Fractional-order boundary value problems with Katugampola fractional integral conditions

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1 Mahmudov and Emin Advances in Difference Equations :8 R E S E A R C H Open Access Fractional-order boundary value problems with Katugampola fractional integral conditions Nazim I. Mahmudov * and Sedef Emin * Correspondence: nazim.mahmudov@emu.edu.tr Department of Mathematics, Eastern Mediterranean University, Gazimagusa, Turkish Republic of Northern Cyprus Abstract In this paper, we study existence uniqueness of solutions for nonlinear fractional differential equations with Katugampola fractional integral conditions. Several fixed point theorems are used for sufficient conditions of existence uniqueness solutions of nonlinear differential equations such as Banach s contraction principle, the Leray Schauder nonlinear alternative, and Krasnoselskii s fixed point theorem. Applications of the main results are also presented. Keywords: Fractional differential equations; Katugampola derivative; Integral boundary conditions Introduction In recent years, boundary value problems for nonlinear fractional differential equations have been studied by several researchers. In fact, fractional differential equations have played an important role in physics, chemical technology, biology, economics, control theory, signal and image processing, see [ 25] and the references cited therein. Boundary value problems of fractional differential equations and inclusions involve different kinds of boundary conditions such as nonlocal, integral, and multipoint boundary conditions. The fractional integral boundary conditions were introduced lately in [26]and nonlocal conditions were presented by Bitsadze, see [3]. In [4] authors gave sufficient criteria for existence of solutions for the following Caputo fractional differential equation: D q xt=f t, xt, 0<t < T, subject to nonlocal generalized Riemann Liouville fractional integral boundary conditions of the form ξ x0 = γ α s xs Ɣα t s ds := γ I α xξ, α xt=δ β Ɣβ 0 ε 0 s xs ξ s β ds := δ I β xε, 0 < ξ, ε < T, The Authors 208. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original authors and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Mahmudovand Emin Advances in Difference Equations :8 Page 2 of 7 where D q denotes the Caputo fractional derivative of order q, I z, z {α, β}, isthegeneralized Riemann Liouville fractional integral of order z >0, >0,ξ, ε arbitrary, with ξ, ε 0, T, γ, δ R and f :[0,T] R R is a continuous function. In [5] authors established the existence of solutions for the following nonlinear Riemann Liouville fractional differential equation subject to nonlocal Erdelyi Kober fractional integral conditions: D q xt=f t, xt, t 0, T, x0 = 0, αxt= m i= β i I γ i,δ i η i xξ i, where < q 2, D q is the standard Riemann Liouville fractional derivative of order q, I γ i,δ i η i is the Erdelyi Kober fractional integral of order δ i >0withη i >0andγ i R, i =,2,...,m, f :[0,T] R R is a continuous function and α, β i R, ξ i 0, T, i =,2,...,m, are given constants. Motivated by the above papers, in this paper, we study the sufficient conditions of existence uniqueness solutions of nonlocal boundary conditions for the following nonlinear fractional differential equation of order α 2, 3]: c D α xt=f t, xt, t [0, T], xt=β I q xɛ, 0 < ɛ < T, x T=γ I q x η, 0 < η < T, x T=δ I q x ζ, 0 < ζ < T, where D α is the Caputo fractional derivative. I q is the Katugampola integral of q >0, >0,f :[0,T] R R is a continuous function. The rest of the paper is organized as follows. In Sect. 2, we recall some definitions and lemmas that we need in the sequel. In Sect. 3, several fixed point theorems are used to give sufficient conditions for existence uniqueness of solutions of such as Banach s contraction principle, Krasnoselskii s fixed point theorem, and the Leray Schauder nonlinear alternative. In Sect. 4,someillustratingexamplesaregiven. 2 Preliminaries In this section, we recall some basic definitions of fractional calculus [, 2, 27] andsome auxiliary lemmas which we need later. Definition [] The Riemann Liouville fractional integral of order p > 0 of a continuous function f :0, R is defined by J p f t= Ɣp t 0 t s p f s ds, provided the right-hand side is point-wise defined on 0,, where Ɣ is the gamma function defined by Ɣp= 0 e s s p ds.

3 Mahmudovand Emin Advances in Difference Equations :8 Page 3 of 7 Definition 2 [] The Riemann Liouville fractional derivative of order p > 0 of a continuous function f :0, R is defined by D p 0 f t= d n t t s n p f s ds, n <p < n, Ɣn p dt 0 where n =[p],[p] denotes the integer part of a real number p. Definition 3 The Caputo derivative of order p for a function f :[0, R can be written as n c D p f t=d p t k 0 f t k! f k 0, t >0,n <p < n. k=0 Remark 4Iff t C n [0,, then c D p f t= t f n s Ɣn p 0 t s ds = p n In p f n t, t >0,n <p < n. Definition 5 [28] Katugampola integral of order q >0and > 0, of a function f t, for all 0<t <,isdefinedas t I q f t= q s f s ds, Ɣq 0 t s q provided the right-hand side is point-wise defined on 0,. Remark 6[28] The above definition corresponds to the one for Riemann Liouville fractional integral of order q >0when =, while the famous Hadamard fractional integral follows for 0, that is, lim I q f t= 0 Ɣq t 0 log t q f s ds. s s Lemma 7 [4] Let, q >0and p >0be the given constants. Then the following formula holds: I q t p = p Ɣ t pq Ɣ pq. q Lemma 8 [] For q >0and x C0, T L0, T. Then the fractional differential equation c D q xt=0has a unique solution xt=c 0 c t c n t n, and the following formula holds: I q D q xt=xtc 0 c t c n t n, where c i R, i =0,,...,n,and n q < n.

4 Mahmudovand Emin Advances in Difference Equations :8 Page 4 of 7 Lemma 9 Let 2<α 3 and β, γ, δ R. Then, for any y C[0, T], R, x is a solution of the following nonlinear fractional differential equation with Katugampola fractional integral conditions: c D α xt=yt, t [0, T], xt=β I q xɛ, 0 < ɛ < T, x T=γ I q x η, 0 < η < T, x T=δ I q x ζ, 0 < ζ < T, 2 if and only if where xt=j α yt β I q J α yɛ J α yt ϖ β, ɛ ϖ2 β, ɛ γ ϖ γ, η ϖ β, ɛ t I q J α yη J α yt ϖ3 β, ɛ ϖ δ, ζ 2ϖ β, ɛ ϖ 2β, ɛϖ 2 γ, η ϖ β, ɛϖ γ, η ϖ 2γ, ηt ϖ γ, η t2 J α 2 yt δ I q J α 2 yζ, 3 2 ϖ α, ξ= α ξ q q ϖ 2 α, ξ= T α ξ q Ɣq Ɣ q q Ɣ ϖ 3 α, ξ= T 2 α ξ q2 2 Ɣ 2q q Ɣ 0, 4, 5. 6 Proof Using Lemma 8, the general solution of the nonlinear fractional differential equation in 2canberepresentedas xt=c 0 c t c 2 t 2 J α yt, c 0, c, c 2 R. 7 By using the first integral condition of problem 2 and applying thekatugampolaintegral on 7, we obtain c 0 c T c 2 T 2 J α yt ɛ q = βc 0 q Ɣq βc ɛ q ɛ q2 2 Ɣ βc 2 q Ɣ 2q Ɣ q q Ɣ β I q J α yɛ.

5 Mahmudovand Emin Advances in Difference Equations :8 Page 5 of 7 After collecting the similar terms in one part, we have the following equation: c 0 β ɛq ɛ c q T q Ɣ βc Ɣq q c 2 T 2 ɛ q2 2 Ɣ βc 2 q Ɣ 2q Ɣ q = β I q J α yɛ J α yt. 8 Rewriting equation 8 by using4, 5, and 6, we obtain c 0 ϖ β, ɛc ϖ 2 β, ɛc 2 ϖ 3 β, ɛ=β I q J α yɛ J α yt. 9 Then, taking the derivative of 7 and using the second integral condition of 2, we get x T=c 2c 2 T J α yt. 0 Now, applying the Katugampola integral on 0, we have c 2c 2 T J α yt η q = γ c q Ɣq 2c 2 γ ηq Ɣ q Ɣ q The above equation implies that c γ ηq 2c q 2 T γ ηq Ɣ Ɣq q γ I q J α yη. Ɣ q = γ I q J α yη J α yt. 2 Also, by using 4and5, equation 2canbewrittenas c ϖ γ, η2c 2 ϖ 2 γ, η = γ I q J α yη J α yt. 3 By using the last integral condition of 2 and applying Katugampola integral operator on the second derivative of 0, we have 2c 2 J α 2 yt=2δc 2 ζ q q Ɣq δ I q J α 2 yζ. Hence, we obtain the following equation: 2c 2 δ ζ q = δ I q J α 2 yζ J α 2 yt. 4 q Ɣq

6 Mahmudovand Emin Advances in Difference Equations :8 Page 6 of 7 By using 4, equation 4can be written as 2c 2 ϖ δ, ζ =δ I q J α 2 yζ J α 2 yt. 5 Moreover, equation 5implies that c 2 = δ I q J α 2 yζ J α 2 yt. 6 2ϖ δ, ζ Substituting the values of 6in3, we get c = γ I q J α yη J α yt ϖ γ, η ϖ 2 γ, η δ I q J α 2 yζ J α 2 yt. 7 ϖ γ, ηϖ δ, ζ Now, substituting the values of 6 and7 in9, we obtain c 0 = β I q J α yɛ J α yt ϖ β, ɛ ϖ 2 β, ɛ γ I q J α yη J α yt ϖ β, ɛϖ γ, η ϖ 3 β, ɛ δ I q J α 2 yζ J α 2 yt 2ϖ β, ɛϖ δ, ζ ϖ 2 β, ɛϖ 2 γ, η δ I q J α 2 yζ J α 2 yt. 8 ϖ β, ɛϖ γ, ηϖ δ, ζ Finally, substituting the valuesof 8, 7, and 6inequation7, we obtain the general solution of problem 2 whichis3. Converse is also true by using the direct computation. 3 Main results Let us denote by C = C[0, T], R the Banach space of all continuous functions from [0, T] R endowed with a topology of uniform convergence with the norm defined by x = sup{ xt : t [0, T]}. We define an operator H : C C on problem as Hxt=J α f s, xs t β I q J α f s, xs ɛ ϖ β, ɛ ϖ2 β, ɛ ϖ β, ɛ t J α f s, xs T ϖ γ, η γ I q J α f s, xs η J α f s, xs T ϖ δ, ζ ϖ 2γ, ηt ϖ γ, η t2 2 ϖ3 β, ɛ 2ϖ β, ɛ ϖ 2β, ɛϖ 2 γ, η ϖ β, ɛϖ γ, η J α 2 f s, xs T δ I q J α 2 f s, xs ζ. 9

7 Mahmudovand Emin Advances in Difference Equations :8 Page 7 of 7 Also, we define the notations T α = Ɣα ϖ β, ɛ Ɣα Ɣ α β ɛ αq T α Ɣ αq ϖ γ, η Ɣα Ɣ α γ Ɣ α q q ϖ δ, ζ Ɣα ϖ2 β, ɛ ϖ β, ɛ T η α q T α q ϖ3 β, ɛ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T ϖ γ, η Ɣ α 2 ζ α 2q δ Ɣ α 2q T α 2 q T and β = ϖ β, ɛ Ɣα Ɣ α ɛ αq Ɣ αq q γ ϖ2 β, ɛ Ɣ α ϖ γ, η Ɣα ϖ β, ɛ T η α q Ɣ α q q δ ϖ3 β, ɛ ϖ δ, ζ Ɣα 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T ϖ γ, η T 2 2 Ɣ α 2 Ɣ α 2q ζ α 2q q. 2 In the following subsections, we prove existence uniqueness results for the boundary value problem by using Banach s fixed point theorem, the Leray Schauder nonlinear alternative, and Krasnoselskii s fixed point theorem. 3. Existence and uniqueness result Theorem 0 Let f :[0,T] R R be a continuous function. Assume that: S f t, x f t, y L x y for all t [0, T], L >0, x, y R; S 2 L <, where is defined by 20. Then the boundary value problem has a unique solution on [0, T]. Proof Byusingthe operator H,which isdefinedby 9, we obtain Hxt Hyt J α f s, xs f s, ys T β I q J α f s, xs f s, ys ɛ ϖ β, ɛ

8 Mahmudovand Emin Advances in Difference Equations :8 Page 8 of 7 ϖ β, ɛ Jα f s, xs f s, ys T γ ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ T I q J α f s, xs f s, ys η ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ T J α f s, xs f s, ys T ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 J α 2 f s, xs f s, ys T δ ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 I q J α 2 f s, xs f s, ys ζ { L x y J α T β I q J α ɛ ϖ β, ɛ J α T ϖ2 β, ɛ γ ϖ γ, η ϖ β, ɛ T I q J α η J α T ϖ δ, ζ ϖ 2γ, η T ϖ γ, η J α 2 T } ϖ3 β, ɛ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η T 2 δ I q J α 2 ζ 2 { T α L Ɣα ϖ β, ɛ Ɣα β Ɣ α ɛ αq Ɣ αq q ϖ γ, η Ɣα Ɣ α γ Ɣ α q ϖ δ, ζ Ɣα T α ϖ2 β, ɛ ϖ β, ɛ T η α q T α q ϖ3 β, ɛ 2 ϖ β, ɛ

9 Mahmudovand Emin Advances in Difference Equations :8 Page 9 of 7 ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 Ɣ α 2 ζ α 2q } δ Ɣ α 2q T α 2 x y q = L x y for any x, y C and for each t [0, T]. This implies that Hx Hy L x y.asl <, the operator H : C C is a contraction mapping. As a result, the boundary value problem hasauniquesolutionon[0,t]. 3.2 Existence results Lemma Let E be a Banach space, Cbeaclosed, convex subset of E, Ubeanopensubset of C and 0 U. Suppose that A : U C is a continuous, compact map. Then either i A has a fixed point in U, or ii there are x U the boundary of U in C and μ 0, with x = μax. Theorem 2 Let f :[0,T] R R be a continuous function. Assume that: S 3 There exist a nonnegative function C[0, T], R and a nondecreasing function :[0, 0, such that f t, u t u for any t, u [0, T] R; S 4 There exists a constant M >0such that M M >, where in 20. Then problem has at least one solution on [0, T]. Proof Let B d = {x C : x d} be a closed bounded subset in C[0, T], R. Notice that problem is equivalent to the problem of finding a fixed point of H. As a first step, we show that the operator H, which is defined by 9, maps bounded sets into bounded sets in C[0, T], R. Then, for t [0, T], we have Hxt J α f s, xs T β I q J α f s, xs ɛ ϖ β, ɛ J α f s, xs T ϖ γ, η ϖ2 β, ɛ ϖ β, ɛ T γ J α f s, xs ηj α f s, xs T ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2

10 Mahmudovand Emin Advances in Difference Equations :8 Page 0 of 7 J α 2 f s, xs T δ I q J α 2 f s, xs ζ x J α st x β I q J α sɛ ϖ β, ɛ J α st x ϖ γ, η ϖ2 β, ɛ ϖ β, ɛ T γ I q J α sηj α st x ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 J α 2 st δ I q J α 2 sζ { T α d Ɣα ϖ β, ɛ Ɣα β Ɣ α ɛ αq Ɣ αq q ϖ γ, η Ɣα Ɣ α γ Ɣ α q ϖ δ, ζ Ɣα T α ϖ2 β, ɛ ϖ β, ɛ T η α q T α q ϖ3 β, ɛ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 Ɣ α 2 ζ α 2q } δ Ɣ α 2q T α 2 q = d, which leads to Hx d.bys 4 thereexistsd >0suchthat d < d. Next, we show that the map H : C[0, T], R C[0, T], R is completely continuous. Therefore, to prove that the map H is completely continuous, we show that H is a map from bounded sets into equicontinuous sets of C[0, T], R. Let us choose t, t 2 from the interval [0, T]andalsot < t 2.Thenwehave Hxt2 Hxt J α f s, xs t 2 J α f s, xs t t 2 t γ q J α f s, xs ηj α f s, xs T ϖ γ, η ϖ2 γ, η ϖ δ, ζ ϖ γ, η t 2 t t2 2 t2 2

11 Mahmudovand Emin Advances in Difference Equations :8 Page of 7 J α 2 f s, xs T δ I q J α 2 f s, xs ζ x s Ɣα [ t x t 2 t ϖ γ, η ϖ2 γ, η ϖ δ, ζ t 2 t 2 [ t 0 [ t2 s α t s α ] ds t2 γ J α sηj α st t ] t 2 s α ds ϖ γ, η J α 2 st δ I q J α 2 sζ ϖ δ, ζ [ t2 s α t s α ] ds t2 d Ɣα 0 t 2 t d { T α γ ϖ γ, η Ɣα ϖ2 γ, η Ɣα ϖ δ, ζ t 2 t 2 T α 2 δ Ɣ α 2 Ɣ α 2q ζ α 2q q t Ɣ α ] t 2 s α ds η α q Ɣ α q q ϖ γ, η ϖ δ, ζ }. 22 It is clear that the right-hand side of 22 isindependentofx. Therefore, as t 2 t 0, inequality 22tendstozero.ThatmeansH is equicontinuous, and by the Arzelà Ascoli theorem, the operator H : C[0, T], R C[0, T], R is completely continuous. In the last step we show that the operator H has a fixed point. Let Hx=x be a solution. Then, for t [0, T], Hx = x x, which implies that x x. In view of S 4, there exists positive M such that x M.Letusset U = { x C [0, T], R : x < M }. Then the operator H : U C[0, T], R is continuous and completely continuous. From thechoiceofu,thereisnox U such that x = μhx for some μ 0,. It can be proved by using contraction. Assume that there exists x U such that x = μhx for some μ 0,. Then x = μhx Hx x, x x. This contradicts x x >.

12 Mahmudovand Emin Advances in Difference Equations :8 Page 2 of 7 Consequently, by the nonlinear alternative of Leray Schauder type, we conclude that H has a fixed point x U, whichisasolutionofproblem. This completes the proof. Theorem 3 [29] Let M be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A, B be the operators such that a Ax By Mwheneverx, y M; bais compact and continuous;c B is a contraction mapping. Then there exists z M such that z = Az Bz. Theorem 4 Let f :[0,T] R R be a continuous function, and let condition S hold. In addition, the function f satisfiesthe assumptions: S 5 There exists a nonnegative function C[0, T], R such that f t, u t for any t, u [0, T] R. S 6 L <, where is defined by 2. Then the boundary value problem has at least one solution on [0, T]. Proof We first define the new operators H and H 2 as and H xt=j α f s, xs t ϖ β, ɛ Jα f s, xs T ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ t J α f s, xs T ϖ3 β, ɛ ϖ δ, ζ 2ϖ β, ɛ ϖ 2β, ɛϖ 2 γ, η ϖ β, ɛϖ γ, η ϖ 2γ, ηt ϖ γ, η t2 2 J α 2 f s, xs T 23 β H 2 xt= I q J α f s, xs ɛ ϖ β, ɛ γ ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ t I q J α f s, xs η δ ϖ3 β, ɛ ϖ δ, ζ 2ϖ β, ɛ ϖ 2β, ɛϖ 2 γ, η ϖ β, ɛϖ γ, η ϖ 2γ, ηt ϖ γ, η t2 2 I q J α 2 f s, xs ζ. 24 Then we consider a closed, bounded, convex, and nonempty subset of the Banach space X as B d = { x C : x < d } with d, where is defined by 20. Now, we show that H x H 2 y B d for any x, y B d,whereh

13 Mahmudovand Emin Advances in Difference Equations :8 Page 3 of 7 and H 2 are denoted by 23and24, respectively. H x H 2 y J α f s, xs T J α f s, xs T β I q J α f s, xs ɛ ϖ β, ɛ ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ T J α f s, xs T γ I q J α f s, xs η ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T ϖ γ, η T 2 2 J α 2 f s, xs T δ I q J α 2 f s, xs ζ J α st J α st β I q J α sɛ ϖ β, ɛ ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ T J α st γ I q J α sη ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 J α 2 st δ I q J α 2 sζ { T α Ɣα ϖ β, ɛ Ɣα T α β ϖ γ, η Ɣα T α γ Ɣ α Ɣ αq ϖ δ, ζ Ɣα ɛ αq q ϖ2 β, ɛ ϖ β, ɛ T Ɣ α η α q Ɣ α q q ϖ3 β, ɛ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 Ɣ α 2 T α 2 ζ α 2q } δ Ɣ α 2q q = d. Therefore, it is clear that H x H 2 y d.hence,h x H 2 y B d.

14 Mahmudovand Emin Advances in Difference Equations :8 Page 4 of 7 The next step is related to the compactness and continuity of the operator H.Theproof issimilartothatoftheorem2. Finally, we show that the operator H 2 is a contraction. By using assumption S, H 2 x H 2 y β I q J α f s, xs f s, ys ɛ ϖ β, ɛ γ ϖ γ, η ϖ2 β, ɛ ϖ β, ɛ T I q J α f s, xs f s, ys η δ ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 ϖ γ, η 2 I q J α 2 f s, xs f s, ys ζ { β L x y I q J α ɛ ϖ β, ɛ γ ϖ2 β, ɛ ϖ γ, η ϖ β, ɛ T I q J α η δ ϖ3 β, ɛ ϖ δ, ζ 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η ϖ 2γ, η T T 2 } I q J α 2 ζ ϖ γ, η 2 { β Ɣ α L ɛ αq ϖ β, ɛ Ɣα Ɣ αq q γ ϖ2 β, ɛ ϖ γ, η Ɣα ϖ β, ɛ T α Ɣ η α q Ɣ α q q δ ϖ3 β, ɛ ϖ δ, ζ Ɣα ϖ 2γ, η T T 2 ϖ γ, η 2 α 2 Ɣ Ɣ α 2q = L x y, ζ α 2q q 2 ϖ β, ɛ ϖ 2β, ɛ ϖ 2 γ, η ϖ β, ɛ ϖ γ, η } x y which means H 2 x H 2 y L x y.asl <, the operator H 2 is a contraction. For this reason, problem hasat leastonesolution on [0,T]. 4 Examples In this section, some examples are illustrated to show our results.

15 Mahmudovand Emin Advances in Difference Equations :8 Page 5 of 7 Example Consider the following nonlinear fractional differential equation with Katugampola fractional integral conditions: c D 5/2 xt= sin2 πt e t 0 xt xt, t [0, 2 ], x 2 = 5 2 I 3 x3/8, x 2 = 5 2 I 3 x/3, x 2 = 5 2 I 3 x2/5. 25 Here, α = 5/2, T = 2, β = /2, γ = /2, δ = /2, ɛ = 3/8, η = /3, ζ = 2/5, =5,q = 3,and f t, x= sin2 πt x e t 0 x. Hence, we have f t, x f t, y 0 x y. Then, assumption S issatisfiedwithl = 0. By using the Matlab program, ω 2, 3 8 = 0.936, ω 2, 3 = , ω 2, 2 5 = , ω 2 2, 3 8 = , ω 2 2, 3 = , ω 3 2, 3 = , and =.226 are found. Therefore, L = <, which implies that assumption S 2 holds true. By using Theorem 0, 8 the boundary value problem 25hasauniquesolutionon[0, 2 ]. Example 2 Consider the following nonlinear fractional differential equation with Katugampola fractional integral conditions: D 5/2 xt= t2 0 x2 t xt xt 2 xt 2, t [0, 2 ], x 2 = 5 2 I 3 x3/8, x 2 = 5 2 I 3 x/3, x 2 = 5 2 I 3 x2/5, 26 where α = 5/2, T = 2, β = /2, γ = /2, δ = /2, ɛ = 3/8, η = /3, ζ = 2/5, =5,q = 3. Moreover, f t, u = t 2 u 2 u 0 u 2 u t2 u. 2 0 By using assumption S 3, it is easy to see that t= t2 and u = u.moreover, 0 = and =.226 which was found in the previous example. Now, we need to show 8 that there exists M >0suchthat M M >, and such M >0existsif >0. By using direct computation = <, assumption S 4 issatisfied.hence,by using Theorem 2,theboundaryvalueproblem26 has at least one solution on [0, 2 ].

16 Mahmudovand Emin Advances in Difference Equations :8 Page 6 of 7 Example 3 Consider the following nonlinear fractional differential equation with Katugampola fractional integral conditions: c D 5/2 xt= 9 sin2 πt e t xt 0 xt, t [0, 2 ], x 2 = 5 2 I 3 x3/8, x 2 = 5 2 I 3 x/3, x 2 = 5 2 I 3 x2/5. 27 Here, α = 5/2, T = 2, β = /2, γ = /2, δ = /2, ɛ = 3/8, η = /3, ζ = 2/5, =5,q = 3,and f t, x= 9 sin2 πt x e t 0 x. Since f t, x f t, y 9 0 x y,thenitimpliesthatl = 9 0 means S issatisfiedbuts 2, which is L <, is not satisfied. [L =.0358 >.] Therefore, we consider S 5 whichis f t, x 9 x e t 0 x 8 e t 0 = t. By using 2, = is found. It is obvious that L = <. So, S 6 issatisfied. Hence, by using Theorem 4, the boundary value problem27 has at least one solution on [0, 2 ]. Acknowledgements The authors would like to thank the anonymous referees and the editor for their constructive suggestions on improving the presentation of the paper. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 29 November 207 Accepted: 25 February 208 References. Kilbas, A.A., Srivastava, H.M., Trujillo, J.J.: Theory and Applications of Fractional Differential Equations. North-Holland Mathematics Studies, vol Elsevier, Amsterdam Samko, S.G., Kilbas, A.A., Marichev, O.I.: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, Yverdon Bitsadze, A.V.: On the theory of nonlocal boundary value problems. Dokl. Akad. Nauk SSSR 277, Ahmad, B., Ntouyas, S.K., Tariboon, J.: Nonlocal fractional-order boundary value problems with generalized Riemann Liouville integral boundary conditions. J. Comput. Anal. Appl. 237, Thongsalee, N., Ntouyas, S.K., Tariboon, J.: Nonlinear Riemann Liouville fractional differential equations with nonlocal Erdelyi Kober fractional integral conditions. Fract. Calc. Appl. Anal. 92, Ahmad, B.: Existence of solutions for fractional differential equations of order q 2, 3] with anti-periodic boundary conditions. J. Appl. Math. Comput. 34, Ahmad, B.: On nonlocal boundary value problems for nonlinear integro-differential equations of arbitrary fractional order. Results Math. 63, Ahmad, B., Alsaedi, A.: Existence and uniqueness of solutions for coupled systems of higher order nonlinear fractional differential equations. Fixed Point Theory Appl. 200,ArticleID Ahmad, B., Nieto, J.J.: Existence results for nonlinear boundary value problems of fractional integro differential equations with integral boundary conditions. Bound. Value Probl. 2009,Article ID Ntouyas, S.K.: Boundary value problems for nonlinear fractional differential equations and inclusions with nonlocal and fractional integral boundary conditions. Opusc. Math. 33, Ahmad, B., Nieto, J.J.: Existence results for a coupled system of nonlinear fractional differential equations with three-point boundary conditions. Comput. Math. Appl. 58,

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