EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6034 ISSN 1307-5543 – ejpam.com Published by New York Business Global The Conformable Double Laplace-Sawi Transform Raed R. Abu Awwad1, Monther Al-Momani2, Baha’ Abughazaleh3,∗, Ali Jaradat4, Abdulkarim Farah3 1 Department of Mathematics, University of Petra, Amman, Jordan 2 Department of Basic Sciences, Al-Ahliyya Amman University, Amman, Jordan 3 Department of Mathematics, Isra University, Amman, Jordan 4 Department of Mathematics, Amman Arab University, Amman, Jordan Abstract. In this study, we introduce the conformable double Laplace-Sawi transform, a method for solving fractional partial differential equations that appear in various physical and engineering models. These models use derivatives and integrals based on the newly defined conformable deriva- tive. The study first explores key properties of the conformable double Laplace-Sawi transform. Then, as an application, the method is applied to solving the conformable telegraph equation, the conformable heat equation, and the conformable Klein-Gordon equation, which are widely used in scientific and engineering fields. 2020 Mathematics Subject Classifications: 44A05, 44A10 Key Words and Phrases: Laplace transform, Sawi transform, double Laplace-Sawi transform, the conformable double Laplace-Sawi transform 1. Introduction Fractional partial differential equations are important in modeling real-world problems in physics, electrical circuits, fluid dynamics, optics, and mathematical biology. One useful concept introduced in [1] is the conformable fractional derivative, which keeps many familiar properties of standard derivatives. Recently, researchers have developed various methods to solve conformable fractional partial differential equations, including the conformable double Laplace transform see [2], [3] and the conformable double Sumudu transform see [4]. More recently, researchers have developed a new technique called Double Laplace-Sawi transform [5], which has been successfully applied to different types of partial differential equations. for more details about integral transform see [6–12]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6034 Email addresses: rabuawwad@uop.edu.jo (R. Abu Awwad), montheralmomani72@gmail.com (M. Al-Momani), baha.abughazaleh@iu.edu.jo (B. Abughazaleh), a.jaradat@aau.edu.jo (A. Jaradat), karim.farah@iu.edu.jo (A. Farah) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 2 of 17 In this study, we introduce the conformable double Laplace-Sawi transform (CLSW) as a new approach to analyzing conformable partial differential equations. We first explore its fundamental properties, including the conditions for its existence and its behavior with differentiation. We explore new methods for solving conformable partial differential equations, providing a new viewpoint that may lead to more advances in math and real- world uses. 2. Conformable Fractional Derivative In this section, we present fundamental definitions and theorems related to conformable fractional derivatives. Definition 1. [1] “Let 0 < α ≤ 1 and ω : (0,∞) → R. The conformable fractional derivative of order α is defined as: dα dγα ω(γ) = lim τ→0 ω(γ + τγ1−α)− ω(γ) τ where γ > 0, and ∂α ∂γα is referred to as the fractional derivative of order α.” Definition 2. [13] “Let 0 < α1, α2 ≤ 1 and ω(γ, η) : (0,∞) × (0,∞) → R. The con- formable partial derivatives of orders α1 and α2 of the function ω(γ, η) are defined as: ∂α1 ∂γα1 ω(γ, η) = lim τ→0 ω(γ + τγ1−α1 , η)− ω(γ, η) τ ∂α2 ∂ηα2 ω(γ, η) = lim τ→0 ω(γ, η + τη1−α2)− ω(γ, η) τ where γ, η > 0, ∂α1 ∂γα1 and ∂α2 ∂ηα2 are referred to as fractional derivatives of orders α1 and α2, respectively.” Theorem 1. [14]Suppose that ω(γ, η) is differentiable at a point γ, η > 0, 0 < α1, α2 ≤ 1, then: ∂α1ω ∂γα1 = γ1−α1 ∂ω ∂γ , ∂α2ω ∂ηα2 = η1−α2 ∂ω ∂η . 3. The Conformable Double Laplace-Sawi transform This section serves to introduce the CLSW. We commence by delineating its fundamen- tal properties, encompassing aspects like linearity. Subsequently, we reveal a novel result associated with partial derivatives. Ultimately, we illustrate how these insights enable us to compute the CLSW for various essential functions. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 3 of 17 Definition 3. Let ω(γ, η) be a continuous function on (0,∞)× (0,∞). Then 1- The Conformable Laplace transformation (CL) of ω(γ, η), denoted by Lα γ [ω(γ, η)], is defined as: H (µ) = Lα γ (ω(γ, η)) = ∞∫ 0 e−µ γα α ω(γ, η)γα−1dγ, µ ∈ C 2- The Conformable Sawi transformation (CSW) of ω(γ, η), denoted by Lα η [ω(γ, η)], is defined as: S (τ) = Wα η (ω(γ, η)) = 1 τ2 ∞∫ 0 e− ηα ταω(γ, η)ηα−1dη, τ ∈ C 3- The Conformable Laplace Sawi transformation (CLSW) of ω(γ, η), denoted by Lα1 γ Wα2 η [ω(γ, η)], is defined as: Ω(µ, τ) = Lα1 γ Wα2 η [ω(γ, η)] = 1 τ2 ∫ ∞ 0 ∫ ∞ 0 e − ( µ γα1 α1 + ηα2 τα2 ) ω(γ, η)γα1−1ηα2−1dγdη. Theorem 2. Assume that ω : (0,∞)×(0,∞) → R such that Ω(µ, τ) = Lα1 γ Wα2 η [ω(γ α1 α1 , η α2 α2 )] exist, then Lα1 γ Wα2 η [ω( γα1 α1 , ηα2 α2 )] = LγWη[ω(γ, η)], where LγWη[ω(γ, η)] = 1 τ2 ∫ ∞ 0 ∫ ∞ 0 e−(µγ+ η τ )ω(γ, η) dγ dη. Lemma 1. Lα1 γ Wα2 η (ω(γ, η)) is a linear transformation. Proof. for nonzero constants λ and ν, we have Lα1 γ Wα2 η (λω1(γ, η)+νω2(γ, η)) = 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) (λω1(γ, η) + νω2(γ, η))γ α1−1ηα2−1dγdη, = λ 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) ω1(γ, η)γ α1−1ηα2−1dγdη + ν 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) ω2(γ, η)γ α1−1ηα2−1dγdη = λLα1 γ Wα2 η (ω1(γ, η)) + νLα1 γ Wα2 η (ω2(γ, η)). If ω(γ, η) can be written as ω(γ, η) = p(γ)q(η) for some continuous functions p and q, then Lα1 γ Wα2 η (ω(γ, η)) = Lα1 γ (p(γ))Wα2 η (q(η)). In fact Lα1 γ Wα2 η (ω(γ, η)) = Lα1 γ Wα2 η (p(γ)q(η)) R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 4 of 17 = 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) p(γ)q(η)γα1−1ηα2−1dγdη = ∞∫ 0 e −µ γα1 α1 p(γ)γα1−1dγ  1 τ2 ∞∫ 0 e − ηα2 τα2 q(η)ηα2−1dη  = Lα1 γ (p(γ))Wα2 η (q(η)). 3.1. The Conformable Double Laplace-Sawi transform for some basic functions (i) Lα1 γ Wα2 η [c] = LγWη[c] = c µτ , c ∈ R, (ii) Lα1 γ Wα2 η [( γα1 α1 )λ(ηα2 α2 )ν ] = LγWη[γ λην ] = τν−1 µλ+1 Γ(λ+ 1)Γ(ν + 1), Re(µ) > 0 and Re(λ) > −1, (iii) Lα1 γ Wα2 η [ e λ γα1 α1 +ν ηα2 α2 ] = LγWη[e λγ+νη] = 1 τ (µ− λ) (1− ντ) ,Re(µ) > Re(λ). 3.2. Existence condition for the Conformable Double Laplace-Sawi trans- form Definition 4. Let 0 < α1, α2 ≤ 1. Then a function ω(γ, η) is said to be of conformable exponential orders λ and ν on 0 < γ < ∞ and 0 < η < ∞. If there exist K,X, Y > 0 such that |ω(γ, η)| ≤ Ke λ γα1 α1 +ν ηα2 α2 , for all γα1 α1 > X, ηα2 α2 > Y. Theorem 3. Let 0 < α1, α2 ≤ 1 and ω(γ, η) be a continuous function on the region (0,∞)×(0,∞) of conformable exponential orders λ and ν. Then Ω(µ, τ) = Lα1 γ Wα2 η [ω(γ, η)] exists for µ, τ whenever Re (µ) > λ and Re ( 1 τ ) > ν. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 5 of 17 Proof. We have |Ω(µ, τ)| = ∣∣∣∣∣∣ 1τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) ω(γ, η)γα1−1ηα2−1 dγdη ∣∣∣∣∣∣ ≤ 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) |ω(γ, η)| γα1−1ηα2−1dγdη ≤ K 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) e λ γα1 α1 +ν ηα2 α2 γα1−1ηα2−1dγdη = K ∞∫ 0 e −(µ−λ) γ α1 α1 γα1−1dγ  1 τ2 ∞∫ 0 e −( 1 τ −ν) η α2 α2 ηα2−1dη  = K τ (µ− λ) (1− ντ) , where Re (µ) > λ and Re ( 1 τ ) > ν. 3.3. Derivatives properties Now, we present some basic properties of the CLSW Let Ω(µ, τ) = Lα1 γ Wα2 η (ω(γ, η)) where ω(γ, η) is a continuous function on (0,∞) × (0,∞). Then (i) Lα1 γ Wα2 η ( ∂α1ω(γ, η) ∂γα1 ) = µΩ(µ, τ)−Wα2 η (ω(0, η)), (1) (ii) Lα1 γ Wα2 η ( ∂2α1ω(γ, η) ∂γ2α1 ) = µ2Ω(µ, τ)− µWα2 η (ω(0, η))−Wα2 η ( ∂α1ω(0, η) ∂γα1 ), (2) (iii) Lα1 γ Wα2 η ( ∂α2ω(γ, η) ∂ηα2 ) = 1 τ Ω(µ, τ)− 1 τ2 Lα1 γ (ω(γ, 0)), (3) (iv) Lα1 γ Wα2 η ( ∂2α2ω(γ, η) ∂η2α2 ) = 1 τ2 Ω(µ, τ)− 1 τ3 Lα1 γ (ω(γ, 0))− 1 τ2 Lα1 γ ( ∂α2ω(γ, 0) ∂ηα2 ). (4) R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 6 of 17 Proof. Proof of Equation 1 Lα1 γ Wα2 η ( ∂α1ω(γ,η) ∂γα1 ) = 1 τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) ∂α1ω(γ,η) ∂γα1 γα1−1ηα2−1dγdη. By Theorem 1, we have ∂α1ω(γ,η) ∂γα1 = γ1−α1 ∂ω(γ,η) ∂γ . So, Lα1 γ Wα2 η ( ∂α1ω(γ,η) ∂γα1 ) = 1 τ2 ∞∫ 0 e − ηα2 τα2 ηα2−1 ∞∫ 0 e −µ γα1 α1 ∂ω(γ,η) ∂γ dγdη. By integrating by parts, we get Lα1 γ Wα2 η ( ∂α1ω(γ,η) ∂γα1 ) = 1 τ2 ∞∫ 0 e − ηα2 τα2 ηα2−1 ( −ω(0, η) + µ ∞∫ 0 e −µ γα1 α1 ω(γ, η)γα1−1 dγ ) dη = − 1 τ2 ∞∫ 0 e − ηα2 τα2 ω(0, η)ηα2−1dη + µ τ2 ∞∫ 0 ∞∫ 0 e − ( µ γα1 α1 + ηα2 τα2 ) ω(γ, η)γα1−1ηα2−1dγdη = µΩ(µ, τ)−Wα2 η (ω(0, η)). The proof of Equations 2, 3 and 4 can be obtained in the same manner. In Table 1, we have the CLSW of some basic functions. Table 1: Table of CLSW ω(γ, η) Lα1 γ Wα2 η (ω(γ, η)) c c µτ , Re(µ) > 0( γα1 α1 )λ ( ηα2 α2 )ν τν−1 µλ+1Γ(λ+ 1)Γ(ν + 1), Re(µ) > 0 and Re(λ) > −1 e λ γα1 α1 +ν ηα2 α2 1 τ(µ−λ)(1−ντ) , Re(µ) > Re(λ) e i ( λ γα1 α1 +ν ηα2 α2 ) i τ(µ−iλ)(i+ντ) , Im(λ) + Re(µ) > 0 sin ( λγα1 α1 + ν ηα2 α2 ) λ+µτν τ(µ2+λ2)(1+ν2τ2) , |Im(λ)| < Re(µ) cos ( λγα1 α1 + ν ηα2 α2 ) µ−τλν τ(µ2+λ2)(1+ν2τ2) , |Im(λ)| < Re(µ) sinh ( λγα1 α1 + ν ηα2 α2 ) λ+µτν τ(µ2−λ2)(1−ν2τ2) , Re(µ) > Re(λ) and Re(µ) + Re(λ) > 0 cosh ( λγα1 α1 + ν ηα2 α2 ) µ+τλν τ(µ2−λ2)(1−ν2τ2) , Re(µ) > Re(λ) and Re(µ) + Re(λ) > 0 p(γ)q(η) Lα1 γ (p(γ))Wα2 η (q(η)) 4. Applications In this section, we use the CLSW for solving conformable partial differential equations Example 1. Consider the conformable telegraph equation ∂2α1ω(γ, η) ∂γ2α1 − 2 ∂2α2ω(γ, η) ∂η2α2 + ∂α1ω(γ, η) ∂γα1 = 4ω(γ, η), where γ, η > 0 (5) With initial conditions (ICs) ω(γ, 0) = e 2 γα1 α1 , ∂α2ω(γ,0) ∂ηα2 = −e 2 γα1 α1 , R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 7 of 17 and boundary conditions (BCs) ω (0, η) = e − ηα1 α1 , ∂α1ω(0,η) ∂γα1 = 2e − ηα1 α1 . Solution 1. By applying the CL to the ICs and the CSW to the BCs, we get Lα1 γ ( e 2 γα1 α1 ) = 1 µ−2 , L α1 γ ( −e 2 γα1 α1 ) = −1 µ−2 , W α2 η ( e − ηα1 α1 ) = 1 τ(1+τ) , W α2 η ( 2e − ηα1 α1 ) = 2 τ(1+τ) Apply the CLSW to Equation 5, we get µ2Ω− µ τ (1 + τ) − 2 τ (1 + τ) − 2 τ2 Ω+ 2 τ3 (µ− 2) − 2 τ2 (µ− 2) + µΩ− 1 τ (1 + τ) = 4Ω So, Ω(µ, τ) = µ+3 τ(1+τ) − 2 τ3(µ−2) + 2 τ2(µ−2) µ2 − 2 τ2 + µ− 4 = τ2(µ−2)(µ+3)−2(1+τ)+2τ(1+τ) τ3(µ−2)(1+τ) µ2τ2+µτ2−4τ2−2 τ2 By simplify, Ω(µ, τ) = 1 τ (µ− 2) (1 + τ) . So, ω(γ, η) = ( Lα1 γ )−1 ( Wα2 η )−1 ( 1 τ (µ− 2) (1 + τ) ) = e 2 γα1 α1 − ηα2 α2 . The following figures show the 3D representation of the solution at α1 = α2 = 0.5, 1. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 8 of 17 The following two figures illustrate the 2D graph of the solution with respect to γ and η at α1 = α2 = 0.7, 0.9, 1. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 9 of 17 Example 2. Consider the conformable heat equation 2 ∂α2ω(γ, η) ∂ηα2 + ∂2α1ω(γ, η) ∂γ2α1 = ω(γ, η) + 2 γα1 α1 , where γ, η > 0 (6) R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 10 of 17 With IC ω(γ, 0) = cos ( γα1 α1 ) − 2γα1 α1 , and BCs ω (0, η) = e ηα1 α1 , ∂α1ω(0,η) ∂γα1 = −2. Solution 2. By applying the CL to the IC and the CSW to the BCs, we get Lα1 γ ( cos ( γα1 α1 ) − 2γα1 α1 ) = µ µ2+1 − 2 µ2 , W α2 η ( e ηα1 α1 ) = 1 τ(1−τ) , W α2 η (−2) = −2 τ Apply the CLSW to Equation 6, we get 2 τ Ω− 2µ τ2 (µ2 + 1) + 4 µ2τ2 + µ2Ω− µ τ (1− τ) + 2 τ = Ω+ 2 µ2τ So, Ω(µ, τ) = 2µ τ2(µ2+1) − 4 µ2τ2 + µ τ(1−τ) − 2 τ + 2 µ2τ 2 τ + µ2 − 1 By simplify, Ω(µ, τ) = µ τ (µ2 + 1) (1− τ) − 2 µ2τ . So, ω(γ, η) = ( Lα1 γ )−1 ( Wα2 η )−1 ( µ τ (µ2 + 1) (1− τ) − 2 µ2τ ) = e ηα2 α2 cos ( γα1 α1 ) − 2 γα1 α1 . The following figures show the 3D representation of the solution at α1 = α2 = 0.7, 1. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 11 of 17 The following two figures illustrate the 2D graph of the solution with respect to γ and η at α1 = α2 = 0.5, 0.8, 1. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 12 of 17 Example 3. Consider the conformable Klein-Gordon equation ∂2α1ω(γ, η) ∂γ2α1 + 2 ∂2α2ω(γ, η) ∂η2α2 = ω(γ, η), where γ, η > 0 (7) R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 13 of 17 With ICs ω(γ, 0) = sin ( γα1 α1 ) , ∂α2ω(0,η) ∂ηα2 = 0, and BCs ω (0, η) = 0, ∂α1ω(0,η) ∂γα1 = cosh ( ηα2 α2 ) . Solution 3. By applying the CL to the ICs and the CSW to the BCs, we get Lα1 γ ( sin ( γα1 α1 )) = 1 µ2+1 , Lα1 γ (0) = 0, Wα2 η (0) = 0, Wα2 η ( cosh ( ηα2 α2 )) = 1 τ(1−τ2) Apply the CLSW to Equation 7, we get µ2Ω− 1 τ (1− τ2) + 2 τ2 Ω− 2 τ3 (µ2 + 1) = Ω So, Ω(µ, τ) = 1 τ(1−τ2) + 2 τ3(µ2+1) µ2 + 2 τ2 − 1 . By simplify, Ω(µ, τ) = 1 τ (µ2 + 1) (1− τ2) . So, ω(γ, η) = ( Lα1 γ )−1 ( Wα2 η )−1 ( 1 τ (µ2 + 1) (1− τ2) ) = sin ( γα1 α1 ) cosh ( ηα2 α2 ) . The following figures show the 3D representation of the solution at α1 = α2 = 0.6, 1. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 14 of 17 The following two figures illustrate the 2D graph of the solution with respect to γ and η at α1 = α2 = 0.4, 0.7, 1. R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 15 of 17 R. Abu Awwad et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6034 16 of 17 5. Conclusion In this study, we introduced the conformable double Laplace-Sawi transform and ex- plored its application to conformable fractional partial derivatives. We demonstrated its effectiveness by solving fractional partial differential equations. Since the conformable double Laplace-Sawi transform is a newly defined approach, there remain many open problems and potential areas for further research. 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