Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 8s (2025) 829 https://internationalpubls.com Fractional Difference Equations with Initial Time Difference D. Naga Purnima 1*, G.V.S.R. Deekshitulu 2 , G. V. Ramana3 and M. Bala Prabhakar4 1Assistant Professor, Department of Mathematics, Aditya University, Surampalem-533 437, A. P., India. Mail: nagapurnimad@adityauniversity.in Professor, Department of Mathematics, JNTUK Kakinada, Surampalem-533 437, A. P., India. Mail: dixitgvsr@gmail.com 3Associate Professor, Department of Mathematics, Aditya University, Surampalem-533 437, A. P., India. Mail: ramanaginjala9@gmail.com 4Associate Professor, Department of Mathematics, Aditya University, Surampalem-533 437, A. P., India. Mail: prabhakar_mb@yahoo.co.in * Corresponding Author : D. Naga Purnima Article History: Received: 09-11-2024 Revised:23-12-2024 Accepted:08-01-2025 Abstract: In this paper, we consider non-linear fractional difference equations at different initial times and establish the existence of solutions using monotone iterative technique. Keywords: fractional order, monotone iterative technique 1. Introduction. Fractional calculus gained importance during the past three decades due to its applicability in diverse fields of science and engineering. The notions of fractional calculus may be traced back to the works of Euler, but the idea of fractional difference is very recent. G.V.S.R. Deekshitulu and J. Jagan Mohan [7] modified the definition of fractional difference given by Nagai [9] and discussed some basic inequalities, comparison theorems and qualitative properties of the solutions of fractional difference equations [2,3,4,5,7]. In the study of initial and boundary value problems most of the times, it is assumed that the independent variable is unchanged and the dependent variable or space variable is perturbed [10,11,12]. But in real world problems, it is almost impossible to measure the initial time or initial value of space variable with out at any error. It is difficult to compare any two solutions if the initial times are different. This has attracted many mathematicians to study the corresponding problems. Though some literature is available on fractional differential equations with initial time difference, not much of work has been yet done in discrete case. In this paper, using lower solutions and upper solutions starting at different initial times, comparison and existence results for fractional difference equations are established. 2. Fractional difference equations with initial time difference In this section, we consider the following Initial value problem (IVP) ∇μρ(n + 1) = f(n, ρ(n)), ρ(n0) = ρ0, (2.1) mailto:dixitgvsr@gmail.com mailto:ramanaginjala9@gmail.com mailto:prabhakar_mb@yahoo.co.in Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 8s (2025) 830 https://internationalpubls.com For n ∈Nn0 + ,n0 ≥ 0. Theorem 2.1. [6] Let ξ, ω : N0 + → R be L.S and U.S of (2.1). Further assume that f (n, r), for r∈ R is strictly non decreasing in r for each n. Then ξ(n) ≤ ω(n) provided ξ (0) ≤ ω (0). Theorem 2.2.[6] Suppose that ∇µm (n + 1) ≤ −Mm(n) for M>0 and m (0) ≤ m(N+1), then m(n) ≤ 0. Theorem 2.3. Assume that 1.ξ(n) and ω(n) be such that ∇μξ(n + 1) ≤ f(n, ξ(n)), ξ(n0) ≤ ρ0, n ϵ Nn0 + ∇μω(n + 1) ≥ f(n, ω(n)), ω(s0) ≥ ρ0, n ϵ Ns0 + with ξ(n0) ≤ ω(s0) 2. For x≥y, M>0, f (n, x)-f (n, y) ≤ - M(x-y). 3. s0 > n0 and f (n, ρ) is nondecreasing in n for each ρ. Then a. ξ(n) ≤ ω (n + η), n ≥ n0, b. ξ (n − η) ≤ ω(n), n ≥ s0, and η = s0 − n0. Proof. (a) Let ω̃ (n) = ω (n + η), n ≥ Nn0 + , ω̃(n0) = ω (n0 + η) = ω(s0) ≥ ρ0. Since f is nondecreasing in n for each ρ. Also ∇μω̃ (n + 1) = ∇μω̃ (n + η + 1) = f (n + η, ω (n + η)) ≥ f (n, ω̃ (n)). Consolidating the above, ∇μω̃ (n + 1) ≥ f (n, ω̃ (n)), ω̃ (n0) ≥ρ0. ∇μξ (n + 1) ≤ f (n, ξ(n)), ξ (n0) ≤ ρ0. Using Theorem 2.1, clearly ξ(n) ≤ ω̃ (n) = ω (n + η). (b) Let ξ̃(n) = ξ(n − η), n ≥ Nn0 + For n = s0, ξ̃(s0) = ξ (s0 − η) = ξ (n0) ≤ ρ0. Now, since f is non decreasing in n, ∇μξ̃ (n + 1) = f (n − η, ξ (n − η)) ≤ f(n, ξ̃(n)). Consolidating the above, ∇μξ̃ (n + 1) ≤ f (n, ξ̃ (n)), ξ̃ (s0) ≤ρ0, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 8s (2025) 831 https://internationalpubls.com ∇μω(n + 1) ≥ f (n, ω(n)), ω (n0) ≥ ρ0. By using Theorem 2.1, ξ (n − η) = ξ̃(n) ≤ ω(n), for n ≥s0 . Here we shall discuss the monotone iterative method to obtain extremal solutions for Fractional difference equations with initial time difference. Theorem 2.4. Assume that 1. ∇μξ (n + 1) ≤ f (n, ξ(n)), ξ (n0) = ρ0, n ∈ Nn0 + , n0 ≥ 0, ∇μω (n + 1) ≥ f (n, ω(n)), ω (s0) =ρ0, n ∈ Ns0 + , such that ξ0 ≤ ω0, 2. f (n, x) – f (n, y) ≥ −M (x − y), where M > 0 and y ≤ x, 3. s0 > n0 > 0 and f (n, ρ(n)) is nondecreasing n for each ρ(n) and ξ(n) ≤ ω (n + η), η = s0 − n0. Then there exit monotone sequences ξ̃n, ω̃n which converge uniformly and monotonically for n ∈ Nn0 + such that ξ̃m → ξ̃ and ω̃m → ω̃ for m → ∞. Moreover ξ̃ and ω̃ are minimal and maximal solutions of (2.1). Proof: Let ω̃0(n) = ω (n + η) and ξ̃0(n) = ξ(n) for n ∈ Nn0 + where η = s0 − n0. Let Θ: Nn0 + → R be such that ξ̃0(n) ≤ Θ ≤ ξ̃0(n), and the following linear fractional difference equation for 0 < µ < 1, ∇μρ (n + 1) = f (n, Θ(n)) − M[ρ(n) − Θ(n)], ρ (n0) =ρ0. (2.2) It is clear that, if Θ(n) = ρ(n), ρ(n) is the unique solution of (3.1) on n ∈ N0 + . If Θ(n) ≠ ρ(n), the non-homogeneous fractional difference equation (2.2) has unique solution ρn [1]. In order to construct monotone sequences ξ̃ (n), ω̃ (n), a mapping A: R → R is defined such that AΘ(n) = ρ(n). Now the following properties of A are proved below. (a). ξ̃0(n) ≤ Aξ̃0(n), ω̃0(n) ≥ Aω̃0(n). (b). A is monotone operator on [ξ̃0, ω̃0] = {ρ(n)/ ξ̃0 ≤ ρ(n) ≤ ω̃0}. To prove (a), set Aξ̃0 (n) = ξ̃1 (n), where ξ̃1(n) is the unique solution of (2.2). with Θ(n) = ξ̃0 (n). Set p (n) = ξ̃0(n) − ξ̃1(n). Consider ∇μp (n + 1) = ∇μ [ξ̃0(n + 1) − ξ̃1(n + 1)] ≤ f (n, ξ̃0(n)) – f (n, ξ̃0(n)) + M[ξ̃1(n) − ξ̃0(n)] = −M[ξ̃0(n) − ξ̃1(n)] or ∇μ p (n + 1) ≤ −M p(n). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 8s (2025) 832 https://internationalpubls.com Also, ρ (n0) = ξ̃0(n0) −ξ̃1(n0) ≤ ρ0 − ρ0 ≤ 0. By using Theorem 2.2, p(n) ≤ 0 or ξ̃0(n) ≤ ξ̃1(n) = Aξ̃0(n). Similarly, it can be proved that ω̃0(n) ≥ A{ω̃0(n)}. To prove (b), Let k ∈ N0 + and Θk and Θk+1∈ [ξ̃0 (n), ω̃0 (n)] such that Θk ≤ Θk+1. Suppose A{Θk(n)} = ξ̃k (n) and A {Θk+1(n)} = ξ̃k+1 (n). Take q(n) = ξ̃k − ξ̃k+1 . Consider ∇μ q (n + 1) = ∇μ [ξ̃k (n + 1) − ξ̃k+1 (n + 1)] = f (n, Θk(n)) – M [ξ̃k (n) − Θk (n)] – f (n, Θk+1 (n)) + M[ξ̃k+1 (n) − Θk+1 (n)] = M [Θk (n) − Θk+1 (n)] + M[ξ̃k+1 (n) − ξ̃k (n)] + f (n, Θk) – f (n, Θk+1) ≤ −M q(n) Also, q (n0) = ξ̃k (n0) − ξ̃k+1 (n0) ≤ ρ0 −ρ0 ≤ 0. By using Theorem 2.2, hence q(n) ≤ 0 or ξ̃k (n) ≤ ξ̃k+1 (n) = Aξ̃k (n). Similarly, it can be proved that ωk(n) ≥ ωk+1(n). that means, a sequence can be constructed as ξ̃0 (n)≤ ξ̃1 (n) ≤ ξ̃2 (n) ≤ .......... ≤ ξ̃m (n) ≤ ω̃m(n) ≤ ............. ≤ ω̃2(n) ≤ ω̃1(n) ≤ ω̃0(n) for n ∈ Nn0 + . By Dini’s theorem as m → ∞, ξ̃m → ξ̃ and ω̃m→ ω̃ where ξ̃ and ω̃ are any two functions defined for n ∈ Nn0 + . Also ξ̃m (n) and ω̃m(n) satisfy ∇μξ̃m (n + 1) = f (n, ξ̃m−1 (n) − M [ ξ̃m (n) −ξ̃m−1 (n) ], ξ̃m (n0)= ρ0, (2.3) ∇μω̃m (n + 1) = f (n, ω̃m−1 (n)) − M[ω̃m(n)− ω̃m−1(n)], ω̃m (n0)= ρ0 (2.4) As m → ∞, ξ̃(n) → ξ̃ and ω̃(n) → ω̃, Hence ∇μξ̃ = f(n, ξ̃), ξ̃(n0)= ρ0 , (2.5) ∇μω̃ = f(n, ω̃), ω̃(n0)= ρ0 . (2.6) It implies that the functions ξ̃ and ω̃ defined on n ∈ Nn0 + are solutions of (2.1). Now to prove that ξ̃ and ω̃ are minimal and maximal solutions of (2.1) respectively, it is sufficient to prove that ξ̃ ≤ ρ(n) ≤ ω̃ (2.7) If ρ(n) is any solution of (2.1) such that ξ̃0 ≤ ρ ≤ ω̃0. (2.8) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 8s (2025) 833 https://internationalpubls.com Let the statement ξ̃k ≤ ρ(n) ≤ ω̃k. (2.9) be true. Let ξ̃k satisfy (2.2) and consider r(n) = ξ̃k+1 (n) − ρ(n). Consider ∇μr (n + 1) = ∇μ [ξ̃k+1 (n + 1) – ρ (n + 1)] = f (n, ξ̃k (n)) – f (n, ρ(n) − M[ξ̃k+1 (n) − ξ̃k (n)] ≤ −M[ξ̃k (n) − ρ(n)] − M[ξ̃k+1 (n) −ξ̃k (n)] = −Mr(n). Also, r (n0) = ξ̃k+1(n0)− ρ (n0) ≤ ρ0 − ρ0 ≤ 0. By using Theorem 2.2, we have ξ̃k+1 (n) ≤ ρ(n). Similarly, it can be proved that ρ(n) ≤ ωk+1(n). Therefore ξ̃k+1 (n) ≤ ρ(n) ≤ ωk+1(n). Hence the statement (2.9) is true for m = k + 1. By the principle of mathematical induction, the statement (2.9) is true for every m ∈ N0 + Thus ξ̃m (n) ≤ ρ(n) ≤ ω̃m (n). Taking the limit as m → ∞, we get ξ̃ ≤ ρ(n) ≤ ω̃ such that ξ̃0 (n) ≤ ρ(n) ≤ ω̃0. Hence the functions ξ̃ and ω̃ defined for n ∈ Nn0 + are minimal and maximal solutions of (2.1) respectively. Theorem 2.5. In addition to the hypothesis of Theorem (2.4), if we assume that f(n, x) − f(n, y) ≥ −M(x − y), (2.10) for ξ ≤ y ≤ x ≤ ω and M ≥ 0. Then ξ = ω = ρ is the unique solution of (2.1). Proof: Since ξ(n) ≤ ω(n), it is enough to prove that ξ(n) ≥ ω(n). Take s(n) = ξ(n) − ω(n). Consider ∇μ s (n + 1) = ∇μ [ξ (n + 1) – ω (n + 1)] = − (f (n, ω(n)) – f (n, ξ(n))) ≤ M[ω(n) − ξ(n)] = −Ms(n). Also r (n0) = ξ (n0) – ρ (n0) ≤ ρ0 − ρ0 ≤ 0. By using Theorem2.2, hence s(n) ≤ 0 for n ∈ Nn0 + . Thus ω(n) ≤ ξ(n). Hence ξ(n) = ω(n) = ρ(n) is the unique solution of (2.1). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 8s (2025) 834 https://internationalpubls.com Conclusion In this chapter, using monotone iterative technique on solutions of fractional difference equations at different initial times the convergence of monotone sequences to maximum and minimum solutions is established. Using lower and upper solutions, existence and uniqueness of solution to fractional difference equations at different initial times is also obtained. Future work: This work can be extended onto the systems of Fractional Difference equations This work may make better some processes concerned with signal processing and image processing, system trajectories and various other fields. Acknowledgement: The authors thank their respective college managements for their continuous support and constant encouragement. 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