Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 258 https://internationalpubls.com Extorial in RL Circuits and Heat Flows T. Sathinathan1*, S. John Borg2, G. Britto Antony Xavier3 1,2,3Department of Mathematics, Sacred Heart College (Autonomous), Tirupattur 635601, Tamilnadu, India. ∗Corresponding author: sathithoma@gmail.com Article History: Received: 21-09-2024 Revised: 24-11-2024 Accepted: 05-12-2024 Abstract: The newly defined ℓ-extorial function was used in this paper to generate the solution of the RL circuit. When an inductor and a resistor are linked across a voltage source, the resulting circuit is known as an RL circuit. Depending on how the resistor and inductor are connected, this circuit may be in series or parallel. One of the solutions to RL’s mathematical difference problem is the extorial function. To solve the RL circuit, we thus develop the theory of the extorial function and employ it. Keywords: ℓ-extorial function, ℓ-Delta operator, RL circuit, Heat equation. 1. Introduction A difference equation is an equation that contains sequence differences. There are various types of difference equations namely ordinary, delay, advanced, neutral, quasilinear, half linear, etc. These equations occur in numerous settings and forms, both in mathematics itself and its applications to Biology, Computer Science, Digital Signal Processing, Economics, Statistics and other fields. The fractional sum of a function 𝑓 (or 𝜈𝑡ℎ order delta integration) is defined by (Δ𝑎 −𝜈𝑢)(𝜅) = 1 Γ(𝜈) ∑𝜅−𝜈 𝑠=𝑎 Γ(𝜅−𝑠) Γ(𝜅−𝑠−(𝜈−1)) 𝑢(𝑠), (1) where 𝜈 > 0, 𝑓 is defined for 𝑠 = 𝑎 𝑚𝑜𝑑(1) and Δ −𝜈𝑓 is defined for 𝜅 = 𝑎 + 𝜈 𝑚𝑜𝑑(1). The basic theory of difference equations is based on the difference operator Δ defined as Δ𝑢(𝜅) = 𝑢(𝜅 + 1) − 𝑢(𝜅), where {𝑢(𝜅)} is a sequence or a function of 𝜅 of numbers. Many authors ([7],[9]) have suggested the definition of generalized difference operator Δℓ on real valued function u defined on ℝ = (−∞,∞) as Δℓ𝑢(𝜅) = 𝑢(𝜅 + ℓ) − 𝑢(𝜅), 𝜅 ∈ ℝ, ℓ > 0. (2) E. Thandapani, M.Maria Susai Manuel, G.B.A Xavier [8] considered the definition of Δℓ as given in (2) and developed the theory of difference equations in a different direction. If there exists a function v such that Δℓ𝑣(𝜅) = 𝑢(𝜅), then we call this function 𝑣 as Δℓ −1𝑣. Hence, for 𝜅 ∈ ℝ = ∪ 0≤𝑗<ℓ ℕℓ(𝑗), if Δℓ𝑣(𝜅) = 𝑢(𝜅), then 𝑣(𝜅) = Δℓ −1𝑢(𝜅) + 𝑐𝑗 , (3) where 𝑐𝑗 is constant for all 𝜅 in each ℕℓ(𝑗) = {𝑗, 𝑗 + ℓ, 𝑗 + 2ℓ, … }, 𝑗 = 𝜅 − [ 𝜅 ℓ ]ℓ. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 259 https://internationalpubls.com In 1989, Miller and Rose introduced the discrete analogue of the Riemann-Liouville fractional derivative and proved some properties of the fractional difference operator. In 1984, Jerzy Popenda [4] introduced a particular type of difference operator on 𝑢 as Δ𝛼𝑢(𝜅) = 𝑢(𝜅 + 1) − 𝛼𝑢(𝜅), In 2011, M.Maria Susai Manuel, et.al, [5] extended the operator Δ𝛼 to generalized 𝛼 −difference operator as Δ 𝛼(ℓ) 𝑣(𝜅) = 𝑣(𝜅 + ℓ) − 𝛼𝑣(𝜅) for real valued function 𝑣. In 2014, the authors in [1] have applied the q-difference operator defined by Δ𝑞𝑣(𝜅) = 𝑣(𝑞𝜅) − 𝑣(𝜅) and delta operator Δ 𝜅(ℓ) with variable coefficients defined by Δ 𝜅(ℓ) 𝑣(𝜅) = 𝑣(𝜅 + ℓ) − 𝜅𝑣(𝜅), ℓ ≠ 0 ∈ ℝ. Also the generalized difference operator with 𝑛-shift values 𝑙 = (ℓ1, ℓ2, ℓ3, . . . , ℓ𝑛) ≠ 0 on a real valued function 𝑣: ℝ𝑛 → ℝ is defined as Δ (ℓ) 𝑣(𝜅) = 𝑣(𝜅1 + ℓ1, 𝜅2 + ℓ2, . . . , 𝜅𝑛 + ℓ𝑛) − 𝑣(𝜅1, 𝜅2, . . . , 𝜅𝑛). (4) This operator Δ (ℓ) becomes generalized partial difference operator if some ℓ𝑖 = 0. The equations involving Δ (ℓ) with atleast one ℓ𝑖 = 0 is called generalized partial difference equation. for one shift value, we take Δ(ℓ) as Δℓ. By defining the inverse Δℓ −1 , many interesting results on sum of partial sums of higher power of arithmetic and geometric functions and applications in numerical methods (see [8, 6, 2, 3, 10, 11, 12, 13]) are obtained. The difference operator defined in (2) becomes the usual difference operator Δ when ℓ = 1. We obtain several results on factorial function by applying Δℓ −1 . The fractional sum of a function 𝑓 (or 𝜈𝑡ℎ order delta integration) is defined by (Δ𝑎 −𝜈𝑢)(𝜅) = 1 Γ(𝜈) ∑𝜅−𝜈 𝑠=𝑎 Γ(𝜅−𝑠) Γ(𝜅−𝑠−(𝜈−1)) 𝑢(𝑠), (1) where 𝜈 > 0, 𝑓 is defined for 𝑠 = 𝑎 𝑚𝑜𝑑(1) and Δ −𝜈𝑓 is defined for 𝜅 = 𝑎 + 𝜈 𝑚𝑜𝑑(1). The basic theory of difference equations is based on the difference operator Δ defined as Δ𝑢(𝜅) = 𝑢(𝜅 + 1) − 𝑢(𝜅), where {𝑢(𝜅)} is a sequence or a function of 𝜅 of numbers. Many authors ([7],[9]) have suggested the definition of generalized difference operator Δℓ on real valued function u defined on ℝ = (−∞,∞) as Δℓ𝑢(𝜅) = 𝑢(𝜅 + ℓ) − 𝑢(𝜅), 𝜅 ∈ ℝ, ℓ > 0. (2) E. Thandapani, M.Maria Susai Manuel, G.B.A Xavier [8] considered the definition of Δℓ as given in (2) and developed the theory of difference equations in a different direction. If there exists a function v such that Δℓ𝑣(𝜅) = 𝑢(𝜅), then we call this function 𝑣 as Δℓ −1𝑣. Hence, for 𝜅 ∈ ℝ = ∪ 0≤𝑗<ℓ ℕℓ(𝑗), if Δℓ𝑣(𝜅) = 𝑢(𝜅), then 𝑣(𝜅) = Δℓ −1𝑢(𝜅) + 𝑐𝑗 , (3) where 𝑐𝑗 is constant for all 𝜅 in each ℕℓ(𝑗) = {𝑗, 𝑗 + ℓ, 𝑗 + 2ℓ, … }, 𝑗 = 𝜅 − [ 𝜅 ℓ ]ℓ. In 1989, Miller and Rose introduced the discrete analogue of the Riemann-Liouville fractional derivative and proved some properties of the fractional difference operator. In 1984, Jerzy Popenda [4] introduced a particular type of difference operator on 𝑢 as Δ𝛼𝑢(𝜅) = 𝑢(𝜅 + 1) − 𝛼𝑢(𝜅), In 2011, M.Maria Susai Manuel, et.al, [5] extended the operator Δ𝛼 to generalized 𝛼 − difference operator as Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 260 https://internationalpubls.com Δ 𝛼(ℓ) 𝑣(𝜅) = 𝑣(𝜅 + ℓ) − 𝛼𝑣(𝜅) for real valued function 𝑣. In 2014, the authors in [1] have applied the q-difference operator defined by Δ𝑞𝑣(𝜅) = 𝑣(𝑞𝜅) − 𝑣(𝜅) and delta operator Δ 𝜅(ℓ) with variable coefficients defined by Δ 𝜅(ℓ) 𝑣(𝜅) = 𝑣(𝜅 + ℓ) − 𝜅𝑣(𝜅), ℓ ≠ 0 ∈ ℝ. Also the generalized difference operator with 𝑛-shift values 𝑙 = (ℓ1, ℓ2, ℓ3, . . . , ℓ𝑛) ≠ 0 on a real valued function 𝑣: ℝ𝑛 → ℝ is defined as Δ (ℓ) 𝑣(𝜅) = 𝑣(𝜅1 + ℓ1, 𝜅2 + ℓ2, . . . , 𝜅𝑛 + ℓ𝑛) − 𝑣(𝜅1, 𝜅2, . . . , 𝜅𝑛). (4) This operator Δ (ℓ) becomes generalized partial difference operator if some ℓ𝑖 = 0. The equations involving Δ (ℓ) with atleast one ℓ𝑖 = 0 is called generalized partial difference equation. for one shift value, we take Δ(ℓ) as Δℓ. By defining the inverse Δℓ −1 , many interesting results on sum of partial sums of higher power of arithmetic and geometric functions and applications in numerical methods (see [8, 6, 2, 3, 10, 11, 12, 13]) are obtained. The difference operator defined in (2) becomes the usual difference operator Δ when ℓ = 1. We obtain several results on factorial function by applying Δℓ −1 . 2. The 𝓵 - Extorial function and its Properties The ℓ-Extorial function is arrived by replacing the polynomial 𝜅𝑛 by polynomial factorial function 𝜅ℓ (𝑛) in the exponential function 𝑒𝜅. The formal definition of extorial function is given below. Definition 2.1. The ℓ-extorial function denoted as 𝑒(𝜅ℓ (𝑛) ) is defined as 𝑒(𝜅ℓ (𝑛) ) = 1 + 𝜅ℓ (𝑛) 1! + 𝜅ℓ (2𝑛) 2! + 𝜅ℓ (3𝑛) 3! + ⋯ + ∞, (5) where |ℓ| ≤ 1 and 𝑛, 𝜅 ∈ ℝ. Definition 2.2. For ℓ ∈ (−1,1) and 𝜅 ∈ ℝ, the 𝑛𝑡ℎ order ℓ-extorial function denoted as 𝑒𝑛(𝜅ℓ) is defined as 𝑒𝑛(𝜅ℓ) = 1 + 𝜅ℓ (𝑛) 𝑛! + 𝜅ℓ (2𝑛) (2𝑛)! + 𝜅ℓ (3𝑛) (3𝑛)! + ⋯ + ∞. (6) From the definition of extorial function, we obtain following lemma. Lemma 2.3. For any real 𝜅 and ℓ, 𝑛 ∈ ℕ, we have (i) 𝑒𝑛(−𝜅ℓ) = 𝑒𝑛(𝜅−ℓ) if n 𝑖𝑠 𝑒𝑣𝑒𝑛 & 1 − 𝜅(−ℓ) (𝑛) 𝑛! + 𝜅(−ℓ) (2𝑛) (2𝑛)! − 𝜅(−ℓ) (3𝑛) (3𝑛)! + ⋯ if n 𝑖𝑠 𝑜𝑑𝑑 and (ii) 𝑒𝑛(−𝜅(−ℓ)) = 𝑒𝑛(𝜅ℓ) if n 𝑖𝑠 𝑒𝑣𝑒𝑛 & 1 − (𝜅)ℓ (𝑛) 𝑛! + (𝜅)ℓ (2𝑛) 2𝑛! − (𝜅)ℓ (3𝑛) 3𝑛! + ⋯ if n 𝑖𝑠 𝑜𝑑𝑑 Lemma 2.4. Let 𝜅 ∈ ℝ and 𝑛, ℓ ∈ ℕ. Then, we have Δℓ𝑒𝑛(𝜅ℓ) = ℓ∑∞ 𝑚=1 𝜅ℓ (𝑚𝑛−1) (𝑚𝑛−1)! , 𝑛𝑚 ≠ 1. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 261 https://internationalpubls.com Proof. We shall prove this by induction method 𝑒2(𝜅ℓ) = 1 + 𝜅ℓ (2) 2! + 𝜅ℓ (4) 4! + 𝜅ℓ (6) 6! + ⋯ + ∞ Δℓ𝑒2(𝜅ℓ) = Δℓ 𝜅ℓ (2) 2! + Δℓ 𝜅ℓ (4) 4! + Δℓ 𝜅ℓ (6) 6! + ⋯ + ∞ = ℓ [ 𝜅ℓ (1) 1! + 𝜅ℓ (3) 3! + 𝜅ℓ (5) 5! + ⋯ ] 𝑒3(𝜅ℓ) = 1 + 𝜅ℓ (3) 3! + 𝜅ℓ (6) 6! + 𝜅ℓ (9) 9! + ⋯ + ∞ Δℓ𝑒3(𝜅ℓ) = Δℓ 𝜅ℓ (3) 3! + Δℓ 𝜅ℓ (6) 6! + Δℓ 𝜅ℓ (9) 9! + ⋯ + ∞ = ℓ [ 𝜅ℓ (2) 2! + 𝜅ℓ (5) 5! + 𝜅ℓ (8) 8! + ⋯ ] In general, we find for 𝑛 ≥ 1 Δℓ𝑒𝑛(𝜅ℓ) = ℓ [ 𝜅ℓ (𝑛−1) (𝑛−1)! + 𝜅ℓ (2𝑛−1) (2𝑛−1)! + 𝜅ℓ (3𝑛−1) (3𝑛−1)! + ⋯ ] = ℓ∑∞ 𝑚=1 𝜅ℓ (𝑚𝑛−1) (𝑚𝑛−1)! Lemma 2.5. For any positive integer 𝑚, we have Δℓ 𝑚𝑒𝑚(𝜅ℓ) = ℓ 𝑚𝑒𝑚(𝜅ℓ). Proof. Δℓ𝑒1(𝜅ℓ) = 0 + Δℓ 𝜅ℓ (1) 1! + Δℓ 𝜅ℓ (2) 2! + Δℓ 𝜅ℓ (3) 3! + ⋯ = ℓ𝑒1(𝜅ℓ). Δℓ𝑒2(𝜅ℓ) = 0 + Δℓ 𝜅ℓ (2) 2! + Δℓ 𝜅ℓ (4) 4! + Δℓ 𝜅ℓ (6) 6! + ⋯ = 2ℓ𝜅ℓ(1) 2! + 4ℓ𝜅ℓ(3) 4! + 6ℓ𝜅ℓ(5) 6! + ⋯ Δℓ 2𝑒2(𝜅ℓ) = 2ℓ(ℓ𝜅ℓ (0) ) 2! + 4ℓ(3ℓ𝜅ℓ (2) ) 4! + 6ℓ(5ℓ𝜅ℓ (4) ) 6! + ⋯ = ℓ 2𝑒2(𝜅ℓ), which yields Δℓ 𝑚𝑒𝑚(𝜅ℓ) = ℓ 𝑚𝑒𝑚(𝜅ℓ). Lemma 2.6. For positive 𝑚 and real 𝜅, we have Δℓ (−𝑚) 𝑒𝑚(𝜅ℓ) = 𝑒𝑚(𝜅ℓ) ℓ𝑚 , ℓ ∈ ℕ. Proof. From the lemma 5, we find Δℓ 𝑚𝑒𝑚(𝜅ℓ) = ℓ 𝑚𝑒𝑚(𝜅ℓ). Taking Δℓ −𝑚 on both sides, we get Δℓ −𝑚(Δℓ 𝑚𝑒𝑚(𝜅ℓ)) = Δℓ −𝑚(ℓ𝑚𝑒𝑚(𝜅ℓ)), which gives Δℓ (−𝑚) 𝑒𝑚(𝜅ℓ) = 𝑒𝑚(𝜅ℓ) ℓ𝑚 . Definition 2.7. For |ℓ| < 1, and 𝑛 ∈ ℕ, 𝑒(−𝑛)(𝑘ℓ) is defined as 𝑒(−𝑛)(𝜅ℓ) = 1 + 1 𝑛! 1 𝜅 ℓ (𝑛) + 1 (2𝑛)! 1 𝜅 ℓ (2𝑛) + 1 (3𝑛)! 1 𝜅 ℓ (3𝑛) + ⋯ + ∞. (7) Lemma 2.8. For ℓ ∈ (−1,1) and positive 𝜅, we have Δℓ𝑒(−𝑛)(𝜅ℓ) = −ℓ[ 1 (𝑛−1)! 1 (𝜅+ℓ) ℓ (𝑛+1) + 1 (2𝑛−1)! 1 (𝜅+ℓ) ℓ (2𝑛+1) + 1 (3𝑛−1)! 1 (𝜅+ℓ) ℓ (3𝑛+1) + ⋯ ] Proof. Putting 𝑛 = 1 in (7), we get Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 262 https://internationalpubls.com 𝑒(−1)(𝜅ℓ) = 1 + 1 1! 1 𝜅 ℓ (1) + 1 2! 1 𝜅 ℓ (2) + 1 3! 1 𝜅 ℓ (3) + ⋯ + ∞Δℓ𝑒(−1)(𝜅ℓ) = 1 + Δℓ 1 1! 1 𝜅 ℓ (1) + Δℓ 1 2! 1 𝜅 ℓ (2) + Δℓ 1 3! 1 𝜅 ℓ (3) + ⋯ + ∞ = −ℓ [ 1 (𝜅 + ℓ) ℓ (2) + 1 1! 1 (𝜅 + ℓ) ℓ (3) + 1 2! 1 (𝜅 + ℓ) ℓ (4) + ⋯ ]. Putting 𝑛 = 2 in (7), we get 𝑒(−2)(𝜅ℓ) = 1 + 1 2! 1 𝜅 ℓ (2) + 1 4! 1 𝜅 ℓ (4) + 1 6! 1 𝜅 ℓ (6) + ⋯ + ∞Δℓ𝑒(−2)(𝜅ℓ) = 1 + Δℓ 1 2! 1 𝜅 ℓ (2) + Δℓ 1 4! 1 𝜅 ℓ (4) + Δℓ 1 6! 1 𝜅 ℓ (6) + ⋯ + ∞ = −ℓ [ 1 1! 1 (𝜅 + ℓ) ℓ (3) + 1 3! 1 (𝜅 + ℓ) ℓ (5) + 1 5! 1 (𝜅 + ℓ) ℓ (7) + ⋯ ]. Putting 𝑛 = 3 in (7), we get 𝑒(−3)(𝜅ℓ) = 1 + 1 3! 1 𝜅 ℓ (3) + 1 6! 1 𝜅 ℓ (6) + 1 9! 1 𝜅 ℓ (9) + ⋯ + ∞Δℓ𝑒(−3)(𝜅ℓ) = 1 + Δℓ 1 3! 1 𝜅 ℓ (3) + Δℓ 1 6! 1 𝜅 ℓ (6) + Δℓ 1 9! 1 𝜅 ℓ (9) + ⋯ + ∞ = −ℓ [ 1 2! 1 (𝜅 + ℓ) ℓ (4) + 1 5! 1 (𝜅 + ℓ) ℓ (7) + 1 8! 1 (𝜅 + ℓ) ℓ (10) + ⋯ ]. In general, Δℓ𝑒(−𝑛)(𝜅ℓ) = −ℓ[ 1 (𝑛−1)! 1 (𝜅+ℓ) ℓ (𝑛+1) + 1 (2𝑛−1)! 1 (𝜅+ℓ) ℓ (2𝑛+1) + 1 (3𝑛−1)! 1 (𝜅+ℓ) ℓ (3𝑛+1) + ⋯ ]. 3. Current Flows in RL Circuit Consider a RL circuit by using the Kirchhoff’s circuit rule. The differential equation connecting voltage V, resistance R, current I and induction L in series is given by first order linear difference equation 𝑉 = 𝑅𝐼(𝜅) + 𝐿 𝑑𝐼(𝜅) 𝑑𝜅 . (8) The discrete analogue of (8) is assumed by replacing 𝑑𝐼(𝜅) = Δ𝐼(𝜅), where Δ𝐼(𝜅) = 𝐼(𝜅 + 1) − 𝐼(𝜅) and 𝑑𝜅 = 1 in (8). The corresponding difference equation for the current flows in RL series circuit in the discrete case takes the form, at time 𝜅 𝑣(𝜅) = 𝑅𝐼(𝜅) + 𝐿Δ𝐼(𝜅). (9) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 263 https://internationalpubls.com Due to the resistance of conductor, heat temperature may be raised in the RL circuit. In that case, we need to modify the difference equation (9). In that case the difference equation (9) becomes fractional difference equation as 𝑉 = 𝑅𝐼(𝜅) + 𝐿Δ 𝜈𝐼(𝜅), (0 < 𝜈 < 1). (10) Which can be expressed as 𝑉(𝜅)−𝑅𝐼(𝜅) 𝐿 = Δ 𝜈𝐼(𝜅) The corresponding discrete integral equations Δ −𝜈 ( 𝑉(𝜅)−𝑅𝐼(𝜅) 𝐿 ) = 𝐼(𝜅) By applying 𝛾𝑡ℎ order delta sum given by (1), 1 𝐿Γ(𝜈) ∑𝜅−𝜈 𝑠=0 Γ(𝜅−𝑠) Γ(𝜅−𝑠−(𝜈−1)) (𝑉(𝑠) − 𝑅𝐼(𝑠)) = 𝐼(𝜅), (11) The corresponding fractional difference equation for de-energizing in RL circuit is obtained by putting 𝑣(𝜅) = 0. In this case, we get 0 = 𝑅𝐼(𝜅) + 𝐿Δ 𝜈𝐼(𝜅). (12) Which is the same as Δ 𝜈𝐼(𝜅) = − 𝑅𝐼(𝜅) 𝐿 . 𝐼(𝜅) = Δ −𝜈 (− 𝑅𝐼(𝜅) 𝐿 ) = − 𝑅 𝐿 Δ −𝜈𝐼(𝜅) By applying fractional order delta integration (1), we obtain 𝐼(𝜅) = 𝑅 𝐿Γ(𝜈) ∑𝜅−𝜈 𝑠=0 Γ(𝜅−𝑠) Γ(𝜅−𝑠−(𝜈−1)) 𝐼(𝑠), (13) The solution (11) and (13) are summation forms. Through our research, we identifies that these fractional difference equations have exact type solutions, when the initial time a is taken as zero. We obtain exact solution for the equations (9) and (10) using our newly defined extorial functions. 4. Extorial Type Solution of RL Circuit In this section, we find solution of equation (9) after arriving at some basic results of extorial functions. This extorial function is easily obtained by replacing polynomial 𝜅𝑛 into factorial polynomial in the expansion of exponential function 𝑒𝜅. This function is useful to arrive at solutions for fractional difference equation. Consider the extorial function 𝑒1((𝑚𝜅)(𝑚)) is defined by 𝑒1((𝑚𝜅)(𝑚)) = 1 + (𝑚𝜅)𝑚 (1) 1! + (𝑚𝜅)𝑚 (2) 2! + ⋯ + ∞ = ∑∞ 𝑟=0 (𝑚𝜅)𝑚 (𝑟) 𝑟! , (14) where (𝑚𝜅)𝑚 (𝑟) = (𝑚𝜅)(𝑚𝜅 − 𝑚) ⋯ (𝑚𝜅 − (𝑟 − 1)𝑚) for positive integer 𝑟, is a falling polynomial factorial. In general, for real index 𝜈, we have Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 264 https://internationalpubls.com 𝑒(𝜈)((𝑚𝜅)(𝑚)) = 1 + (𝑚𝜅)𝑚 (𝜈) 1! + (𝑚𝜅)𝑚 (2𝜈) 2! + ⋯ + ∞ = ∑∞ 𝑟=0 (𝑚𝜅)𝑚 (𝑟𝜈) 𝑟! , (15) where (𝑚𝜅)𝑚 (𝑟𝜈) = (𝑚)(𝑟𝜈) Γ(𝜅+1) Γ(𝐾+1−𝑟𝜈) and Γ(. ) is the gamma function. Lemma 4.1. If 𝑒1((𝑚𝜅)(𝑚)) is an extorial function, then we have Δ𝑒1((𝑚𝜅)(𝑚)) = (𝑚)𝑒1((𝑚𝜅)(𝑚)). (16) Proof. Applying Δ on the extorial function 𝑒1((𝑚𝜅)(𝑚)), we arrive at Δ𝑒1((𝑚𝜅)(𝑚)) = Δ(1) + Δ (𝑚𝜅)𝑚 (1) 1! + Δ (𝑚𝜅)𝑚 (2) 2! + ⋯ + ∞ = 0 + (𝑚(𝜅+1))𝑚 (1) 1! + (𝑚𝜅)𝑚 (1) 1! + 1 2! [(𝑚(𝜅 + 1))𝑚 (2) − (𝑚𝜅)𝑚 (2) ] + ⋯ + ∞ = (𝑚) 1! + 1 2! [(𝑚𝜅 + 𝑚)(𝑚𝜅) − (𝑚𝜅)(𝑚𝜅 − 𝑚)] + ⋯ + ∞ = 𝑚 + 2𝑚 2! (𝑚𝜅)𝑚 (1) + 3𝑚 3! (𝑚𝜅)𝑚 (2) + ⋯ + ∞ = 𝑚[1 + (𝑚𝜅)𝑚 (𝜅) 1! + (𝑚𝜅)𝑚 (2) 2! + ⋯ + ∞] Δ𝑒(𝑚𝜅)𝑚 = (𝑚)𝑒1((𝑚𝜅)(𝑚)). Lemma 4.2. The extorial function 𝑢(𝜅) = 𝑒1((𝑚𝜅)(𝑚)) is a solution of equation (𝐴Δ 2 + 𝐵Δ + 𝐶)𝑢(𝜅) = 0, (17) if 𝑚 is a root of the auxiliary equation 𝐴𝑚2 + 𝐵𝑚 + 𝐶 = 0. Proof. If we try 𝑢(𝜅) = 𝑒1((𝑚𝜅)(𝑚)) as a solution of equation(17), then it should satisfy the equation 𝐴Δ 2𝑒1((𝑚𝜅)(𝑚)) + 𝐵Δ𝑒1((𝑚𝜅)(𝑚)) + 𝐶𝑒1((𝑚𝜅)(𝑚)) = 0. (18) By linear property of Δ and the expansion of 𝑒1((𝑚𝜅)(𝑚)), we arrive at Δℓ𝑒1((𝑚𝜅)(𝑚)) = 0 + (𝑚) (𝑚𝜅)(𝑚) (0) 1! + 2𝑚(𝑚𝜅)(𝑚) (1) 2! + 3𝑚(𝑚𝜅)(𝑚) (2) 3! +. .. i.e, Δℓ𝑒1((𝑚𝜅)(𝑚)) = 𝑚[1 + (𝑚𝜅)(𝑚) (1) 1! + (𝑚𝜅)(𝑚) (2) 2! +. . . ] = 𝑚𝑒1((𝑚𝜅)(𝑚)), which yields Δℓ 2𝑒1((𝑚𝜅)(𝑚)) = (𝑚)Δℓ𝑒1((𝑚𝜅)(𝑚)) = (𝑚)2𝑒1((𝑚𝜅)(𝑚)). Applying the values of Δ𝑒1((𝑚𝜅)(𝑚)) and Δℓ 2𝑒1((𝑚𝜅)(𝑚)) in (18), we obtain (𝐴𝑚2 + 𝐵𝑚 + 𝐶)𝑒1((𝑚𝜅)(𝑚)) = 0, Since 𝑒1((𝑚𝜅)𝑚) ≠ 0, we get 𝐴𝑚2 + 𝐵𝑚 + 𝐶 = 0. (19) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 265 https://internationalpubls.com Hence 𝑢(𝜅) = 𝑒1((𝑚𝜅)(𝑚)) is a solution of (17) when 𝑚 is a root of (18). Remark 4.3. The above lemma can be extended to higher order linear difference equation with constant coefficients. Theorem 4.4. Let 𝐼0 be initial value of 𝐼(𝜅) and 𝜈 = 1. The de-energizing difference equation (12) for 𝜈 = 1 has a solution of the form 𝐼(𝜅) = 𝐼0𝑒1 (( −𝑅 𝐿 𝜅) ( −𝑅 𝐿 ) ) (20) where 𝑒1 denotes the extorial function. Proof. Consider the first order difference equation 𝐿Δ𝐼(𝜅) + 𝑅𝐼(𝜅) = 0 which is obtained from (12) by taking 𝜈 = 1 and its Auxillary equation ML+R=0. The auxiliary equation 𝑚𝐿 + 𝑅 = 0 has an unique solution 𝑚 = −𝑅 𝐿 , 𝐿 ≠ 0. Applying Lemma 4 for first order difference equation by taking 𝐴 = 0, 𝐼(𝑡) = 𝐼0𝑒1(( −𝑅 𝐿 𝜅) ( −𝑅 𝐿 ) ), which is a solution of the equation (12) for 𝜈 = 1. Theorem 4.5. For 𝜈 = 1, the energizing difference equation (10) has a solution 𝐼(𝜅) = 𝑉 𝐿(𝑒𝑠−1)+𝑅 + 𝐼0𝑒1 (( −𝑅 𝐿 𝜅) ( −𝑅 𝐿 ) ), (21) where s is a constant. Proof. Let 𝐼(𝜅) = 𝑉 𝑐 𝑒𝑠𝜅 be a solution of equation (10) for 𝜈 = 1, where c is to be determined. Since 𝑠 is a constant, we get Δ𝑒𝑠𝜅 = 𝑒𝑠(𝜅+1) − 𝑒𝑠𝜅 = (𝑒𝑠 − 1)𝑒𝑠𝜅 and Δ𝐼(𝜅) = 𝑉 𝑐 Δ𝑒𝑠𝜅 = 𝑉 𝑐 𝑒𝑠𝜅(𝑒𝑠 − 1). Substituting 𝜈 = 1, 𝐼(𝜅) and Δ𝐼(𝜅) in (10), we arrive 𝐼(𝜅)𝑅 + 𝐿Δ𝐼(𝜅) = 𝑅 𝑉 𝑐 𝑒𝑠𝜅 + 𝐿[ 𝑉 𝑐 𝑒𝑠𝜅(𝑒𝑠 − 1)], which yields [𝑅 + 𝐿Δ]𝐼 = 𝑉 𝑐 [𝐿(𝑒𝑠 − 1 + 𝑅)]𝑒𝑠𝜅. Hence, taking 𝑐 = 𝐿(𝑒𝑠 − 1 + 𝑅), we find a particular solution of equation (10) when 𝜈 = 1, as 𝐼(𝜅) = 𝑉 𝐿(𝑒𝑠−1+𝑅) 𝑒𝑠𝜅 . (22) Now (21) follows by adding (20) and (22) the proof is complete. Corollary 4.6. If 𝐼0 = −𝑉 𝑅 , then the extorial solution of difference equation (9) of the RL circuit is 𝐼(𝜅) = 𝑉 𝑅 − 𝑉 𝑅 𝑒1 (( −𝑅 𝐿 𝜅) ( −𝑅 𝐿 ) ). Proof. The proof follows by taking 𝑠 = 0 in (21). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 266 https://internationalpubls.com Finding the solutions of integer order difference equation is comparatively easier than the fractional order difference equation. 5. Extorial Energizing for RL Circuit In this section, we derive at the solution of RL circuit model with extorial energizing. Here, we deal with fractional order difference equation also. Theorem 5.1. The flow of current in the RL circuit creates chaos due to increase of temperature of heat. In this case, the difference equation of RL circuit becomes 𝑉𝑒1((𝑠𝜅)𝑠) = 𝑅𝐼(𝜅) + 𝐿Δ 𝜈𝐼(𝜅), (0 < 𝜈 < 1). (23) Equation (23) is 𝜈𝑡ℎ order fractional difference equation. When there is no choas in RL circuit, the parameter 𝜈 takes integer value. Theorem 5.2. For 𝜈 = 1, the difference equation (23) has a extorial solution 𝐼(𝜅) = 𝑉𝑒1((𝑠𝜅)𝑠ℓ) 𝐿(𝑒𝑠−1)+𝑅 + 𝐼0𝑒1(( −𝑅 𝐿 𝜅) ( −𝑅 𝐿 ) ). (24) Proof. Let 𝐼(𝜅) = 𝑉 𝑐 𝑒1((𝑠𝜅)𝑠) be a solution of equation (23) (𝜈 = 1), where 𝑐 is to be determined. Since 𝑠 is a constant, from (24), we get Δ𝑒1((𝑠𝜅)𝑠) = 𝑒1((𝑠(𝜅 + 1))(𝑠)) − 𝑒1((𝑠𝜅)𝑠). This gives Δ𝐼(𝜅) = 𝑉 𝑐 Δ𝑒1((𝑠𝜅)𝑠) = 𝑉 𝑐 𝑒1((𝑠𝜅)(𝑠))(𝑒1((1)(𝑠)) − 1). Substituting 𝐼(𝜅) and Δ𝐼(𝜅) in the above equation, we arrive 𝐼(𝜅)𝑅 + 𝐿Δ𝐼(𝜅) = 𝑅 𝑉 𝑐 𝑒1((𝑠𝜅)𝑠) + 𝐿[ 𝑉 𝑐 𝑒(𝑠𝜅)(𝑒1(1(𝑠)) − 1)], which yields [𝑅 + 𝐿Δ]𝐼 = 𝑉 𝑐 [𝐿(𝑒1(ℓ(𝑠)) − 1 + 𝑅)]𝑒1((𝑠𝜅)𝑠). Hence taking 𝑐 = 𝐿(𝑒1(1(𝑠)) − 1 + 𝑅), we find 𝐼(𝜅) = 𝑉 𝐿(𝑒1(1(𝑠))−1+𝑅) 𝑒𝑠𝜅 is a particular solution of equation when 𝜈 = 1 and (24) follows. Theorem 5.3. For For 0 < 𝜈 < 1, the energizing fractional difference equation 𝑉𝑒1((𝑠𝜅)(𝑠)) = 𝐼(𝜅)𝑅 + Δ 𝜈𝐼(𝜅), (25) has an extorial solution of the form 𝑉𝑒1((𝑠𝜅)(𝑠)) 𝐿(𝑒1(ℓ(𝑠))−1)𝜈+𝑅 + 𝐼0𝑒1(( −𝑅 𝐿 ) 1 𝜈𝜅 ( −𝑅 𝐿 ) 1 𝜈 ). (26) Proof. We try 𝐼(𝜅) = 𝑉𝑐𝑒1((𝑠𝜅)(𝑠)) as a solution of equation (25), where c is to be determined. Δ𝐼(𝜅) = 𝑉𝑐(𝑒1(1(𝑠)) − 1)𝑒1((𝑠𝜅)(𝑠)), Δ 2𝐼(𝜅) = 𝑉𝑐(𝑒1(1(𝑠)) − 1)2𝑒1((𝑠𝜅)(𝑠)) ⋯, Δ 𝜈𝐼(𝜅) = 𝑉𝑐(𝑒1(1(𝑠)) − 1)𝜈𝑒1((𝑠𝜅)(𝑠)) is obtained from Δ𝐼(𝜅) = 𝐼(𝜅 + 1) − 𝐼(𝜅). Substituting 𝐼 and Δ 𝜈𝐼 in (25), we find Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 267 https://internationalpubls.com 𝐼(𝜅)𝑅 + 𝐿 ℓ Δ 𝜈𝐼(𝜅) = 𝑅𝑉𝑐𝑒1((𝑠𝜅)(𝑠)) + 𝐿[𝑉𝑐(𝑒1(1(𝑠)) − 1)𝜈𝑒1((𝑠𝜅)(𝑠))] = 𝑉𝑐[ 𝐿 ℓ (𝑒1(1(𝑠)) − 1)𝜈 + 𝑅]𝑒1((𝑠𝜅)(𝑠)). Hence 𝐼(𝜅) = 𝑉 𝐿(𝑒1(1(𝑠))−1)𝜈+𝑅 𝑒1((𝑠𝜅)(𝑠)) is a particular solution of (26). Thus extorial function is used to obtain the solution to RL-circuit difference equation. Also we have obtained solution of RL circuit of chaos situation represented by fractional order difference equation. 6. Fractional Difference Heat Equation Model In this section, we apply the alpha and Fibonacci difference operators and obtain new model of heat equations. The solution of these equations are expressed in terms of extorial functions. The materials up to three dimensions i.e., rod, thin plate and medium are taken for study and the transfer of heat is examined. The two operators (alpha and Fibonacci) are used for the study of transfer of heat and are defined accordingly. Let 𝛼 ≠ 0, 𝑙 = (1,1,1, . . . ,1), 𝜅 = (𝜅1, 𝜅2, ⋯ , 𝜅𝑛) ∈ ℝ𝑛 and 𝑣(𝜅) be a real valued n-variable function defined on ℝ𝑛. The n-variable 𝛼-difference operator, denoted as Δ𝛼, on 𝑣(𝜅) is defined by Δ 𝛼 𝑣(𝜅) = 𝑣(𝜅1 + 1, 𝜅2 + 1, . . . , 𝜅𝑛 + 1) − 𝛼𝑣(𝜅1, 𝜅2, . . . , 𝜅𝑛). (27) This operator becomes partial 𝛼-difference operator if we replace by 𝜅𝑖 + 1 in centain component i. Thus the above definition of the alpha and Fibonacci difference operators and its equations are employed in the forthcoming sections and solutions are derived for heat equations. Also we present solutions of partial fractional alpha difference equation with polynomial factorial and extorial functions. We also apply these type of solutions to heat flows. In the following lemma, some identities related to alpha difference operator on extorial function are given. Lemma 6.1. Let 𝜅(𝑟𝑛) ≠ 0, 𝑛 ∈ 𝑁. Then we have the following identities with extorial function: (i). Δ𝛼𝑒1(𝜅) = 𝑒1(𝜅)[1 + 1 − 𝛼], (ii). Δ𝛼𝑒(−1)(𝜅) = 𝑒(−1)(𝜅)[𝑒(−1) − 𝛼](−1), (iii). Δ𝛼𝑒1((−𝜅)) = 𝑒1((−𝜅))[1 + 1 − 𝛼], 𝜅 > 0. Proof. (i). By (27), and applying Δ𝛼 on 𝑒1(𝜅), we arrive Δ𝛼𝑒1(𝜅) = 𝑒1((𝜅 + 1)) − 𝛼𝑒1(𝜅) = 𝑒1(𝜅). 𝑒1 − 𝛼𝑒1(𝜅) = 𝑒1(𝜅)[𝑒1(1) − 𝛼] = 𝑒1(𝜅)[1 + 1 1! + 1 2! + ⋯ − 𝛼] = 𝑒1(𝜅)[1 + 1 − 𝛼]. (ii). By (27), and applying Δ𝛼 on 𝑒(𝜅(−1)), we arrive Δ𝛼𝑒(−1)(𝜅) = 𝑒(−1)((𝜅 + 1)) − 𝛼𝑒(−1)(𝜅) = 𝑒(−1)(𝜅). 𝑒(−1) − 𝛼𝑒(−1)(𝜅). = 𝑒(−1)(𝜅)[𝑒(−1) − 𝛼](−1). (iii). follows from (ii) by replacing 𝜅 as −𝜅. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 268 https://internationalpubls.com Theorem 6.2. If 𝑣(𝜅1, 𝜅2) = 𝑒1((𝜅1)). 𝑒1((𝜅2)) then we have the identities: (𝑖)Δ𝛼𝑣(𝜅1, 𝜅2) = 𝑒1((𝜅1)). 𝑒1((𝜅2))[𝑒1(1) − 𝛼], (𝑖𝑖)Δ𝛼𝑣(𝜅1, 𝜅2) = 𝑒1((𝜅2)). 𝑒1((𝜅1))[𝑒1((1)) − 𝛼]. Proof. (𝑖)Δ𝛼𝑣(𝜅1, 𝜅2) = 𝑒1((𝜅1))[Δ𝛼𝑒1((𝜅2))] = 𝑒1((𝜅1))[𝑒1((𝜅2 + 1)) − 𝛼𝑒1((𝜅2))] = 𝑒1((𝜅1))𝑒1((𝜅2))[𝑒1(1) − 𝛼]. In the similar way, the proof of (ii) follows. Assume that 𝑣(𝜅1, 𝜅2) be the temperature of a rod at position 𝜅1 at time 𝜅2, ℓ1 and ℓ2 be shift values of 𝜅1 and 𝜅2 respectively and 𝛾 be the rate of conductivity of rod. When considering impact of external climate change on the rod, the partial 𝛼 - difference equation of heat flow in the rod becomes fractional 𝛼- difference equation Δ 𝜈 𝛼 𝑣(𝜅1, 𝜅2) = 𝛾[Δ𝜈 𝛼 𝑣(𝜅1, 𝜅2) + Δ 𝜈 𝛼 𝑣(𝜅1, 𝜅2)]. (28) Theorem 6.3. If 𝛾 = [𝑒1(1) − 𝛼/𝑒1(±(1)) − 𝛼], then the function 𝑣(𝜅1, 𝜅2) = 𝑒1((𝜅1)). 𝑒1((𝜅2)) is the exact solution of the 𝛼- difference equation (28). Proof. By applying the Theorem 6, we get the proof. Corollary 6.4. The fractional partial 𝛼-difference heat equation (28) has a solution of the form 𝑣(𝜅1, 𝜅2) = 𝑒1((𝜅1)). 𝑒1((𝜅2)) if 𝛾 = [(𝑒1(1) − 𝛼)𝜈/(𝑒1(±(1)) − 𝛼)𝜈]. Assume that 𝑣(𝜅1, 𝜅2, 𝜅3) be the temperature of a thin plate at position (𝜅1, 𝜅2) at time 𝜅3. Let (1,1,1) be the shift values of (𝜅1, 𝜅2) and 𝜅3 and 𝛾 be the rate of conductivity of thin plate. The fractional partial 𝛼-difference heat equation of thin plate is given by Δ 𝜈 𝛼 𝑣(𝜅1, 𝜅2, 𝜅3) = 𝛾 {Δ 𝜈 𝛼 𝑣(𝜅1, 𝜅2, 𝜅3) + Δ 𝜈 𝛼 𝑣(𝜅1, 𝜅2, 𝜅3)}. (29) Corollary 6.5. If 𝛾 = [1 + 1 − 𝛼)𝜈/(𝑒1(±(1)1) − 𝛼)𝜈], then the function𝑣(𝜅) = ∏3 𝑖=1 𝑒1(𝜅𝑖(1𝑖) ) is an exact solution of the fractional partial heat equation (29). Assume that 𝑣(𝜅1, 𝜅2, 𝜅3, 𝜅4, 𝜅5) be the temperature of a medium at position (𝜅1, 𝜅2, 𝜅3) at time 𝜅4 and at density 𝜅5. Let (1,1,1,1,1) be the shift values of (𝜅1, 𝜅2, 𝜅3), 𝜅4 and 𝜅5 and 𝛾 be the rate of conductivity of medium. The fractional partial 𝛼-difference equation of heat flow in medium is Δ 𝜈 𝛼 𝑣(𝜅1, 𝜅2, 𝜅3, 𝜅4, 𝜅5) = 𝛾 { Δ 𝜈 𝛼(±1) 𝑣(𝜅1, 𝜅2, 𝜅3, 𝜅4, 𝜅5)}. (30) Corollary 6.6. If 𝛾 = [1 + 1 − 𝛼)𝜈 + 𝑒1(1 + 1 − 𝛼)𝜈/(𝑒1(±(1)1) − 𝛼)𝜈], then 𝑣(𝜅) = ∏5 𝑖=1 𝑒1(𝜅𝑖(1𝑖) ) is a closed form solution of the fractional partial 𝛼-difference equation (30). 7. Conclusion In conclusion, the resistor and inductor, as fundamental linear and passive circuit elements, form the basis of RL circuits, which can be configured in series or parallel. The mathematical analysis of RL Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 4s (2025) 269 https://internationalpubls.com circuits involves differential equations, with solutions often expressed in terms of extorial functions. By developing the theory of extorial functions, we successfully applied it to derive solutions for RL circuits and extended its utility to solving problems in wave motion, demonstrating its versatility and practical significance. References [1] Britto Antony Xavier.G, Gerly.T.G and Nasira Begum.H, Finite Series of Polynomials and Polynomial Factorials arising from Generalized q-Difference operator, Far East Journal of Mathematical Sciences,94(1)(2014), 47-63. [2] Britto Antony Xavier.G, John Borg. S, Meganathan. M, Discrete heat equation model with shift values, Applied Mathematics, 2017, 8, 1343-1350. [3] Weisstein, Eric. Harmonic Series MathWorld. Web. 24 August 2014. 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