International Journal of Analysis and Applications Volume 17, Number 6 (2019), 940-957 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-940 NONLINEAR COUPLED FRACTIONAL ORDER SYSTEMS WITH INTEGRO-MULTISTRIP-MULTIPOINT BOUNDARY CONDITIONS BASHIR AHMAD1,∗, AHMED ALSAEDI2, SOTIRIS K. NTOUYAS1,2 1Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 2Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece ∗Corresponding author: bashirahmad qau@yahoo.com Abstract. We study the existence and uniqueness of solutions for a nonlinear system of coupled fractional differential equations equipped with nonlocal coupled integro-multistrip-multipoint boundary conditions. Our results are new in the sense that the given boundary conditions connect the values of the known functions over the given domain with the ones described on different sub-segments and different nonlocal positions within the given domain. We make use of Banach contraction mapping principle, Leray-Schauder alternative and Krasnoselskii fixed point theorem to prove the desired results for the problem at hand. An example illustrating the existence and uniqueness result is also presented. 1. Introduction Fractional calculus is found to be more practical and effective than the classical calculus in the mathe- matical modeling of several real world phenomena. The topic of fractional differential equations has evolved as an important and significant area of investigation in view of its numerous applications in viscoelasticity, electroanalytical chemistry, and many physical problems [1]- [4]. In recent years, many works have been devoted to the study of the mathematical aspects of fractional order differential equations. Many advanced Received 2019-03-22; accepted 2019-04-10; published 2019-11-01. 2010 Mathematics Subject Classification. 26A33, 34B15. Key words and phrases. boundary value problem; fractional derivative; fractional integral; fixed point theorem. c©2019 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 940 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-940 Int. J. Anal. Appl. 17 (6) (2019) 941 and efficient methods have been applied to develop the existence theory for fractional differential equations. One of the powerful tools for developing the existence crireria for solutions to such equations is based on the fixed point theory. Many authors applied fixed point theorems to establish the existence theory for nonlinear fractional differential equations; for example, see [5]- [13] and the references cited therein. On the other hand, the coupled systems of nonlinear fractional differential equations also received con- siderable attention. Such systems appear in various disciplines of applied nature, for instance, see [14, 15]. The tools of the fixed point theory also played a key role in developing the existence theory for the coupled systems of fractional differential equations [18]- [20]. In [21], the authors investigated a coupled system of nonlinear fractional differential equations:  cDαx(t) = f(t,x(t),y(t),Dγy(t)), t ∈ [0,T], 1 < α ≤ 2, 0 < γ < 1, cDβy(t) = g(t,x(t),Dδx(t),y(t)), t ∈ [0,T], 1 < β ≤ 2, 0 < δ < 1, equipped with coupled nonlocal and integral boundary conditions of the form:  x(0) = h(y), ∫ T 0 y(s)ds = µ1x(η), y(0) = φ(x), ∫ T 0 x(s)ds = µ2y(ξ), η,ξ ∈ (0,T), where cDi denote the Caputo fractional derivatives of order i, i = α,β,γ,δ, f,g : [0,T] × R × R × R → R, h,φ : C([0,T],R) → R are given continuous functions, and µ1,µ2 are real constants. In [22], the existence of solutions for the following boundary value problem of coupled system of nonlinear fractional differential equations was discussed:  cDqx(t) = f(t,x(t),y(t),c Dσ1y(t)), 1 < q ≤ 2, 0 < σ1 < 1, t ∈ [0, 1], cDpy(t) = g(t,x(t),c Dσ2x(t),y(t)), 1 < p ≤ 2, 0 < σ2 < 1, t ∈ [0, 1],   x(0) = ψ1(y), x(1) = a1 ∫ ξ 0 y(s)ds + b1 m−2∑ i=1 αiy(ηi), y(0) = ψ2(x), y(1) = a2 ∫ ξ 0 x(s)ds + b2 m−2∑ i=1 βix(ηi), 0 < ξ < η1 < η2 < · · · < ηm−2 < 1, m ≥ 3, where cDj (j = p,q,σ1,σ2) denote the Caputo fractional derivative of order j, f,g : [0, 1] × R × R × R → R, ψ1,ψ2 : C([0, 1],R) → R are given appropriate functions, a1,a2,b1 and b2 are real constants and αi, βi, i = 1, 2, . . . ,m− 2, are positive real constants. In this paper, we are concerned with existence of solutions for a nonlinear system of coupled fractional differential equations: Dσx(t) = f(t,x(t),y(t)), n− 1 < σ < n, n ≥ 3, t ∈ J := [0, 1], Dφx(t) = g(t,x(t),y(t)), m− 1 < φ < m, m ≥ 3, t ∈ J := [0, 1], (1.1) Int. J. Anal. Appl. 17 (6) (2019) 942 subject to integro-multistrip-multipoint boundary conditions: x( î )(0) = 0, î = 0, 1, 2, . . . ,n− 2,∫ 1 0 x(s)ds = p∑ i=2 βi−1 ∫ ηi ηi−1 y(s)ds + q∑ j=1 γj y(ρj), βi−1 > 0,γj > 0, y( ĵ )(0) = 0, ĵ = 0, 1, 2, . . . ,m− 2,∫ 1 0 y(s)ds = µ∑ i=2 β′i−1 ∫ θi θi−1 x(s)ds + λ∑ j=1 γ′j x(ζj), β ′ i−1 > 0,γ ′ j > 0, (1.2) where Dσ,Dφ are the standard Riemann-Liouville fractional derivatives of order σ and φ respectively, f,g : J × R × R → R are continuous functions and 0 < η1 < η2 < ... < ηp < ρ1 < ρ2 < ... < ρq < 1, 0 < θ1 < θ2 < ... < θµ < ζ1 < ζ2 < ... < ζλ < 1 with p,q,µ,λ ∈ N. The rest of the paper is organized as follows. In Section 2, we recall some basic definitions of fractional calculus and present an auxiliary lemma, which plays a pivotal role in obtaining the main results presented in Section 3. We also discuss an example for illustration of the existence-uniqueness result. 2. Preliminaries First of all, we recall some basic definitions of fractional calculus [2]. Definition 2.1. The fractional integral of order r with the lower limit zero for a function f is defined as Irf(t) = 1 Γ(r) ∫ t 0 f(s) (t−s)1−r ds, t > 0, r > 0, provided the right hand-side is point-wise defined on [0,∞), where Γ(·) is the gamma function, which is defined by Γ(r) = ∫∞ 0 tr−1e−tdt. Definition 2.2. The Riemann-Liouville fractional derivative of order r > 0, n − 1 < r < n, n ∈ N, is defined as Dr0+f(t) = 1 Γ(n−r) ( d dt )n ∫ t 0 (t−s)n−r−1f(s)ds, where the function f has absolutely continuous derivatives upto order (n− 1). The following lemma is of great importance in the proof of our main results. Lemma 2.1. Let h,k ∈ C(J,R) and Ω = 1 σφ − Λ1Λ2 6= 0, where Λ1 = 1 φ p∑ i=2 βi−1(η φ i −η φ i−1) + q∑ j=1 γjρ φ−1 j , Int. J. Anal. Appl. 17 (6) (2019) 943 Λ2 = 1 σ µ∑ i=2 β′i−1(θ σ i −θ σ i−1) + λ∑ j=1 γ′jζ σ−1 j . Then the solution of the linear fractional differential system Dσx(t) = h(t), n− 1 < σ < n, t ∈ J := [0, 1], Dφy(t) = k(t), m− 1 < φ < m, t ∈ J := [0, 1], (2.1) supplemented with the boundary conditions (1.2) is equivalent to the system of integral equations x(t) = 1 Γ(σ) ∫ t 0 (t−s)σ−1h(s)ds + tσ−1 Ω { 1 φ ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1k(s)dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1k(s)ds− 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1h(s)dsdt ) (2.2) +Λ1 ( µ∑ i=2 β ′ i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1h(τ)dτds + λ∑ j=1 γ ′ j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1h(s)ds− 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1k(s)dsdt )} , and y(t) = 1 Γ(φ) ∫ t 0 (t−s)φ−1k(s)ds + tφ−1 Ω { 1 σ ( µ∑ i=2 β ′ i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1h(s)dτds + λ∑ j=1 γ ′ j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1h(s)ds− 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1h(s)dsdt ) (2.3) +Λ2 ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1k(τ)dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1k(s)ds− 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1h(s)dsdt )} . Proof. As argued in [2], the general solution of the equations Dσx(t) = h(t), n− 1 < σ < n and Dφy(t) = k(t), m− 1 < φ < m, can be written as x(t) = b1t σ−1 + b2t σ−2 + · · · + bntσ−n + 1 Γ(σ) ∫ t 0 (t−s)σ−1h(s)ds, (2.4) y(t) = d1t φ−1 + d2t φ−2 + · · · + dmtφ−m + 1 Γ(φ) ∫ t 0 (t−s)φ−1k(s)ds, (2.5) where bi, i = 1, 2, . . . ,n, and dj,j = 1, . . . ,m are arbitrary constants. Using the conditions x ( î )(0) = 0, î = 0, 1, 2, . . . ,n − 2, and y( ĵ )(0) = 0, ĵ = 0, 1, 2, . . . ,m − 2, we find that b2 = b3 = · · · = bn = 0 and d1 = d2 = . . . = dm = 0. Thus (2.4) and (2.5) become x(t) = b1t σ−1 + 1 Γ(σ) ∫ t 0 (t−s)σ−1g(s)ds, (2.6) y(t) = d1t φ−1 + 1 Γ(φ) ∫ t 0 (t−s)φ−1g(s)ds. (2.7) Int. J. Anal. Appl. 17 (6) (2019) 944 Using the conditions ∫ 1 0 x(s)ds = ∑p i=2 βi−1 ∫ηi ηi−1 y(s)ds + ∑q j=1 γjy(ρj) and∫ 1 0 y(s)ds = ∑µ i=2 β ′ i−1 ∫ θi θi−1 x(s)ds + ∑λ j=1 γ ′ jx(ζj) in (2.6) and (2.7), we get 1 σ b1 − Λ1d1 = p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1k(τ)dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1k(s)ds− 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1h(s)dsdt, −Λ2b1 + 1 φ d1 = µ∑ i=2 β′i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1h(τ)dτds + λ∑ j=1 γ′j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1h(s)ds− 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1k(s)dsdt. Solving the above system for b1 and d1, we find that b1 = 1 Ω ( 1 φ [ p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1k(τ)dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1k(s)ds− 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1h(s)dsdt ] +Λ1 [ µ∑ i=2 β′i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1h(τ)dτds + λ∑ j=1 γ′j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1h(s)ds− 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1k(s)dsdt ]) , and d1 = 1 Ω ( 1 σ [ µ∑ i=2 β′i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1h(τ)dτds + λ∑ j=1 γ′j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1h(s)ds− 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1k(s)dsdt ] +Λ2 [ p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1k(τ)dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1k(s)ds− 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1h(s)dsdt ]) . Inserting the above values of b1 and d1 in (2.6) and (2.7) leads to the solutions (2.2) and (2.3). The converse follows by direct computation. The proof is completed. � 3. Main Results Let us introduce the space X = {x(t)|x(t) ∈ C([0, 1],R)} endowed with the norm ‖x‖ = sup{|x(t)|, t ∈ [0, 1]}. Obviously (X,‖ · ‖) is a Banach space. Then the product space (X × X,‖(x,y)‖) is also a Banach space equipped with norm ‖(x,y)‖ = ‖x‖ + ‖y‖. Int. J. Anal. Appl. 17 (6) (2019) 945 In view of Lemma 2.1, we define an operator T : X ×X → X ×X by T(x,y)(t) =   T1(x,y)(t) T2(x,y)(t)   , where T1(x,y)(t) = 1 Γ(σ) ∫ t 0 (t−s)σ−1f(s,x(s),y(s))ds + tσ−1 Ω { 1 φ ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1g(s,x(s),y(s))dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1g(s,x(s),y(s))ds (3.1) − 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1f(s,x(s),y(s))dsdt ) +Λ1 ( µ∑ i=2 β ′ i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1f(τ,x(τ),y(τ))dτds + λ∑ j=1 γ ′ j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1f(s,x(s),y(s))ds − 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1g(s,x(s),y(s))dsdt )} , and T2(x,y)(t) = 1 Γ(φ) ∫ t 0 (t−s)φ−1g(s,x(s),y(s))ds + tφ−1 Ω { 1 σ ( µ∑ i=2 β ′ i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1f(τ,x(τ),y(τ))dτds + λ∑ j=1 γ ′ j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1f(s,x(s),y(s))ds (3.2) − 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1g(s,x(s),y(s))dsdt ) +Λ2 ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1g(τ,x(τ),y(τ))dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1g(s,x(s),y(s))ds − 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1f(s,x(s),y(s))dsdt )} . For the sake of computational convenience, we define Q1 = 1 Γ(σ + 1) + 1 |Ω| [ 1 φ 1 Γ(σ + 2) + |Λ1| ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) + λ∑ j=1 |γ′j| ζσj Γ(σ + 1) )] , (3.3) Q2 = 1 |Ω| [ 1 φ ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) ) + |Λ1| 1 Γ(φ + 2) ] , (3.4) Int. J. Anal. Appl. 17 (6) (2019) 946 Q3 = 1 |Ω| [ 1 σ ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) + λ∑ j=1 |γ′j| ζσj Γ(σ + 1) ) + |Λ2| 1 Γ(σ + 2) ] , (3.5) Q4 = 1 Γ(φ + 1) + 1 |Ω| [ 1 σ 1 Γ(φ + 2) + |Λ2| ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) )] . (3.6) In the first result, we prove the existence and uniqueness of solutions for the system (1.1)-(1.2) via Banach contraction mapping principle. Theorem 3.1. Assume that: (H1): f,g : [0, 1] × R × R → R are continuous functions and there exist positive constants `1 and `2 such that for all t ∈ [0, 1] and xi,yi ∈ R, i = 1, 2, we have |f(t,x1,x2) −f(t,y1,y2)| ≤ `1(|x1 −y1| + |x2 −y2|), |g(t,x1,x2) −g(t,y1,y2)| ≤ `2(|x1 −y1| + |x2 −y2|). If (Q1 + Q3)`1 + (Q2 + Q4)`2 < 1, where Qi, i = 1, 2, 3, 4 are given by (3.3)-(3.6), then the system (1.1)-(1.2) has a unique solution on [0, 1]. Proof. Define supt∈[0,1] f(t, 0, 0) = N1 < ∞ and supt∈[0,1] g(t, 0, 0) = N2 < ∞ and r > 0 such that r > (Q1 + Q3)N1 + (Q2 + Q4)N2 1 − (Q1 + Q3)`1 − (Q2 + Q4)`2 . We show that TBr ⊂ Br, where Br = {(x,y) ∈ X ×X : ‖(x,y)‖≤ r}. By the assumption (H1), for (u,v) ∈ Br, t ∈ [0, 1], we have |f(t,x(t),y(t))| ≤ |f(t,x(t),y(t)) −f(t, 0, 0)| + |f(t, 0, 0)| ≤ `1(|x(t)| + |y(t)|) + N1 ≤ `1(‖x‖ + ‖y‖) + N1 ≤ `1r + N1, and |g(t,x(t),y(t))| ≤ `2(‖x‖ + ‖y‖) + N2 ≤ `2r + N2, which lead to |T1(x,y)(t)| ≤ 1 Γ(σ) ∫ t 0 (t−s)σ−1(`1r + N1)ds + 1 |Ω| { 1 φ ( p∑ i=2 |βi−1| 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1(`2r + N2)dτds Int. J. Anal. Appl. 17 (6) (2019) 947 + q∑ j=1 |γj| 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1(`2r + N2)ds + 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1(`1r + N1)dsdt ) +|Λ1| ( µ∑ i=2 |β′i−1| 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1(`1r + N1)dτds + λ∑ j=1 |γ′j| 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1(`1r + N1)ds + 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1(`2r + N2)dsdt )} = (Q1`1 + Q2`2)r + Q1N1 + Q2N2. Hence ‖T1(x,y)‖≤ (Q1`1 + Q2`2)r + Q1N1 + Q2N2. In a similar manner, one can find that ‖T2(x,y)‖≤ (Q3`1 + Q4`2)r + Q3N1 + Q4N2. Consequently, we have ‖T(x,y)‖≤ [(Q1 + Q3)`1 + (Q2 + Q4)`2]r + (Q1 + Q3)N1 + (Q2 + Q4)N2 ≤ r. Now, for (x2,y2), (x1,y1) ∈ X ×X, and for any t ∈ [0, 1], we get |T1(x2,y2)(t) −T1(x1,y1)(t)| ≤ 1 Γ(σ) ∫ t 0 (t−s)σ−1`1(‖x2 −x1‖ + ‖y2 −y1‖)ds + 1 |Ω| { 1 φ ( p∑ i=2 |βi−1| 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1`2(‖x2 −x1‖ + ‖y2 −y1‖)dτds + q∑ j=1 |γj| 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1`2(‖x2 −x1‖ + ‖y2 −y1‖)ds + 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1`1(‖x2 −x1‖ + ‖y2 −y1‖)dsdt ) +|Λ1| ( µ∑ i=2 |β′i−1| 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1`1(‖x2 −x1‖ + ‖y2 −y1‖)dτds + λ∑ j=1 |γ′j| 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1`1(‖x2 −x1‖ + ‖y2 −y1‖)ds + 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1`2(‖x2 −x1‖ + ‖y2 −y1‖)dsdt )} ≤ (Q1`1 + Q2`2)(‖x2 −x1‖ + ‖y2 −y1‖), Int. J. Anal. Appl. 17 (6) (2019) 948 which implies that ‖T1(x2,y2) −T1(x1,y1)‖≤ (Q1`1 + Q2`2)(‖x2 −x1‖ + ‖y2 −y1‖). (3.7) Similarly, we obtain ‖T2(x2,y2)(t) −T2(x1,y1)‖≤ (Q3`1 + Q4`2)(‖x2 −x1‖ + ‖y2 −y1‖). (3.8) Thus it follows from (3.7) and (3.8) that ‖T(x2,y2) −T(x1,y1)‖≤ [(Q1 + Q3)`1 + (Q2 + Q4)`2](‖x2 −x1‖ + ‖y2 −y1‖). Since (Q1 + Q3)`1 + (Q2 + Q4)`2 < 1, therefore, T is a contraction. So, by Banach fixed point theorem, the operator T has a unique fixed point, which corresponds to a unique solution of problem (1.1)-(1.2). This completes the proof. � In the following theorem, we prove the existence of solutions for the system (1.1)-(1.2) by means of Leray-Schauder alternative. Lemma 3.1. (Leray-Schauder alternative) ( [23] p. 4.) Let F : E → E be a completely continuous operator (i.e., a map that restricted to any bounded set in E is compact). Let E(F) = {x ∈ E : x = λF(x) for some 0 < λ < 1}. Then either the set E(F) is unbounded, or F has at least one fixed point. Theorem 3.2. Assume that: (H3): f,g : [0, 1] × R × R → R are continuous functions and there exist real constants ki,γi ≥ 0, (i = 0, 1, 2) and k0 > 0,γ0 > 0 such that ∀xi ∈ R (i = 1, 2), |f(t,x1,x2)| ≤ k0 + k1|x1| + k2|x2|, |g(t,x1,x2)| ≤ γ0 + γ1|x1| + γ2|x2|. If (Q1 + Q3)k1 + (Q2 + Q4)γ1 < 1 and (Q1 + Q3)k2 + (Q2 + Q4)γ2 < 1, where Mi (i = 1, 2, 3, 4) are given by (3.3)-(3.6), then there exists at least one solution to the system (1.1)- (1.2) on [0, 1]. Proof. First we show that the operator T : X × X → X × X is completely continuous. By continuity of functions f and g, it is easy to show that the operator T is continuous. Let Ω ⊂ X ×X be bounded. Then there exist positive constants L1 and L2 such that |f(t,x(t),y(t))| ≤ L1, |g(t,x(t),y(t))| ≤ L2, ∀(x,y) ∈ Ω. Int. J. Anal. Appl. 17 (6) (2019) 949 Then, for any (x,y) ∈ Ω, we have |T1(x,y)(t)| ≤ 1 Γ(σ) ∫ t 0 (t−s)σ−1L1ds + 1 |Ω| { 1 φ ( p∑ i=2 |βi−1| 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1L2dτds + q∑ j=1 |γj| 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1L2ds + 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1L1dsdt ) +|Λ1| ( µ∑ i=2 |β′i−1| 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1L1dτds + λ∑ j=1 |γ′j| 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1L1ds + 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1L2dsdt )} ≤ Q1L1 + Q2L2, which implies that ‖T1(x,y)‖≤ Q1L1 + Q2L2. In a similar manner, we can get ‖T2(x,y)‖≤ Q3L1 + Q4L2. Thus, it follows from the above inequalities that the operator T is uniformly bounded, since ‖T(x,y)|| ≤ (Q1 + Q3)L1 + (Q2 + Q4)L2. Next, we show that T is equicontinuous. Let t1, t2 ∈ [0, 1] with t1 < t2. Then we have |T1(x(t2),y(t2)) −T1(x(t1),y(t1))| ≤ L1 { 1 Γ(σ) ∫ t1 0 [(t2 −s)σ−1 − (t1 −s)σ−1]ds + 1 Γ(σ) ∫ t2 t1 (t2 −s)σ−1ds } + |tσ−12 − t σ−1 1 | |Ω| {[ 1 φ 1 Γ(σ + 2) + |Λ1| ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) )] L1 + [ 1 φ ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) ) + |Λ1| 1 Γ(φ + 2) ] L2 } ≤ L1 Γ(σ + 1) [2(t2 − t1)σ + |tσ2 − t σ 1 |] + |tσ−12 − t σ−1 1 | |Ω| {[ 1 φ 1 Γ(σ + 2) + |Λ1| ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) )] L1 + [ 1 φ ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) ) + |Λ1| 1 Γ(φ + 2) ] L2 } . Analogously, we can obtain |T2(x(t2),y(t2)) −T2(x(t1),y(t1))| Int. J. Anal. Appl. 17 (6) (2019) 950 ≤ L2 Γ(φ + 1) [2(t2 − t1)φ + |t2 − t φ 1 |] + |tφ−12 − t φ−1 1 | |Ω| {[ 1 σ ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) + λ∑ j=1 |γ′j| ζσj Γ(σ + 1) ) + |Λ2| 1 Γ(σ + 2) ] L1 + [ 1 σ 1 Γ(φ + 2) + |Λ2| ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) )] L2 } . From the preceding inequalities, we deduce that the operator T(x,y) is equicontinuous, and thus the operator T(x,y) is completely continuous. Finally, it will be verified that the set E = {(x,y) ∈ X ×X|(x,y) = λT(x,y), 0 ≤ λ ≤ 1} is bounded. Let (x,y) ∈E, with (x,y) = λT(x,y). For any t ∈ [0, 1], we have x(t) = λT1(x,y)(t), y(t) = λT2(x,y)(t). Then |x(t)| ≤ Q1(k0 + k1|x| + k2|y|) + Q2(γ0 + γ1|x| + γ2|y|) = Q1k0 + Q2γ0 + (Q1k1 + Q2γ1)|x| + (Q1k2 + Q2γ2)|y|, and |y(t)| ≤ Q3(k0 + k1|x| + k2|y|) + Q4(γ0 + γ1|x| + γ2|y|) = Q3k0 + Q4γ0 + (Q3k1 + Q4γ1)|x| + (Q3k2 + Q4γ2)|y|. Hence we have ‖x‖≤ Q1k0 + Q2γ0 + (Q1k1 + Q2γ1)‖x‖ + (Q1k2 + Q2γ2)‖y|| and ‖y‖≤ Q3k0 + Q4γ0 + (Q3k1 + Q4γ1)‖x‖ + (Q3k2 + Q4γ2)‖y‖, which imply that ‖x‖ + ‖y‖ ≤ (Q1 + Q3)k0 + (Q2 + Q4)γ0 + [(Q1 + Q3)k1 + (Q2 + Q4)γ1]‖x‖ +[(Q1 + Q3)k2 + (Q2 + Q4)γ2)]‖y‖. Consequently, ‖(x,y)‖≤ (Q1 + Q3)k0 + (Q2 + Q4)γ0 M0 , where M0 = min{1 − [(Q1 + Q3)k1 + (Q2 + Q4)γ1], 1 − [(Q1 + Q3)k2 + (Q2 + Q4)γ2)]}, which proves that E is bounded. Thus, by Lemma 3.1, the operator T has at least one fixed point. Hence the system (1.1)-(1.2) has at least one solution. The proof is complete. � Our last result is based on Krasnoselskii fixed point theorem [24]. Int. J. Anal. Appl. 17 (6) (2019) 951 Lemma 3.2. (Krasnoselskii) Let M be a closed, bounded, convex and nonempty subset of a Banach space X. Let A,B be operators mapping M to X such that (a) Ax + By ∈ M where x,y ∈ M; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists z ∈ M such that z = Az + Bz. Theorem 3.3. Assume that f,g : [0, 1] × R → R are continuous functions satisfying assumption (H1) in Theorem 3.1. In addition we suppose that there exist two positive constants L1,L2 such that for all t ∈ [0, 1] and xi,yi ∈ R, i = 1, 2, |f(t,x1,x2)| ≤ L1 and |g(t,x1,x2)| ≤ L2. (3.9) If `1 Γ(σ + 1) + `2 Γ(φ + 1) < 1, (3.10) then the problem (1.1)-(1.2) has at least one solution on [0, 1]. Proof. In order to verify the hypotheses of Lemma 3.2, we decompose the operator T into four operators T1,1,T1,2,T2,1 and T2,2 on Bδ = {(x,y) ∈ X ×Y : ‖(x,y)‖≤ δ} as follows: T1,1(x,y)(t) = tσ−1 Ω { 1 φ ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1g(s,x(s),y(s))dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1g(s,x(s),y(s))ds − 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1f(s,x(s),y(s))dsdt ) +Λ1 ( µ∑ i=2 β ′ i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1f(τ,x(τ),y(τ))dτds + λ∑ j=1 γ ′ j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1f(s,x(s),y(s))ds − 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1g(s,x(s),y(s))dsdt )} , T1,2(x,y)(t) = 1 Γ(σ) ∫ t 0 (t−s)σ−1f(s,x(s),y(s))ds, and T2,1(x,y)(t) = tφ−1 Ω { 1 σ ( µ∑ i=2 β ′ i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1f(τ,x(τ),y(τ))dτds + λ∑ j=1 γ ′ j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1f(s,x(s),y(s))ds − 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1g(s,x(s),y(s))dsdt ) Int. J. Anal. Appl. 17 (6) (2019) 952 +Λ2 ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1g(τ,x(τ),y(τ))dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1g(s,x(s),y(s))ds − 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1f(s,x(s),y(s))dsdt )} , T2,2(x,y)(t) = 1 Γ(φ) ∫ t 0 (t−s)φ−1g(s,x(s),y(s))ds. Notice that T1(x,y)(t) = T1,1(x,y)(t) + T1,2(x,y)(t) and T2(x,y)(t) = T2,1(x,y)(t) + T2,2(x,y)(t) on Bδ and that the ball Bδ is a closed, bounded and convex subset of the Banach space X × X. Let us select δ ≥ max{Q1L1 + Q2L2,Q3L1 + Q4L2} and show that TBδ ⊂ Bδ for verifying condition (a) of Lemma 3.2. Setting x = (x1,x2),y = (y1,y2) ∈ Bδ, and using condition (3.9), we obtain |T1,1(x,y)(t) + T1,2(x,y)(t)| ≤ 1 Γ(σ) ∫ t 0 (t−s)σ−1L1ds + 1 |Ω| { 1 φ ( p∑ i=2 |βi−1| 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1L2dτds + q∑ j=1 |γj| 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1L2ds + 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1L1dsdt ) +|Λ1| ( µ∑ i=2 |β′i−1| 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1L1dτds + λ∑ j=1 |γ′j| 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1L1ds + 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1L2dsdt )} = Q1L1 + Q2L2 ≤ δ. Likewise, we can find that |T2,1(x̂, ŷ)(t) + T2,2(x̂, ŷ)(t)| ≤ Q3L1 + Q4L2 ≤ δ. Clearly the above two inequalities lead to the fact that T1(x,y) + T2(x̂, ŷ) ∈ Bδ. Now we establish that the operator (T1,2,T2,2) is a contraction satisfying condition (c) of Lemma 3.2. For (x1,y1), (x2,y2) ∈ Bδ, we have |T1,2(x1,y1)(t) −T1,2(x2,y2)(t)| = 1 Γ(σ) ∫ t 0 (t−s)σ−1|f(s,x1(s),y1(s)) −f(s,x2(s),y2(s))|ds ≤ `1 Γ(σ + 1) (‖x1 −y1‖ + ‖x2 −y2‖), (3.11) and |T2,2(x1,y1)(t) −T2,2(x2,y2)(t)| Int. J. Anal. Appl. 17 (6) (2019) 953 = 1 Γ(φ) ∫ t 0 (t−s)φ−1|g(s,x1(s),y1(s)) −g(s,x2(s),y2(s))|ds ≤ `2 Γ(φ + 1) (‖x1 −y1‖ + ‖x2 −y2‖). (3.12) It follows from (3.11) and (3.12) that ‖(T1,2,T2,2)(x1,y1) − (T1,2,T2,2)(x2,y2)‖ ≤ [ `1 Γ(σ + 1) + `2 Γ(φ + 1) ] (‖x1 −y1‖ + ‖x2 −y2‖), which is a contraction by (3.10). Therefore, the condition (c) of Lemma 3.2 is satisfied. Next we will show that the operator (T1,1,T2,1) satisfies the condition (b) of Lemma 3.2. By applying the continuity of the functions f,g on [0, 1]×R×R, we can conclude that the operator (T1,1,T2,1) is continuous. For each (x,y) ∈ Bδ, we have |T1,1(x,y)(t)| ≤ 1 |Ω| { 1 φ ( p∑ i=2 |βi−1| 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1L2dτds + q∑ j=1 |γj| 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1L2ds + 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1L1dsdt ) +|Λ1| ( µ∑ i=2 |β′i−1| 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1L1dτds + λ∑ j=1 |γ′j| 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1L1ds + 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1L2dsdt )} = P∗, and |T2,1(x,y)(t)| ≤ tφ−1 |Ω| { 1 σ ( µ∑ i=2 β′i−1 1 Γ(σ) ∫ θi θi−1 ∫ s 0 (s− τ)σ−1L1dτds + λ∑ j=1 γ′j 1 Γ(σ) ∫ ζj 0 (ζj −s)σ−1L1ds + 1 Γ(φ) ∫ 1 0 ∫ t 0 (t−s)φ−1L2dsdt ) +Λ2 ( p∑ i=2 βi−1 1 Γ(φ) ∫ ηi ηi−1 ∫ s 0 (s− τ)φ−1L2dτds + q∑ j=1 γj 1 Γ(φ) ∫ ρj 0 (ρj −s)φ−1L2ds + 1 Γ(σ) ∫ 1 0 ∫ t 0 (t−s)σ−1L1dsdt )} = Q∗, Int. J. Anal. Appl. 17 (6) (2019) 954 which lead to the fact that ‖(T1,1,T2,1)(x,y)‖≤ P∗ + Q∗. Thus the set (T1,1,T2,1)Bδ is uniformly bounded. In the next step, we will show that the set (T1,1,T2,1)Bδ is equicontinuous. For t1, t2 ∈ [0, 1] with t1 < t2 and for any (x,y) ∈ Bδ, we obtain |T1,1(x,y)(t2) −T1,1(x,y)(t1)| ≤ |tσ−12 − t σ−1 1 | |Ω| {[ 1 φ 1 Γ(σ + 2) + |Λ1| ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) )] L1 + [ 1 φ ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) ) + |Λ1| 1 Γ(φ + 2) ] L2 } ≤ + |tσ−12 − t σ−1 1 | |Ω| {[ 1 φ 1 Γ(σ + 2) + |Λ1| ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) )] L1 + [ 1 φ ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) ) + |Λ1| 1 Γ(φ + 2) ] L2 } . Analogously, we can obtain |T2,1(x,y)(t2) −T2,1(x,y)(t1)| ≤ |tφ−12 − t φ−1 1 | |Ω| {[ 1 σ ( µ∑ i=2 |β′i−1| θσ+1i −θ σ+1 i−1 Γ(σ + 2) + λ∑ j=1 |γ′j| ζσj Γ(σ + 1) ) + |Λ2| 1 Γ(σ + 2) ] L1 + [ 1 σ 1 Γ(φ + 2) + |Λ2| ( p∑ i=2 |βi−1| η φ+1 i −η φ+1 i−1 Γ(φ + 2) + q∑ j=1 |γj| ρ φ j Γ(φ + 1) )] L2 } . Thus |(T1,1,T2,1)(x,y)(t2) − (T1,1,T2,1)(x,y)(t1)| tends to zero as t1 → t2 independent of (x,y) ∈ Bδ. Therefore the set (T1,1,T2,1)Bδ is equicontinuous. Thus it follows by the Arzelá-Ascoli theorem that the operator (T1,1,T2,1) is compact on Bδ. By the conclusion of Lemma 3.2, we deduce that the problem (1.1)- (1.2) has at least one solution on [0, 1]. This completes the proof. � Example 3.1. Consider the following system of fractional boundary value problems  cD3/2x(t) = 1 4(t + 2)2 |x(t)| 1 + |x(t)| + 1 + 1 32 sin2 y(t) + 1 √ t2 + 1 , t ∈ [0, 1], cD5/2y(t) = 1 32π sin(2πx(t)) + |y(t)| 16(1 + |y(t)|) + 1 2 , t ∈ [0, 1], ∫ 1 0 x(s)ds = 5 2 ∫ 1/6 1/7 y(s)ds + 4 ∫ 1/5 1/6 y(s)ds + 3 2 y (1 4 ) + 2y (1 3 ) , ∫ 1 0 y(s)ds = 2 ∫ 1/2 1/3 x(s)ds + 4 ∫ 2/3 1/2 x(s)ds + x (1 3 ) + 1 2 x (2 3 ) . (3.13) Here σ = 3/2,φ = 3/2,β1 = 5/2,β2 = 4,η1 = 1/7,η2 = 1/6,η3 = 1/5,γ1 = 3/2,γ2 = 2,ρ1 = 1/4,ρ2 = 1/3,β′1 = 2,β ′ 2 = 4,γ ′ 1 = 1,γ ′ 2 = 1/2,θ1 = 1/3,θ2 = 1/2,θ3 = 2/3,ζ1 = 1/3,ζ2 = 2/3, f(t,x,y) = 1 4(t + 2)2 |x| 1 + |x| + 1 + 1 32 sin2 y, and g(t,x,y) = 1 32π sin(2πx) + |y| 16(1 + |y|) + 1 2 . With the given data, we find Int. J. Anal. Appl. 17 (6) (2019) 955 that Q1 ≈ 1.618712, Q2 ≈ 0.096199, Q3 ≈ 1.276613, Q4 ≈ 0.199029. Note that |f(t,x1,x2) −f(t,y1,y2)| ≤ 1 16 |x1−x2|+ 1 16 |y1−y2|, |g(t,x1,x2)−g(t,y1,y2)| ≤ 1 16 |x1−x2|+ 1 16 |y1−y2|, and (Q1 +Q3)`1 +(Q2 +Q4)`2 ≈ 0.217861 < 1. Thus all the conditions of Theorem 3.1 are satisfied and consequently, its conclusion applies to the problem (3.13). 4. Conclusions In this paper we introduced and solved a new boundary value problem consisting of nonlinear coupled fractional differential equations and nonlocal coupled integro-multistrip-multipoint boundary conditions. Assuming different conditions on the nonlinear functions involved in the given problem, we have presented the criteria ensuring the existence of solutions for the problem at hand by applying Banach contraction mapping principle, Leray-Schauder alternative and Krasnoselskii fixed point theorem. The obtained results are of quite general nature and lead to several interesting special cases (new results) by fixing the values of the parameters involved in the problem appropriately. For example, if we take all βi−1 = 0, i = 2, . . . ,p in the results of this paper, we obtain the ones associated with the boundary conditions of the form: x ( î ) (0) = 0, ∫ 1 0 x(s)ds = q∑ j=1 γj y(ρj), î = 0, 1, 2, . . . ,n− 2, y ( ĵ ) (0) = 0, ∫ 1 0 y(s)ds = µ∑ i=2 β ′ i−1 ∫ θi θi−1 x(s)ds + λ∑ j=1 γ ′ j x(ζj), ĵ = 0, 1, 2, . . . ,m− 2. Letting all β′i−1 = 0, i = 2, . . . ,p, our results correspond to the ones with the boundary conditions: x ( î ) (0) = 0, ∫ 1 0 x(s)ds = p∑ i=2 βi−1 ∫ ηi ηi−1 y(s)ds + q∑ j=1 γj y(ρj), î = 0, 1, 2, . . . ,n− 2, y ( ĵ ) (0) = 0, ∫ 1 0 y(s)ds = λ∑ j=1 γ ′ j x(ζj), ĵ = 0, 1, 2, . . . ,m− 2. In case all βi−1 = 0, i = 2, . . . ,p and β ′ i−1 = 0, i = 2, . . . ,µ, the results of this paper lead to the ones for the integro-multipoint boundary conditions: x ( î ) (0) = 0, ∫ 1 0 x(s)ds = q∑ j=1 γj y(ρj); y ( ĵ ) (0) = 0, ∫ 1 0 y(s)ds = λ∑ j=1 γ ′ j x(ζj), where î = 0, 1, 2, . . . ,n− 2 and ĵ = 0, 1, 2, . . . ,m− 2. Fixing all γj = 0,j = 1, . . . ,q and γ ′ j = 0, j = 1, . . . ,λ in the results of this paper, we get the ones associated with coupled integro-multistrip conditions of the form: x( î )(0) = 0, ∫ 1 0 x(s)ds = p∑ i=2 βi−1 ∫ ηi ηi−1 y(s)ds, î = 0, 1, 2, . . . ,n− 2, y( ĵ )(0) = 0, ∫ 1 0 y(s)ds = µ∑ i=2 β′i−1 ∫ θi θi−1 x(s)ds, ĵ = 0, 1, 2, . . . ,m− 2. Int. J. Anal. Appl. 17 (6) (2019) 956 References [1] I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, 1999. [2] A.A. Kilbas, H.M. Srivastava, J.J. 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Nauk 10 (1955), 123-127. 1. Introduction 2. Preliminaries 3. Main Results 4. Conclusions References