IHJPAS. 53 (4)2022 220 This work is licensed under a Creative Commons Attribution 4.0 International License. On the Double of the Emad - Falih Transformation and Its Properties with Applications Abstract In this paper, we have generalized the concept of one dimensional Emad - Falih integral transform into two dimensional, namely, a double Emad - Falih integral transform. Further, some main properties and theorems related to the double Emad - Falih transform are established. To show the proposed transform's efficiency, high accuracy, and applicability, we have implemented the new integral transform for solving partial differential equations. Many researchers have used double integral transformations in solving partial differential equations and their applications. One of the most important uses of double integral transformations is how to solve partial differential equations and turning them into simple algebraic ones. The most important partial differential equations are Laplace, Poisson, wave, heat, telegraph, and other equations. A new Double Emad -Falih integral transform denoted by the operator 𝐃𝐸𝐹{.}, the transform form is as follows: 𝐃𝐸𝐹[𝑓(π‘₯,𝑑)] = 𝐓(𝑒,𝑣) = 1 𝑒𝑣 ∫ β€Š ∞ 0 ∫ β€Š ∞ 0 𝑓(π‘₯,𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ Keywords: Emad - Falih Transform, Partial Differential Equation, Double Integral Transform. 1. Introduction An integral transformation maps a function from its main function space into a new function space via integration, with properties of the main function that might be more characterized and manipulated than in the main function space. The transformed function can generally be mapped back to the main function space by applying the inverse integral transformation. Many researchers have applied double transforms for solving partial differential equations and their applications [1- 4]. Many applications and problems in most applied nature science and engineering fields encounter double integral transforms or partial differential equations describing the physical phenomena. Solving such equations using single transforms is more complicated than applying the double integral transformation. Doi: 10.30526/35.4.2938 Article history: Received 20 July 2022, Accepted 21 August 2022, Published in October 2022. Ibn Al-Haitham Journal for Pure and Applied Sciences Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Saed M. Turq Teacher at the Ministry of Education Hebron, Palestine. saedturq@gmail.com Emad A. Kuffi Al-Qadisiyah University, College of Engineering, Al-Qadisiyah, Iraq. emad.abbas@qu.edu.iq https://creativecommons.org/licenses/by/4.0/ mailto:saedturq@gmail.com mailto:emad.abbas@qu.edu.iq IHJPAS. 53 (4)2022 221 There are many types of integral transformations, such as: Fourier transformation [5], Laplace transformation [6], Sumudu transformation [7], Natural transformation [8], Elzaki tranformation [9] ,and so on. These types of transformations have wide variety applications in various areas in applied mathematics, physics, engineering, and in most of other sciences [10]. Emad - Falih integral transformation of a single variable is a new integral transform which has recently been introduced by Emad A. Kuffi and Sara F. in (2021) [11]. 2. The Double Emad - Falih Integral Transform Definition 2.1. Emad- Falih Integral Transform: The Emad- Falih Integral Transform of 𝑔(𝑑) is defined as: 𝐄𝐅[𝑔(𝑑)] = 𝐓(𝑣) = 1 𝑣 ∫ β€Š ∞ 0 𝑔(𝑑)π‘’βˆ’π‘£ 2𝑑𝑑𝑑 3. The Double Emad- Falih Integral Transform of Some Famous Functions This Section introduces the new double Emad- Falih integral transform of some famous functions: 1. Let 𝑓(π‘₯,𝑑) = 1, then 𝐃𝐸𝐹[1] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š1π‘’βˆ’(𝑒 𝛼π‘₯+𝑣𝛼𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Š1π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯ β‹… 1 𝑣 ∫ β€Š ∞ 0 β€Š1π‘’βˆ’π‘£ 2𝑑𝑑𝑑 = 𝐄𝐅[1] ⋅𝐄𝐅[1], = 1 𝑒3 β‹… 1 𝑣3 = 1 (𝑒𝑣)3 2 Let 𝑓(π‘₯,𝑑) = π‘₯π‘›π‘‘π‘š, then 𝐃𝐸𝐹[π‘₯ π‘›π‘‘π‘š] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘₯π‘›π‘‘π‘šπ‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘₯π‘›π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯ β‹… 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘‘π‘šπ‘’βˆ’π‘£ 2𝑑𝑑𝑑 = 𝐄𝐅[π‘₯𝑛] β‹… 𝐄𝐅[𝑑𝑛] = 𝑛! 𝑒2𝑛+3 β‹… π‘š! 𝑣2π‘š+3 3 Let 𝑓(π‘₯,𝑑) = π‘’π‘Žπ‘₯+𝑏𝑑, then IHJPAS. 53 (4)2022 222 𝐃𝐸𝐹[𝑒 π‘Žπ‘₯+𝑏𝑑] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’π‘Žπ‘₯+π‘π‘‘π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’π‘Žπ‘₯π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯ β‹… 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’π‘π‘‘π‘’βˆ’π‘£ 2𝑑𝑑𝑑 = 𝐄𝐅[π‘’π‘Žπ‘₯] β‹… 𝐄𝐅[𝑒𝑏𝑑] = 1 𝑒(𝑒2 βˆ’ π‘Ž) β‹… 1 𝑣(𝑣2 βˆ’ 𝑏) = 1 𝑒𝑣(𝑒2 βˆ’ π‘Ž)(𝑣2 βˆ’ 𝑏) 4 Let 𝑓(π‘₯,𝑑) = π‘’βˆ’(π‘Žπ‘₯+𝑏𝑑), then 𝐃𝐸𝐹[𝑒 βˆ’(π‘Žπ‘₯+𝑏𝑑)] = 1 (𝑒𝑣) ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’βˆ’(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘Žπ‘₯π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯ β‹… 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘π‘‘π‘’βˆ’π‘£ 2𝑑𝑑𝑑 = 𝐄𝐅[π‘’βˆ’π‘Žπ‘₯] β‹… 𝐄𝐅[π‘’βˆ’π‘π‘‘] = 1 𝑒(𝑒2 + π‘Ž) β‹… 1 𝑣(𝑣2 + 𝑏) = 1 𝑒𝑣(𝑒2 + π‘Ž)(𝑣2 + 𝑏) 5 Let 𝑓(π‘₯,𝑑) = 𝑒𝑖(π‘Žπ‘₯+𝑏𝑑), then 𝐃𝐸𝐹[𝑒 𝑖(π‘Žπ‘₯+𝑏𝑑)] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’π‘–(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘₯(𝑒 2βˆ’π‘–π‘Ž)𝑑π‘₯ β‹… 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘‘(𝑣 2βˆ’π‘–π‘)𝑑𝑑 = 1 𝑒 1 βˆ’(𝑒2 βˆ’ π‘–π‘Ž) π‘’βˆ’π‘₯(𝑒 2βˆ’π‘–π‘Ž)| 0 ∞ 1 𝑣 1 βˆ’(𝑣2 βˆ’ 𝑖𝑏) π‘’βˆ’π‘‘(𝑣 2βˆ’π‘–π‘)| 0 ∞ = 1 𝑒 1 βˆ’(𝑒2 βˆ’ π‘–π‘Ž) [0βˆ’ 1] 1 𝑣 1 βˆ’(𝑣2 βˆ’ 𝑖𝑏) [0 βˆ’ 1] = 1 𝑒(𝑒2 βˆ’ π‘–π‘Ž) 1 𝑣(𝑣2 βˆ’ 𝑖𝑏) , = 1 𝑒𝑣(𝑒2 βˆ’ π‘–π‘Ž)(𝑣2 βˆ’ 𝑖𝑏) . 6 Let 𝑓(π‘₯,𝑑) = π‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑), then 𝐃𝐸𝐹[𝑒 βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑)] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘₯(𝑒 2+π‘–π‘Ž)𝑑π‘₯ β‹… 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘‘(𝑣 2+𝑖𝑏)𝑑𝑑 = 1 𝑒(𝑒2 + π‘–π‘Ž) 1 𝑣(𝑣2 + 𝑖𝑏) = 1 𝑒𝑣(𝑒2 + π‘–π‘Ž)(𝑣2 + 𝑖𝑏) IHJPAS. 53 (4)2022 223 7 Let 𝑓(π‘₯,𝑑) = sin (π‘Žπ‘₯ + 𝑏𝑑), then 𝐃𝐸𝐹[sin (π‘Žπ‘₯ + 𝑏𝑑)] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šsin (π‘Žπ‘₯ + 𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š[ 𝑒𝑖(π‘Žπ‘₯+𝑏𝑑) βˆ’ π‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑) 2𝑖 ]π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 2𝑖 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’π‘–(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ βˆ’ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ ] = 1 2𝑖 [𝐃𝐸𝐹[𝑒 𝑖(π‘Žπ‘₯+𝑏𝑑)] βˆ’ 𝐃𝐸𝐹[𝑒 βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑)]] = 1 2𝑖 [ 1 𝑒𝑣(𝑒𝛼 βˆ’π‘–π‘Ž)(𝑣2 βˆ’ 𝑖𝑏) βˆ’ 1 𝑒𝑣(𝑒2 + π‘–π‘Ž)(𝑣2 + 𝑏) ] = 𝑏𝑒2 + π‘Žπ‘£2 𝑒𝑣(𝑒4 + π‘Ž2)(𝑣4 + 𝑏2) β‹… 8 Let 𝑓(π‘₯,𝑑) = cos (π‘Žπ‘₯ + 𝑏𝑑), then 𝐃𝐸𝐹[cos (π‘Žπ‘₯ + 𝑏𝑑)] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šcos (π‘Žπ‘₯ + 𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š[ 𝑒𝑖(π‘Žπ‘₯+𝑏𝑑) + π‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑) 2 ]π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’π‘–(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ + 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ ] = 1 2 [𝐃𝐸𝐹[𝑒 𝑖(π‘Žπ‘₯+𝑏𝑑)] + 𝐃𝐸𝐹[𝑒 βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑)]] = 1 2 [ 1 𝑒𝑣(𝑒𝛼 βˆ’ π‘–π‘Ž)(𝑣2 βˆ’ 𝑖𝑏) + 1 𝑒𝑣(𝑒2 + π‘–π‘Ž)(𝑣2 + 𝑏) ] = 𝑒2𝑣2 βˆ’ π‘Žπ‘ 𝑒𝑣(𝑒4 + π‘Ž2)(𝑣4 + 𝑏2) β‹… 9 Let 𝑓(π‘₯,𝑑) = sinh (π‘Žπ‘₯ + 𝑏𝑑), then IHJPAS. 53 (4)2022 224 𝐃𝐸𝐹[sinh (π‘Žπ‘₯ + 𝑏𝑑)] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šsinh (π‘Žπ‘₯ + 𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š[ 𝑒(π‘Žπ‘₯+𝑏𝑑) βˆ’π‘’βˆ’(π‘Žπ‘₯+𝑏𝑑) 2 ]π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ βˆ’ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’βˆ’(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ ] = 1 2 [𝐃𝐸𝐹[𝑒 (π‘Žπ‘₯+𝑏𝑑)] βˆ’ 𝐃𝐸𝐹[𝑒 βˆ’(π‘Žπ‘₯+𝑏𝑑)]] = 1 2 [ 1 𝑒𝑣(𝑒2 βˆ’ π‘Ž)(𝑣2 βˆ’ 𝑏) βˆ’ 1 𝑒𝑣(𝑒2 + π‘Ž)(𝑣2 + 𝑏) ] = π‘Žπ‘£2 + 𝑏𝑒2 𝑒𝑣(𝑒4 βˆ’ π‘Ž2)(𝑣4 βˆ’ 𝑏2) 10 Let 𝑓(π‘₯,𝑑) = cosh (π‘Žπ‘₯ + 𝑏𝑑), then 𝐃𝐸𝐹[cosh (π‘Žπ‘₯ + 𝑏𝑑)] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šcosh (π‘Žπ‘₯ + 𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š[ 𝑒(π‘Žπ‘₯+𝑏𝑑) + π‘’βˆ’(π‘Žπ‘₯+𝑏𝑑) 2 ]π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ + 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘’βˆ’(π‘Žπ‘₯+𝑏𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ ] = 1 2 [𝐃𝐸𝐹[𝑒 (π‘Žπ‘₯+𝑏𝑑)] + 𝐃𝐸𝐹[𝑒 βˆ’(π‘Žπ‘₯+𝑏𝑑)]] = 1 2 [ 1 𝑒𝑣(𝑒2 βˆ’ π‘Ž)(𝑣2 βˆ’ 𝑏) + 1 𝑒𝑣(𝑒2 + π‘Ž)(𝑣2 +𝑏) ] = 𝑒2𝑣2 + π‘Žπ‘ 𝑒𝑣(𝑒4 βˆ’ π‘Ž2)(𝑣4 βˆ’ 𝑏2) β‹… 4. Summarization The new double Emad- Falih integral transform of some basic functions is shown in the following table: Table 1: The new double Emad - Falih integral transform of some basic functions. 𝑓(π‘₯,𝑑) 𝐃𝐸𝐹[Cosh (π‘Žπ‘₯ + 𝑏𝑑)] 1 1 (𝑒𝑣)3 π‘₯π‘›π‘‘π‘š 𝑛! 𝑒2𝑛+3 β‹… π‘š! 𝑣2π‘š+3 π‘’π‘Žπ‘₯+𝑏𝑑 1 𝑒𝑣(𝑒2 βˆ’ π‘Ž)(𝑣2 βˆ’π‘) π‘’βˆ’(π‘Žπ‘₯+𝑏𝑑) 1 𝑒𝑣(𝑒2 + π‘Ž)(𝑣2 +𝑏) 𝑒𝑖(π‘Žπ‘₯+𝑏𝑑) 1 𝑒𝑣(𝑒2 βˆ’ π‘–π‘Ž)(𝑣2 βˆ’π‘–π‘) IHJPAS. 53 (4)2022 225 π‘’βˆ’π‘–(π‘Žπ‘₯+𝑏𝑑) 1 𝑒𝑣(𝑒2 + π‘–π‘Ž)(𝑣2 +𝑖𝑏) Sin (π‘Žπ‘₯ + 𝑏𝑑) 𝑏𝑒2 + π‘Žπ‘£2 𝑒𝑣(𝑒4 +π‘Ž2)(𝑣4 + 𝑏2) Cos (π‘Žπ‘₯ + 𝑏𝑑) 𝑒2𝑣2 βˆ’ π‘Žπ‘ 𝑒𝑣(𝑒4 +π‘Ž2)(𝑣4 + 𝑏2) Sinh (π‘Žπ‘₯ + 𝑏𝑑) π‘Žπ‘£2 + 𝑏𝑒2 𝑒𝑣(𝑒4 βˆ’π‘Ž2)(𝑣4 βˆ’ 𝑏2) Cosh (π‘Žπ‘₯ + 𝑏𝑑) 𝑒2𝑣2 + π‘Žπ‘ 𝑒𝑣(𝑒4 βˆ’π‘Ž2)(𝑣4 βˆ’ 𝑏2) 5. Theorems And Proof Theorem πŸ“.𝟏. 𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ ] = βˆ’ 1 𝑒 𝐓(0,𝑣)+ 𝑒2𝐓(𝑒,𝑣) Proof. 𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ ] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’(𝑒 𝛼π‘₯+𝑣𝛼𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [ 1 𝑒 ∫ β€Š ∞ 0 βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯⏟ ]𝑑𝑑 πΌπ‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘” 𝑏𝑦 π‘π‘Žπ‘Ÿπ‘‘π‘  = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [ 1 𝑒 (π‘’βˆ’π‘’ 2π‘₯𝑓(π‘₯,𝑑)| 0 ∞ +𝑒2 ∫ β€Š ∞ 0 β€Šπ‘“(π‘₯,𝑑)π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯)]𝑑𝑑 = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [βˆ’ 𝑓(0,𝑑) 𝑒 + π‘’βˆ« β€Š ∞ 0 β€Šπ‘“(π‘₯,𝑑)π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯]𝑑𝑑, = βˆ’ 1 𝑒 [ 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑𝑓(0,𝑑)𝑑𝑑] + 𝑒2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘“(π‘₯,𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯] = βˆ’ 1 𝑒 𝐓(0,𝑣) +𝑒2𝐓(𝑒,𝑣). Theorem 5.2. 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯2 ] = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘₯ βˆ’ 𝑒𝐓(0,𝑣) +𝑒4𝐓(𝑒,𝑣) Proof. IHJPAS. 53 (4)2022 226 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯2 ] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯2 π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑[ 1 𝑒 ∫ β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯2 π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯ ⏟ Integration by parts ]𝑑𝑑, Let 𝜁 = π‘’βˆ’π‘’ 2π‘₯ π‘‘πœ‚ = βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯2 𝑑π‘₯ π‘‘πœ = βˆ’π‘’2π‘’βˆ’π‘’ 2π‘₯ πœ‚ = βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [ 1 𝑒 (π‘’βˆ’π‘’ 2π‘₯ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ | 0 ∞ + 𝑒2 ∫ β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯)]𝑑𝑑, = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [ 1 𝑒 π‘’βˆ’π‘’ 2π‘₯ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ | 0 ∞ + π‘’βˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯]𝑑𝑑 = βˆ’ 1 𝑒 [ 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 βˆ‚π‘“(0,𝑑) βˆ‚π‘₯ 𝑑𝑑] + (𝑒)2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ )π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯] , = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘₯ + 𝑒2𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ ]. = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘₯ + 𝑒2 [βˆ’ 1 𝑒 𝐓(0,𝑣)+ 𝑒2𝐓(𝑒,𝑣)] ; = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘₯ βˆ’ 𝑒𝐓(0,𝑣) + 𝑒4𝐓(𝑒,𝑣). In general 𝐃𝐸𝐹 [ βˆ‚π‘›π‘“(π‘₯,𝑑) βˆ‚π‘₯𝑛 ] = 𝑒2𝑛𝐓(𝑒,𝑣)βˆ’ [π‘’βˆ’1 βˆ‚π‘›βˆ’1𝐓(0,𝑣) βˆ‚π‘₯π‘›βˆ’1 + 𝑒1 βˆ‚π‘›βˆ’2𝐓(0,𝑣) βˆ‚π‘₯π‘›βˆ’2 + 𝑒3 βˆ‚π‘›βˆ’3𝐓(0,𝑣) βˆ‚π‘₯π‘›βˆ’3 +β‹―+ 𝑒2(π‘›βˆ’1)βˆ’3 βˆ‚π“(0,𝑣) βˆ‚π‘₯ + 𝑒2π‘›βˆ’3𝐓(0,𝑣)] or = 𝑒2𝑛𝐓(𝑒,𝑣)βˆ’ βˆ‘β€Š 𝑛 π‘˜=1 β€Šπ‘’2π‘˜βˆ’3 βˆ‚π‘›βˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘›βˆ’π‘˜ Theorem 5.3. 𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ ] = βˆ’ 1 𝑣 𝐓(𝑒,0) + 𝑣2𝐓(𝑒,𝑣) Proof. IHJPAS. 53 (4)2022 227 𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ ] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’(𝑒 𝛼π‘₯+𝑣𝛼𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯[ 1 𝑣 ∫ β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’π‘£ 2𝑑𝑑𝑑]𝑑π‘₯ ⏟ Integration by parts = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [ 1 𝑣 (π‘’βˆ’π‘£ 2𝑑𝑓(π‘₯,𝑑)| 0 ∞ +𝑣2 ∫ β€Š ∞ 0 β€Šπ‘“(π‘₯,𝑑)π‘’βˆ’π‘£ 2𝑑𝑑𝑑)]𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [βˆ’ 𝑓(π‘₯,0) 𝑣 + π‘£βˆ« β€Š ∞ 0 β€Šπ‘“(π‘₯,𝑑)π‘’βˆ’π‘£ 2𝑑𝑑𝑑]𝑑π‘₯ = βˆ’ 1 𝑣 [ 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯𝑓(π‘₯,0)𝑑π‘₯]+ 𝑣2[ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Šπ‘“(π‘₯,𝑑)π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯] = βˆ’ 1 𝑣 𝐓(𝑒,0)+ 𝑣2𝐓(𝑒,𝑣). Theorem 5.4. 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘2 ] = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘‘ βˆ’ 𝑣𝐓(0,𝑣) + 𝑣4𝐓(𝑒,𝑣) Proof. 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘2 ] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘2 π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯[ 1 𝑣 ∫ β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘2 π‘’βˆ’π‘£ 2𝑑𝑑𝑑 ⏟ Integration by parts ]𝑑π‘₯, Let 𝜁 = π‘’βˆ’π‘£ 2𝑑 π‘‘πœ‚ = βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘2 𝑑𝑑 π‘‘πœ = βˆ’π‘£2π‘’βˆ’π‘£ 2𝑑 πœ‚ = βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [ 1 𝑣 (π‘’βˆ’π‘£ 2𝑑 βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ | 0 ∞ + 𝑣2 ∫ β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’π‘£ 2𝑑𝑑𝑑)]𝑑π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [ 1 𝑣 π‘’βˆ’π‘£ 2𝑑 βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ | 0 ∞ + π‘£βˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’π‘£ 2𝑑𝑑𝑑]𝑑π‘₯, = βˆ’ 1 𝑣 [ 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ βˆ‚π‘“(π‘₯,0) βˆ‚π‘‘ 𝑑𝑑] + (𝑣)2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯], = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘‘ + 𝑣2𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ ] . = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘‘ + 𝑣2 [βˆ’ 1 𝑣 𝐓(𝑒,0)+ 𝑣2𝐓(𝑒,𝑣)] ; = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘‘ βˆ’ 𝑣𝐓(0,𝑣) + 𝑣4𝐓(𝑒,𝑣). In general IHJPAS. 53 (4)2022 228 𝐃𝐸𝐹 [ βˆ‚π‘›π‘“(π‘₯,𝑑) βˆ‚π‘‘π‘› ] = 𝑣2𝑛𝐓(𝑒,𝑣) βˆ’ [π‘£βˆ’1 βˆ‚π‘›βˆ’1𝐓(𝑒,0) βˆ‚π‘‘π‘›βˆ’1 + 𝑣1 βˆ‚π‘›βˆ’2𝐓(𝑒,0) βˆ‚π‘‘π‘›βˆ’2 + 𝑣3 βˆ‚π‘›βˆ’3𝐓(𝑒,0) βˆ‚π‘‘π‘›βˆ’3 +β‹―+ 𝑣2(π‘›βˆ’1)βˆ’3 βˆ‚π“(𝑒,0) βˆ‚π‘‘ + 𝑒2π‘›βˆ’3𝐓(𝑒,0)] , or = 𝑣2𝑛𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 𝑛 π‘˜=1 β€Šπ‘£2π‘˜βˆ’3 βˆ‚π‘›βˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘›βˆ’π‘˜ Theorem πŸ“.πŸ“. 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘βˆ‚π‘₯ ] = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘₯ βˆ’ 𝑣2 𝑒 𝐓(0,𝑣) + 𝑣2𝑒2𝐓(𝑒,𝑣) Proof. 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘βˆ‚π‘₯ ] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘βˆ‚π‘₯ π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯, = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯[ 1 𝑣 ∫ β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘βˆ‚π‘₯ π‘’βˆ’π‘£ 2𝑑𝑑𝑑 ⏟ Integration by parts ]𝑑π‘₯, Let 𝜁 = π‘’βˆ’π‘£ 2𝑑 π‘‘πœ‚ = βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘βˆ‚π‘₯ 𝑑𝑑 π‘‘πœ = βˆ’π‘£2π‘’βˆ’π‘£ 2𝑑 πœ‚ = βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [ 1 𝑣 (π‘’βˆ’π‘£ 2𝑑 βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ | 0 ∞ + 𝑣2 ∫ β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’π‘£ 2𝑑𝑑𝑑)]𝑑π‘₯, = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [ 1 𝑣 π‘’βˆ’π‘£ 2𝑑 βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ | 0 ∞ + π‘£βˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’π‘£ 2𝑑𝑑𝑑]𝑑π‘₯, = 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ [βˆ’ 1 𝑣 βˆ‚π‘“(π‘₯,0) βˆ‚π‘₯ + π‘£βˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’π‘£ 2𝑑𝑑𝑑]𝑑π‘₯, = βˆ’ 1 𝑣 [ 1 𝑒 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘’ 2π‘₯ βˆ‚π‘“(π‘₯,0) βˆ‚π‘₯ ]+ 𝑣2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯], = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘₯ + 𝑣2𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ ] = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘₯ + 𝑣2 [βˆ’ 1 𝑒 𝐓(0,𝑣) +𝑒2𝐓(𝑒,𝑣)] , = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘₯ βˆ’ 𝑣2 𝑒 𝐓(0,𝑣) + 𝑣2𝑒2𝐓(𝑒,𝑣) Theorem πŸ“.πŸ”. 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯βˆ‚π‘‘ ] = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘‘ βˆ’ 𝑒2 𝑣 𝐓(0,𝑣) + 𝑣2𝑒2𝐓(𝑒,𝑣) Proof. IHJPAS. 53 (4)2022 229 𝐃𝐸𝐹 [ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯βˆ‚π‘‘ ] = 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯βˆ‚π‘‘ π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯, = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑[ 1 𝑒 ∫ β€Š ∞ 0 β€Š βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯βˆ‚π‘‘ π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯ ⏟ Integration by parts ]𝑑𝑑, Let 𝜁 = π‘’βˆ’π‘’ 2π‘₯ π‘‘πœ‚ = βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯βˆ‚π‘‘ 𝑑π‘₯ π‘‘πœ = βˆ’π‘’2π‘’βˆ’π‘’ 2π‘₯ πœ‚ = βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [ 1 𝑒 (π‘’βˆ’π‘’ 2π‘₯ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ | 0 ∞ + 𝑒2 ∫ β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’π‘’ 2π‘₯𝑑𝑑)]𝑑𝑑, = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [ 1 𝑒 π‘’βˆ’π‘’ 2π‘₯ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ | 0 ∞ + π‘’βˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯]𝑑𝑑 = 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 [βˆ’ 1 𝑒 βˆ‚π‘“(π‘₯,0) βˆ‚π‘‘ + π‘’βˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ π‘’βˆ’π‘’ 2π‘₯𝑑π‘₯]𝑑𝑑, = βˆ’ 1 𝑒 [ 1 𝑣 ∫ β€Š ∞ 0 β€Šπ‘’βˆ’π‘£ 2𝑑 βˆ‚π‘“(0,𝑑) βˆ‚π‘‘ ] + 𝑒2 [ 1 𝑒𝑣 ∫ β€Š ∞ 0 β€Šβˆ« β€Š ∞ 0 β€Š βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ π‘’βˆ’(𝑒 2π‘₯+𝑣2𝑑)𝑑𝑑𝑑π‘₯], = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘‘ + 𝑒2𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ ] , = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘‘ + 𝑒2 [βˆ’ 1 𝑣 𝐓(𝑒,0)+ 𝑣2𝐓(𝑒,𝑣)] , = βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘‘ βˆ’ 𝑒2 𝑣 𝐓(0,𝑣)+ 𝑣2𝑒2𝐓(𝑒,𝑣) Table 2. Summarization 𝑓(π‘₯,𝑑) 𝐃𝐸𝐹[𝑓(π‘₯,𝑑)] = 𝐓(𝑒,𝑣) βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ βˆ’ 1 𝑒 𝐓(0,𝑣) + 𝑒2𝐓(𝑒,𝑣) βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯2 βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘₯ βˆ’ 𝑒𝐓(0,𝑣) + 𝑒4𝐓(𝑒,𝑣) βˆ‚n𝑓(π‘₯,𝑑) βˆ‚π‘₯n 𝑒2𝑛𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 𝑛 π‘˜=1 β€Šπ‘’2π‘˜βˆ’3 βˆ‚π‘›βˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘›βˆ’π‘˜ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ βˆ’ 1 𝑣 𝐓(𝑒,0) + 𝑣2𝐓(𝑒,𝑣) βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘2 βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘‘ βˆ’ 𝑣𝐓(0,𝑣) + 𝑣4𝐓(𝑒,𝑣) βˆ‚π‘›π‘“(π‘₯,𝑑) βˆ‚π‘‘π‘› 𝑣2𝑛𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 𝑛 π‘˜=1 β€Šπ‘£2π‘˜βˆ’3 βˆ‚π‘›βˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘›βˆ’π‘˜ βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘‘βˆ‚π‘₯ βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘₯ βˆ’ 𝑣2 𝑒 𝐓(0,𝑣) + 𝑣2𝑒2𝐓(𝑒,𝑣) βˆ‚2𝑓(π‘₯,𝑑) βˆ‚π‘₯βˆ‚π‘‘ βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘‘ βˆ’ 𝑒2 𝑣 𝐓(0,𝑣) + 𝑣2𝑒2𝐓(𝑒,𝑣) IHJPAS. 53 (4)2022 230 Theorem 5.7. Let 𝐅𝑐(𝑒,𝑣) be the new double Emad- Falih transform of 𝑓(π‘₯,𝑑)(𝐓(𝑒,𝑣) = 𝐃𝐸𝐹[𝑓(π‘₯,𝑑)]), then 𝐃𝐸𝐹 [ βˆ‚π‘›π‘“(π‘₯,𝑑) βˆ‚π‘₯𝑛 ] = 𝑒2𝑛𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 𝑛 π‘˜=1 𝑒2π‘˜βˆ’3 βˆ‚π‘›βˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘›βˆ’π‘˜ (1) And 𝐃𝐸𝐹 [ βˆ‚π‘›π‘“(π‘₯,𝑑) βˆ‚π‘‘π‘› ] = 𝑣2𝑛𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 𝑛 π‘˜=1 𝑣2π‘˜βˆ’3 βˆ‚π‘›βˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘›βˆ’π‘˜ (2) Proof. Firstly, we prove (1) by the mathematical induction: 1 for 𝑛 = 1 𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘₯ ] = 𝑒2(1)𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 1 π‘˜=1 β€Šπ‘’2π‘˜βˆ’3 βˆ‚1βˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯1βˆ’π‘˜ = 𝑒2𝐓(𝑒,𝑣) βˆ’ 1 𝑒 𝐓(0,𝑣). Thus, true for 𝑛 = 1 2 Assume true for 𝑛 = π‘š, to get: 𝐃𝐸𝐹 [ βˆ‚π‘šπ‘“(π‘₯,𝑑) βˆ‚π‘₯π‘š ] = 𝑒2π‘šπ“(𝑒,𝑣) βˆ’ βˆ‘ β€Šπ‘šπ‘˜=1 𝑒 2π‘˜βˆ’3 βˆ‚ π‘šβˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘šβˆ’π‘˜ . 3 We want to prove (1) for 𝑛 = π‘š + 1: IHJPAS. 53 (4)2022 231 𝐃𝐸𝐹 [ βˆ‚π‘š+1𝑓(π‘₯,𝑑) βˆ‚π‘₯π‘š+1 ] = 𝐃𝐸𝐹 [ βˆ‚ βˆ‚π‘₯ [ βˆ‚π‘šπ‘“(π‘₯,𝑑) βˆ‚π‘₯π‘š ]] = 𝑒2𝐃𝐸𝐹 [ βˆ‚π‘šπ‘“(π‘₯,𝑑) βˆ‚π‘₯π‘š ] βˆ’ 1 𝑒 βˆ‚π‘šπ“(0,𝑣) βˆ‚π‘₯π‘š , = 𝑒2 [𝑒2π‘šπ“(𝑒,𝑣) βˆ’ βˆ‘β€Š π‘š π‘˜=1 β€Šπ‘’2π‘˜βˆ’3 βˆ‚π‘šβˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘šβˆ’π‘˜ ] βˆ’ 1 𝑒 βˆ‚π‘šπ“(0,𝑣) βˆ‚π‘₯π‘š , = 𝑒2(π‘š+1)𝐓(𝑒,𝑣) βˆ’π‘’2 βˆ‘β€Š π‘š π‘˜=1 β€Šπ‘’2π‘˜βˆ’3 βˆ‚π‘šβˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘šβˆ’π‘˜ βˆ’ 1 𝑒 βˆ‚π‘šπ“(0,𝑣) βˆ‚π‘₯π‘š = 𝑒2(π‘š+1)𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š π‘š π‘˜=1 𝑒2π‘˜βˆ’1 βˆ‚π‘šβˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘šβˆ’π‘˜ βˆ’ 1 𝑒 βˆ‚π‘šπ“(0,𝑣) βˆ‚π‘₯π‘š , = 𝑒2(π‘š+1)𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š π‘š π‘˜=0 β€Šπ‘’2π‘˜βˆ’1 βˆ‚π‘šβˆ’π‘˜π“(0,𝑣) βˆ‚π‘₯π‘šβˆ’π‘˜ , = 𝑒2(π‘š+1)𝐓(𝑒,𝑣) βˆ’ βˆ‘ β€Š π‘š+1 π‘˜=1 β€Šπ‘’2(π‘˜βˆ’1)βˆ’1 βˆ‚π‘šβˆ’(π‘˜βˆ’1)𝐓(0,𝑣) βˆ‚π‘₯π‘šβˆ’(π‘˜βˆ’1) , = 𝑒2(π‘š+1)𝐓(𝑒,𝑣) βˆ’ βˆ‘ β€Š π‘š+1 π‘˜=1 β€Šπ‘’2π‘˜βˆ’3 βˆ‚(π‘š+1βˆ’π‘˜)𝐓(0,𝑣) βˆ‚π‘₯(π‘š+1βˆ’π‘˜) , = 𝐃𝐸𝐹 [ βˆ‚π‘š+1𝑓(π‘₯,𝑑) βˆ‚π‘₯π‘š+1 ] . So theorem is true for 𝑛 ∈ β„•. Finally, we want prove (2) by the mathematical induction: 1 for 𝑛 = 1 𝐃𝐸𝐹 [ βˆ‚π‘“(π‘₯,𝑑) βˆ‚π‘‘ ] = 𝑣2(1)𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š 1 π‘˜=1 β€Šπ‘£2π‘˜βˆ’3 βˆ‚1βˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘1βˆ’π‘˜ = 𝑣2𝐓(𝑒,𝑣) βˆ’ 1 𝑣 𝐓(𝑒,0). Thus, true for 𝑛 = 1 2 Assume true for 𝑛 = π‘š, to get: 𝐃𝐸𝐹 [ βˆ‚π‘šπ‘“(π‘₯,𝑑) βˆ‚π‘‘π‘š ] = 𝑣2π‘šπ“(𝑒,𝑣) βˆ’ βˆ‘β€Š π‘š π‘˜=1 𝑣2π‘˜βˆ’3 βˆ‚π‘šβˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘šβˆ’π‘˜ . 3 We want to prove (1) for 𝑛 = π‘š + 1: IHJPAS. 53 (4)2022 232 𝐃𝐸𝐹 [ βˆ‚π‘š+1𝑓(π‘₯,𝑑) βˆ‚π‘‘π‘š+1 ] = 𝐃𝐸𝐹 [ βˆ‚ βˆ‚π‘₯ [ βˆ‚π‘šπ‘“(π‘₯,𝑑) βˆ‚π‘‘π‘š ]] = 𝑣2𝐃𝐸𝐹 [ βˆ‚π‘šπ‘“(π‘₯,𝑑) βˆ‚π‘‘π‘š ] βˆ’ 1 𝑣 βˆ‚π‘šπ“(𝑒,0) βˆ‚π‘‘π‘š , = 𝑣2 [𝑣2π‘šπ“(𝑒,𝑣) βˆ’ βˆ‘β€Š π‘š π‘˜=1 β€Šπ‘£2π‘˜βˆ’3 βˆ‚π‘šβˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘šβˆ’π‘˜ ] βˆ’ 1 𝑣 βˆ‚π‘šπ“(𝑒,0) βˆ‚π‘‘π‘š = 𝑣2(π‘š+1)𝐓(𝑒,𝑣)βˆ’ 𝑣2 βˆ‘β€Š π‘š π‘˜=1 β€Šπ‘£2π‘˜βˆ’3 βˆ‚π‘šβˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘šβˆ’π‘˜ βˆ’ 1 𝑣 βˆ‚π‘šπ“(𝑒,0) βˆ‚π‘‘π‘š = 𝑣2(π‘š+1)𝐓(𝑒,𝑣) βˆ’ βˆ‘β€Š π‘š π‘˜=1 β€Šπ‘£2π‘˜βˆ’1 βˆ‚π‘šβˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘šβˆ’π‘˜ βˆ’ 1 𝑣 βˆ‚π‘šπ“(𝑒,0) βˆ‚π‘‘π‘š , = 𝑣2(π‘š+1)𝐓(𝑒,𝑣)βˆ’ βˆ‘β€Š π‘š π‘˜=0 β€Šπ‘£2π‘˜βˆ’1 βˆ‚π‘šβˆ’π‘˜π“(𝑒,0) βˆ‚π‘‘π‘šβˆ’π‘˜ , = 𝑣2(π‘š+1)𝐓(𝑒,𝑣)βˆ’ βˆ‘ β€Š π‘š+1 π‘˜=1 β€Šπ‘£2(π‘˜βˆ’1)βˆ’1 βˆ‚π‘šβˆ’(π‘˜βˆ’1)𝐓(𝑒,0) βˆ‚π‘‘π‘šβˆ’(π‘˜βˆ’1) , = 𝑣2(π‘š+1)𝐓(𝑒,𝑣)βˆ’ βˆ‘ β€Š π‘š+1 π‘˜=1 β€Šπ‘£2π‘˜βˆ’3 βˆ‚(π‘š+1βˆ’π‘˜)𝐓(𝑒,0) βˆ‚π‘‘(π‘š+1βˆ’π‘˜) , = 𝐃𝐸𝐹 [ βˆ‚π‘š+1𝑓(π‘₯,𝑑) βˆ‚π‘‘π‘š+1 ] . So theorem is true for 𝑛 ∈ β„•. 6. Applications In this section, we introduce the solution of the linear partial differential equation (Telegraph equation). Application 6.1. Consider the linear telegraph equation: βˆ‚2𝜁 βˆ‚π‘₯2 = βˆ‚2𝜁 βˆ‚π‘‘2 + 2 βˆ‚πœ βˆ‚π‘‘ + 𝜁. Subject to initial conditions: 𝜁(π‘₯,0) = 𝑒π‘₯, 𝜁(0,𝑑) = π‘’βˆ’2𝑑 πœπ‘‘(π‘₯,0) = βˆ’2𝑒 π‘₯, 𝜁π‘₯(0,𝑑) = 𝑒 βˆ’2𝑑 Solution: Applying Emad- Falih transform, 𝐃𝐸𝐹 [ βˆ‚2𝜁 βˆ‚π‘₯2 = βˆ‚2𝜁 βˆ‚π‘‘2 + 2 βˆ‚πœ βˆ‚π‘‘ + 𝜁], IHJPAS. 53 (4)2022 233 to obtain the following: βˆ’ 1 𝑒 βˆ‚π“(0,𝑣) βˆ‚π‘₯ βˆ’ 𝑒𝐓(0,𝑣) + 𝑒4𝐓(𝑒,𝑣) = βˆ’ 1 𝑣 βˆ‚π“(𝑒,0) βˆ‚π‘‘ βˆ’ 𝑣𝐓(0,𝑣)+ 𝑣4𝐓(𝑒,𝑣) +2[βˆ’ 1 𝑣 𝐓(𝑒,0) + 𝑣2𝐓(𝑒,𝑣)] + 𝑇(𝑒,𝑣). Applying the Emad- Falih transform to the initial conditions, we have: βˆ’ 1 𝑒 ( 1 𝑣(𝑣2 + 2) )βˆ’ 𝑒( 1 𝑣(𝑣2 + 2) )+ 𝑒4𝐓(𝑒,𝑣) = βˆ’ 1 𝑣 ( βˆ’2 𝑒(𝑒2 βˆ’1) ) βˆ’ 𝑣( 1 𝑒(𝑒2 βˆ’ 1) )+ 𝑣4𝐓(𝑒,𝑣) βˆ’2 1 𝑣 ( 1 𝑒(𝑒2 βˆ’ 1) )+ 2𝑣2𝐓(𝑒,𝑣) + 𝑇(𝑒,𝑣). We get: (𝑒4 βˆ’ 𝑣4 βˆ’ 2𝑣2 βˆ’ 1)𝐓(𝑒,𝑣) = 𝑒4 βˆ’ 𝑣4 βˆ’2𝑣2 βˆ’ 1 𝑒𝑣(𝑣2 + 2)(𝑒2 βˆ’ 1) , 𝐓(𝑒,𝑣) = 1 𝑒𝑣(𝑣2 + 2)(𝑒2 βˆ’ 1) . Then 𝜁(π‘₯,𝑑) = 𝑒π‘₯βˆ’2𝑑 7.Conclusions In this research, we have extended the double Emad - Falih integral transform. First, we have proved the fundamental properties and theorems using the new double Emad - Falih integral transform which have also been introduced and presented. Second, the new double integral transform to solve the telegraph partial differential equation is applied. References 1. Dhunde, R. R. ; Waghmare, G. L. Double Laplace Transform Method in mathematical Physics. International Journal of Theoretical and Mathematical Physics, 2017,7(1),14–20. 2. Eltayeb, H. ; KilicΒΈman. A. On Double Sumudu Transform And Double Laplace Transform. Malaysian Journal of Mathematical Sciences, 2010, 4(1),17–30. 3. Hassan, M. A. Partial Differential Equation. Journal of Applied Mathematics and Physics, 2020,8(8), 1463-1471. 4. Meddahi M. ; Jafari, H. ; Yang, X. 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