EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6384 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Conformable Double Sumudu-Shehu Transform and its Properties with Applications 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 introduce a new transform called the conformable double Sumudu-Shehu transform. It helps solve fractional partial differential equations that appear in physical and engineering prob- lems. The transform uses the conformable derivative idea. We explain its basic properties and show how it works. Then we apply it to some well-known equations like the wave and Klein-Gordon equations. 2020 Mathematics Subject Classifications: 44A05 Key Words and Phrases: Conformable derivatives, Sumudu transform, Shehu transform, double transform, the conformable double Sumudu-Shehu transform. 1. Introduction Fractional partial differential equations are used in many real-life problems in physics, circuits, fluids, optics, and biology. One important idea used to deal with such equations is the conformable derivative, introduced in [1], which keeps most of the key features of classical derivatives. Several approaches were proposed to solve these equations. The conformable double Laplace transform was discussed in [2], [3], and the conformable double Sumudu transform appeared in [4]. More results about these transforms are found in [5] and [6]. Later, a method called the double Sumudu-Shehu transform was introduced in [7]. It was successfully applied to different equations. More studies on integral transforms appear in [8–15]. In this paper, we define the conformable double Sumudu-Shehu transform (CD-SSH). We explain when the transform exists and how it behaves with derivatives. Then we show how it can be used to solve some well-known conformable equations. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6384 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), 6384 2 of 14 2. Preliminaries In this section, we give the main definitions and results related to conformable frac- tional derivatives. Definition 1. [1] Let 0 < θ ≤ 1 and χ : (0,∞) → R. The conformable fractional derivative of order θ is defined as: dθ dλθ χ(λ) = lim η→0 χ(λ+ ηλ1−θ)− χ(λ) η , where λ > 0, and ∂θ ∂λθ is referred to as the fractional derivative of order θ. Definition 2. [16] Let 0 < θ1, θ2 ≤ 1 and χ(λ, δ) : (0,∞)× (0,∞) → R. The conformable partial derivatives of orders θ1 and θ2 of the function χ(λ, δ) are defined as: ∂θ1 ∂λθ1 χ(λ, δ) = lim η→0 χ(λ+ ηλ1−θ1 , δ)− χ(λ, δ) η , ∂θ2 ∂δθ2 χ(λ, δ) = lim η→0 χ(λ, δ + ηδ1−θ2)− χ(λ, δ) η , where λ, δ > 0, ∂θ1 ∂λθ1 and ∂θ2 ∂δθ2 are referred to as fractional derivatives of orders θ1 and θ2, respectively. Theorem 1. [17]Suppose that χ(λ, δ) is differentiable at a point λ, δ > 0, 0 < θ1, θ2 ≤ 1, then: ∂θ1χ ∂λθ1 = λ1−θ1 ∂χ ∂λ , ∂θ2χ ∂δθ2 = δ1−θ2 ∂χ ∂δ . 3. The Conformable Double Sumudu-Shehu transform In this section, we introduce the CD-SSH transform and explain its main properties such as linearity. We also present a new result related to partial derivatives. Finally, we show how these ideas help in finding the CD-SSH of some basic functions. Definition 3. Let χ(λ, δ) be a continuous function on (0,∞)× (0,∞). Then 1- The Conformable Sumudu transformation (C-S) of χ(λ, δ), denoted by Sθ λ[χ(λ, δ)], is defined as: Φ (ρ) = Sθ λ(χ(λ, δ)) = 1 ρ ∞∫ 0 e −λθ ρθ χ(λ, δ)λθ−1dλ, ρ ∈ C. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 3 of 14 2- The Conformable Shehu transformation (C-SH) of χ(λ, δ), denoted by Hθ δ [χ(λ, δ)], is defined as: Ω (ϵ, η) = Hθ δ (χ(λ, δ)) = ∞∫ 0 e −ϵ δ θ ηθχ(λ, δ)δθ−1dδ, ϵ, η ∈ C. 3- The Conformable Sumudu-Shehu transformation (CD-SSH) of χ(λ, δ), denoted by Sθ1 λ Hθ2 δ [χ(λ, δ)], is defined as: Ψ(ρ, ϵ, η) = Sθ1 λ Hθ2 δ [χ(λ, δ)] = 1 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) χ(λ, δ)λθ1−1δθ2−1dλdδ. Theorem 2. Assume that χ : (0,∞)×(0,∞) → R such that Ψ(ρ, ϵ, η) = Sθ1 λ Hθ2 δ [χ(λ θ1 θ1 , δ θ2 θ2 )] exist, then Sθ1 λ Hθ2 δ [χ( λθ1 θ1 , δθ2 θ2 )] = SλHδ[χ(λ, δ)], where SλHδ[χ(λ, δ)] = 1 ρ ∞∫ 0 ∞∫ 0 e − ( λ ρ + ϵδ η ) χ(λ, δ) dλ dδ. Lemma 1. Sθ1 λ Hθ2 δ (χ(λ, δ)) is a linear transformation. Proof. for nonzero constants a1 and a2, we have Sθ1 λ Hθ2 δ (a1χ1(λδ)+a2χ2(λ, δ)) = 1 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) (a1χ1(λ, δ) + a2χ2(λ, δ))λ θ1−1δθ2−1dλdδ = a1 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) χ1(λ, δ)λ θ1−1δθ2−1dλdδ + a2 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) χ2(λ, δ)λ θ1−1δθ2−1dλdδ = a1S θ1 λ Hθ2 δ (χ1(λ, δ)) + a2S θ1 λ Hθ2 δ (χ2(λ, δ)). If χ(λ, δ) can be written as χ(λ, δ) = p(λ)q(δ) for some continuous functions p and q, then Sθ1 λ Hθ2 δ (χ(λ, δ)) = Sθ1 λ (p(λ))Hθ2 δ (q(δ)). In fact Sθ1 λ Hθ2 δ (χ(λ, δ)) = Sθ1 λ Hθ2 δ (p(λ)q(δ)) M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 4 of 14 = 1 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) p(λ)q(δ)λθ1−1δθ2−1dλdδ = 1 ρ ∞∫ 0 e −λθ1 ρθ1 p(λ)λθ1−1dλ ∞∫ 0 e −ϵ δ θ2 ηθ2 q(δ)δθ2−1dδ  = Sθ1 λ (p(λ))Hθ2 δ (q(δ)). 3.1. The Conformable Double Sumudu-Shehu transform for some basic functions (i) Sθ1 λ Hθ2 δ [c] = Sθ1 λ Hθ2 δ [c] = cη ϵ , c ∈ R, (ii) Sθ1 λ Hθ2 δ [ e a1 λθ1 θ1 +a2 δθ2 θ2 ] = Sθ1 λ Hθ2 δ [ea1λ+a2δ] = η (1− a1ρ) (ϵ− a2η) ,Re( 1 ρ ) > Re(a1), (iii) Sθ1 λ Hθ2 δ [( λθ1 θ1 )a1 (δθ2 θ2 )a2] = Sθ1 λ Hθ2 δ [λa1δa2 ] = ρa1ηa2+1 ϵa2+1 Γ(a1 + 1)Γ(a2 + 1), Re( 1 ρ ) > 0 and Re(a1) > −1. 3.2. Existence condition for the Conformable Double Sumudu-Shehu trans- form Definition 4. Let 0 < θ1, θ2 ≤ 1. Then a function χ(λ, δ) is said to be of conformable exponential orders a1 and a2 on 0 < λ < ∞ and 0 < δ < ∞. If there exist A,B,C > 0 such that |χ(λ, δ)| ≤ Ae a1 λθ1 θ1 +a2 δθ2 θ2 , for all λθ1 θ1 > B, δθ2 θ2 > C. Theorem 3. Let 0 < θ1, θ2 ≤ 1 and χ(λ, δ) be a continuous function on the region (0,∞)× (0,∞) of conformable exponential orders a1 and a2. Then Ψ(ρ, ϵ, η) = Sθ1 λ Hθ2 δ [χ(λ, δ)] exists for ρ, η whenever Re(1ρ) > a1 and Re ( ϵ η ) > a2. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 5 of 14 Proof. We have |Ψ(ρ, ϵ, η)| = ∣∣∣∣∣∣1ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) χ(λ, δ)λθ1−1δθ2−1 dλdδ ∣∣∣∣∣∣ ≤ 1 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) |χ(λ, δ)|λθ1−1δθ2−1dλdδ ≤ A ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) e a1 λθ1 θ1 +a2 δθ2 θ2 λθ1−1δθ2−1dλdδ = A 1 ρ ∞∫ 0 e −( 1 ρ −a1) λθ1 θ1 λθ1−1dλ ∞∫ 0 e −( ϵ η −a2) δθ2 θ2 δθ2−1dδ  = Aη (1− a1ρ) (ϵ− a2η) , where Re(1ρ) > a1 and Re ( ϵ η ) > a2. 3.3. Derivatives properties Now, we present some basic properties of the CD-SSH Let Ψ(ρ, ϵ, η) = Sθ1 λ Hθ2 δ (χ(λ, δ)) where χ(λ, δ) is a continuous function on (0,∞) × (0,∞). Then (i) Sθ1 λ Hθ2 δ ( ∂θ1χ(λ, δ) ∂λθ1 ) = 1 ρ Ψ(ρ, ϵ, η)− 1 ρ Hθ2 δ (χ(0, δ)), (1) (ii) Sθ1 λ Hθ2 δ ( ∂2θ1χ(λ, δ) ∂λ2θ1 ) = 1 ρ2 Ψ(ρ, ϵ, η)− 1 ρ2 Hθ2 δ (χ(0, δ))− 1 ρ Hθ2 δ ( ∂θ1χ(0, δ) ∂λθ1 ), (2) (iii) Sθ1 λ Hθ2 δ ( ∂θ2χ(λ, δ) ∂δθ2 ) = ϵ η Ψ(ρ, ϵ, η)− Sθ1 λ (χ(λ, 0)), (3) (iv) Sθ1 λ Hθ2 δ ( ∂2θ2χ(λ, δ) ∂δ2θ2 ) = ϵ2 η2 Ψ(ρ, ϵ, η)− ϵ η Sθ1 λ (χ(λ, 0))− Sθ1 λ ( ∂θ2χ(λ, 0) ∂δθ2 ). (4) M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 6 of 14 Proof. Proof of Equation 1 Sθ1 λ Hθ2 δ ( ∂θ1χ(λ, δ) ∂λθ1 ) = 1 ρ ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) ∂θ1χ(λ, δ) ∂λθ1 λθ1−1δθ2−1dλdδ. By Theorem 1, we have ∂θ1χ(λ,δ) ∂λθ1 = λ1−θ1 ∂χ(λ,δ) ∂λ . So, Sθ1 λ Hθ2 δ ( ∂θ1χ(λ, δ) ∂λθ1 ) = 1 ρ ∞∫ 0 e −ϵ δ θ2 ηθ2 δθ2−1 ∞∫ 0 e −λθ1 ρθ1 ∂χ(λ, δ) ∂λ dλdδ. By integrating by parts, we get Sθ1 λ Hθ2 δ ( ∂θ1χ(λ, δ) ∂λθ1 ) = 1 ρ ∞∫ 0 e −ϵ δ θ2 ηθ2 δθ2−1 −χ(0, δ) + 1 ρ ∞∫ 0 e −λθ1 ρθ1 χ(λ, δ)λθ1−1 dλ  dδ = −1 ρ ∞∫ 0 e −ϵ δ θ2 ηθ2 χ(0, δ)δθ2−1dδ + 1 ρ2 ∞∫ 0 ∞∫ 0 e − ( λθ1 ρθ1 +ϵ δ θ2 ηθ2 ) χ(λ, δ)λθ1−1δθ2−1dλdδ = 1 ρ Ψ(ρ, ϵ, η)− 1 ρ Hθ2 δ (χ(0, δ)). The proof of Equations 2, 3 and 4 follow using the same steps. In Table 1, we have the CD-SSH of some basic functions. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 7 of 14 Table 1: Table of CD-SSH χ(λ, δ) Sθ1 λ Hθ2 δ (χ(λ, δ)) c cη ϵ , Re(ρ) > 0( λθ1 θ1 )a1 ( δθ2 θ2 )a2 ρa1ηa2+1 ϵa2+1 Γ(a1 + 1)Γ(a2 + 1), Re(ρ) > 0 and Re(a1) > −1 e a1 λθ1 θ1 +a2 δθ2 θ2 η (1−a1ρ)(ϵ−a2η) , Re(1ρ) > Re(a1) e i ( a1 λθ1 θ1 +a2 δθ2 θ2 ) iη (i+a1ρ)(ϵ−ia2η) , Im(a1) + Re(1ρ) > 0 sin ( a1 λθ1 θ1 + a2 δθ2 θ2 ) η(ρϵa1+ηa2) (1+a21ρ 2)(ϵ2+a22η 2) , |Im(a1)| < Re(1ρ) cos ( a1 λθ1 θ1 + a2 δθ2 θ2 ) η(ϵ−ρηa1a2) (1+a21ρ 2)(ϵ2+a22η 2) , |Im(a1)| < Re(1ρ) sinh ( a1 λθ1 θ1 + a2 δθ2 θ2 ) η(ρϵa1+ηa2) (ρ2a21−1)(ϵ2−a22η 2) , Re(1ρ) > Re(a1) and Re(1ρ) + Re(a1) > 0 cosh ( a1 λθ1 θ1 + a2 δθ2 θ2 ) η(ϵ−ρηa1a2) (ρ2a21−1)(ϵ2−a22η 2) , Re(1ρ) > Re(a1) and Re(1ρ) + Re(a1) > 0 p(λ)q(δ) Sθ1 λ (p(λ))Hθ2 δ (q(δ)) 4. Applications In this section, we use the CD-SSH for solving conformable partial differential equations Example 1. Consider the conformable wave equation ∂2θ1χ(λ, δ) ∂λ2θ1 + 9 ∂2θ2χ(λ, δ) ∂δ2θ2 = 18, where λ, δ > 0. (5) With initial conditions (ICs) χ(λ, 0) = cosh ( 3λθ1 θ1 ) , ∂θ2χ(λ,0) ∂δθ2 = 0, and boundary conditions (BCs) χ (0, δ) = cos ( δθ2 θ2 ) + ( δθ2 θ2 )2 , ∂θ1χ(0,δ) ∂λθ1 = 0. Solution 1. By applying the C-S to the ICs and the C-SH to the BCs, we get Sθ1 λ ( cosh ( 3λθ1 θ1 )) = 1 1−9ρ2 , Sθ1 λ (0) = 0, Hθ2 δ ( cos ( δθ2 θ2 ) + ( δθ2 θ2 )2 ) = ϵη ϵ2+η2 + 2η3 ϵ3 , Hθ2 δ (0) = 0. Apply the CD-SSH to Equation 5, we get 1 ρ2 Ψ− ϵη ρ2 (ϵ2 + η2) − 2η3 ρ2ϵ3 + 9ϵ2 η2 Ψ− 9ϵ η (1− 9ρ2) = 18η ϵ . So, Ψ(ρ, ϵ, η) = ϵη ρ2(ϵ2+η2) + 2η3 ρ2ϵ3 + 9ϵ η(1−9ρ2) + 18η ϵ 1 ρ2 + 9ϵ2 η2 M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 8 of 14 = ϵη2−9ρ2ϵη2+9ρ2ϵ3+9ρ2ϵη2 ρ2η(1−9ρ2)(ϵ2+η2) + 2η3+18ρ2ϵ2η ρ2ϵ3 η2+9ρ2ϵ2 ρ2η2 . By simplify, Ψ(ρ, ϵ, η) = ϵη (1− 9ρ2) (ϵ2 + η2) + 2η3 ϵ3 . So, χ(λ, δ) = ( Sθ1 λ )−1 ( Hθ2 δ )−1 ( ϵη (1− 9ρ2) (ϵ2 + η2) + 2η3 ϵ3 ) = cosh ( 3 λθ1 θ1 ) cos ( δθ2 θ2 ) + ( δθ2 θ2 )2 . The following figures show the 3D representation of the solution at θ1 = θ2 = 0.6, 1. The following two figures illustrates the 2D graph of the solution with respect to λ and δ at θ1 = θ2 = 0.7, 0.85, 1. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 9 of 14 Example 2. Consider the conformable Klein-Gordon equation 2 ∂2θ1χ(λ, δ) ∂λ2θ1 + ∂2θ2χ(λ, δ) ∂δ2θ2 = χ(λ, δ)− 3 ( λθ1 θ1 )( δθ2 θ2 ) , where λ, δ > 0. (6) M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 10 of 14 With ICs χ(λ, 0) = 0, ∂θ2χ(λ,0) ∂δθ2 = e 2−λθ1 θ1 + 3λθ1 θ1 , and BCs χ (0, δ) = e2 sin ( δθ2 θ2 ) , ∂θ1χ(0,δ) ∂λθ1 = −e2 sin ( δθ2 θ2 ) + 3 δθ2 θ2 . Solution 2. By applying the C-S to the ICs and the C-SH to the BCs, we get Sθ1 λ (0) = 0, Sθ1 λ ( e 2−λθ1 θ1 + 3λθ1 θ1 ) = e2 1+ρ + 3ρ, Hθ2 δ ( e2 sin ( δθ2 θ2 )) = e2η2 ϵ2+η2 , Hθ2 δ ( −e2 sin ( δθ2 θ2 ) + 3 δθ2 θ2 ) = −e2η2 ϵ2+η2 + 3η2 ϵ2 . Apply the CD-SSH to Equation 6, we get 2 ρ2 Ψ− 2e2η2 ρ2 (ϵ2 + η2) + 2e2η2 ρ (ϵ2 + η2) − 6η2 ρϵ2 + ϵ2 η2 Ψ− e2 1 + ρ − 3ρ = Ψ− 3ρη2 ϵ2 . So, Ψ(ρ, ϵ, η) = 2e2η2 ρ2(ϵ2+η2) − 2e2η2 ρ(ϵ2+η2) + 6η2 ρϵ2 + e2 1+ρ + 3ρ− 3ρη2 ϵ2 2 ρ2 + ϵ2 η2 − 1 . By simplify, Ψ(ρ, ϵ, η) = e2η2 (ρ+ 1) (ϵ2 + η2) + 3ρη2 ϵ2 . So, χ(λ, δ) = ( Sθ1 λ )−1 ( Hθ2 δ )−1 ( e2η2 (ρ+ 1) (ϵ2 + η2) + 3ρη2 ϵ2 ) = e 2−λθ1 θ1 sin ( δθ2 θ2 ) + 3 ( λθ1 θ1 )( δθ2 θ2 ) . The following figures show the 3D representation of the solution at θ1 = θ2 = 0.4, 1. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 11 of 14 The following two figures illustrates the 2D graph of the solution with respect to λ and δ at θ1 = θ2 = 0.5, 0.75, 1. M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 12 of 14 M. Al-Momani, B. Abughazaleh / Eur. J. Pure Appl. Math, 18 (4) (2025), 6384 13 of 14 5. Conclusion We introduced the CD-SSH and showed its main properties. We applied it to solve several conformable equations. The results show that the method is useful and works well. It may help with other equations in future work. 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