γ2 )] exist, then Lγ1 λ Hγ2 µ [r( λγ1 γ1 , µγ2 γ2 )] = LλHµ[r(λ, µ)], where LλHµ[r(λ, µ)] = ∫ ∞ 0 ∫ ∞ 0 e−(ςλ+ τµ υ )r(λ, µ) dλ dµ. Proof. Lγ1 λ Hγ2 µ [r( λγ1 γ1 , µγ2 γ2 )] = ∫ ∞ 0 ∫ ∞ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) r( λγ1 γ1 , µγ2 γ2 )λγ1−1µγ2−1dλdµ (1) Substitute z = λγ1 γ1 and w = µγ2 γ2 in Equation 1, we have Lγ1 λ Hγ2 µ [r( λγ1 γ1 , µγ2 γ2 )] = ∫ ∞ 0 ∫ ∞ 0 e−(ςz+ τw υ )r(z, w)dzdw = ∫ ∞ 0 ∫ ∞ 0 e−(ςλ+ τµ υ )r(λ, µ) dλ dµ = LλHµ[r(λ, µ)] Lemma 1. Lγ1 λ Hγ2 µ (r(λ, µ)) is a linear transformation. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 4 of 12 Proof. for nonzero constants α and β, we have Lγ1 λ W γ2 µ (αr1(λ, µ)+βr2(λ, µ)) = ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) (αr1(λ, µ) + βr2(λ, µ))λ γ1−1µγ2−1dλdµ, = α ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) r1(λ, µ)λ γ1−1µγ2−1dλdµ+ β ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) r2(λ, µ)λ γ1−1µγ2−1dλdµ = αLγ1 λ W γ2 µ (r1(λ, µ)) + βLγ1 λ W γ2 µ (r2(λ, µ)). If r(λ, µ) can be written as r(λ, µ) = p(λ)q(µ) for some continuous functions p and q, then Lγ1 λ Hγ2 µ (r(λ, µ)) = Lγ1 λ (p(λ))Hγ2 µ (q(µ)). In fact Lγ1 λ Hγ2 µ (r(λ, µ)) = Lγ1 λ Hγ2 µ (p(λ)q(µ)) = ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) p(λ)q(µ)λγ1−1µγ2−1dλdµ = ∞∫ 0 e −ς λγ1 γ1 p(λ)λγ1−1dλ ∞∫ 0 e −τ µγ2 υγ2 q(µ)µγ2−1dµ  = Lγ1 λ (p(λ))Hγ2 µ (q(µ)). Definition 4. Let 0 < γ1, γ2 ≤ 1. Then a function r(λ, µ) is said to be of conformable exponential orders α and β on 0 ≤ λ < ∞ and 0 ≤ µ < ∞. If there exist K,N,M > 0 such that |r(λ, µ)| ≤ Ke αλγ1 γ1 +β µγ2 γ2 , for all λγ1 γ1 > N , µγ2 γ2 > M. Theorem 3. Let 0 < γ1, γ2 ≤ 1 and r(λ, µ) be a continuous function on the region [0,∞)×[0,∞) of conformable exponential orders α and β. Then R(ς, υ) = Lγ1 λ Hγ2 µ [r(λ, µ)] exists for ς, υ whenever Re (ς) > α and Re ( τ υ ) > β. We have |R(ς, υ)| = ∣∣∣∣∣∣ ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) r(λ, µ)λγ1−1µγ2−1 dλdµ ∣∣∣∣∣∣ ≤ ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) |r(λ, µ)|λγ1−1µγ2−1dλdµ ≤ K ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) e αλγ1 γ1 +β µγ2 γ2 λγ1−1µγ2−1dλdµ M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 5 of 12 = K ∞∫ 0 e −(ς−α)λ γ1 γ1 λγ1−1dλ ∞∫ 0 e −( 1 υ −β)µ γ2 γ2 µγ2−1dµ  = Kυ (ς − α) (τ − βυ) , where Re (ς) > α and Re ( τ υ ) > β. 3.2. The conformable double Laplace-Shehu transform for some basic functions (i) Lγ1 λ Hγ2 µ [c] = LλHµ[c] = cυ ςτ , c ∈ R, (ii) Lγ1 λ Hγ2 µ [( λγ1 γ1 )α(µγ2 γ2 )β ] = LλHµ[λ αµβ ] = υβ+1 ςα+1τβ+1 Γ(α+ 1)Γ(β + 1), Re(ς) > 0 and Re(α) > −1, (iii) Lγ1 λ Hγ2 µ [ e αλγ1 γ1 +β µγ2 γ2 ] = LλHµ[e αλ+βµ] = υ (ς − α) (τ − βυ) ,Re(ς) > Re(α). 3.3. Derivatives properties Now, we present some basic properties of the CL-SH Let R(ς, υ) = Lγ1 λ Hγ2 µ (r(λ, µ)) where r(λ, µ) is a continuous function on [0,∞)×[0,∞). Then (i) Lγ1 λ Hγ2 µ ( ∂γ1r(λ, µ) ∂λγ1 ) = ςR(ς, υ)−Hγ2 µ (r(0, µ)), (2) (ii) Lγ1 λ Hγ2 µ ( ∂2γ1r(λ, µ) ∂λ2γ1 ) = ς2R(ς, υ)− ςHγ2 µ (r(0, µ))−Hγ2 µ ( ∂γ1r(0, µ) ∂λγ1 ), (3) M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 6 of 12 (iii) Lγ1 λ Hγ2 µ ( ∂γ2r(λ, µ) ∂µγ2 ) = τ υ R(ς, υ)− Lγ1 λ (r(λ, 0)), (4) (iv) Lγ1 λ Hγ2 µ ( ∂2γ2r(λ, µ) ∂µ2γ2 ) = τ2 υ2 R(ς, υ)− τ υ Lγ1 λ (r(λ, 0))− Lγ1 λ ( ∂γ2r(λ, 0) ∂µγ2 ). (5) Proof. Proof of Equation 2 Lγ1 λ W γ2 µ ( ∂γ1r(λ,µ) ∂λγ1 ) = ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) ∂γ1r(λ,µ) ∂λγ1 λγ1−1µγ2−1dλdµ. By Theorem 1, we have ∂γ1r(λ,µ) ∂λγ1 = λ1−γ1 ∂r(λ,µ) ∂λ . So, Lγ1 λ W γ2 µ ( ∂γ1r(λ,µ) ∂λγ1 ) = ∞∫ 0 e −τ µγ2 υγ2 µγ2−1 ∞∫ 0 e −ς λγ1 γ1 ∂r(λ,µ) ∂λ dλdµ. By integrating by parts, we get Lγ1 λ W γ2 µ ( ∂γ1r(λ,µ) ∂λγ1 ) = ∞∫ 0 e −τ µγ2 υγ2 µγ2−1 ( −r(0, µ) + ς ∞∫ 0 e −ς λγ1 γ1 r(λ, µ)λγ1−1 dλ ) dµ = − ∞∫ 0 e −τ µγ2 υγ2 r(0, µ)µγ2−1dµ+ ς ∞∫ 0 ∞∫ 0 e − ( ς λγ1 γ1 +τ µγ2 υγ2 ) r(λ, µ)λγ1−1µγ2−1dλdµ = ςR(ς, υ)−W γ2 µ (r(0, µ)). The proof of Equations 3, 4 and 5 can be obtained in the same manner. In Table 1, we have the CL-SH of some basic functions. Table 1: Table of conformable double Laplace-Shehu transform r(λ, µ) Lγ1 λ Hγ2 µ (r(λ, µ)) c cυ ςτ , Re(ς) > 0( λγ1 γ1 )α ( µγ2 γ2 )β υβ+1 ςα+1τβ+1Γ(α+ 1)Γ(β + 1), Re(ς) > 0 and Re(α) > −1 e αλγ1 γ1 +β µγ2 γ2 υ (ς−α)(τ−βυ) , Re(ς) > Re(α) e i ( αλγ1 γ1 +β µγ2 γ2 ) iυ (ς−iα)(τ−iβυ) , Im(α) + Re(ς) > 0 sin ( αλγ1 γ1 + β µγ2 γ2 ) υ(τα+ςυβ) (ς2+α2)(τ2+β2υ2) , |Im(α)| < Re(ς) cos ( αλγ1 γ1 + β µγ2 γ2 ) υ(ςτ−υαβ) (ς2+α2)(τ2+β2υ2) , |Im(α)| < Re(ς) sinh ( αλγ1 γ1 + β µγ2 γ2 ) υ(τα+ςυβ) (ς2−α2)(τ2−β2υ2) , Re(ς) > Re(α) and Re(ς) + Re(α) > 0 cosh ( αλγ1 γ1 + β µγ2 γ2 ) υ(ςτ−υαβ) (ς2−α2)(τ2−β2υ2) , Re(ς) > Re(α) and Re(ς) + Re(α) > 0 p(λ)q(µ) Lγ1 λ (p(λ))Hγ2 µ (q(µ)) 4. Applications M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 7 of 12 In this section, we apply the CL-SH transform to solve some conformable partial differential equations. Example 1. Consider the conformable wave equation ∂2γ1r(λ, µ) ∂λ2γ1 + 4 ∂2γ2r(λ, µ) ∂µ2γ2 = 6 λγ1 γ1 , where λ, µ ≥ 0 (6) With initial conditions (ICs) r(λ, 0) = ( λγ1 γ1 )3 , ∂γ2r(λ,0) ∂µγ2 = cos ( 2λγ1 γ1 ) , and boundary conditions (BCs) r (0, µ) = sinh ( µγ2 γ2 ) , ∂γ1r(0,µ) ∂λγ1 = 0. Solution 1. By applying the CL to the ICs and the CSH to the BCs, we get Lγ1 λ (( λγ1 γ1 )3 ) = 6 ς4 , Lγ1 λ ( cos ( 2λγ1 γ1 )) = ς ς2+4 , Hγ2 µ ( sinh ( µγ2 γ2 )) = υ2 τ2−υ2 , H γ2 µ (0) = 0. Apply the CL-SH to Equation 6, we get ς2R− ςυ2 τ2 − υ2 + 4τ2 υ2 R− 24τ ς4υ − 4ς ς2 + 4 = 6υ ς2τ So, R(ς, υ) = ςυ2 τ2−υ2 + 24τ ς4υ + 4ς ς2+4 + 6υ ς2τ ς2 + 4τ2 υ2 = ς3υ2+4ςτ2 (ς2+4)(τ2−υ2) + 24τ2+6ς2υ2 ς4τυ ς2υ2+4τ2 υ2 By simplify, R(ς, υ) = ςυ2 (ς2 + 4) (τ2 − υ2) + 6υ ς4τ . So, r(λ, µ) = ( Lγ1 λ )−1 ( Hγ2 µ )−1 ( ςυ2 (ς2 + 4) (τ2 − υ2) + 6υ ς4τ ) = cos ( 2 λγ1 γ1 ) sinh ( µγ2 γ2 ) + ( λγ1 γ1 )3 . The following figures show the 3D representation of the solution at γ1 = γ2 = 0.5, 1. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 8 of 12 The following two figures illustrates the 2D graph of the solution with respect to λ and µ at γ1 = γ2 = 0.3, 0.7, 1. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 9 of 12 Example 2. Consider the conformable heat equation ∂γ1r(λ, µ) ∂λγ1 = ∂2γ2r(λ, µ) ∂µ2γ2 − 3, where λ, µ ≥ 0 (7) With IC r(λ, 0) = −3λγ1 γ1 , ∂γ2r(λ,0) ∂µγ2 = e −λγ1 γ1 , and BCs r (0, µ) = sin ( µγ2 γ2 ) . Solution 2. By applying the CL to the IC and the CSH to the BCs, we get Lγ1 λ ( −3λγ1 γ1 ) = −3 ς2 , Lγ1 λ ( e −λγ1 γ1 ) = 1 ς+1 , H γ2 µ ( sin ( µγ2 γ2 )) = υ2 τ2+υ2 . Apply the CL-SH to Equation 7, we get ςR− υ2 τ2 + υ2 = τ2 υ2 R+ 3τ ς2υ − 1 ς + 1 − 3υ ςτ . So, R(ς, υ) = υ2 τ2+υ2 + 3τ ς2υ − 1 ς+1 − 3υ ςτ ς − τ2 υ2 . By simplify, R(ς, υ) = υ2 (ς + 1) (τ2 + υ2) − 3υ ς2τ . So, r(λ, µ) = ( Lγ1 λ )−1 ( Hγ2 µ )−1 ( υ2 (ς + 1) (τ2 + υ2) − 3υ ς2τ ) = e −λγ1 γ1 sin ( µγ2 γ2 ) − 3 λγ1 γ1 . M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 10 of 12 The following figures show the 3D representation of the solution at γ1 = γ2 = 0.8, 1. The following two figures illustrates the 2D graph of the solution with respect to λ and µ at γ1 = γ2 = 0.5, 0.75, 1. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 11 of 12 5. Conclusion We introduced a new double transform based on the conformable approach. We showed that it works well for solving fractional differential equations. The examples we gave proved that the method is simple and gives correct results. This shows that the CL-SH is a helpful tool in this area. Future work may include other equations and systems. References [1] R. Khalil, M. Al Horani, A. Yousef, and M. Sababheh. A new definition of fractional derivative. Journal of Computational and Applied Mathematics, 264:65–70, 2014. [2] F. S. Silva, D. M. Moreira, and M. A. Moret. Conformable laplace transform of fractional differential equations. Axioms, 7(3):55, 2018. [3] O. Özkan and A. Kurt. On conformable double laplace transform. Optical and Quantum Electronics, 50:1–9, 2018. [4] S. Alfaqeih, G. Bakıçıerler, and E. Misirli. Conformable double sumudu transform with applications. Journal of Applied and Computational Mechanics, 7(2):578–586, 2021. [5] R. Abu Awwad, M. Al-Momani, B. Abughazaleh, A. Jaradat, and A. Farah. The conformable double laplace-sawi transform. European Journal of Pure and Applied Mathematics, 18(2):6034, 2025. [6] M. Al-Momani, A. Jaradat, B. Abughazaleh, and A. Farah. The conformable dou- ble laplace-sawi transform. European Journal of Pure and Applied Mathematics, 18(2):6099, 2025. [7] M. Al-Momani and B. Abughazaleh. The conformable double sumudu-shehu trans- M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6340 12 of 12 form and its properties with applications. European Journal of Pure and Applied Mathematics, 18(3):6384, 2025. [8] M. Hunaiber and A. Al-Aati. On double laplace-shehu transform and its properties with applications. Turkish Journal of Mathematics and Computer Science, 15(2):218– 226, 2023. [9] M. Al-Momani, A. Jaradat, and B. Abughazaleh. Double laplace-sawi transform. European Journal of Pure and Applied Mathematics, 18(1):5619, 2025. [10] M. Mahgoub and M. Mohand. The new integral transform “sawi transform”. Advances in Theoretical and Applied Mathematics, 14(1):81–87, 2019. [11] S. Khan, A. Ullah, M. De la Sen, and S. Ahmad. Double sawi transform: Theory and applications to boundary values problems. Symmetry, 15(4):921, 2023. [12] B. Abughazaleh, M. A. Amleh, A. Al-Natoor, and R. Saadeh. Double mellin-ara transform. Springer Proceedings in Mathematics and Statistics, 466:383–394, 2024. [13] R. Abu Awwad, M. Al-Momani, B. Abughazaleh, A. Jaradat, and A. Farah. The double sumudu-sawi transform. European Journal of Pure and Applied Mathematics, 18(2):5967, 2025. [14] R. Abu Awwad, M. Al-Momani, A. Jaradat, B. Abughazaleh, and A. Al-Natoor. The double ara-sawi transform. European Journal of Pure and Applied Mathematics, 18(1):5807, 2025. [15] M. Al-Momani, A. Jaradat, B. Abughazaleh, and A. Farah. Solving partial differential equations via the double sumudu-shehu transform. European Journal of Pure and Applied Mathematics, 18(2):5898, 2025. [16] M. Al-Momani, B. Abughazaleh, and A. Farah. The double sawi-shehu transform. European Journal of Pure and Applied Mathematics, 18(3):6890, 2025. [17] H. Thabet and S. Kendre. Analytical solutions for conformable space-time fractional partial differential equations via fractional differential transform. Chaos, Solitons & Fractals, 109:238–245, 2018. [18] H. Eltayeb and S. Mesloub. A note on conformable double laplace transform and singular conformable pseudoparabolic equations. Journal of Function Spaces, 2020(1):8106494, 2020.