International Journal of Analysis and Applications ISSN 2291-8639 Volume 15, Number 1 (2017), 18-22 http://www.etamaths.com POSITIVE SOLUTIONS FOR MULTI-ORDER NONLINEAR FRACTIONAL SYSTEMS A. GUEZANE-LAKOUD AND R. KHALDI∗ Abstract. In this paper, we study the existence of positive solutions for a class of multi-order systems of fractional differential equations with nonlocal conditions. The main tool used is Schauder fixed point theorem and upper and lower solutions method. The results obtained are illustrated by a numerical example. 1. Introduction Recently, the investigation of fractional differential equations attracted more attention since it has many applications in several fields of sciences such as in engineering, physics, chemistry, biology, etc ... [8], [10]. In this work, we use the method of upper and lower solutions to prove the existence of positive solutions for a system of multi-order fractional differential equations with nonlocal boundary conditions, where each equation has an order that may be different from the order of the other equations, that is : (P) { Dα 0+ u (t) + f (t,u (t)) = 0, 0 < t < 1, u (0) = u′ (0) = 0, Au (1) = Bu′ (1) , where the function u = (u1,u2, ...,un), ui : [0, 1] → R, Dα0+u (t) = ( Dα1 0+ u1 (t) ,D α2 0+ u2 (t) , ...,D αn 0+ un (t) ) , Dαi 0+ denotes the Reimann-Liouville fractional derivative of order αi, 2 < αi < 3, i ∈ {1, ..,n} , n ≥ 2, the function f is such that f (t,u) = (f1 (t,u) , ...,fn (t,u)) , u = (u1,u2, ...,un) , fi ∈ C ([0, 1] ×Rn,R+) , A = (a1, ...,an) , B = (b1, ...,bn) ∈ Rn. Fractional differential systems can arise from sciences problems such population problems, dielectric polarization, electromagnetic waves,...see [3]. Many methods are used for the investigation of fractional differential equations, such fixed point theory, lower and upper solutions method, Mawhin theory,...see [1], [2], [4], [5], [6], [7], [9], [11]. This paper is organized as follows: in the second Section, we state some preliminary materials that will be used later. In section three, we use the upper and lower solutions method to prove the existence of positive solutions for problem (P). Finally, we give an example illustrating the obtained results. 2. Preliminaries In this section, we recall the basic definitions and lemmas from fractional calculus theory and the details can be found in [7], [10]. Received 30th April, 2017; accepted 11th July, 2017; published 1st September, 2017. 2010 Mathematics Subject Classification. 34B10, 26A33, 34B15. Key words and phrases. fractional Rieman-Liouville derivative; fractional differential equation; upper and lower solutions method. c©2017 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 18 POSITIVE SOLUTIONS FOR MULTI-ORDER NONLINEAR FRACTIONAL SYSTEMS 19 Definition 2.1. The Riemann-Liouville fractional integrals of order α of a function h is defined as Iα0+h (t) = 1 Γ (α) ∫ t 0 h (s) (t−s)1−α ds, where Γ (α) = ∫∞ 0 e−ttα−1dt is the Gamma function, α > 0. Definition 2.2. The Riemann-Liouville derivative of fractional order α > 0 for a function h is defined as Dα0+h (t) = 1 Γ (n−α) ( d dt )n ∫ t 0 (t−s)n−α−1 h (s) ds, where n = [α] + 1 ([α] denotes the integer part of the real number α). Lemma 2.1. For α > 0, the solution of the homogeneous equation Dα0+h (t) = 0, is given by h(t) = c1t α−1 + c2t α−2 + ··· + cntα−n, where ci, i = 1, 2, ...,n, are real constants. Lemma 2.2. Let p, q ≥ 0, h ∈ L1 [0, 1]. Then I p 0+ I q 0+ h (t) = I p+q 0+ h (t) = I q 0+ I p 0+ h (t) . 3. Main results Lemma 3.1. Let yi ∈ C ([0, 1] ,R) , i ∈{1, ..,n}. Assume that ai > 0 and bi < 0, then for i ∈{1, ..,n}, the linear nonhomogeneous problem (Si) = Dαi 0+ ui (t) = −yi (t) , 0 < t < 1, ui (0) = u ′ i (0) = 0, aiui (1) = biu ′ i (1) , (3.1) has the following solution ui (t) = ∫ 1 0 Gi (t,s) yi (s) ds, 0 ≤ t ≤ 1,∀i ∈{1, ..,n} (3.2) where Gi (t,s) = −(t−s)αi−1 Γ (αi) + tαi−1 (1 −s)αi−2 (ai − bi (αi − 1)) Γ (αi − 1) ( ai (1 −s) αi − 1 − bi ) , s ≤ t, tαi−1 (1 −s)αi−2 (ai − bi (αi − 1)) Γ (αi − 1) ( ai (1 −s) αi − 1 − bi ) ,s ≥ t. Proof. By assuming that ui is a solution of the fractional boundary value problem (P) and using Lemma 2.1, we obtain ui (t) = −Iαi0+yi (t) + c1t αi−1 + c2t αi−2 + c3t αi−3, (3.3) According to conditions ui (0) = 0 and u ′ i (0) = 0, we obtain c2 = c3 = 0. Using the nonlocal condition aiui (1) = biu ′ i (1), it yields c1 = 1 ai − bi (αi − 1) ( aiI αi 0+ y (1) − biIαi−10+ y (1) ) . (3.4) Substituting c1 in Equation 3.3, we get what follows ui (t) = ∫ 1 0 Gi (t,s) yi (s) ds, (3.5) where Gi is given above. � Lemma 3.2. If ai > 0 and bi < 0, i ∈ {1, ..,n} , then the functions Gi are nonnegative, continuous and 0 ≤ Gi (t,s) ≤ 1 Γ (αi) , 0 ≤ s,t ≤ 1,∀i ∈{1, ..,n} , 20 GUEZANE-LAKOUD, KHALDI Proof. The proof is direct, we omit it. � Let X be the Banach space C ([0, 1] ,R) × ...×C ([0, 1] ,R)︸ ︷︷ ︸ n times , equipped with the norm ‖u‖ = ∑i=n i=1 maxt∈[0,1] |ui (t)| . Define the integral operator T : X → X by Tu = (T1u,T2u,...,Tnu) where Tiu (t) = ∫ 1 0 Gi (t,s) fi (s,u (s)) ds, ∀i ∈{1, ..,n} (3.6) Let C = (c1, ...,cn) , D = (d1, ...,dn) ∈ Rn+ such that D > C. We recall that for x = (x1, ...,xn) , y = (y1, ...,yn) then x ≤ y means xi ≤ yi, for all i ∈{1, ..,n} and [C,D] = {x = (x1, ...,xn) ,ci ≤ xi ≤ di,∀i ∈{1, ..,n}} . We define the upper and lower control operators U = ( U1, ...,Un ) , U = (U1, ...,Un) respectively by Ui (t,x) = sup{fi (t,y) ,C ≤ y ≤ x} , Ui (t,x) = inf {fi (t,y) ,x ≤ y ≤ D} , 0 ≤ t ≤ 1. From the definition of Ui and Ui we have Ui (t,x) ≤ fi (t,x) ≤ Ui (t,x) , x ∈ [C,D] , 0 ≤ t ≤ 1, i ∈{1, ..,n} . Lemma 3.3. The function u ∈ X is a solution of the system (P) if and only if Tiu (t) = ui (t), for all t ∈ [0, 1] , ∀i ∈{1, ...,n} . Consequently, the existence of solutions for system (P) can be turned into a fixed point problem in X for the operator T. Define the cone K = {u ∈ X,u(t) ≥ 0, 0 ≤ t ≤ 1} . Let us make the following hypothesis: (H) There exist two functions θ = ( θ1, ...,θn ) , θ = (θ1, ...,θn) ∈ K, such that C ≤ θ (t) ≤ θ (t) ≤ D, 0 ≤ t ≤ 1 and { θi (t) ≥ ∫ 1 0 Gi (t,s) Ui ( s,θ (s) ) ds,i ∈{1, ..,n} θi(t) ≤ ∫ 1 0 Gi (t,s) Ui (s,θ (s)) ds,i ∈{1, ..,n} The functions θ and θ are called respectively upper and lower solutions for problem (P). Now we are ready to give the main result for problem (P). Theorem 3.1. Assume that hypothesis (H) holds and f (t, 0) 6= 0, 0 ≤ t ≤ 1, then the fractional boundary value problem (P) has at least one positive solution u ∈ K satisfying θ (t) ≤ u (t) ≤ θ (t) , 0 ≤ t ≤ 1. Proof. Clearly, the continuity of the operator T follows from the continuity of f. Set Ω = { u ∈ K : θ (t) ≤ u (t) ≤ θ (t) , 0 ≤ t ≤ 1 } , then Ω is a nonempty, closed and convex subset of X. Firstly, we show that T (Ω) ⊂ Ω. In fact, let u ∈ Ω, then by the definition of the control functions and hypothesis (H), it yields Tiu (t) = ∫ 1 0 Gi (t,s) fi (s,u (s)) ds ≤ ∫ 1 0 Gi (t,s) Ui ( s,θ (s) ) ds ≤ θi (t) , i ∈{1, ..,n} thus Tu (t) ≤ θ (t) , 0 ≤ t ≤ 1. Similarly, we get Tiu (t) = ∫ 1 0 Gi (t,s) fi (s,u (s)) ds ≥ ∫ 1 0 Gi (t,s) Ui (s,θ (s)) ds ≥ θi (t) , i ∈{1, ..,n} POSITIVE SOLUTIONS FOR MULTI-ORDER NONLINEAR FRACTIONAL SYSTEMS 21 from which follows Tu (t) ≥ θ (t) , 0 ≤ t ≤ 1 thus T (Ω) ⊂ Ω. Now, we prove that T : Ω → X is completely continuous operator. Set Mi = max{fi (t,u(t)) , 0 ≤ t ≤ 1,u ∈ Ω} , then we have |Tiu (t)| ≤ ∫ 1 0 Gi (t,s) fi (s,u (s)) ds ≤ Mi Γ (αi) . Taking the supremum over [0, 1], then summing the obtained inequalities according to i from 1 to n, we get ‖Tu‖≤ n∑ i=1 Mi Γ (αi) , which implies that T (Ω) is uniformly bounded. Let us show that (Tu) is equicontinuous. Indeed, let u ∈ Ω and 0 ≤ t1 < t2 ≤ 1, then |Tiu (t1) −Tiu (t2)| ≤ ∫ 1 0 |Gi (t1,s) −Gi (t2,s)|fi (s,u (s)) ds ≤ Mi [∫ t1 0 |Gi (t1,s) −Gi (t2,s)|ds + ∫ t2 t1 |Gi (t1,s) −Gi (t2,s)|ds + ∫ 1 t2 |Gi (t1,s) −Gi (t2,s)|ds ] by computation, we get |Tiu (t1) −Tiu (t2)| ≤ Mi ( (t2 − t1) (αi − 1) Γ (αi) + (t2 − t1) αi−1 Γ (αi) + 3 ( tαi−12 − t αi−1 1 ) ai − bi (αi − 1) ( ai Γ (αi) + bi Γ (αi − 1) )) . As t1 → t2, the right-hand side of the above inequality tends to zero. By Ascoli-Arzela theorem, we conclude that the operator T : Ω → Ω is completely continuous. Finally, Schauder fixed point theorem implies that T has at least one fixed point u ∈ Ω and then problem (P) has at least one positive solution in Ω. As direct consequence of Theorem 3.1, we get the following Corollary. � Corollary 3.1. Assume that fi are continuous, nonnegative, fi (t, 0) 6= 0, 0 ≤ t ≤ 1 and there exist two positive constants li and Li such that 0 < li ≤ fi (t,x) ≤ Li, x ≥ 0, 0 ≤ t ≤ 1, i ∈{1, ..,n} , (3.7) then problem (P) has at at least one positive solution u ∈ X. Furthermore the solution satisfies 0 < li ∫ 1 0 Gi (t,s) ds ≤ ui (t) ≤ Li ∫ 1 0 Gi (t,s) ds, 0 ≤ t ≤ 1, ∀i ∈{1, ..,n} . 22 GUEZANE-LAKOUD, KHALDI Proof. From equation 3.7 we have Ui (t,x) ≤ Li, Ui (t,x) ≥ li, 0 ≤ t ≤ 1, x ≥ 0. Let us choose θi (t) = Li ∫ 1 0 Gi (t,s) ds = Li tαi−1 αiΓ (αi) ( 1 + 1 + 1 αi ) ≥ ∫ 1 0 Gi (t,s) Ui ( s,θ (s) ) ds, i ∈{1, ..,n} θi(t) = li ∫ 1 0 Gi (t,s) ds = li tαi−1 αiΓ (αi) ( 1 + 1 + 1 αi ) ≤ ∫ 1 0 Gi (t,s) Ui (s,θ (s)) ds, i ∈{1, ..,n} , then the conclusion follows from Theorem 3.1. � Now, we give an examples to illustrate the usefulness of our main results. Example 3.1. Consider the following two-dimensional fractional order system (S) = D 5 2 u1 (t) + ( 1 + 1 1+u1+u2 ) = 0, D 8 3 u2 (t) + (1 + e −u1 ) = 0, u1 (0) = 0, u ′ 1 (0) = 0, u1 (1) −u′1 (0) = 0, u2 (0) = 0, u ′ 2 (0) = 0, u2 (1) −u′2 (0) = 0. We have α = ( 5 2 , 8 3 ) , a1 = a2 = 1, b1 = b2 = −1, f1 (t,u1,u2) = 1 + 11+u1+u2 , f2 (t,u1,u2) = 1 + e −u1 , fi ∈ C ( [0, 1] ×R2,R+ ) , fi (t, 0) 6= 0, and 1 ≤ fi (t,u1,u2) ≤ 2. From Corollary 3.1, we conclude the system (S) has at at least one positive solution u ∈ X. 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