CUBO, A Mathematical Journal Vol.22, No¯ 01, (125–136). April 2020 http: // doi. org/ 10. 4067/ S0719-06462020000100125 On Katugampola fractional order derivatives and Darboux problem for differential equations Djalal Boucenna1, Abdellatif Ben Makhlouf2 and Mohamed Ali Hammami3 1 Laboratory of Advanced Materials, Faculty of Sciences, Badji Mokhtar-Annaba University, P.O. Box 12, Annaba, 23000, Algeria. 2 Department of Mathematics, College of Science, Jouf University, Aljouf, Saudi Arabia Department of Mathematics, Faculty of Sciences of Sfax, Route Soukra, BP 1171, 3000 Sfax, Tunisia benmakhloufabdellatif@gmail.com 3 Department of Mathematics, Faculty of Sciences of Sfax, Route Soukra, BP 1171, 3000 Sfax, Tunisia ABSTRACT In this paper, we investigate the existence and uniqueness of solutions for the Darboux problem of partial differential equations with Caputo-Katugampola fractional deriva- tive. RESUMEN En este art́ıculo investigamos la existencia y unicidad de soluciones para el problema de Darboux de ecuaciones diferenciales parciales con derivada fraccional de Caputo- Katugampola. Keywords and Phrases: Darboux problem, Fractional differential equations, Caputo-Katugampola derivative. 2010 AMS Mathematics Subject Classification: 34A34, 34A08, 65L20. http://doi.org/10.4067/S0719-06462020000100125 126 Djalal Boucenna, Abdellatif Ben Makhlouf & Mohamed Ali Hammami CUBO 22, 1 (2020) 1 Introduction To investigate many different fields of science and engineering, the fractional calculus represents a powerful tool, with many applications in mathematical physics, hydrology, finance, astrophysics, thermodynamics, statistical mechanics, biophysics, control theory, cosmology, bioengineering and so on, [5, 6]. In recent years, there has been an important works in ordinary and partial fractional differential equations. For the Caputo fractional-order ordinary differential equations case, see Kilbas et al. [7], Miller and Ross [8]. In addition, Yunru Bai and Hua Kong have treated the existence of solution for nonlinear Caputo-Hadamard fractional differential equations in [9]. For the Caputo fractional-order partial differential equations case, see the work of Tian Liang Guo and KanJian Zhang in [10]. Furthermore, Xianmin Zhang has investigated the Caputo-Hadamard partial frac- tional differential equations in [11]. The choice of an appropriate fractional derivative (or integral) depends on the considered system, and for this reason there are a large number of works devoted to different fractional operators. Recently, U. Katugampola presented new types of fractional operators, which generalize both the Riemann-Liouville and Hadamard fractional operators [4]. Although the Katugampola frac- tional integral operator is an Erdélyi-Kober type operator [13] author in [14] argued that is not possible to obtain Hadamard equivalence operators from Erdélyi-Kober type operators. In this sense, Almeida, Malinowska and Odzijewicz [2] introduced a new fractional operator, called the Caputo-Katugampola derivative, which generalizes the concept of Caputo and Caputo-Hadamard fractional derivatives. It turns out that, the new operator is the left inverse of the Katugam- pola fractional integral and keeps some of the fundamental properties of the Caputo and Caputo- Hadamard fractional derivatives. Such derivative is the generalization of the Caputo and Caputo- Hadamard fractional derivative. The existence and uniqueness of the solution of the ordinary Caputo-Katugampola differential equations is given in [3]. A. Cernea in [12] studied a Darboux problem associated to a fractional hyperbolic integro-differential inclusion defined by Caputo- Katugampola fractional derivative and several existence results for this problem are proved. In this paper, we study the existence and uniqueness of solutions of the following partial differential equation with Caputo-Katugampola fractional derivative CD α,ρ a+ u (x,y) = f (x,y,u (x,y)) ,(x,y) ∈ J = [a1,b1] × [a2,b2] , (1.1) u (x,a2) = ϕ(x) , x ∈ [a1,b1] , u (a1,y) = ψ (y) , y ∈ [a2,b2] , ϕ(a1) = ψ (a2) , (1.2) CUBO 22, 1 (2020) On Katugampola fractional order derivatives and Darboux . . . 127 where f : J × R → R, ϕ : [a1,b1] → R and ψ : [a2,b2] → R are given continuous functions. The rest of the paper is organized as follows. Some definitions and preliminaries are presented in Sect. 2. Finally, the existence and uniqueness results, is given in Sect. 3. 2 Preliminaries In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. Definition 1. [2, 3, 4] Given α > 0, ρ > 0 and an interval [a,b] of R, where 0 < a < b. The Katugampola fractional integral of a function u ∈ L1([a,b]) is defined by I α,ρ a+ u (t) = ρ1−α Γ (α) t ∫ a sρ−1u (s) (tρ − sρ) 1−α ds, where Γ is the Gamma function. Definition 2. [2, 3, 4] Given α > 0, ρ > 0 and an interval [a,b] of R, where 0 < a < b. The Katugampola fractional derivative is defined by D α,ρ a+ u (t) = ρα Γ (1 − α) t 1−ρ d dt t ∫ a sρ−1u (s) (tρ − sρ) α ds. Definition 3. [2, 3, 4] Given 0 < α < 1, ρ > 0 and an interval [a,b] of R, where 0 < a < b. The Caputo-Katugampola fractional derivative is defined by C D α,ρ a+ u (t) =D α,ρ a+ [u (t) − u (a)] = ρα Γ (1 − α) t1−ρ d dt t ∫ a sρ−1[u (s) − u(a)] (tρ − sρ) α ds. Definition 4. Let 0 < ai < bi, i = 1,2 reals numbers, a = (a1,a2) and u : [a1,b1] × [a2,b2] → R be an integrable function. The mixed Katugampola fractional integrals of order α = (α1,α2) , and parameter ρ = (ρ1,ρ2) is defined by I α,ρ a+ u (x,y) = ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 sρ1−1tρ2−1 (xρ1 − sρ1) 1−α1 (yρ2 − tρ2) 1−α2 u (s,t)dtds. where α1,α2,ρ1 and ρ2 are strictly positives. Definition 5. Let 0 < ai < bi, i = 1,2 reals numbers, a = (a1,a2) and u : [a1,b1] × [a2,b2] → R be a function. The mixed Katugampola fractional derivative of order α = (α1,α2) , and parameter 128 Djalal Boucenna, Abdellatif Ben Makhlouf & Mohamed Ali Hammami CUBO 22, 1 (2020) ρ = (ρ1,ρ2) is defined by D α,ρ a+ u (x,y) = x1−ρ1y1−ρ2D2x,yI 1−α,ρ a+ u (x,y) = x1−ρ1y1−ρ2ρ α1 1 ρ α2 2 Γ (1 − α1) Γ (1 − α2) D 2 x,y ∫ x a + 1 ∫ y a + 2 sρ1−1tρ2−1 (xρ1 − sρ1) α1 (yρ2 − tρ2) α2 ×u (s,t)dtds. Where (α1,α2) ∈ (0,1) 2 , D2x,y = ∂ 2 ∂x∂y and ρ1, ρ2 are strictly positives. Definition 6. Let 0 < ai < bi, i = 1,2 reals numbers, a = (a1,a2) and u : [a1,b1] × [a2,b2] → R be a function. The mixed Caputo-Katugampola fractional derivative of order α = (α1,α2) , and parameter ρ = (ρ1,ρ2) is defined by CD α,ρ a+ u (x,y) = D α,ρ a+ (u (x,y) − u (x,a2) − u (a1,y) + u (a1,a2)) where (α1,α2) ∈ (0,1) 2 and ρ1, ρ2 are strictly positives. Lemma 2.1. Let 0 < ai < bi, i = 1,2 reals numbers, a = (a1,a2) and u : [a1,b1] × [a2,b2] → R is an absolutely continuous function. The mixed Caputo-Katugampola fractional derivative of order α = (α1,α2) , and parameter ρ = (ρ1,ρ2) is given by C D α,ρ a+ u (x,y) = I 1−α,ρ a+ ( x 1−ρ1y 1−ρ2D 2 x,yu (x,y) ) = ρ α1 1 ρ α2 2 Γ (1 − α1) Γ (1 − α2) ∫ x a + 1 ∫ y a + 2 D2s,tu (s,t) (xρ1 − sρ1) α1 (yρ2 − tρ2) α2 dtds almost everywhere, where (α1,α2) ∈ (0,1) 2 , D2s,t = ∂ 2 ∂s∂t and ρ1,ρ2 are strictly positives. Lemma 2.2. Let 0 < ai < bi, i = 1,2 reals numbers, a = (a1,a2) and u : [a1,b1] × [a2,b2] → R be an integrable function. Then I α,ρ a+ I β,ρ a+ u (x,y) = I α+β,ρ a+ u (x,y) (2.1) almost everywhere, where α = (α1,α2) , β = (β1,β2) and parameter ρ = (ρ1,ρ2). If additionally u is a continuous function, then the identity (2.1) holds everywhere. CUBO 22, 1 (2020) On Katugampola fractional order derivatives and Darboux . . . 129 Proof. Using Fubini’s Theorem we get I α,ρ a+ I β,ρ a+ u (x,y) = ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 s ρ1−1 1 s ρ2−1 2 I β,ρ a+ u (s1,s2) (xρ1 − s ρ1 1 ) 1−α1 (yρ2 − s ρ2 2 ) 1−α2 ds2ds1 = ρ 1−β1 1 ρ 1−β2 2 Γ (β1) Γ (β2) ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 s ρ1−1 1 s ρ2−1 2 (xρ1 − s ρ1 1 ) 1−α1 (yρ2 − s ρ2 2 ) 1−α2 × ∫ s1 a + 1 ∫ s2 a + 2 t ρ1−1 1 t ρ2−1 2 (s ρ1 1 − t ρ1 1 ) 1−β1 (s ρ2 2 − t ρ2 2 ) 1−β2 u (t1, t2)dt2dt1ds2ds1 = ρ 1−β1 1 ρ 1−β2 2 Γ (β1) Γ (β2) ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 t ρ1−1 1 t ρ2−1 2 u (t1, t2) × (2.2) ∫ x t1 ∫ y t2 s ρ1−1 1 s ρ2−1 2 ds2ds1dt2dt1 (xρ1 − s ρ1 1 ) 1−α1 (yρ2 − s ρ2 2 ) 1−α2 (s ρ1 1 − t ρ1 1 ) 1−β1 (s ρ2 2 − t ρ2 2 ) 1−β2 . Using the change of variables x = (s ρ1 1 − t ρ1 1 ) 1−β1 (xρ1 − t ρ1 1 ) 1−α1 and y = (s ρ2 2 − t ρ2 2 ) 1−β2 (yρ2 − t ρ2 2 ) 1−α2 , we get ∫ x t1 ∫ y t2 s ρ1−1 1 s ρ2−1 2 (xρ1 − s ρ1 1 ) 1−α1 (yρ2 − s ρ2 2 ) 1−α2 1 (s ρ1 1 − t ρ1 1 ) 1−β1 (s ρ2 2 − t ρ2 2 ) 1−β2 ds2ds1 = ∫ x t1 s ρ1−1 1 (xρ1 − s ρ1 1 ) 1−α1 (s ρ1 1 − t ρ1 1 ) 1−β1 ds1 × ∫ y t2 s ρ2−1 2 (yρ2 − s ρ2 2 ) 1−α2 (s ρ2 2 − t ρ2 2 ) 1−β2 ds2 = (xρ1 − t ρ1 1 ) ρ1 (yρ2 − t ρ2 2 ) ρ2 ∫ 1 0 (1 − x) α1−1 x β1dx ∫ 1 0 (1 − y) α1−1 y β1dy = (xρ1 − t ρ1 1 ) ρ1 (yρ2 − t ρ2 2 ) ρ2 B (α1,β1)B (α2,β2) = (xρ1 − t ρ1 1 ) ρ1 (yρ2 − t ρ2 2 ) ρ2 Γ (α1) Γ (β1) Γ (α1 + β1) Γ (α2) Γ (β2) Γ (α2 + β2) . (2.3) From (2.2) and (2.3) we obtain (2.1). Lemma 2.3. Let 0 < ai < bi, i = 1,2 reals numbers, a = (a1,a2) and u : [a1,b1] × [a2,b2] → R be an integrable function. Then D α,ρ a+ I α,ρ a+ u (x,y) = u (x,y) almost everywhere, where α = (α1,α2) ∈ (0,1) 2 and parameter ρ = (ρ1,ρ2). Proof. From Lemma (2.2), we get D α,ρ a+ I α,ρ a+ u (x,y) = x1−ρ1y1−ρ2D2x,yI 1−α,ρ a+ I α,ρ a+ u (x,y) = x1−ρ1y1−ρ2D2x,yI 1,ρ a+ u (x,y) = u (x,y) . 130 Djalal Boucenna, Abdellatif Ben Makhlouf & Mohamed Ali Hammami CUBO 22, 1 (2020) 3 Existence and uniqueness results For the existence and uniqueness of solutions for the problem (1.1)-(1.2) we need the following lemma. Lemma 3.1. The function u ∈ C (J) is a solution of fractional order problem (1.1)-(1.2) if and only if u (x,y) = ϕ(x) + ψ (y) − ϕ(a1) + I α,ρ a+ f (x,y,u (x,y)) . (3.1) Proof. First suppose that u is a solution of the integral equation (3.1). Applied CD α,ρ a+ and using Lemma 2.3 we obtain that u solves the the equation (1.1). Since the integral is zero when x = a1, or y = a2, then the initial conditions in (1.2) are satisfied. Hence u solves the problem (1.1)-(1.2). Conversly, if u is a solution of the problem (1.1)-(1.2). Let h(x,y) = f (x,y,u (x,y)) = D α,ρ a+ (u (x,y) − u (x,a2) − u (a1,y) + u (a1,a2)) = x1−ρ1y1−ρ2D2x,yI 1−α,ρ a+ [u (x,y) − u (x,a2) − u (a1,y) + u (a1,a2)] . (3.2) Applying the operator I 1,ρ a+ to (3.2), we get I 1,ρ a+ h(x,y) = I 1−α,ρ a+ [u (x,y) − u (x,a2) − u (a1,y) + u (a1,a2)] . Applying the operator D 1−α,ρ a+ to this equation we find [u (x,y) − u (x,a2) − u (a1,y) + u (a1,a2)] = D 1−α,ρ a+ I 1,ρ a+ h(x,y) = ( x1−ρ1y1−ρ2 ) D2x,yI α,ρ a+ I 1,ρ a+ h(x,y) = I α,ρ a+ h(x,y) . Hence, the proof is complete. 3.1 Existence of solutions In this subsection we study the existence of solutions for the problem (1.1)-(1.2). Theorem 3.1. Let k > 0,h∗1 > a1 and h ∗ 2 > a2. Define G = {(x,y,u) : (x,y) ∈ [a1,h ∗ 1] × [a2,h ∗ 2] , |u − ϕ(x) − ψ (y) + ϕ(a1)| ≤ k} , M = sup (x,y,u)∈G |f (x,y,u)| CUBO 22, 1 (2020) On Katugampola fractional order derivatives and Darboux . . . 131 and (h1,h2) =      (h∗1,h ∗ 2) if M = 0, ( min ( h∗1, ( k 1 2 ρ α1 1 Γ(α1+1) M 1 2 ) 1 α1 ) ,min ( h∗2, ( k 1 2 ρ α2 2 Γ(α2+1) M 1 2 ) 1 α2 )) otherwise. Then, there exists a function u ∈ C [a1,h1] × [a2,h2] that solves the problem (1.1)-(1.2). Proof. If M = 0 then f (x,y,u) = 0, for all (x,y,u) ∈ G. In this case it is clear that the function u : [a1,h1] × [a2,h2] → R with u (x,y) = ϕ(x) + ψ (y) − ϕ(a1) is a solution of the problem (1.1)- (1.2). For M 6= 0, using Lemma 3.1 we obtain that the problem (1.1)-(1.2) is equivalent to the Volterra integral equation (3.1). Define the function T by T (x,y) = ϕ(x) + ψ (y) − ϕ(a1) . (3.3) and the set U by U = {u ∈ C ([a1,h1] × [a2,h2]) ,‖u − T‖∞ ≤ k} . (3.4) The set U is nonempty since T ∈ U. It is clear that U is a closed and convex subset of the Banach space of all continuous functions on [a1,h1] × [a2,h2]. We define the operator A on this set U by (Au) (x,y) = T (x,y) + ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 sρ1−1tρ2−1f (s,t,u (s,t)) (xρ1 − sρ1) 1−α1 (yρ2 − tρ2) 1−α2 dtds. (3.5) We have to show that A has a fixed point. This is done through the Schauder’s Fixed Point Theorem. It is easy to see that A is continuous. Now we show that A is defined to U into itself, let u ∈ U and (x,y) ∈ [a1,h1] × [a2,h2] then |(Au) (x,y) − T (x,y)| = ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 sρ1−1tρ2−1 |f (s,t,u (s,t))| (xρ1 − sρ1) 1−α1 (yρ2 − tρ2) 1−α2 dtds ≤ Mρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x a + 1 ∫ y a + 2 sρ1−1tρ2−1 (xρ1 − sρ1) 1−α1 (yρ2 − tρ2) 1−α2 dtds ≤ M Γ (α1 + 1) Γ (α2 + 1) ( xρ1 − a ρ1 1 ρ1 )α1 ( yρ2 − a ρ2 2 ρ2 )α2 ≤ M ρ α1 1 ρ α2 2 Γ (α1 + 1) Γ (α2 + 1) h ρ1α1 1 h ρ2α2 2 ≤ M ρ α1 1 ρ α2 2 Γ (α1 + 1) Γ (α2 + 1) h α1 1 h α2 2 ≤ M ρ α1 1 ρ α2 2 Γ (α1 + 1) Γ (α2 + 1) kρ α1 1 ρ α2 2 Γ (α1 + 1) Γ (α2 + 1) M ≤ k. 132 Djalal Boucenna, Abdellatif Ben Makhlouf & Mohamed Ali Hammami CUBO 22, 1 (2020) Thus, we have Au ∈ U if u ∈ U. We will now show that AU = {Au : u ∈ U} is relatively compact. This is done by the using Arzela-Ascoli Theorem. Firstly, we show that A(U) is uniformly bounded. Indeed, let u ∈ U and (x,y) ∈ [a1,h1] × [a2,h2] and from the previous step we get ‖Au‖ ∞ ≤ ‖T‖ ∞ + k. Secondly, we show that A(U) is equicontinuous. Indeed, let (x1,y1) ∈ [a1,h1] × [a2,h2] ,(x2,y2) ∈ [a1,h1] × [a2,h2] such that x1 < x2 and y1 < y2, we have |(Au) (x1,y1) − (Au) (x2,y2)| ≤ |T (x1,y1) − T (x2,y2)| + Mρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x1 a + 1 ∫ y1 a + 2 sρ1−1tρ2−1 (x ρ1 1 − s ρ1) 1−α1 (y ρ2 1 − t ρ2) 1−α2 − sρ1−1tρ2−1 (x ρ1 2 − s ρ1) 1−α1 (y ρ2 2 − t ρ2) 1−α2 dtds + Mρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x1 a + 1 ∫ y2 y1 sρ1−1tρ2−1 (x ρ1 2 − s ρ1) 1−α1 (y ρ2 2 − t ρ2) 1−α2 dtds + Mρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x2 x1 ∫ y1 a + 2 sρ1−1tρ2−1 (x ρ1 2 − s ρ1) 1−α1 (y ρ2 2 − t ρ2) 1−α2 dtds + Mρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ x2 x1 ∫ y2 y1 sρ1−1tρ2−1 (x ρ1 2 − s ρ1) 1−α1 (y ρ2 2 − t ρ2) 1−α2 dtds ≤ |T (x1,y1) − T (x2,y2)| + 3M ρ α1 1 ρ α2 2 Γ (α1) Γ (α2) [ (x ρ1 2 − a ρ1 1 ) α1 (y ρ2 2 − y ρ2 1 ) α2 + (y ρ2 2 − a ρ2 2 ) α2 (x ρ1 2 − x ρ1 1 ) α1 ] Hence, A(U) is equicontinous, since T is uniformly continuous in [a1,h1]×[a2,h2]. As a consequence of the Schauder’s Fixed Point Theorem, we deduce that A has a fixed point u in U. This fixed point is the required solution of the problem (1.1)-(1.2). Hence, the proof is complete. 3.2 Uniqueness of solutions In this subsection we discuss the uniqueness results for the problem (1.1)-(1.2). Let u1,u2 ∈ C ([a1,h1] × [a2,h2]) , and (x,y) ∈ [a1,h1] × [a2,h2]. Suppose there exists a constant L > 0 independent of x,y,u1, and u2 such that |f (x,y,u1) − f (x,y,u2)| ≤ L |u1 − u2| , (3.6) then we have ‖(Au1) − (Au2)‖C([a1,x]×[a2,y]) ≤ L ‖u1 − u2‖C([a1,x]×[a2,y]) Γ (α1 + 1) Γ (α2 + 1) ( xρ1 ρ1 )α1 ( yρ2 ρ2 )α2 . (3.7) CUBO 22, 1 (2020) On Katugampola fractional order derivatives and Darboux . . . 133 Indeed, let u1,u2 ∈ C ([a1,h1] × [a2,h2]) , (x,y) ∈ [a1,h1] × [a2,h2] and (v,w) ∈ [a1,x] × [a2,y], we have |(Au1) (v,w) − (Au2) (v,w)| = ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ v a + 1 ∫ w a + 2 sρ1−1tρ2−1 |f (s,t,u1 (s,t)) − f (s,t,u2 (s,t))| (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 dtds ≤ Lρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ v a + 1 ∫ w a + 2 sρ1−1tρ2−1 (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 |u1 (s,t) − u2 (s,t)|dtds ≤ Lρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ‖u1 − u2‖C([a1,x]×[a2,y]) ∫ v a + 1 ∫ w a + 2 sρ1−1tρ2−1 (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 dtds ≤ L Γ (α1 + 1) Γ (α2 + 1) ‖u1 − u2‖C([a1,x]×[a2,y]) ( vρ1 ρ1 )α1 ( wρ2 ρ2 )α2 ≤ L Γ (α1 + 1) Γ (α2 + 1) ‖u1 − u2‖C([a1,x]×[a2,y]) ( xρ1 ρ1 )α1 ( yρ2 ρ2 )α2 . From the above inequality we get (3.7). ‖(Au1) − (Au2)‖C([a1,x]×[a2,y]) ≤ L ‖u1 − u2‖C([a1,x]×[a2,y]) Γ (α1 + 1) Γ (α2 + 1) ( xρ1 ρ1 )α1 ( yρ2 ρ2 )α2 . Next, we have the following result Theorem 3.2. Suppose that the assumptions of Theorem 3.1 are satisfied. Also let j ∈ N,(x,y) ∈ [a1,h1]×[a2,h2] and u1,u2 ∈ U. Suppose f satisfies the Lipschitz condition with respect to the third variable with the Lipschitz constant L. Then ∥ ∥Aju1 − A ju2 ∥ ∥ C([a1,x]×[a2,y]) ≤ ( x ρ1 ρ1 )α1j ( y ρ2 ρ2 )α2j Γ (1 + α1j) Γ (1 + α2j) ‖u1 − u2‖C([a1,x]×[a2,y]) . (3.8) Proof. We will prove (3.8) by induction. In the case j = 0, the inequality holds. Assume (3.8) is 134 Djalal Boucenna, Abdellatif Ben Makhlouf & Mohamed Ali Hammami CUBO 22, 1 (2020) true for j − 1 ∈ N0 then for all (x,y) ∈ [a1,h1] × [a2,h2] and (v,w) ∈ [a1,x] × [a2,y] we have ∣ ∣ ( Aju1 ) (v,w) − ( Aju2 ) (v,w) ∣ ∣ = ∣ ∣ ( AA j−1 u1 ) (v,w) − ( AA j−1 u2 ) (v,w) ∣ ∣ = ρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ v a + 1 ∫ w a + 2 sρ1−1tρ2−1 ∣ ∣f ( s,t,Aj−1u1 (s,t) ) − f ( s,t,Aj−1u2 (s,t) ) ∣ ∣ (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 dtds ≤ Lρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ v a + 1 ∫ w a + 2 sρ1−1tρ2−1 ∣ ∣Aj−1u1 (s,t) − A j−1u2 (s,t) ∣ ∣ (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 dtds ≤ Lρ 1−α1 1 ρ 1−α2 2 Γ (α1) Γ (α2) ∫ v a + 1 ∫ w a + 2 sρ1−1tρ2−1 ∥ ∥Aj−1u1 − A j−1u2 ∥ ∥ C([a1,s]×[a2,t]) (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 dtds ≤ Ljρ 1−α1j 1 ρ 1−α2j 2 Γ (α1) Γ (α2) Γ (1 + α1 (j − 1)) Γ (1 + α2 (j − 1)) ‖u1 − u2‖C([a1,x]×[a2,y]) ∫ v a + 1 ∫ w a + 2 sρ1+α1ρ1(j−1)−1tρ2+α2ρ2(j−1)−1 (vρ1 − sρ1) 1−α1 (wρ2 − tρ2) 1−α2 dtds ≤ Ljρ 1−α1j 1 ρ 1−α2j 2 Γ (α1) Γ (α2) Γ (1 + α1 (j − 1)) Γ (1 + α2 (j − 1)) ‖u1 − u2‖C([a1,x]×[a2,y]) × Γ (α1) Γ (α2) Γ (1 + α1 (j − 1)) Γ (1 + α2 (j − 1)) Γ (1 + α1j) Γ (1 + α2j) xρ1α1j ρ1 yρ2α2j ρ2 ≤ ( x ρ1 ρ1 )α1j ( y ρ2 ρ2 )α2j Γ (1 + α1j) Γ (1 + α2j) ‖u1 − u2‖C([a1,x]×[a2,y]) . Hence, the proof is complete. Theorem 3.3. Let k,h∗1 and h ∗ 1 are positive numbers, define the set G as in Theorem 3.1 and assume that the function f : G → R satisfies a Lipschitz condition with respect to the third variable with the Lipschitz constant L. Then, there exists a unique solution u ∈ C ([a1,h1] × [a2,h2]) for the problem (1.1)-(1.2). Where h1,h2 are the same as in Theorem 3.1. Proof. According to Theorem 3.1, the problem (1.1)-(1.2) has a solution. To prove the uniqueness, we adopt Theorem 3.2, we use the operato A as defined in (3.5), the function T as defined in (3.3) and the set U as defined in (3.4). We will apply Weissinger’s Fixed Point Theorem to prove that A has a unique fixed point. Let j ∈ N and u1,u2 ∈ C ([a1,h1] × [a2,h2]) . From (3.8) and taking the norms on [a1,h1]×[a2,h2], we get ∥ ∥Aj−1u1 − A j−1u2 ∥ ∥ C([a1,h1]×[a2,h2]) ≤ ( x ρ1 ρ1 )α1j ( y ρ2 ρ2 )α2j Γ (1 + α1j) Γ (1 + α2j) ‖u1 − u2‖C([a1,h1]×[a2,h2]) . CUBO 22, 1 (2020) On Katugampola fractional order derivatives and Darboux . . . 135 Let ωj = ( x ρ1 ρ1 ) α1j ( y ρ2 ρ2 ) α2j Γ(1+α1j)Γ(1+α2j) . It is clear that ∞ ∑ j=0 ωj = ∞ ∑ j=0 (( x ρ1 ρ1 )α1 ( y ρ2 ρ2 )α2 )j Γ (1 + α1j) Γ (1 + α2j) = E ( (αi,1)i=1,2 ; ( ( xρ1 ρ1 )α1 ( yρ2 ρ2 )α2 ) ) , hence the series converges. This completes the proof. 4 Conclusion Here we have studied the existence and uniqueness of the solutions for the Darboux problem of partial differential equations with Caputo-Katugampola fractional derivative. 136 Djalal Boucenna, Abdellatif Ben Makhlouf & Mohamed Ali Hammami CUBO 22, 1 (2020) References [1] H . J. Haubold, A. M. Mathai, and R. K. Saxena, Mittag-Leffler Functions and Their Applica- tions, Jour of App Math Volume 2011, Article ID 298628 , 51 pages. [2] Almeida, R., Malinowska, A. B., Odzijewicz, T., “Fractional Differential Equations With Dependence on the Caputo-Katugampola Derivative,” J. Comput. Nonlinear Dynam, Vol. 11, pp. 061017 (2016). 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Introduction Preliminaries Existence and uniqueness results Existence of solutions Uniqueness of solutions Conclusion