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On Nonlinear Implicit Neutral Generalized Hilfer Fractional Differential Equations with Terminal Conditions and Delay

On Nonlinear Implicit Neutral Generalized Hilfer Fractional Differential Equations with Terminal... Topol. Algebra Appl. 2022; 10:77–93 Research Article Open Access Soufyane Bouriah, Abdelkrim Salim*, and Mouak Benchohra On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations with Terminal Conditions and Delay https://doi.org/10.1515/taa-2022-0115 Received 30 May, 2022; accepted 14 June, 2022 Abstract: In this paper, we establish the existence of solutions for a class of nonlinear implicit neutral frac- tional dierential equations with terminal condition and Hilfer-Katugampola fractional derivative. The Ba- nach contraction principle and Krasnoselskii’s xed point theorem are used to support the arguments. An illustration is provided to demonstrate the relevance of our results. Keywords: Hilfer-Katugampola fractional derivative, implicit neutral fractional dierential equations, exis- tence, uniqueness, xed point MSC: 26A33, 34A08, 34K37 1 Introduction Fractional calculus is a generalization of ordinary dierentiation and integration to arbitrary order (non- integer). For some details and recent publications on the subject, see the books [1–4], the papers [5–23] and the references therein. In [13], Benchohra et al. considered the following terminal value problem with Hilfer-Katugampola type fractional derivative: λ,ζ λ,ζ ρ ρ D x (t) = f t, x(t), D x (t) , for each t 2 (a, T ], a > 0, + + a a x(T ) = c 2 R, λ,ζ where D is the Hilfer-Katugampola fractional derivative of order λ 2 (0, 1) and type ζ 2 [0, 1] and f : (a, T ] × R × R ! R is a given function. Their reasoning is mainly based upon dierent types of classical xed point theorems such as the Banach contraction principle and Krasnoselskii’s xed point theorem. Motivated by the works mentioned above, we establish in this paper existence and uniqueness results to the terminal value problem of the following Hilfer-Katugampola type fractional dierential equation: λ,ζ D x(t) − g(t, x ) = f t, x , t 2 (0, T ], 0 < T < ∞, (1) + t ( t ) Soufyane Bouriah: Department of Mathematics, Faculty of Exact Sciences and Informatics, Hassiba Benbouali University, P.O. Box 151 Chlef 02000, Algeria, E-mail: s.bouriah@univ-chlef.dz, bouriahsouane@yahoo.fr Laboratory of Mathematics and Applications Hassiba Benbouali University of Chlef Hay Salem, P.O. Box 151 Chlef Algeria *Corresponding Author: Abdelkrim Salim: Faculty of Technology, Hassiba Benbouali University, P.O. Box 151 Chlef 02000, Algeria, E-mail: salim.abdelkrim@yahoo.com, a.salim@univ-chlef.dz Laboratory of Mathematics, Djillali Liabes University of Sidi Bel-Abbes, P.O. Box 89 Sidi Bel Abbes 22000, Algeria Mouak Benchohra: Laboratory of Mathematics, Djillali Liabes University of Sidi Bel-Abbes, P.O. Box 89 Sidi Bel Abbes 22000, Algeria, E-mail: benchohra@yahoo.com Open Access. © 2022 Soufyane Bouriah et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution 4.0 License. 78 Ë S. Bouriah et al. x(T ) = c 2 R, (2) x(t) = (t), t 2 [−θ, 0], θ > 0, (3) λ,ζ 1−η ρ ρ where D , I are the Hilfer-Katugampola fractional derivative of order λ 2 (0, 1) and type ζ 2 [0, 1] + + 0 0 and Katugampola fractional integral of order 1− η, (η = λ + ζ − λζ ) respectively, f : (0, T ]× C([−θ, 0],R) ! R, and g : (0, T ] × C([−θ, 0],R) ! R are two given functions and 2 C([−θ, 0],R). For each function x dened on [−θ, T ] and for any t 2 [0, T ], we denote by x the element of C([−θ, 0],R) dened by: x (θ) = x(t + θ), θ 2 [−θ, 0], The following is how the current paper is arranged. In Section 2, we present certain notations and review some preliminary notions concerning the Hilfer-Katugampola fractional derivative and auxiliary ndings. Section 3 presents two solutions results to problem (1)-(3). In the nal part, we provide an illustration to demonstrate the application of our key results. 2 Preliminaries This part introduces the preliminary information that will be utilized throughout the study. Let 0 < T, J = [0, T ]. By C([−θ, 0],R), C(J,R) we denote the Banach spaces of all continuous functions from [−θ, 0] into R (resp from [0, T ] into R) with the norms: kxk = supjx(t)j, t 2 [−θ, 0] andkxk = supjx(t)j, t 2 [0, T ] (respectively). C ∞ We consider the weighted spaces of continuous functions C (J) = x : (0, T ] ! R : x(t) 2 C(J,R) , 0 ≤ η < 1, η,ρ and n o n n−1 (n) C (J) = x 2 C (J) : x 2 C (J) , n 2 N, η,ρ η,ρ C (J) = C (J), η,ρ η,ρ with the norms kxk = sup x(t) , η,ρ t2J and n−1 (k) (n) kxk n = kx k +kx k . C C η,ρ η,ρ k=0 Consider the space X (a, T ), (ϑ 2 R, 1 ≤ p ≤ ∞) of those complex-valued Lebesgue measurable functions f on [a, T ] for whichkfk < ∞, where the norm is dened by 0 1 dt ϑ p @ A kfk p = jt f (t)j , (1 ≤ p < ∞, ϑ 2 R). ϑ t p p In particular, when ϑ = , the space X (a, T ) coincides with the L (a, T ) space: X (a, T ) = L (a, T ). p 1 p p ϑ p On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 79 Denition 2.1 (Katugampola fractional integral [24, 25]). Let λ 2 R , ϑ 2 R and h 2 X (a, T ). The Katugam- pola fractional integral of order λ is given by λ−1 ρ ρ t − s h(s) ρ λ ρ−1 I +h (t) = s ds, t > a, ρ > 0, ρ Γ(λ) λ−1 −t where Γ(·) is the Euler gamma function dened by Γ(λ) = t e dt, λ > 0. Denition 2.2 (Katugampola fractional derivative [24, 25]). Let λ 2 R \N and ρ > 0. The Katugampola frac- ρ λ tional derivative D of order λ is dened by ρ λ n ρ n−λ + + D h (t) = δ ( I h)(t) a ρ a n n−λ−1 ρ ρ d t − s h(s) 1−ρ ρ−1 = t s ds, t > a, ρ > 0, dt ρ Γ(n − λ) n 1−ρ where n = [λ] + 1 and δ = t . dt Theorem 2.3 ([25]). Let λ > 0, ζ > 0, a < T < ∞ and ρ, ϑ 2 R, ρ ≥ ϑ. Then, for h 2 X (a, T ) the semigroup property is valid, i.e. ζ λ+ζ ρ λ ρ ρ I + I h (t) = I h (t). + + a a ρ λ Lemma 2.4 ([24–26]). Let λ > 0, and 0 ≤ η < 1. Then, I is bounded from C (J) into C (J). + η,ρ η,ρ ρ λ Lemma 2.5 ([26]). Let a < T < ∞, λ > 0, 0 ≤ η < 1 and x 2 C (J). If λ > η, then I x is continuous on J and η,ρ + ρ λ ρ λ I +x (a) = lim I +x (t) = 0. a a t!a Lemma 2.6 ([27]). For λ ≥ 0 and ζ > 0, we have " # ζ−1 λ+ζ−1 ρ ρ ρ ρ s − a Γ(ζ ) t − a ρ λ I + (t) = ρ ρ Γ(λ + ζ ) " # λ−1 ρ ρ s − a ρ λ D (t) = 0, 0 < λ < 1. Lemma 2.7 ([26]). Let λ > 0, 0 ≤ η < 1 and h 2 C [a, T ]. Then, ρ λ ρ λ + + D I h (t) = h(t), for all t 2 (a, T ]. a a ρ 1−λ 1 Lemma 2.8 ([26]). Let 0 < λ < 1, 0 ≤ η < 1. If h 2 C [a, T ] and I + h 2 C [a, T ], then η,ρ η,ρ ρ 1−λ λ−1 I + h (a) ρ ρ t − a ρ λ ρ λ I + D +h (t) = h(t) − , for all t 2 (a, T ]. a a Γ(λ) ρ Denition 2.9 ([26]). Let λ and ζ verify n − 1 < λ < n and 0 ≤ ζ ≤ 1, with n 2 N. The Hilfer-Katugampola fractional derivative to of a function h 2 C [a, T ], is given by 1−η,ρ λ,ζ ζ (n−λ) (1−ζ )(n−λ) ρ ρ ρ−1 ρ D h (t) = I t I h (t) + + + a a a dt ζ (n−λ) (1−ζ )(n−λ) ρ n ρ = I δ I h (t). + + a a 80 Ë S. Bouriah et al. λ,ζ Property 2.10 ([26]). The operator D can be written as ρ λ,ζ ρ ζ (1−λ) ρ 1−η ρ ζ (1−λ) ρ η D = I δ I = I D , η = λ + ζ − λζ . + + ρ + + + a a a a a Consider the following parameters λ, ζ , η satisfying η = λ + ζ − λζ , 0 < λ, ζ , η < 1. Thus, we dene the spaces n o λ,ζ λ,ζ C (J) = x 2 C (J), D x 2 C (J) 1−η,ρ 1−η,ρ 1−η,ρ a and η η C (J) = x 2 C (J), D x 2 C (J) . 1−η,ρ + 1−η,ρ 1−η,ρ a ρ λ,ζ ρ η(1−λ) ρ η Since D x = I D x, it follows from Lemma 2.4 that + + + a a a η λ,ζ C (J)  C (J)  C (J). 1−η,ρ 1−η,ρ 1−η,ρ Lemma 2.11 ([26]). Let 0 < λ < 1, 0 ≤ ζ ≤ 1 and η = λ + ζ − λζ . If x 2 C (J), then 1−η,ρ ρ η ρ η ρ λ ρ λ,ζ I D x = I + D x + + + a a a and η ζ (1−λ) ρ ρ λ ρ D I +x = D x. + + a a 3 Existence of Solutions Let Λ be the Banach space dened by: Λ = fx : [−θ, T ] ! R : xj 2 C([−θ, 0],R) and xj 2 C (J)g, [−θ,0] (0,T ] 1−η,ρ with the norm kxk = kxk +kxk . Λ C C 1−η,ρ We consider the following problem: λ,ζ D x(t) − g(t, x ) = ϖ(t), t 2 (0, T ], (4) x(T ) = c, c 2 R, (5) x(t) = (t), t 2 [−θ, 0]. (6) where ϖ(·) 2 C (J) and g(·, x ) 2 C (J). The following theorem shows that (4)-(6) is equivalent to the 1−η,ρ 1−η,ρ integral equation: > (t), if t 2 [−θ, 0], 2 3 > T > 1−η λ−1 η−1 < ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 c − g(T, x ) − s ϖ(s)ds x(t) = T (7) ρ ρ ρ Γ(λ) > Z λ−1 > ρ ρ 1 t − s ρ−1 +g(t, x ) + s ϖ(s)ds, t 2 (0, T ]. > t Γ(λ) ρ 0 On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 81 Theorem 3.1. Let η = λ + ζ − λζ , where 0 < λ < 1 and 0 ≤ ζ ≤ 1. If ϖ : (0, T ] ! R is a function where ϖ(·) 2 C (J), then x 2 Λ satises the problem (4)-(6) if and only if it satises Equation (7). 1−η,ρ Proof. ()) Let x 2 Λ, if t 2 [−θ, 0], then we have x(t) = (t). On the other hand, for t 2 (0, T ], we have x 2 C (J) a solution of the equations (4) and (5). We prove that 1−η,ρ x is also a solution of (7). From the denition of C (J), Lemma 2.4, and using Denition 2.2, we have 1−η,ρ ρ 1−η I x(·) − g(·, x ) 2 C(J,R) + · and η 1−η ρ ρ D x(·) − g(·, x ) = δ I x(·) − g(·, x ) 2 C (J). (8) + · ρ + · 1−η,ρ 0 0 By the denition of the space C (J), it follows that 1−η,ρ ρ 1−η 1 I x(·) − g(·, x ) 2 C (J). + · 1−η,ρ Using Lemma 2.8, with λ = η, we obtain: η η ρ ρ I D x(s) − g(s, x )) (t) = x(t) − g(t, x ) (9) + + s 0 0 1−η η−1 I x(s) − g(s, x ) (0) s ρ − , Γ(η) where t 2 (0, T ]. By hypothesis, x 2 C (J), using Lemma 2.11 with Equation (4), we have 1−η,ρ η η λ,ζ ρ ρ ρ λ ρ I D x(s) − g(s, x ) (t) = I + D x(s) − g(s, x ) (t) (10) + + s + s 0 0 0 ρ λ = I +ϖ (t). Comparing (9) and (10), we see that 1−η η−1 I x(s) − g(s, x ) (0) ρ + s ρ λ x(t) = + g(t, x ) + I +ϖ (t). (11) Γ(η) ρ Using Equation (5) we obtain: 2 3 1−η λ−1 η−1 ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 x(t) = c − g(T, x ) − s ϖ(s)ds ρ Γ(λ) ρ ρ λ−1 ρ ρ 1 t − s ρ−1 + g(t, x ) + s ϖ(s)ds, Γ(λ) where t 2 (0, T ], that is x(·) satises Equation (7). (() Let x 2 Λ, satisfying Equation (7). We show that x also satises the problem (4)-(6). If t 2 [−θ, 0], then x(t) = (t), so the condition 6 is satised. ( ) η η If t 2 (0, T ], then x 2 C (J). Apply operator D on both sides of Equation (7). Then, from Lemmas 2.6 1−η,ρ 0 and 2.11 we get η ζ (1−λ) ρ ρ D x(s) − g(s, x ) (t) = D ϖ (t). (12) + s + 0 0 By Equation (8) we have D x(·) − g(·, x ) 2 C (J). Then, Equation (12) implies + · 1−η,ρ ρ η ρ 1−ζ (1−λ) ρ ζ (1−λ) D x(s) − g(s, x ) (t) = δ I ϖ (t) = D ϖ (t) 2 C (J). (13) + s ρ + + 1−η,ρ 0 0 0 As ϖ(·) 2 C (J) and from Lemma 2.4, it follows that 1−η,ρ 1−ζ (1−λ) I ϖ 2 C (J). (14) 1−η,ρ 0 82 Ë S. Bouriah et al. From Equations (13) and (14) and by the Denition of the space C (J), we obtain 1−η,ρ 1−ζ (1−λ) ρ 1 I ϖ 2 C (J). 1−η,ρ ζ (1−λ) Applying operator I on both sides of Equation (13) and using Lemmas 2.5 and 2.8, we have 1−ζ (1−λ) ζ (1−λ)−1 I ϖ(t) (0) + ρ ρ ζ (1−λ) ρ η I D x(t) − g(t, x ) = ϖ(t) + , + + 0 0 Γ(ζ (1 − λ)) which implies that: λ,ζ D x(t) − g(t, x ) = ϖ(t), + t that is, Equation(4) holds. Clearly, if x 2 C (J) satises Equation(7), then it also satises Equation(5). 1−η,ρ Suppose that the function f : (0, T ] × C([−θ, 0],R) ! R is continuous and veries the requirements: (B1)Set the functions f : (0, T ] × C([−θ, 0],R) ! R and g : (0, T ] × C([−θ, 0],R) ! R such that: ζ (1−λ) η f (·, u ) 2 C (J) and g(·, u ) 2 C (J) for any u 2 C (J). · · 1−η,ρ 1−η,ρ 1−η,ρ (B2)There exist constants β > 0 and 0 < β < 1 such that: 1 2 jf (t, u) − f (t, u ¯ )j ≤ β ku − u ¯k and jg(t, w) − g(t, w ¯ )j ≤ β kw − w ¯k 2 C for any u, w, u ¯ , w ¯ 2 C([−θ, 0],R) and t 2 (0, T ]. As a consequence of Theorem 3.1, we have Theorem 3.2. Theorem 3.2. Let η = λ + ζ − λζ where 0 < λ < 1 and 0 ≤ ζ ≤ 1; let f : (0, T ] × C([−θ, 0],R) ! R be a function ζ (1−λ) η such that f (·, x ) 2 C (J)  C (J) and g : (0, T ] × C([−θ, 0],R) ! R where g(·, x ) 2 C (J) for any · · 1−η,ρ 1−η,ρ 1−η,ρ x 2 C (J). 1−η,ρ If x 2 Λ, then x satises the problem (1)−(3) if and only if x is the xed point of the operator G : Λ ! Λ dened by: (t) if t 2 [−θ, 0] 2 3 > 1−η λ−1 η−1 < ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 c − g(T, x ) − s h(s)ds Gx(t) = T (15) ρ ρ ρ Γ(λ) > Z λ−1 > ρ ρ 1 t − s ρ−1 +g(t, x ) + s h(s)ds, t 2 (0, T ], > t Γ(λ) ρ where: h : (0, T ] ! R be a function verifying h(t) = f (t, x ). Clearly, h 2 C (J). 1−η,ρ The following proof is to show that for any x 2 Λ then Gx 2 Λ, i.e. the operator G is well dened. Proof. Let x 2 Λ, if t 2 [−θ, 0] then Gx(t) = (t). If t 2 (0, T ] then we have: 2 3 1−η λ−1 η−1 ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 Gx(t) = c − g(T, x ) − s h(s)ds ρ ρ ρ Γ(λ) λ−1 ρ ρ 1 t − s ρ−1 + g(t, x ) + s h(s)ds. Γ(λ) ρ 0 On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 83 Applying D to both sides and by Lemmas 2.6 and 2.11, we have: η η η ρ ρ ρ ρ λ D Gx(t) = D g(t, x ) + D I +f (s, x ) (t) + + t + s 0 0 0 η ζ (1−λ) ρ ρ = D g(t, x ) + D f (s, x ) (t). + t + s 0 0 ζ (1−λ) η Since η ≥ λ, and f (·, x ) 2 C (J), the right hand side is in C (J) and thus D Gx 2 C (J), which · + 1−η,ρ 1−η,ρ 1−η,ρ 0 implies that Gx 2 C (J). We can conclude that Gx 2 Λ. 1−η,ρ Based on Banach’s xed point, we now present and verify our existence result for problem (1)-(3). Theorem 3.3. Suppose that (B1) and (B2) are met. If ( ) λ 1−η 1−η+λ ρ ρ ρ β Γ(η) T T β T 1 1 1 max β + ; β + < , (16) 2 2 Γ(λ + η) ρ ρ Γ(λ + 1) ρ 2 then the problem (1)−(3) has unique solution in Λ. Proof. We show that the operator G dened in Equation (15) has a unique xed point x in Λ. Let x, u 2 Λ. If t 2 [−θ, 0], then jGx(t) − Gu(t)j = 0. For t 2 (0, T ], we have jGx(t) − Gu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ 1 T t T − s ρ−1 ≤ s jh(s) − ϖ(s)jds ρ ρ ρ Γ(λ) λ−1 ρ ρ 1 t − s ρ−1 + jg(t, x ) − g(t, u )j + s jh(s) − ϖ(s)jds t t Γ(λ) ρ 1−η η−1 ρ ρ T t + jg(T, x ) − g(T, u )j, T T ρ ρ where h, ϖ 2 C (J) such that 1−η,ρ h(t) = f (t, x ) ϖ(t) = f (t, u ). By (B2), we have jh(t) − ϖ(t)j = jf (t, x ) − f (t, u )j t t ≤ β kx − u k . 1 t t C On the other hand, we havekx − u k = sup jx (θ) − u (θ)j, θ 2 [−θ, 0] , then there exist θ 2 [−θ, 0] such t t C t t that * * * kx − u k = jx (θ ) − u (θ )j = jx(t + θ ) − u(t + θ )j. t t C t t * If t + θ 2 [−θ, 0], then kx − u k ≤ kx − uk = kx − uk , t t C C Λ 84 Ë S. Bouriah et al. which implies that for each t 2 (0, T ], we have jGx(t) − Gu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s ρ ρ ρ Γ(λ) λ−1 ρ ρ β t − s 1 ρ−1 + β kx − u k + s kx − u k ds 2 t t s s C C Γ(λ) ρ 1−η η−1 ρ ρ T t + β kx − u k 2 T T C ρ ρ 1−η η−1 λ ρ ρ ρ β T t T ≤ kx − uk Γ(λ + 1) ρ ρ ρ β t + β kx − uk + kx − uk Λ Λ Γ(λ + 1) ρ 1−η η−1 ρ ρ T t + β kx − uk . ρ ρ Hence: 1−η 1−η+λ 1−η ρ ρ ρ t β T t Gx(t) − Gu(t) ≤ + β ρ Γ(λ + 1) ρ ρ 1−η+λ 1−η ρ ρ β t T + + β kx − uk Γ(λ + 1) ρ ρ " # 1−η 1−η+λ ρ ρ T β T ≤ 2 β + kx − uk , 2 Λ ρ Γ(λ + 1) ρ which implies that " # 1−η 1−η+λ ρ ρ T β T kGx − Guk = kGx − Guk ≤ 2 β + kx − uk . C Λ 2 Λ 1−η,ρ ρ ρ Γ(λ + 1) If t + θ 2 [0, T ], then we have η−1 1−η η−1 η−1 ρ ρ ρ ρ t t t t kx − u k = kx − u k ≤ kx − uk ≤ kx − uk . t t C t t C C Λ 1−η,ρ ρ ρ ρ ρ Hence, for each t 2 (0, T ] we obtain jGx(t) − Gu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s Γ(λ) ρ ρ ρ λ−1 ρ ρ β t − s 1 ρ−1 + β kx − u k + s kx − u k ds s s 2 t t C C Γ(λ) 1−η η−1 ρ ρ T t + β kx − u k 2 T T C ρ ρ 1−η η−1 η−1 ρ ρ ρ T t s ρ λ ≤ β kx − uk I + (T ) 1 C 0 1−η,ρ ρ ρ ρ η−1 η−1 ρ ρ t s + 2β kx − uk + β I + (t)kx − uk . 2 C 1 C 1−η,ρ 0 1−η,ρ ρ ρ By Lemma 2.6, we have λ η−1 ρ ρ β Γ(η) T t jGx(t) − Gu(t)j ≤ Γ(λ + η) ρ ρ On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 85 η−1 λ+η−1 ρ ρ t β Γ(η) t + 2β + kx − uk , 2 C 1−η,ρ ρ Γ(λ + η) ρ hence 1−η Gx(t) − Gu(t) " # λ λ ρ ρ β Γ(η) T β Γ(η) t 1 1 ≤ + 2β + kx − uk 1−η,ρ Γ(λ + η) ρ Γ(λ + η) ρ " # β Γ(η) T ≤ 2 β + kx − uk , 2 Λ Γ(λ + η) ρ which implies that " # β Γ(η) T kGx − Guk = kGx − Guk ≤ 2 β + kx − uk . C Λ Λ 1−η,ρ Γ(λ + η) ρ By Equations (16), the operator G is a contraction. Hence, by Banach’s contraction principle, G has a unique xed point x 2 Λ. As a consequence of Theorem 3.2, we can conclude that the problem (1)−(3) have a unique solution in Λ. We present now the second result, which is based on Krasnoselskii xed point theorem [28]. Theorem 3.4. Assume (B1) and (B2) hold. If: " # " # 1−η λ λ+1−η ρ ρ ρ T 2β Γ(η) T T β 1 + + + < 1, ρ Γ(λ + η) ρ ρ and λ λ+1−η 1−η ρ ρ ρ β Γ(η) T β T T 1 1 max + 2β ; + 2β 2 2 ρ ρ ρ Γ(λ + η) Γ(λ + 1) := max { ;{ < 1, (17) ( ) 2 1 then the problem (1)−(3) have at least one solution. Proof. Consider the set B * = fx 2 Λ : jjxjj ≤ ξ g, ξ Λ where 8 9 1−η * λ ρ ρ < 2Γ(η)f = T * T jcj+2g + ρ Γ(λ+η) ρ ξ ≥ max ; sup j (t)j , 1−η λ λ+1−η ρ ρ ρ 2β Γ(η) : T 1 T T ; 1− β 1+ + + t2[−θ,0] ρ Γ(λ+η) ρ ρ 1−η f = sup jf (t, 0)j, t2J and 1−η g = sup jg(t, 0)j. t2J We dene the operators H and F on B * by: (t), if t 2 [−θ, 0] 2 3 1−η λ−1 η−1 Hx(t) = ρ ρ ρ ρ (18) T 1 T − s t ρ−1 4 5 > c − g(T, x ) − s h(s)ds > ρ ρ ρ Γ(λ) +g(t, x ), t 2 (0, T ], t 86 Ë S. Bouriah et al. (t), if t 2 [−θ, 0] Fx(t) = (19) λ−1 ρ ρ > 1 t − s ρ−1 s h(s)ds, t 2 (0, T ]. : Γ(λ) ρ Then the fractional integral equation (15) can be written as the operator equation: Gx(t) = Hx(t) + Fx(t), t 2 [−θ, T ]. The proof will be given in several steps: Step 1: We prove that Hx + Fu 2 B * for any x, u 2 B * . ξ ξ 1 1 If t 2 [−θ, 0] thenjHx(t)j ≤ sup j (t)j andjFu(t)j ≤ sup j (t)j, 2 2 t2[−θ,0] t2[−θ,0] which implies that 1 1 kHx + Fuk ≤ sup j (t)j + sup j (t)j 2 2 t2[−θ,0] t2[−θ,0] = sup j (t)j t2[−θ,0] ≤ ξ . 1−η If t 2 (0, T ] then multiplying both sides of Equation (18) by , we have 2 3 1−η 1−η λ−1 ρ ρ ρ ρ t T 1 T − s ρ−1 4 5 Hx(t) = c − g(T, x ) − s h(s)ds ρ ρ ρ Γ(λ) 1−η +g(t, x ) . Then 1−η 1−η λ−1 ρ ρ ρ ρ t T 1 T − s ρ−1 Hx(t) ≤ s jh(s)jds ρ ρ Γ(λ) ρ 1−η +jcj +jg(T, x )j + jg(t, x ) − g(t, 0)j +jg(t, 0)j T t 2 3 1−η λ−1 ρ ρ ρ T 1 T − s ρ−1 4 5 ≤ s jh(s)jds +jcj ρ Γ(λ) ρ " # η−1 1−η ρ ρ t t * * + g + β kx k + g . 2 t ρ ρ Thus 2 3 1−η 1−η λ−1 ρ ρ ρ ρ t T 1 T − s ρ−1 4 5 Hx(t) ≤ jcj + s jh(s)jds ρ ρ Γ(λ) ρ (20) 1−η + 2g + β kx k . 2 C By (B2), we have for each t 2 (0, T ], jh(t)j = jf (t, x ) − f (t, 0) + f (t, 0)j t On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 87 ≤ jf (t, x ) − f (t, 0)j +jf (t, 0)j η−1 ≤ β kx k + f . 1 t C 1−η Multiplying both sides of the above inequality by , we get 1−η 1−η ρ ρ t t h(t) ≤ f + β kx k . 1 t ρ ρ If t + θ 2 [−θ, 0] thenkx k = kxk = kxk , which implies that for each t 2 (0, T ], we have t C C Λ 1−η 1−η ρ ρ t t h(t) ≤ f + β kxk ρ ρ 1−η * * ≤ f + β ξ := Θ . (21) By replacing (21) in the inequality (20) and using Lemma 2.6, we have " # 1−η 1−η λ+η−1 ρ ρ ρ t T Θ Γ(η) T 1 * Hx(t) ≤ jcj + + β ξ ρ ρ ρ Γ(λ + η) +2g . This gives " # 1−η λ+η−1 ρ ρ T Θ Γ(η) T 1 * * jjHxjj ≤ jcj + + β ξ + 2g := Ω . C 2 1 1−η,ρ ρ ρ Γ(λ + η) If t + θ 2 [0, T ], then η−1 1−η ρ ρ t t kx k = kx k t t C C ρ ρ η−1 ≤ kxk . Thus, for each t 2 (0, T ], we have 1−η * * h(t) ≤ f + β ξ := Θ . (22) 1 2 By replacing (22) in the inequality (20) and using Lemma 2.6, we obtain " # 1−η 1−η λ+η−1 ρ ρ ρ t T Θ Γ(η) T Hx(t) ≤ jcj + ρ ρ Γ(λ + η) ρ * * + β ξ + 2g . Hence " # 1−η λ+η−1 ρ ρ T Θ Γ(η) T 2 * * jjHxjj ≤ jcj + + β ξ + 2g := Ω . 2 2 1−η,ρ ρ Γ(λ + η) ρ If t + θ 2 [−θ, 0] with t 2 (0, T ] then using Equation (21) and Lemma 2.6, we have " # 1−η λ+η−1 * * ρ ρ Γ(η)f β Γ(η)ξ T t jF(u)(t)j ≤ + . Γ(λ + η) Γ(λ + η) ρ ρ 88 Ë S. Bouriah et al. Therefore " # 1−η 1−η λ ρ ρ ρ * * t Γ(η)f β Γ(η)ξ T t Fu(t) ≤ + , ρ ρ ρ Γ(λ + η) Γ(λ + η) " # 1−η λ * * ρ ρ Γ(η)f β Γ(η)ξ T T ≤ + . Γ(λ + η) Γ(λ + η) ρ ρ Thus " # 1−η λ ρ ρ * * Γ(η)f β Γ(η)ξ T T kFuk ≤ + := Y . C 1 1−η,ρ Γ(λ + η) Γ(λ + η) ρ ρ If t + θ 2 [0, T ] then for each t 2 (0, T ] and by using Equation (22) and Lemma 2.6, we have λ+η−1 * * ρ Γ(η)f β Γ(η)ξ t jF(u)(t)j ≤ + . Γ(λ + η) Γ(λ + η) Therefore 1−η λ ρ * * ρ t Γ(η)f β Γ(η)ξ t Fu(t) ≤ + , ρ ρ Γ(λ + η) Γ(λ + η) * * ρ Γ(η)f β Γ(η)ξ T ≤ + . Γ(λ + η) Γ(λ + η) ρ Thus * * Γ(η)f β Γ(η)ξ T kFuk ≤ + := Y . 1−η,ρ Γ(λ + η) Γ(λ + η) ρ Consequently, for every x, u 2 B * we obtain kHx + Fuk ≤ kHxk +kFuk C C C 1−η,ρ 1−η,ρ 1−η,ρ ≤ maxfΩ ; Ω g + maxfY ; Y g 1 2 1 2 " # λ λ λ+1−η ρ ρ ρ * * 2Γ(η)f T 2β Γ(η)ξ T T ≤ + + ρ ρ ρ Γ(λ + η) Γ(λ + η) " # 1−η 1−η ρ ρ T T * * + jcj + 2g + β ξ 1 + . ρ ρ Since 1−η λ ρ * ρ T 2Γ(η)f T jcj + 2g + ρ ρ Γ(λ + η) " " # " ## ξ ≥ , 1−η λ λ+1−η ρ ρ ρ T 2β Γ(η) T T 1 − β 1 + + + ρ Γ(λ + η) ρ ρ we have kHx + Fuk = kHx + Fuk ≤ ξ , C Λ 1−η,ρ which implies that for each t 2 [−θ, T ] we get kHx + Fuk ≤ ξ , which infers that Hx + Fu 2 B . Step 2: H is a contraction. Let x, u 2 Λ. If t 2 [−θ, 0]. Then, we have jHx(t) − Hu(t)j = 0. For t 2 (0, T ], we have On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 89 jHx(t) − Hu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ 1 T t T − s ρ−1 ≤ s jh(s) − ϖ(s)jds Γ(λ) ρ ρ ρ 1−η η−1 ρ ρ T t + jg(t, x ) − g(t, u )j + jg(T, x ) − g(T, u )j t t T T ρ ρ 1−η η−1 λ−1 ρ ρ ρ ρ 1 T t T − s ρ−1 ≤ s jh(s) − ϖ(s)jds Γ(λ) ρ ρ ρ 1−η η−1 ρ ρ T t + β kx − u k + β kx − u k . 2 t t C 2 T T C ρ ρ where h, ϖ 2 C (J) such that 1−η,ρ h(t) = f (t, x ), ϖ(t) = f (t, u ). By (B2), we have jh(t) − ϖ(t)j = jf (t, x ) − f (t, u )j t t ≤ β kx − u k . 1 t t If t + θ 2 [−θ, 0] then kx − u k ≤ kx − uk = kx − uk , t t C C Λ which implies that for each t 2 (0, T ] we get jHx(t) − Hu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s ρ ρ ρ Γ(λ) 1−η η−1 ρ ρ T t + β kx − u k + β kx − u k 2 t t C 2 T T C ρ ρ λ+1−η η−1 ρ ρ β T t ≤ kx − uk + β kx − uk Λ Λ Γ(λ + 1) ρ ρ 1−η η−1 ρ ρ T t + β kx − uk . ρ ρ Hence 1−η λ+1−η 1−η ρ ρ ρ t β T t Hx(t) − Hu(t) ≤ + β ρ Γ(λ + 1) ρ ρ 1−η +β kx − uk , 2 Λ which implies that " # λ+1−η 1−η ρ ρ β T T kHx − Huk ≤ + 2β := { kx − uk . 2 1 Λ 1−η,ρ Γ(λ + 1) ρ ρ If t + θ 2 [0, T ] then η−1 η−1 ρ ρ t t kx − u k ≤ kx − uk = kx − uk . t t C C Λ 1−η,ρ ρ ρ 90 Ë S. Bouriah et al. Therefore, for each t 2 (0, T ], we obtain jHx(t) − Hu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s Γ(λ) ρ ρ ρ 1−η η−1 ρ ρ T t + β kx − u k + β kx − u k 2 t t C 2 T T C ρ ρ 1−η η−1 η−1 ρ ρ ρ T t s ρ λ ≤ β kx − uk I + (T ) Λ 0 ρ ρ ρ η−1 + 2β kx − uk . By Lemma 2.6, we have λ η−1 ρ ρ β Γ(η) T t jHx(t) − Hu(t)j ≤ kx − uk ρ ρ Γ(λ + η) η−1 + 2β kx − uk . Hence " # 1−η λ ρ ρ t β Γ(η) T Hx(t) − Hu(t) ≤ + 2β kx − uk , 2 Λ ρ Γ(λ + η) ρ which implies that " # β Γ(η) T kHx − Huk ≤ + 2β kx − uk := { kx − uk . C 2 Λ 2 Λ 1−η,ρ Γ(λ + η) Thus kHx − Huk = kHx − Huk ≤ max ({ ,{ )kx − uk . C Λ 1 2 Λ 1−η,ρ By Equation (17) the operator H is a contraction. Step 3: F is compact and continuous. The continuity of F follows from the continuity of f . Next, we prove that F is uniformly bounded on B * . Let any u 2 B * . If t 2 [−θ, 0]. Then jFu(t)j = j (t)j ≤ j (t)j ≤ sup j (t)j ≤ ξ , t2[−θ,0] which implies that kFuk ≤ ξ . If t 2 (0, T ] then we have kFuk = kFuk ≤ maxfY , Y g . Λ C 1 2 1−η,ρ Consequently, F is uniformly bounded on B . Now, we demonstrate that QB is equicontinuous. Let any * * ξ ξ u 2 B and 0 < t < t ≤ T. Then 1 2 1−η  1−η ρ ρ t t 2 1 F(x)(t ) − F(x)(t ) ≤ 2 1 ρ ρ 1−η Z λ−1 ρ t − s ρ−1 s jh(s)jds Γ(λ) ρ 1 On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 91 1−η λ−1 ρ ρ 1 t t − s ρ−1 2 2 + s ρ ρ Γ(λ) 1−η λ−1 ρ ρ t t − s ρ−1 1 1 − s jh(s)jds ρ ρ 1−η maxfΘ ; Θ g Γ(η) λ+η−1 2 1 ρ ρ ρ t − t 2 1 Γ(λ + η) Z 1−η λ−1 ρ ρ t t − s maxfΘ ; Θ g 2 1 ρ−1 2 2 + s Γ(λ) ρ ρ 1−η λ−1 ρ ρ η−1 t t − s ρ−1 1 1 − s ds. ρ ρ ρ Note that 1−η  1−η ρ ρ t t 2 1 F(x)(t ) − F(x)(t ) ! 0 as t ! t . 2 1 2 1 ρ ρ This shows that F is equicontinuous on [−θ, T ]. Therefore, F is relatively compact on B . By C , type 1−η Arzela–Ascoli Theorem F is compact on B . As a consequence of Krasnoselskii’s xed point theorem [28], we conclude that G has at least a xed point x 2 Λ. Using Lemma 3.2, we conclude that the problem (1)−(3) have at least one solution in the space Λ. 4 An example Consider the following terminal value problem: " # ,0 te jx j 1 ln( t + 1) p p D x − − = (23) + t p p 200 (1 +jx j) 2 t 2 t 2 +jx j + , t 2 (0, 2], −t+3 84e 1 +jx j ( ) x(2) = c 2 R, (24) x(t) = (t), t 2 [−θ, 0], θ > 0. (25) where 2 C([−θ, 0],R). Set 2 +juj ln( t + 1) f (t, u) = + p , t 2 (0, 2], u 2 C([−θ, 0],R), −t+3 84e (1 +juj) 2 t and te jwj 1 g(t, w) = + p , t 2 (0, 2], w 2 C([−θ, 0],R). 200 1 +jwj ( ) 2 t We have ζ (1−λ) C ([0, 2]) = C ([0, 2]) = C 1 1 ([0, 2]), 1 1 1−η,ρ , , 2 2 2 2 1 2 with η = λ = ρ = and ζ = 0. Clearly, the functions f 2 C ([0, 2]) and g 2 C ([0, 2]). Hence condition 1 1 1 1 2 2 2 2 (B1) is satised. For each u, u 2 C([−θ, 0],R) and t 2 (0, 2] we get jf (t, u) − f (t, u ¯ )j ≤ ku − u ¯k −t+3 84e 92 Ë S. Bouriah et al. ≤ ku − u ¯k 84e and jg(t, u) − g(t, u ¯ )j ≤ ku − u ¯k . 1 e Hence condition (B2) is satised with β = and β = . 1 2 84e 100 The condition ( ) λ 1−η 1−η+λ ρ ρ ρ β Γ(η) T T β T 1 1 1 max β + ; β + ≈ 0.1358 < , 2 2 Γ(λ + η) ρ ρ Γ(λ + 1) ρ 2 is satised with T = 2. It follows from Theorem 3.3 that the problem (24)-(25) has a unique solution. Conict of interest: The authors declare that they have no conict of interest. References [1] S. Abbes, M. Benchohra and G M. N’Guérékata, Topics in Fractional Dierential Equations, Springer-Verlag, New York, 2012. [2] S. Abbes, M. Benchohra and G M. N’Guérékata, Advanced Fractional Dierential and Integral Equations, Nova Science Pub- lishers, New York, 2014. [3] A. A. Kilbas, Hari M. Srivastava, and Juan J. Trujillo, Theory and Applications of Fractional Dierential Equations. North- Holland Mathematics Studies, 204. Elsevier Science B.V., Amsterdam, 2006. [4] I. Podlubny, Fractional Dierential Equations, Academic Press, San Diego, 1999. 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Odzijewicz, Fractional dierential equations with dependence on the Caputo- Katugampola derivative, J. Comput. Nonlinear Dynam. 11 (2016), 1-11. [28] Y.Zhou, Basic Theory of Fractional Dierential Equations. World scientic, 2014. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Topological Algebra and its Applications de Gruyter

On Nonlinear Implicit Neutral Generalized Hilfer Fractional Differential Equations with Terminal Conditions and Delay

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Topol. Algebra Appl. 2022; 10:77–93 Research Article Open Access Soufyane Bouriah, Abdelkrim Salim*, and Mouak Benchohra On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations with Terminal Conditions and Delay https://doi.org/10.1515/taa-2022-0115 Received 30 May, 2022; accepted 14 June, 2022 Abstract: In this paper, we establish the existence of solutions for a class of nonlinear implicit neutral frac- tional dierential equations with terminal condition and Hilfer-Katugampola fractional derivative. The Ba- nach contraction principle and Krasnoselskii’s xed point theorem are used to support the arguments. An illustration is provided to demonstrate the relevance of our results. Keywords: Hilfer-Katugampola fractional derivative, implicit neutral fractional dierential equations, exis- tence, uniqueness, xed point MSC: 26A33, 34A08, 34K37 1 Introduction Fractional calculus is a generalization of ordinary dierentiation and integration to arbitrary order (non- integer). For some details and recent publications on the subject, see the books [1–4], the papers [5–23] and the references therein. In [13], Benchohra et al. considered the following terminal value problem with Hilfer-Katugampola type fractional derivative: λ,ζ λ,ζ ρ ρ D x (t) = f t, x(t), D x (t) , for each t 2 (a, T ], a > 0, + + a a x(T ) = c 2 R, λ,ζ where D is the Hilfer-Katugampola fractional derivative of order λ 2 (0, 1) and type ζ 2 [0, 1] and f : (a, T ] × R × R ! R is a given function. Their reasoning is mainly based upon dierent types of classical xed point theorems such as the Banach contraction principle and Krasnoselskii’s xed point theorem. Motivated by the works mentioned above, we establish in this paper existence and uniqueness results to the terminal value problem of the following Hilfer-Katugampola type fractional dierential equation: λ,ζ D x(t) − g(t, x ) = f t, x , t 2 (0, T ], 0 < T < ∞, (1) + t ( t ) Soufyane Bouriah: Department of Mathematics, Faculty of Exact Sciences and Informatics, Hassiba Benbouali University, P.O. Box 151 Chlef 02000, Algeria, E-mail: s.bouriah@univ-chlef.dz, bouriahsouane@yahoo.fr Laboratory of Mathematics and Applications Hassiba Benbouali University of Chlef Hay Salem, P.O. Box 151 Chlef Algeria *Corresponding Author: Abdelkrim Salim: Faculty of Technology, Hassiba Benbouali University, P.O. Box 151 Chlef 02000, Algeria, E-mail: salim.abdelkrim@yahoo.com, a.salim@univ-chlef.dz Laboratory of Mathematics, Djillali Liabes University of Sidi Bel-Abbes, P.O. Box 89 Sidi Bel Abbes 22000, Algeria Mouak Benchohra: Laboratory of Mathematics, Djillali Liabes University of Sidi Bel-Abbes, P.O. Box 89 Sidi Bel Abbes 22000, Algeria, E-mail: benchohra@yahoo.com Open Access. © 2022 Soufyane Bouriah et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution 4.0 License. 78 Ë S. Bouriah et al. x(T ) = c 2 R, (2) x(t) = (t), t 2 [−θ, 0], θ > 0, (3) λ,ζ 1−η ρ ρ where D , I are the Hilfer-Katugampola fractional derivative of order λ 2 (0, 1) and type ζ 2 [0, 1] + + 0 0 and Katugampola fractional integral of order 1− η, (η = λ + ζ − λζ ) respectively, f : (0, T ]× C([−θ, 0],R) ! R, and g : (0, T ] × C([−θ, 0],R) ! R are two given functions and 2 C([−θ, 0],R). For each function x dened on [−θ, T ] and for any t 2 [0, T ], we denote by x the element of C([−θ, 0],R) dened by: x (θ) = x(t + θ), θ 2 [−θ, 0], The following is how the current paper is arranged. In Section 2, we present certain notations and review some preliminary notions concerning the Hilfer-Katugampola fractional derivative and auxiliary ndings. Section 3 presents two solutions results to problem (1)-(3). In the nal part, we provide an illustration to demonstrate the application of our key results. 2 Preliminaries This part introduces the preliminary information that will be utilized throughout the study. Let 0 < T, J = [0, T ]. By C([−θ, 0],R), C(J,R) we denote the Banach spaces of all continuous functions from [−θ, 0] into R (resp from [0, T ] into R) with the norms: kxk = supjx(t)j, t 2 [−θ, 0] andkxk = supjx(t)j, t 2 [0, T ] (respectively). C ∞ We consider the weighted spaces of continuous functions C (J) = x : (0, T ] ! R : x(t) 2 C(J,R) , 0 ≤ η < 1, η,ρ and n o n n−1 (n) C (J) = x 2 C (J) : x 2 C (J) , n 2 N, η,ρ η,ρ C (J) = C (J), η,ρ η,ρ with the norms kxk = sup x(t) , η,ρ t2J and n−1 (k) (n) kxk n = kx k +kx k . C C η,ρ η,ρ k=0 Consider the space X (a, T ), (ϑ 2 R, 1 ≤ p ≤ ∞) of those complex-valued Lebesgue measurable functions f on [a, T ] for whichkfk < ∞, where the norm is dened by 0 1 dt ϑ p @ A kfk p = jt f (t)j , (1 ≤ p < ∞, ϑ 2 R). ϑ t p p In particular, when ϑ = , the space X (a, T ) coincides with the L (a, T ) space: X (a, T ) = L (a, T ). p 1 p p ϑ p On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 79 Denition 2.1 (Katugampola fractional integral [24, 25]). Let λ 2 R , ϑ 2 R and h 2 X (a, T ). The Katugam- pola fractional integral of order λ is given by λ−1 ρ ρ t − s h(s) ρ λ ρ−1 I +h (t) = s ds, t > a, ρ > 0, ρ Γ(λ) λ−1 −t where Γ(·) is the Euler gamma function dened by Γ(λ) = t e dt, λ > 0. Denition 2.2 (Katugampola fractional derivative [24, 25]). Let λ 2 R \N and ρ > 0. The Katugampola frac- ρ λ tional derivative D of order λ is dened by ρ λ n ρ n−λ + + D h (t) = δ ( I h)(t) a ρ a n n−λ−1 ρ ρ d t − s h(s) 1−ρ ρ−1 = t s ds, t > a, ρ > 0, dt ρ Γ(n − λ) n 1−ρ where n = [λ] + 1 and δ = t . dt Theorem 2.3 ([25]). Let λ > 0, ζ > 0, a < T < ∞ and ρ, ϑ 2 R, ρ ≥ ϑ. Then, for h 2 X (a, T ) the semigroup property is valid, i.e. ζ λ+ζ ρ λ ρ ρ I + I h (t) = I h (t). + + a a ρ λ Lemma 2.4 ([24–26]). Let λ > 0, and 0 ≤ η < 1. Then, I is bounded from C (J) into C (J). + η,ρ η,ρ ρ λ Lemma 2.5 ([26]). Let a < T < ∞, λ > 0, 0 ≤ η < 1 and x 2 C (J). If λ > η, then I x is continuous on J and η,ρ + ρ λ ρ λ I +x (a) = lim I +x (t) = 0. a a t!a Lemma 2.6 ([27]). For λ ≥ 0 and ζ > 0, we have " # ζ−1 λ+ζ−1 ρ ρ ρ ρ s − a Γ(ζ ) t − a ρ λ I + (t) = ρ ρ Γ(λ + ζ ) " # λ−1 ρ ρ s − a ρ λ D (t) = 0, 0 < λ < 1. Lemma 2.7 ([26]). Let λ > 0, 0 ≤ η < 1 and h 2 C [a, T ]. Then, ρ λ ρ λ + + D I h (t) = h(t), for all t 2 (a, T ]. a a ρ 1−λ 1 Lemma 2.8 ([26]). Let 0 < λ < 1, 0 ≤ η < 1. If h 2 C [a, T ] and I + h 2 C [a, T ], then η,ρ η,ρ ρ 1−λ λ−1 I + h (a) ρ ρ t − a ρ λ ρ λ I + D +h (t) = h(t) − , for all t 2 (a, T ]. a a Γ(λ) ρ Denition 2.9 ([26]). Let λ and ζ verify n − 1 < λ < n and 0 ≤ ζ ≤ 1, with n 2 N. The Hilfer-Katugampola fractional derivative to of a function h 2 C [a, T ], is given by 1−η,ρ λ,ζ ζ (n−λ) (1−ζ )(n−λ) ρ ρ ρ−1 ρ D h (t) = I t I h (t) + + + a a a dt ζ (n−λ) (1−ζ )(n−λ) ρ n ρ = I δ I h (t). + + a a 80 Ë S. Bouriah et al. λ,ζ Property 2.10 ([26]). The operator D can be written as ρ λ,ζ ρ ζ (1−λ) ρ 1−η ρ ζ (1−λ) ρ η D = I δ I = I D , η = λ + ζ − λζ . + + ρ + + + a a a a a Consider the following parameters λ, ζ , η satisfying η = λ + ζ − λζ , 0 < λ, ζ , η < 1. Thus, we dene the spaces n o λ,ζ λ,ζ C (J) = x 2 C (J), D x 2 C (J) 1−η,ρ 1−η,ρ 1−η,ρ a and η η C (J) = x 2 C (J), D x 2 C (J) . 1−η,ρ + 1−η,ρ 1−η,ρ a ρ λ,ζ ρ η(1−λ) ρ η Since D x = I D x, it follows from Lemma 2.4 that + + + a a a η λ,ζ C (J)  C (J)  C (J). 1−η,ρ 1−η,ρ 1−η,ρ Lemma 2.11 ([26]). Let 0 < λ < 1, 0 ≤ ζ ≤ 1 and η = λ + ζ − λζ . If x 2 C (J), then 1−η,ρ ρ η ρ η ρ λ ρ λ,ζ I D x = I + D x + + + a a a and η ζ (1−λ) ρ ρ λ ρ D I +x = D x. + + a a 3 Existence of Solutions Let Λ be the Banach space dened by: Λ = fx : [−θ, T ] ! R : xj 2 C([−θ, 0],R) and xj 2 C (J)g, [−θ,0] (0,T ] 1−η,ρ with the norm kxk = kxk +kxk . Λ C C 1−η,ρ We consider the following problem: λ,ζ D x(t) − g(t, x ) = ϖ(t), t 2 (0, T ], (4) x(T ) = c, c 2 R, (5) x(t) = (t), t 2 [−θ, 0]. (6) where ϖ(·) 2 C (J) and g(·, x ) 2 C (J). The following theorem shows that (4)-(6) is equivalent to the 1−η,ρ 1−η,ρ integral equation: > (t), if t 2 [−θ, 0], 2 3 > T > 1−η λ−1 η−1 < ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 c − g(T, x ) − s ϖ(s)ds x(t) = T (7) ρ ρ ρ Γ(λ) > Z λ−1 > ρ ρ 1 t − s ρ−1 +g(t, x ) + s ϖ(s)ds, t 2 (0, T ]. > t Γ(λ) ρ 0 On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 81 Theorem 3.1. Let η = λ + ζ − λζ , where 0 < λ < 1 and 0 ≤ ζ ≤ 1. If ϖ : (0, T ] ! R is a function where ϖ(·) 2 C (J), then x 2 Λ satises the problem (4)-(6) if and only if it satises Equation (7). 1−η,ρ Proof. ()) Let x 2 Λ, if t 2 [−θ, 0], then we have x(t) = (t). On the other hand, for t 2 (0, T ], we have x 2 C (J) a solution of the equations (4) and (5). We prove that 1−η,ρ x is also a solution of (7). From the denition of C (J), Lemma 2.4, and using Denition 2.2, we have 1−η,ρ ρ 1−η I x(·) − g(·, x ) 2 C(J,R) + · and η 1−η ρ ρ D x(·) − g(·, x ) = δ I x(·) − g(·, x ) 2 C (J). (8) + · ρ + · 1−η,ρ 0 0 By the denition of the space C (J), it follows that 1−η,ρ ρ 1−η 1 I x(·) − g(·, x ) 2 C (J). + · 1−η,ρ Using Lemma 2.8, with λ = η, we obtain: η η ρ ρ I D x(s) − g(s, x )) (t) = x(t) − g(t, x ) (9) + + s 0 0 1−η η−1 I x(s) − g(s, x ) (0) s ρ − , Γ(η) where t 2 (0, T ]. By hypothesis, x 2 C (J), using Lemma 2.11 with Equation (4), we have 1−η,ρ η η λ,ζ ρ ρ ρ λ ρ I D x(s) − g(s, x ) (t) = I + D x(s) − g(s, x ) (t) (10) + + s + s 0 0 0 ρ λ = I +ϖ (t). Comparing (9) and (10), we see that 1−η η−1 I x(s) − g(s, x ) (0) ρ + s ρ λ x(t) = + g(t, x ) + I +ϖ (t). (11) Γ(η) ρ Using Equation (5) we obtain: 2 3 1−η λ−1 η−1 ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 x(t) = c − g(T, x ) − s ϖ(s)ds ρ Γ(λ) ρ ρ λ−1 ρ ρ 1 t − s ρ−1 + g(t, x ) + s ϖ(s)ds, Γ(λ) where t 2 (0, T ], that is x(·) satises Equation (7). (() Let x 2 Λ, satisfying Equation (7). We show that x also satises the problem (4)-(6). If t 2 [−θ, 0], then x(t) = (t), so the condition 6 is satised. ( ) η η If t 2 (0, T ], then x 2 C (J). Apply operator D on both sides of Equation (7). Then, from Lemmas 2.6 1−η,ρ 0 and 2.11 we get η ζ (1−λ) ρ ρ D x(s) − g(s, x ) (t) = D ϖ (t). (12) + s + 0 0 By Equation (8) we have D x(·) − g(·, x ) 2 C (J). Then, Equation (12) implies + · 1−η,ρ ρ η ρ 1−ζ (1−λ) ρ ζ (1−λ) D x(s) − g(s, x ) (t) = δ I ϖ (t) = D ϖ (t) 2 C (J). (13) + s ρ + + 1−η,ρ 0 0 0 As ϖ(·) 2 C (J) and from Lemma 2.4, it follows that 1−η,ρ 1−ζ (1−λ) I ϖ 2 C (J). (14) 1−η,ρ 0 82 Ë S. Bouriah et al. From Equations (13) and (14) and by the Denition of the space C (J), we obtain 1−η,ρ 1−ζ (1−λ) ρ 1 I ϖ 2 C (J). 1−η,ρ ζ (1−λ) Applying operator I on both sides of Equation (13) and using Lemmas 2.5 and 2.8, we have 1−ζ (1−λ) ζ (1−λ)−1 I ϖ(t) (0) + ρ ρ ζ (1−λ) ρ η I D x(t) − g(t, x ) = ϖ(t) + , + + 0 0 Γ(ζ (1 − λ)) which implies that: λ,ζ D x(t) − g(t, x ) = ϖ(t), + t that is, Equation(4) holds. Clearly, if x 2 C (J) satises Equation(7), then it also satises Equation(5). 1−η,ρ Suppose that the function f : (0, T ] × C([−θ, 0],R) ! R is continuous and veries the requirements: (B1)Set the functions f : (0, T ] × C([−θ, 0],R) ! R and g : (0, T ] × C([−θ, 0],R) ! R such that: ζ (1−λ) η f (·, u ) 2 C (J) and g(·, u ) 2 C (J) for any u 2 C (J). · · 1−η,ρ 1−η,ρ 1−η,ρ (B2)There exist constants β > 0 and 0 < β < 1 such that: 1 2 jf (t, u) − f (t, u ¯ )j ≤ β ku − u ¯k and jg(t, w) − g(t, w ¯ )j ≤ β kw − w ¯k 2 C for any u, w, u ¯ , w ¯ 2 C([−θ, 0],R) and t 2 (0, T ]. As a consequence of Theorem 3.1, we have Theorem 3.2. Theorem 3.2. Let η = λ + ζ − λζ where 0 < λ < 1 and 0 ≤ ζ ≤ 1; let f : (0, T ] × C([−θ, 0],R) ! R be a function ζ (1−λ) η such that f (·, x ) 2 C (J)  C (J) and g : (0, T ] × C([−θ, 0],R) ! R where g(·, x ) 2 C (J) for any · · 1−η,ρ 1−η,ρ 1−η,ρ x 2 C (J). 1−η,ρ If x 2 Λ, then x satises the problem (1)−(3) if and only if x is the xed point of the operator G : Λ ! Λ dened by: (t) if t 2 [−θ, 0] 2 3 > 1−η λ−1 η−1 < ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 c − g(T, x ) − s h(s)ds Gx(t) = T (15) ρ ρ ρ Γ(λ) > Z λ−1 > ρ ρ 1 t − s ρ−1 +g(t, x ) + s h(s)ds, t 2 (0, T ], > t Γ(λ) ρ where: h : (0, T ] ! R be a function verifying h(t) = f (t, x ). Clearly, h 2 C (J). 1−η,ρ The following proof is to show that for any x 2 Λ then Gx 2 Λ, i.e. the operator G is well dened. Proof. Let x 2 Λ, if t 2 [−θ, 0] then Gx(t) = (t). If t 2 (0, T ] then we have: 2 3 1−η λ−1 η−1 ρ ρ ρ ρ T 1 T − s t ρ−1 4 5 Gx(t) = c − g(T, x ) − s h(s)ds ρ ρ ρ Γ(λ) λ−1 ρ ρ 1 t − s ρ−1 + g(t, x ) + s h(s)ds. Γ(λ) ρ 0 On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 83 Applying D to both sides and by Lemmas 2.6 and 2.11, we have: η η η ρ ρ ρ ρ λ D Gx(t) = D g(t, x ) + D I +f (s, x ) (t) + + t + s 0 0 0 η ζ (1−λ) ρ ρ = D g(t, x ) + D f (s, x ) (t). + t + s 0 0 ζ (1−λ) η Since η ≥ λ, and f (·, x ) 2 C (J), the right hand side is in C (J) and thus D Gx 2 C (J), which · + 1−η,ρ 1−η,ρ 1−η,ρ 0 implies that Gx 2 C (J). We can conclude that Gx 2 Λ. 1−η,ρ Based on Banach’s xed point, we now present and verify our existence result for problem (1)-(3). Theorem 3.3. Suppose that (B1) and (B2) are met. If ( ) λ 1−η 1−η+λ ρ ρ ρ β Γ(η) T T β T 1 1 1 max β + ; β + < , (16) 2 2 Γ(λ + η) ρ ρ Γ(λ + 1) ρ 2 then the problem (1)−(3) has unique solution in Λ. Proof. We show that the operator G dened in Equation (15) has a unique xed point x in Λ. Let x, u 2 Λ. If t 2 [−θ, 0], then jGx(t) − Gu(t)j = 0. For t 2 (0, T ], we have jGx(t) − Gu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ 1 T t T − s ρ−1 ≤ s jh(s) − ϖ(s)jds ρ ρ ρ Γ(λ) λ−1 ρ ρ 1 t − s ρ−1 + jg(t, x ) − g(t, u )j + s jh(s) − ϖ(s)jds t t Γ(λ) ρ 1−η η−1 ρ ρ T t + jg(T, x ) − g(T, u )j, T T ρ ρ where h, ϖ 2 C (J) such that 1−η,ρ h(t) = f (t, x ) ϖ(t) = f (t, u ). By (B2), we have jh(t) − ϖ(t)j = jf (t, x ) − f (t, u )j t t ≤ β kx − u k . 1 t t C On the other hand, we havekx − u k = sup jx (θ) − u (θ)j, θ 2 [−θ, 0] , then there exist θ 2 [−θ, 0] such t t C t t that * * * kx − u k = jx (θ ) − u (θ )j = jx(t + θ ) − u(t + θ )j. t t C t t * If t + θ 2 [−θ, 0], then kx − u k ≤ kx − uk = kx − uk , t t C C Λ 84 Ë S. Bouriah et al. which implies that for each t 2 (0, T ], we have jGx(t) − Gu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s ρ ρ ρ Γ(λ) λ−1 ρ ρ β t − s 1 ρ−1 + β kx − u k + s kx − u k ds 2 t t s s C C Γ(λ) ρ 1−η η−1 ρ ρ T t + β kx − u k 2 T T C ρ ρ 1−η η−1 λ ρ ρ ρ β T t T ≤ kx − uk Γ(λ + 1) ρ ρ ρ β t + β kx − uk + kx − uk Λ Λ Γ(λ + 1) ρ 1−η η−1 ρ ρ T t + β kx − uk . ρ ρ Hence: 1−η 1−η+λ 1−η ρ ρ ρ t β T t Gx(t) − Gu(t) ≤ + β ρ Γ(λ + 1) ρ ρ 1−η+λ 1−η ρ ρ β t T + + β kx − uk Γ(λ + 1) ρ ρ " # 1−η 1−η+λ ρ ρ T β T ≤ 2 β + kx − uk , 2 Λ ρ Γ(λ + 1) ρ which implies that " # 1−η 1−η+λ ρ ρ T β T kGx − Guk = kGx − Guk ≤ 2 β + kx − uk . C Λ 2 Λ 1−η,ρ ρ ρ Γ(λ + 1) If t + θ 2 [0, T ], then we have η−1 1−η η−1 η−1 ρ ρ ρ ρ t t t t kx − u k = kx − u k ≤ kx − uk ≤ kx − uk . t t C t t C C Λ 1−η,ρ ρ ρ ρ ρ Hence, for each t 2 (0, T ] we obtain jGx(t) − Gu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s Γ(λ) ρ ρ ρ λ−1 ρ ρ β t − s 1 ρ−1 + β kx − u k + s kx − u k ds s s 2 t t C C Γ(λ) 1−η η−1 ρ ρ T t + β kx − u k 2 T T C ρ ρ 1−η η−1 η−1 ρ ρ ρ T t s ρ λ ≤ β kx − uk I + (T ) 1 C 0 1−η,ρ ρ ρ ρ η−1 η−1 ρ ρ t s + 2β kx − uk + β I + (t)kx − uk . 2 C 1 C 1−η,ρ 0 1−η,ρ ρ ρ By Lemma 2.6, we have λ η−1 ρ ρ β Γ(η) T t jGx(t) − Gu(t)j ≤ Γ(λ + η) ρ ρ On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 85 η−1 λ+η−1 ρ ρ t β Γ(η) t + 2β + kx − uk , 2 C 1−η,ρ ρ Γ(λ + η) ρ hence 1−η Gx(t) − Gu(t) " # λ λ ρ ρ β Γ(η) T β Γ(η) t 1 1 ≤ + 2β + kx − uk 1−η,ρ Γ(λ + η) ρ Γ(λ + η) ρ " # β Γ(η) T ≤ 2 β + kx − uk , 2 Λ Γ(λ + η) ρ which implies that " # β Γ(η) T kGx − Guk = kGx − Guk ≤ 2 β + kx − uk . C Λ Λ 1−η,ρ Γ(λ + η) ρ By Equations (16), the operator G is a contraction. Hence, by Banach’s contraction principle, G has a unique xed point x 2 Λ. As a consequence of Theorem 3.2, we can conclude that the problem (1)−(3) have a unique solution in Λ. We present now the second result, which is based on Krasnoselskii xed point theorem [28]. Theorem 3.4. Assume (B1) and (B2) hold. If: " # " # 1−η λ λ+1−η ρ ρ ρ T 2β Γ(η) T T β 1 + + + < 1, ρ Γ(λ + η) ρ ρ and λ λ+1−η 1−η ρ ρ ρ β Γ(η) T β T T 1 1 max + 2β ; + 2β 2 2 ρ ρ ρ Γ(λ + η) Γ(λ + 1) := max { ;{ < 1, (17) ( ) 2 1 then the problem (1)−(3) have at least one solution. Proof. Consider the set B * = fx 2 Λ : jjxjj ≤ ξ g, ξ Λ where 8 9 1−η * λ ρ ρ < 2Γ(η)f = T * T jcj+2g + ρ Γ(λ+η) ρ ξ ≥ max ; sup j (t)j , 1−η λ λ+1−η ρ ρ ρ 2β Γ(η) : T 1 T T ; 1− β 1+ + + t2[−θ,0] ρ Γ(λ+η) ρ ρ 1−η f = sup jf (t, 0)j, t2J and 1−η g = sup jg(t, 0)j. t2J We dene the operators H and F on B * by: (t), if t 2 [−θ, 0] 2 3 1−η λ−1 η−1 Hx(t) = ρ ρ ρ ρ (18) T 1 T − s t ρ−1 4 5 > c − g(T, x ) − s h(s)ds > ρ ρ ρ Γ(λ) +g(t, x ), t 2 (0, T ], t 86 Ë S. Bouriah et al. (t), if t 2 [−θ, 0] Fx(t) = (19) λ−1 ρ ρ > 1 t − s ρ−1 s h(s)ds, t 2 (0, T ]. : Γ(λ) ρ Then the fractional integral equation (15) can be written as the operator equation: Gx(t) = Hx(t) + Fx(t), t 2 [−θ, T ]. The proof will be given in several steps: Step 1: We prove that Hx + Fu 2 B * for any x, u 2 B * . ξ ξ 1 1 If t 2 [−θ, 0] thenjHx(t)j ≤ sup j (t)j andjFu(t)j ≤ sup j (t)j, 2 2 t2[−θ,0] t2[−θ,0] which implies that 1 1 kHx + Fuk ≤ sup j (t)j + sup j (t)j 2 2 t2[−θ,0] t2[−θ,0] = sup j (t)j t2[−θ,0] ≤ ξ . 1−η If t 2 (0, T ] then multiplying both sides of Equation (18) by , we have 2 3 1−η 1−η λ−1 ρ ρ ρ ρ t T 1 T − s ρ−1 4 5 Hx(t) = c − g(T, x ) − s h(s)ds ρ ρ ρ Γ(λ) 1−η +g(t, x ) . Then 1−η 1−η λ−1 ρ ρ ρ ρ t T 1 T − s ρ−1 Hx(t) ≤ s jh(s)jds ρ ρ Γ(λ) ρ 1−η +jcj +jg(T, x )j + jg(t, x ) − g(t, 0)j +jg(t, 0)j T t 2 3 1−η λ−1 ρ ρ ρ T 1 T − s ρ−1 4 5 ≤ s jh(s)jds +jcj ρ Γ(λ) ρ " # η−1 1−η ρ ρ t t * * + g + β kx k + g . 2 t ρ ρ Thus 2 3 1−η 1−η λ−1 ρ ρ ρ ρ t T 1 T − s ρ−1 4 5 Hx(t) ≤ jcj + s jh(s)jds ρ ρ Γ(λ) ρ (20) 1−η + 2g + β kx k . 2 C By (B2), we have for each t 2 (0, T ], jh(t)j = jf (t, x ) − f (t, 0) + f (t, 0)j t On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 87 ≤ jf (t, x ) − f (t, 0)j +jf (t, 0)j η−1 ≤ β kx k + f . 1 t C 1−η Multiplying both sides of the above inequality by , we get 1−η 1−η ρ ρ t t h(t) ≤ f + β kx k . 1 t ρ ρ If t + θ 2 [−θ, 0] thenkx k = kxk = kxk , which implies that for each t 2 (0, T ], we have t C C Λ 1−η 1−η ρ ρ t t h(t) ≤ f + β kxk ρ ρ 1−η * * ≤ f + β ξ := Θ . (21) By replacing (21) in the inequality (20) and using Lemma 2.6, we have " # 1−η 1−η λ+η−1 ρ ρ ρ t T Θ Γ(η) T 1 * Hx(t) ≤ jcj + + β ξ ρ ρ ρ Γ(λ + η) +2g . This gives " # 1−η λ+η−1 ρ ρ T Θ Γ(η) T 1 * * jjHxjj ≤ jcj + + β ξ + 2g := Ω . C 2 1 1−η,ρ ρ ρ Γ(λ + η) If t + θ 2 [0, T ], then η−1 1−η ρ ρ t t kx k = kx k t t C C ρ ρ η−1 ≤ kxk . Thus, for each t 2 (0, T ], we have 1−η * * h(t) ≤ f + β ξ := Θ . (22) 1 2 By replacing (22) in the inequality (20) and using Lemma 2.6, we obtain " # 1−η 1−η λ+η−1 ρ ρ ρ t T Θ Γ(η) T Hx(t) ≤ jcj + ρ ρ Γ(λ + η) ρ * * + β ξ + 2g . Hence " # 1−η λ+η−1 ρ ρ T Θ Γ(η) T 2 * * jjHxjj ≤ jcj + + β ξ + 2g := Ω . 2 2 1−η,ρ ρ Γ(λ + η) ρ If t + θ 2 [−θ, 0] with t 2 (0, T ] then using Equation (21) and Lemma 2.6, we have " # 1−η λ+η−1 * * ρ ρ Γ(η)f β Γ(η)ξ T t jF(u)(t)j ≤ + . Γ(λ + η) Γ(λ + η) ρ ρ 88 Ë S. Bouriah et al. Therefore " # 1−η 1−η λ ρ ρ ρ * * t Γ(η)f β Γ(η)ξ T t Fu(t) ≤ + , ρ ρ ρ Γ(λ + η) Γ(λ + η) " # 1−η λ * * ρ ρ Γ(η)f β Γ(η)ξ T T ≤ + . Γ(λ + η) Γ(λ + η) ρ ρ Thus " # 1−η λ ρ ρ * * Γ(η)f β Γ(η)ξ T T kFuk ≤ + := Y . C 1 1−η,ρ Γ(λ + η) Γ(λ + η) ρ ρ If t + θ 2 [0, T ] then for each t 2 (0, T ] and by using Equation (22) and Lemma 2.6, we have λ+η−1 * * ρ Γ(η)f β Γ(η)ξ t jF(u)(t)j ≤ + . Γ(λ + η) Γ(λ + η) Therefore 1−η λ ρ * * ρ t Γ(η)f β Γ(η)ξ t Fu(t) ≤ + , ρ ρ Γ(λ + η) Γ(λ + η) * * ρ Γ(η)f β Γ(η)ξ T ≤ + . Γ(λ + η) Γ(λ + η) ρ Thus * * Γ(η)f β Γ(η)ξ T kFuk ≤ + := Y . 1−η,ρ Γ(λ + η) Γ(λ + η) ρ Consequently, for every x, u 2 B * we obtain kHx + Fuk ≤ kHxk +kFuk C C C 1−η,ρ 1−η,ρ 1−η,ρ ≤ maxfΩ ; Ω g + maxfY ; Y g 1 2 1 2 " # λ λ λ+1−η ρ ρ ρ * * 2Γ(η)f T 2β Γ(η)ξ T T ≤ + + ρ ρ ρ Γ(λ + η) Γ(λ + η) " # 1−η 1−η ρ ρ T T * * + jcj + 2g + β ξ 1 + . ρ ρ Since 1−η λ ρ * ρ T 2Γ(η)f T jcj + 2g + ρ ρ Γ(λ + η) " " # " ## ξ ≥ , 1−η λ λ+1−η ρ ρ ρ T 2β Γ(η) T T 1 − β 1 + + + ρ Γ(λ + η) ρ ρ we have kHx + Fuk = kHx + Fuk ≤ ξ , C Λ 1−η,ρ which implies that for each t 2 [−θ, T ] we get kHx + Fuk ≤ ξ , which infers that Hx + Fu 2 B . Step 2: H is a contraction. Let x, u 2 Λ. If t 2 [−θ, 0]. Then, we have jHx(t) − Hu(t)j = 0. For t 2 (0, T ], we have On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 89 jHx(t) − Hu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ 1 T t T − s ρ−1 ≤ s jh(s) − ϖ(s)jds Γ(λ) ρ ρ ρ 1−η η−1 ρ ρ T t + jg(t, x ) − g(t, u )j + jg(T, x ) − g(T, u )j t t T T ρ ρ 1−η η−1 λ−1 ρ ρ ρ ρ 1 T t T − s ρ−1 ≤ s jh(s) − ϖ(s)jds Γ(λ) ρ ρ ρ 1−η η−1 ρ ρ T t + β kx − u k + β kx − u k . 2 t t C 2 T T C ρ ρ where h, ϖ 2 C (J) such that 1−η,ρ h(t) = f (t, x ), ϖ(t) = f (t, u ). By (B2), we have jh(t) − ϖ(t)j = jf (t, x ) − f (t, u )j t t ≤ β kx − u k . 1 t t If t + θ 2 [−θ, 0] then kx − u k ≤ kx − uk = kx − uk , t t C C Λ which implies that for each t 2 (0, T ] we get jHx(t) − Hu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s ρ ρ ρ Γ(λ) 1−η η−1 ρ ρ T t + β kx − u k + β kx − u k 2 t t C 2 T T C ρ ρ λ+1−η η−1 ρ ρ β T t ≤ kx − uk + β kx − uk Λ Λ Γ(λ + 1) ρ ρ 1−η η−1 ρ ρ T t + β kx − uk . ρ ρ Hence 1−η λ+1−η 1−η ρ ρ ρ t β T t Hx(t) − Hu(t) ≤ + β ρ Γ(λ + 1) ρ ρ 1−η +β kx − uk , 2 Λ which implies that " # λ+1−η 1−η ρ ρ β T T kHx − Huk ≤ + 2β := { kx − uk . 2 1 Λ 1−η,ρ Γ(λ + 1) ρ ρ If t + θ 2 [0, T ] then η−1 η−1 ρ ρ t t kx − u k ≤ kx − uk = kx − uk . t t C C Λ 1−η,ρ ρ ρ 90 Ë S. Bouriah et al. Therefore, for each t 2 (0, T ], we obtain jHx(t) − Hu(t)j 1−η η−1 λ−1 ρ ρ ρ ρ β T t T − s 1 ρ−1 ≤ s kx − u k ds s s Γ(λ) ρ ρ ρ 1−η η−1 ρ ρ T t + β kx − u k + β kx − u k 2 t t C 2 T T C ρ ρ 1−η η−1 η−1 ρ ρ ρ T t s ρ λ ≤ β kx − uk I + (T ) Λ 0 ρ ρ ρ η−1 + 2β kx − uk . By Lemma 2.6, we have λ η−1 ρ ρ β Γ(η) T t jHx(t) − Hu(t)j ≤ kx − uk ρ ρ Γ(λ + η) η−1 + 2β kx − uk . Hence " # 1−η λ ρ ρ t β Γ(η) T Hx(t) − Hu(t) ≤ + 2β kx − uk , 2 Λ ρ Γ(λ + η) ρ which implies that " # β Γ(η) T kHx − Huk ≤ + 2β kx − uk := { kx − uk . C 2 Λ 2 Λ 1−η,ρ Γ(λ + η) Thus kHx − Huk = kHx − Huk ≤ max ({ ,{ )kx − uk . C Λ 1 2 Λ 1−η,ρ By Equation (17) the operator H is a contraction. Step 3: F is compact and continuous. The continuity of F follows from the continuity of f . Next, we prove that F is uniformly bounded on B * . Let any u 2 B * . If t 2 [−θ, 0]. Then jFu(t)j = j (t)j ≤ j (t)j ≤ sup j (t)j ≤ ξ , t2[−θ,0] which implies that kFuk ≤ ξ . If t 2 (0, T ] then we have kFuk = kFuk ≤ maxfY , Y g . Λ C 1 2 1−η,ρ Consequently, F is uniformly bounded on B . Now, we demonstrate that QB is equicontinuous. Let any * * ξ ξ u 2 B and 0 < t < t ≤ T. Then 1 2 1−η  1−η ρ ρ t t 2 1 F(x)(t ) − F(x)(t ) ≤ 2 1 ρ ρ 1−η Z λ−1 ρ t − s ρ−1 s jh(s)jds Γ(λ) ρ 1 On Nonlinear Implicit Neutral Generalized Hilfer Fractional Dierential Equations Ë 91 1−η λ−1 ρ ρ 1 t t − s ρ−1 2 2 + s ρ ρ Γ(λ) 1−η λ−1 ρ ρ t t − s ρ−1 1 1 − s jh(s)jds ρ ρ 1−η maxfΘ ; Θ g Γ(η) λ+η−1 2 1 ρ ρ ρ t − t 2 1 Γ(λ + η) Z 1−η λ−1 ρ ρ t t − s maxfΘ ; Θ g 2 1 ρ−1 2 2 + s Γ(λ) ρ ρ 1−η λ−1 ρ ρ η−1 t t − s ρ−1 1 1 − s ds. ρ ρ ρ Note that 1−η  1−η ρ ρ t t 2 1 F(x)(t ) − F(x)(t ) ! 0 as t ! t . 2 1 2 1 ρ ρ This shows that F is equicontinuous on [−θ, T ]. Therefore, F is relatively compact on B . By C , type 1−η Arzela–Ascoli Theorem F is compact on B . As a consequence of Krasnoselskii’s xed point theorem [28], we conclude that G has at least a xed point x 2 Λ. Using Lemma 3.2, we conclude that the problem (1)−(3) have at least one solution in the space Λ. 4 An example Consider the following terminal value problem: " # ,0 te jx j 1 ln( t + 1) p p D x − − = (23) + t p p 200 (1 +jx j) 2 t 2 t 2 +jx j + , t 2 (0, 2], −t+3 84e 1 +jx j ( ) x(2) = c 2 R, (24) x(t) = (t), t 2 [−θ, 0], θ > 0. (25) where 2 C([−θ, 0],R). Set 2 +juj ln( t + 1) f (t, u) = + p , t 2 (0, 2], u 2 C([−θ, 0],R), −t+3 84e (1 +juj) 2 t and te jwj 1 g(t, w) = + p , t 2 (0, 2], w 2 C([−θ, 0],R). 200 1 +jwj ( ) 2 t We have ζ (1−λ) C ([0, 2]) = C ([0, 2]) = C 1 1 ([0, 2]), 1 1 1−η,ρ , , 2 2 2 2 1 2 with η = λ = ρ = and ζ = 0. Clearly, the functions f 2 C ([0, 2]) and g 2 C ([0, 2]). Hence condition 1 1 1 1 2 2 2 2 (B1) is satised. For each u, u 2 C([−θ, 0],R) and t 2 (0, 2] we get jf (t, u) − f (t, u ¯ )j ≤ ku − u ¯k −t+3 84e 92 Ë S. 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Journal

Topological Algebra and its Applications – de Gruyter

Published: Jan 1, 2022

Keywords: Hilfer-Katugampola fractional derivative; implicit neutral fractional differential equations; existence; uniqueness; fixed point; 26A33; 34A08; 34K37

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