120 This work is licensed under a Creative Commons Attribution 4.0 International License. The New Complex Integral Transform "Complex Sadik Transform" and It’s Applications Abstract In this work, we present a novel complex transform namely the "Complex Sadik Transform". The propositions of this transformation are investigated. The complex transform is used to convert the core problem to a simple algebraic equation. Then, the answer to this primary problem can be obtained to find the solution to this equation and apply the inverse of the complex Sadik transform. As well, the complex Sadik transform is applied and used to find the solution of linear higher order ordinary differential equations. As well, we present and discuss, some important real life problems such as pharmacokinetics problems, nuclear physics problems, and Beam problems. Keywords: Complex integral transformation, the inverse of complex transform, Sadik transform, ordinary differential equations. 1. Introduction In (2018), researcher Sadik L. sheikh [11] presented a new integral transformation defined as follows : The Sadik integral transform of 𝑔(𝑑) is defined as : π’π‘Ž[𝑔(𝑑)] = 𝐅(𝑣 𝛼 , 𝛽) = 1 𝑣𝛽 ∫ β€Š ∞ 0 𝑔(𝑑)π‘’βˆ’π‘£ 𝛼𝑑 𝑑𝑑 where 𝑣 ∈ β„‚ , 𝛼 ∈ β„βˆ— , and 𝛽 ∈ ℝ . The following properties of Sadik integral transform [11] : 1 If 𝑔(𝑑) = 𝑑𝑛 , then π’π‘Ž[𝑑 𝑛 ] = 𝑛! 𝑣𝑛𝛼+(𝛼+𝛽) , 𝑛 β‰₯ 0. 2 If 𝑔(𝑑) = π‘’π‘Žπ‘‘ , then π’π‘Ž[𝑒 π‘Žπ‘‘ ] = π‘£βˆ’π›½ π‘£π›Όβˆ’π‘Ž , where π‘Ž is a constant. 3 If 𝑔(𝑑) = sin π‘Žπ‘‘, then π’π‘Ž [sin π‘Žπ‘‘] = π‘Žπ‘£βˆ’π›½ 𝑣2𝛼+π‘Ž2 . Doi: 10.30526/35.3.2850 Ibn Al Haitham Journal for Pure and Applied Science Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index Article history: Received 12 May 2022, Accepted 6 June 2022, Published in July 2022. Saed M. Turq Teacher at the Ministry of Education Hebron,Palestine saedturq@gmail.com Emad A. Kuffi College of Engineering, Al-Qadisiyah University, 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 (3)2022 121 4 If 𝑔(𝑑) = cos π‘Žπ‘‘, then π’π‘Ž [cos π‘Žπ‘‘] = π‘£π›Όβˆ’π›½ 𝑣2𝛼+π‘Ž2 . 5 If 𝑔(𝑑) = sinh π‘Žπ‘‘, then π’π‘Ž[sinh π‘Žπ‘‘] = π‘Žπ‘£βˆ’π›½ 𝑣2π›Όβˆ’π‘Ž2 6 If 𝑔(𝑑) = cosh π‘Žπ‘‘, then π’π‘Ž[cosh π‘Žπ‘‘] = π‘£π›Όβˆ’π›½ 𝑣2π›Όβˆ’π‘Ž2 . 7 π’π‘Ž[𝑔 (𝑛)(𝑑)] = 𝑣𝑛𝛼 𝐅(𝑣) βˆ’ βˆ‘π‘˜=0 π‘›βˆ’1 β€Šπ‘£π‘˜π›Όβˆ’π›½ 𝑔((π‘›βˆ’1)βˆ’π‘˜)(0). Now, the complex Sadik transform is a new complex transform and it is applied and used to find the solution of ordinary differential equation and has applications in domains such as engineering, applied physics, and signed processing [3,4,7]. We analyze functions in the set 𝐂 defined by a novel complex transform defined for functions of exponential order: 𝐂 = {𝑔(𝑑): there exists 𝑀, 𝐿1 and 𝐿2 are greater than zero such that |𝑔(𝑑)| < π‘€π‘’βˆ’π‘–πΏπ‘—|𝑑|, if 𝑑 ∈ (βˆ’1)𝑗 Γ— [0, ∞), 𝑗 = 1,2} where 𝑖 is a complex number. The constant 𝑀 must be a finite number for a particular function 𝑔(𝑑) in the set 𝐂, while 𝐿1 and 𝐿2 are may be finite or infinite. The Complex Sadik Transform (CST) denoted by the operator π’π‘Ž{. }, the transform as follows: π’π‘Ž 𝑐 [𝑔(𝑑)] = 𝐅𝑐 (𝑠𝛼 , 𝛽) = 1 𝑠𝛽 ∫ β€Š ∞ 0 𝑔(𝑑)π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑 where 𝑠 ∈ β„‚ , 𝛼 ∈ β„βˆ— , and 𝛽 ∈ ℝ . The aim of this work (complex transform) to find the solution of higher order ordinary differential equations. Many researchers have proposed new integral transformations for the purpose of solving ordinary and partial differential equations and their applications [2,8,9,12]. 2. A Novel Complex Transform "Complex Sadik Transform" of Important Functions In this section, we present the complex Sadik transform of famous functions: 1 If 𝑓(𝑑) = 𝑑𝑛 , 𝑛 ∈ β„• , then π’π‘Ž 𝑐 {𝑑𝑛 } = (βˆ’π‘–)𝑛+1 𝑛! 𝑠𝑛𝛼+(𝛼+𝛽) , 𝑠 > 0. Proof. Since π’π‘Ž 𝑐 [𝑑𝑛 ] = 1 𝑠𝛽 ∫ β€Š ∞ 0 𝑑𝑛 π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑 Let 𝑒 = 𝑖𝑑 β†’ 𝑑𝑒 = 𝑖𝑑𝑑 or 𝑑𝑒 𝑖 = 𝑑𝑑 or βˆ’π‘–π‘‘π‘’ = 𝑑𝑑 and we know 𝑒 = 𝑖𝑑 β†’ βˆ’π‘–π‘’ = 𝑑, when 𝑑 β†’ 0 then 𝑒 β†’ 0 and when 𝑑 β†’ ∞ then 𝑒 β†’ ∞, that is: π’π‘Ž 𝑐 [𝑑𝑛 ] = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š 𝑑𝑛 π‘’βˆ’π‘–π‘  𝛼 𝑑 𝑑𝑑 = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š (βˆ’π‘–π‘’)π‘›π‘’βˆ’π‘  𝛼𝑒(βˆ’π‘–)𝑑𝑒 = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š (βˆ’π‘–)π‘›π‘’π‘›π‘’βˆ’π‘  𝛼 𝑒(βˆ’π‘–)𝑑𝑒 = (βˆ’π‘–)𝑛+1 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š π‘’π‘›π‘’βˆ’π‘  𝛼𝑒 𝑑𝑒 = (βˆ’π‘–)𝑛+1π’π‘Ž[𝑑 𝑛 ] = (βˆ’π‘–)𝑛+1 𝑛! 𝑠𝑛𝛼+(𝛼+𝛽) β‹… 𝑠 > 0 2 If 𝑓(𝑑) = π‘’π‘Žπ‘‘ , π‘Ž is a constant number, then : π’π‘Ž 𝑐 [π‘’π‘Žπ‘‘ ] = βˆ’1 𝑠𝛽 [ π‘Ž (𝑠2𝛼 + π‘Ž2) + 𝑖 𝑠𝛼 (𝑠2𝛼 + π‘Ž2) ] , 𝑠 > π‘Ž IHJPAS. 53 (3)2022 122 Proof. Since π’π‘Ž 𝑐 [𝑒 π‘Žπ‘‘ ] = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š π‘’π‘Žπ‘‘ π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑 = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š π‘’βˆ’π‘‘(𝑖𝑠 π›Όβˆ’π‘Ž)𝑑𝑑 = 1 𝑠𝛽 1 𝑖𝑠𝛼 βˆ’ π‘Ž , = 1 𝑠𝛽 1 (𝑖𝑠𝛼 βˆ’ π‘Ž) (βˆ’π‘–π‘ π›Ό βˆ’ π‘Ž) (βˆ’π‘–π‘ π›Ό βˆ’ π‘Ž) , = 1 𝑠𝛽 [ (βˆ’1)(π‘Ž + 𝑖𝑠𝛼 ) (𝑠2𝛼 + π‘Ž2) ] , = βˆ’1 𝑠𝛽 [ π‘Ž (𝑠2𝛼 + π‘Ž2) + 𝑖 𝑠𝛼 (𝑠2𝛼 + π‘Ž2) ] , 𝑠 > π‘Ž 3 Let 𝑓(𝑑) = sin (π‘Žπ‘‘), π‘Ž is a constant number, then π’π‘Ž 𝑐 [sin (π‘Žπ‘‘)] = βˆ’π‘Ž 𝑠𝛽 (𝑠2𝛼 βˆ’ π‘Ž2) , 𝑠 > |π‘Ž| Proof. Since π’π‘Ž 𝑐 [sin (π‘Žπ‘‘)] = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Šsin (π‘Žπ‘‘)π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑 = 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š π‘’π‘–π‘Žπ‘‘ βˆ’ π‘’βˆ’π‘–π‘Žπ‘‘ 2𝑖 π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑 after simple computations, we get: π’π‘Ž 𝑐 [sin (π‘Žπ‘‘)] = βˆ’π‘Ž 𝑠𝛽 (𝑠2𝛼 βˆ’ π‘Ž2) , 𝑠 > |π‘Ž| The result will be benefit in determining the complicated transform of: 4 π’π‘Ž 𝑐 [cos (π‘Žπ‘‘)] = βˆ’π‘–π‘ π›Ό 𝑠𝛽(𝑠2π›Όβˆ’π‘Ž2) , 𝑠 > |π‘Ž|. 5 π’π‘Ž 𝑐 [sinh (π‘Žπ‘‘)] = βˆ’π‘Ž 𝑠𝛽(𝑠2𝛼+π‘Ž2) , 𝑠 > 0. 6 π’π‘Ž 𝑐 [cosh (π‘Žπ‘‘)] = βˆ’π‘–π‘ π›Ό 𝑠𝛽(𝑠2𝛼+π‘Ž2) , 𝑠 > 0. 2.1. The Sadik and Complex Sadik Integral Transforms for Some Basic Functions In this section, we will present the Sadik transform and the novel complex transform for some basic functions in the following Table 1: Table 1: Sadik transform and the complex Sadik integral transform for some basic functions Functions 𝑔(𝑑) π’π‘Ž [𝑔(𝑑)] = 𝐅(𝑠) "Sadik Transform" π’π‘Ž 𝑐 [𝑔(𝑑)] = 𝐅𝑐 (𝑠) "Complex Sadik Transform" 𝑑 𝑛, 𝑛 ∈ β„• 𝑛! 𝑠𝑛𝛼+(𝛼+𝛽) (βˆ’π‘–)𝑛+1 𝑛! 𝑠𝑛𝛼+(𝛼+𝛽) 𝑒 π‘Žπ‘‘ , π‘Ž constant 1 𝑠𝛽 (𝑠𝛼 βˆ’ π‘Ž) βˆ’1 𝑠𝛽 [ π‘Ž (𝑠2𝛼 + π‘Ž2) + 𝑖 𝑠𝛼 (𝑠2𝛼 + π‘Ž2) ] sin (π‘Žπ‘‘) π‘Ž 𝑠𝛽 (𝑠2𝛼 + π‘Ž2) βˆ’π‘Ž 𝑠𝛽 (𝑠2𝛼 βˆ’ π‘Ž2) cos (π‘Žπ‘‘) 𝑠 𝛼 𝑠𝛽 (𝑠2𝛼 + π‘Ž2) βˆ’π‘–π‘ π›Ό 𝑠𝛽 (𝑠2𝛼 βˆ’ π‘Ž2) sinh (π‘Žπ‘‘) π‘Ž 𝑠𝛽 (𝑠2𝛼 βˆ’ π‘Ž2) βˆ’π‘Ž 𝑠𝛽 (𝑠2𝛼 + π‘Ž2) IHJPAS. 53 (3)2022 123 cosh (π‘Žπ‘‘) 𝑠 𝛼 𝑠𝛽 (𝑠2𝛼 βˆ’ π‘Ž2) βˆ’π‘–π‘ π›Ό 𝑠𝛽 (𝑠2𝛼 + π‘Ž2) 2.2. The inverse of Sadik Complex Integral Transform : If π’π‘Ž 𝑐 {𝑔(𝑑)} = 𝐅𝑐 (𝑠) is the Sadik complex transform, then 𝑔(𝑑) = (π’π‘Ž 𝑐 )βˆ’1[𝐅𝑐 (𝑠)] is said to be an inverse of the Sadik complex transform. In this section, we present the inverse of Sadik complex integral transform of simple functions: 1 (π’π‘Ž 𝑐 )βˆ’1 [(βˆ’π‘–)𝑛+1 𝑛! 𝑠𝑛𝛼+(𝛼+𝛽) ] = 𝑑𝑛. 2 (π’π‘Ž 𝑐 )βˆ’1 [ βˆ’1 𝑠𝛽 [ π‘Ž (𝑠2𝛼+π‘Ž2) + 𝑖 𝑠𝛼 (𝑠2𝛼+π‘Ž2) ]] = π‘’π‘Žπ‘‘. 3 (π’π‘Ž 𝑐 )βˆ’1 [ βˆ’π‘Ž 𝑠𝛽(𝑠2π›Όβˆ’π‘Ž2) ] = sin (π‘Žπ‘‘). 4 (π’π‘Ž 𝑐 )βˆ’1 [ βˆ’π‘–π‘ π›Ό 𝑠𝛽(𝑠2π›Όβˆ’π‘Ž2) ] = cos (π‘Žπ‘‘). 5 (π’π‘Ž 𝑐 )βˆ’1 [ βˆ’π‘Ž 𝑠𝛽(𝑠2𝛼+π‘Ž2) ] = sinh (π‘Žπ‘‘) 6 (π’π‘Ž 𝑐 )βˆ’1 [ βˆ’π‘–π‘ π›Ό 𝑠𝛽(𝑠2𝛼+π‘Ž2) ] = cosh (π‘Žπ‘‘) 3. Complex Sadik Integral Transform of Derivatives: Let 𝑓(𝑑) be a continuous function and piecewise continuous on any interval, then the complex Sadik transform of first derivative of 𝑓(𝑑) is given by: π’π‘Ž 𝑐 [𝑓 β€²(𝑑)] = 1 𝑠𝛽 ∫ β€Š ∞ 0 𝑓 β€²(𝑑)π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑, integrating by parts. Let 𝑒 = π‘’βˆ’π‘–π‘  𝛼 𝑑 , 𝑑𝑣 = 𝑓 β€²(𝑑)𝑑𝑑 𝑑𝑒 = βˆ’π‘–π‘ π›Ό π‘’βˆ’π‘–π‘  𝛼𝑑 𝑑𝑑, 𝑣 = 𝑓(𝑑) 1 𝑠𝛽 [βˆ’π‘“(0) + 𝑖𝑠𝛼 ∫ β€Š ∞ 0 β€Š π‘’βˆ’π‘–π‘  𝛼𝑑 𝑓(𝑑)𝑑𝑑] βˆ’π‘“(0) 𝑠𝛽 + 𝑖𝑠𝛼 1 𝑠𝛽 ∫ β€Š ∞ 0 β€Š π‘’βˆ’π‘–π‘  𝛼𝑑 𝑓(𝑑)𝑑𝑑 βˆ’π‘“(0) 𝑠𝛽 + 𝑖𝑠𝛼 π’π‘Ž 𝑐 [𝑓(𝑑)] π’π‘Ž 𝑐 [𝑓 β€²(𝑑)] = 1 𝑠𝛽 [βˆ’π‘“(0) + 𝑖𝑠𝛼 ∫ β€Š ∞ 0 β€Š π‘’βˆ’π‘–π‘  𝛼𝑑 𝑓(𝑑)𝑑𝑑] , or π’π‘Ž 𝑐 [𝑓 β€²(𝑑)] = 𝑖𝑠𝛼 𝐅𝑐 (𝑠) βˆ’ 𝑓(0) 𝑠𝛽 Therefore, when substituted 𝑓(𝑑) by 𝑓 β€²(𝑑) and 𝑓 β€²(𝑑) by 𝑓 β€²β€²(𝑑), we get π’π‘Ž 𝑐 [𝑓 β€²β€²(𝑑)] = (𝑖𝑠𝛼 )2𝐅𝑐 (𝑠) βˆ’ 𝑓 β€²(0) 𝑠𝛽 βˆ’ 𝑖𝑠𝛼 𝑓(0) 𝑠𝛽 Similarly, π’π‘Ž 𝑐 [𝑓 β€²β€²β€²(𝑑)] = (𝑖𝑠𝛼 )3𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [𝑓 β€²β€²(0) + 𝑖𝑠𝛼 𝑓 β€²(0) + (𝑖𝑠𝛼 )2𝑓(0)]. In general: π’π‘Ž 𝑐 [𝑓 (𝑛)(𝑑)] = (𝑖𝑠𝛼 )𝑛𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [𝑓 (π‘›βˆ’1)(0) + 𝑖𝑠𝛼 𝑓 (π‘›βˆ’2)(0) + (𝑖𝑠𝛼 )2𝑓 (π‘›βˆ’3)(0) + β‹― + (𝑖𝑠𝛼 )π‘›βˆ’2𝑓 β€²(0) + (𝑖𝑠𝛼 )π‘›βˆ’1𝑓(0)] IHJPAS. 53 (3)2022 124 or π’π‘Ž 𝑐 [𝑓 (𝑛 (𝑑)] = (𝑖𝑠𝛼 )𝑛𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘π‘˜=1 𝑛 β€Š(𝑖𝑠𝛼 )π‘˜βˆ’1𝑓 (π‘›βˆ’π‘˜)(0)]. Theorem 3.1. Let 𝐅𝑐 (𝑠) be the complex Sadik integral transform of 𝑓(𝑑)(𝐅𝑐 (𝑠) = π’π‘Ž 𝑐 [𝑓(𝑑)]), then : π’π‘Ž 𝑐 [𝑓 (𝑛)(𝑑)] = (𝑖𝑠𝛼 )𝑛𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘ β€Š 𝑛 π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜βˆ’1𝑓 (π‘›βˆ’π‘˜)(0)]. Proof. By Mathematical Induction 1 For 𝑛 = 1, π’π‘Ž 𝑐 [𝑓 β€²(𝑑)] = 𝑖𝑠𝛼 𝐅𝑐 (𝑠) βˆ’ 𝑓(0) 𝑠𝛽 Thus true for 𝑛 = 1. 2. Assume that, true for 𝑛 = π‘š that means: π’π‘Ž 𝑐 [𝑓 (π‘š)(𝑑)] = (𝑖𝑠𝛼 )π‘šπ…π‘ (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘ β€Š π‘š π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜βˆ’1𝑓 (π‘šβˆ’π‘˜)(0)] 3 we want to prove for 𝑛 = π‘š + 1 π’π‘Ž 𝑐 [𝑓 (π‘š+1)(𝑑)] = π’π‘Ž 𝑐 [(𝑓 (π‘š)(𝑑)) β€² ] = 𝑖𝑠𝛼 π’π‘Ž 𝑐 [𝑓 (π‘š)(𝑑)] βˆ’ 1 𝑠𝛽 [𝑓 (π‘š)(0)], = 𝑖𝑠𝛼 π’π‘Ž 𝑐 [𝑓 (π‘š)(𝑑)] βˆ’ 𝑓 (π‘š)(0) 𝑠𝛽 , = 𝑖𝑠𝛼 [(𝑖𝑠𝛼 )π‘šπ…π‘ (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘ β€Š π‘š π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜βˆ’1𝑓 (π‘šβˆ’π‘˜)(0)]] βˆ’ 𝑓 (π‘š) 𝑠 (0)] = (𝑖𝑠𝛼 )π‘š+1𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘ β€Š π‘š π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜ 𝑓 (π‘šβˆ’π‘˜)(0)]] βˆ’ 𝑓 (π‘š)(0) 𝑠𝛽 = (𝑖𝑠𝛼 )π‘š+1𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘ β€Š π‘š π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜ 𝑓 (π‘šβˆ’π‘˜)(0)] + 𝑓 (π‘š)(0)] , = (𝑖𝑠𝛼 )π‘š+1𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [βˆ‘ β€Š π‘š π‘˜=0 β€Š (𝑖𝑠𝛼 )π‘˜ 𝑓 (π‘šβˆ’π‘˜)(0)]] , = (𝑖𝑠𝛼 )π‘š+1𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [ βˆ‘ β€Š π‘š+1 π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜βˆ’1𝑓 (π‘šβˆ’(π‘˜βˆ’1))(0)]] = (𝑖𝑠𝛼 )π‘š+1𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [ βˆ‘ β€Š π‘š+1 π‘˜=1 β€Š (𝑖𝑠𝛼 )π‘˜βˆ’1𝑓 (π‘š+1βˆ’π‘˜))(0)]] , So theorem is true for 𝑛 ∈ β„•. 4. Applications of Complex Sadik Integral Transform: In this section, we introduce three real life problems: pharmacokinetics problem, nuclear physics and Beam problems. Example 4.1. For a physical explanation of the present scheme, we consider a problem from the field of "pharmacokinetics" for solving the concentration of the drug at any given time " 𝑑 " in the blood during continuous intravenous injection of drug and find its solution in this application. This IHJPAS. 53 (3)2022 125 application can be written in terms of 1𝑠𝑑 order linear ordinary differential equation with constant coefficients as [1,5,6]. 𝑑𝑔(𝑑) 𝑑𝑑 + πœ†π‘”(𝑑) = 𝛾 volume , where 𝑑 > 0 (1) with initial conditions 𝑔(0) = 0 (2) Here: 𝑔(𝑑) : is the drug concentration in the blood at any time " 𝑑 ". πœ†: is the constant velocity of elimination. 𝛾 : the rate of infusion (in mg/min.) Volume: volume in which drug is distributed. Complex Sadik transform of both sides of equation (1) gives: π’π‘Ž 𝑐 { 𝑑𝑔(𝑑) 𝑑𝑑 } + πœ†π’π‘Ž 𝑐 {𝑔(𝑑)} = 𝛾 volume π’π‘Ž 𝑐 {1} (3) Applying Theorem 3.1, we get: 𝑖𝑠𝛼 𝐅𝑐 (𝑠) βˆ’ 𝑔(0) 𝑠𝛽 + πœ†π…π‘ (𝑠) = 𝛾 volume βˆ’π‘– 𝑠𝛼+𝛽 (4) The use of the initial condition equation (2) in (4) gives: 𝑖𝑠𝛼 𝐅𝑐 (𝑠) + πœ†π…π‘ (𝑠) = βˆ’π‘–π›Ύ volume 𝑠𝛼+𝛽 . 𝐅𝑐 (𝑠) = 𝛾 volume βˆ’π‘–π‘ βˆ’π›½ 𝑠𝛼 (πœ† + 𝑖𝑠𝛼 ) (5) Applying inverse complex Sadik transform in equation (5 ), we get: 𝑔(𝑑) = 𝛾 volume (π’π‘Ž 𝑐 )βˆ’1 [ βˆ’π‘–π‘ βˆ’π›½ 𝑠𝛼 (πœ† + 𝑖𝑠𝛼 ) ] By a fractional fraction, after simple computations, we get: 𝑔(𝑑) = 𝛾 πœ† volume [1 βˆ’ π‘’βˆ’πœ†π‘‘ ]. Which is the required concentration of drug at any given time " 𝑑" in the blood during continuous intravenous injection of a drug. Example 4.2. "Complex Sadik Transform in Nuclear physics": Consider the first order linear differential equation: 𝑑𝑔(𝑑) 𝑑𝑑 = βˆ’πœ†π‘”(𝑑) This differential equation is the fundamental relationship describing radioactive decay, where 𝑔(𝑑) represents the number of un decayed atoms remaining in a sample of radioactive isotope at the time " 𝑑 " and πœ† is the decay constant, [7,10]. We can apply the complex Sadik transform to find the solution to this differential equation. Rearranging the above differential equation, we obtain: 𝑑𝑔(𝑑) 𝑑𝑑 + πœ†π‘”(𝑑) = 0 Taking complex Sadik transform on both sides, we have: π’π‘Ž 𝑐 { 𝑑𝑔(𝑑) 𝑑𝑑 } + πœ†π’π‘Ž 𝑐 {𝑔(𝑑)} = 0 then: IHJPAS. 53 (3)2022 126 𝑖𝑠𝛼 π’π‘Ž 𝑐 {𝑔(π‘₯)} βˆ’ 𝑔(0) 𝑠𝛽 + πœ†π’π‘Ž 𝑐 {𝑔(𝑑)} = 0 (𝑖𝑠𝛼 + πœ†)π’π‘Ž 𝑐 {𝑔(𝑑)} = 𝑔(0) 𝑠𝛽 π’π‘Ž 𝑐 {𝑔(𝑑)} = 𝑔(0) 𝑠𝛽 (𝑖𝑠𝛼 + πœ†) , here 𝑔(0) = 𝑔0 Then π’π‘Ž 𝑐 {𝑔(𝑑)} = 𝑔0 𝑠𝛽 (𝑖𝑠𝛼 + πœ†) Now, we take the inverse complex Sadik transform on both sides, we obtain: 𝑔(𝑑) = 𝑔0(π’π‘Ž 𝑐 )βˆ’1 { 1 𝑠𝛽 (𝑖𝑠𝛼 + πœ†) } = 𝑔0(π’π‘Ž 𝑐 )βˆ’1 { 1 𝑠𝛽 (𝑖𝑠𝛼 + πœ†) πœ† βˆ’ 𝑖𝑠𝛼 πœ† βˆ’ 𝑖𝑠𝛼 } = 𝑔0(π’π‘Ž 𝑐 )βˆ’1 { 1 𝑠𝛽 [ πœ† 𝑠2𝛼 + πœ†2 βˆ’ 𝑖 𝑠𝛼 𝑠2𝛼 + πœ†2 ]} = 𝑔0(π’π‘Ž 𝑐 )βˆ’1 { βˆ’1 𝑠𝛽 [ βˆ’πœ† 𝑠2𝛼 + πœ†2 + 𝑖 𝑠𝛼 𝑠2𝛼 + πœ†2 ]} = 𝑔0𝑒 βˆ’πœ†π‘‘ Which is indeed the correct formula for radioactive decay. Example 4.3. Problem to Beams: A beam that is hinged at its ends, π‘₯ = 0 and π‘₯ = 𝐿 carries a uniform loud 𝑀0 per unit length. Find the deflection at any point 𝑃. Solutions: The ordinary differential equation and boundary conditions are: 𝑑4𝑦 𝑑π‘₯4 = 𝑀0 𝐸1 , 0 < π‘₯ < 𝐿 (6) 𝑦(0) = 𝑦′′(0) = 0, 𝑦(𝐿) = 𝑦′′(𝐿) = 0 (7) where 𝐸 is young's modulus, I is the moment of inertia of the cross section about an axis normal to the plane of bending and 𝐸𝐼 is said to be the flexural rigidity of the beam. Some physical quantities associated with the application are: 𝑦′(π‘₯), 𝑀(π‘₯) = 𝐸𝐼𝑦′′(π‘₯) and 𝑆(π‘₯) = 𝑀′(π‘₯)𝐸𝐼𝑦′′(π‘₯) which respectively represent the "Slope", bending moment, and shear at a point 𝑃. Taking complex Sadik transform of both sides of equation (6) , we get, if 𝐅𝑐 (𝑠) = π’π‘Ž 𝑐 {𝑦(π‘₯)}, (𝑖𝑠𝛼 )4𝐅𝑐 (𝑠) βˆ’ 1 𝑠𝛽 [𝑦′′′(0) + 𝑖𝑠𝛼 𝑦′′(0) + (𝑖𝑠𝛼 )2𝑦′(0) + (𝑖𝑠𝛼 )3𝑦(0)] = 𝑀0 𝐸𝐼 ( βˆ’π‘– 𝑠𝛼+𝛽 ) (𝑖𝑠𝛼 )4𝐅𝑐(𝑠) βˆ’ 1 𝑠𝛽 [𝐢2 + (𝑖𝑠 𝛼 )2𝐢1)] = βˆ’π‘€0𝑖 𝐸𝐼𝑠𝛼+𝛽 , 𝑠4𝛼 𝐅𝑐 (𝑠) = βˆ’π‘€0𝑖 𝐸𝐼𝑠𝛼+𝛽 + 1 𝑠𝛽 [𝐢2 βˆ’ 𝑠 2𝛼 𝐢1)] , π’π‘Ž 𝑐 {𝑦(π‘₯)} = 𝐅𝑐 (𝑠) = βˆ’π‘€0𝑖 𝐸𝐼𝑠5𝛼+𝛽 + 𝐢2 𝑠4𝛼+𝛽 βˆ’ 𝐢1 𝑠2𝛼+𝛽 Inverting to find the solution: 𝑦(π‘₯) = 𝐢1π‘₯ + 𝐢2 π‘₯3 3! + 𝑀0 𝐸𝐼 π‘₯4 4! , or IHJPAS. 53 (3)2022 127 𝑦(π‘₯) = 𝐢1π‘₯ + 𝐢2 π‘₯3 6 + 𝑀0 𝐸𝐼 π‘₯4 24 . From the last two conditions in Equation (7) , we find: 𝐢1 = 𝑀0𝐿 3 24𝐸𝐼 , 𝐢2 = 𝑀0𝐿 2𝐸𝐼 . Thus, the required deflection is : 𝑦(π‘₯) = 𝑀0 24𝐸𝐼 π‘₯(𝐿 βˆ’ π‘₯)(𝐿2 βˆ’ 𝐿π‘₯ βˆ’ π‘₯2). It is possible to calculate the bending moment and shear at any point 𝑃 of the beam, and in particular, at the ends. 5. 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