On high order fractional integro-differential equations including the Caputo–Fabrizio derivative

On high order fractional integro-differential equations including the Caputo–Fabrizio derivative Department of Mathematics, By using the fractional Caputo–Fabrizio derivative, we introduce two types new high Cankaya University, Ankara, Turkey Institute of Space Sciences, order derivations called CFD and DCF. Also, we study the existence of solutions for Bucharest, Romania two such type high order fractional integro-differential equations. We illustrate our Full list of author information is results by providing two examples. available at the end of the article MSC: 34A08; 34A99 Keywords: Caputo–Fabrizio derivative; Fractional integro-differential equation; High order derivation 1 Introduction Fractional integro-differential equations have been studied by many researchers from dif- ferent points of view during the last decades (see for example, [5, 10]and [15–19]). In 2015, a new fractional derivation without singular kernel was introduced by Caputo and Fabrizio ([8]). Some researchers tried to use it for solving different equations (see, for example, [2, 9]and [14]). Recently, approximate solutions of some fractional differential equations have been reviewed (see, for example, [3, 4, 6, 12, 13]and [7]). Also, one is finding some new applications for fractional derivations (see, for example, [3]). In this manuscript we consider b >0, x ∈ H (0, b)and α ∈ (0, 1). The expression of the Caputo–Fabrizio fractional derivative of order α for the function x has the form B(α) –α CF α D x(t)= exp( (t –s))x (s) ds,where t ≥ 0([1, 8]and [9]). B(α)is a normalization 1–α 0 1–α constant (B(1) = B(0) = 1). The fractional integral of order α for the function x is written CF α 1–α α as ([14]) I x(t)= x(t)+ x(s) ds,whenever 0 < α <1. If n ≥ 1and α ∈ [0, 1], then B(α) B(α) 0 CF α+n CF α+n CF α n the fractional derivative D of order n + α is defined by D x := D (D x(t)) ([6] (k) CF α n and [8]). If the function x is such that x =0 for k = 1,2,3,..., n,then D (D x(t)) = n CF α D ( D )x(t)([8]). Here, D is the ordinary derivation. Lemma 1.1 ([1]and [14]) Let 0< α <1. Then the unique solution for the problem CF α 1–α α D x(t)= y(t) is given by x(t)= x(0) + y(t)+ x(s) ds. B(α) B(α) 0 Theorem 1.2 ([11]) Let (X, d) be a complete metric space and F : X → X be a mapping such that ϕ(d(Fx, Fy)) ≤ ϕ(d(x, y)) – φ(d(x, y)), for all x, y ∈ X, where ϕ, φ : [0, 1] → [0, 1] © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 2 of 15 are continuous non-decreasing maps and ϕ(t)= φ(t)=0 if and only if t =0. Then F has a unique fixed point. 2 Main result (n) 1 Let n be a natural number, α ∈ (0, 1) and x ∈ H (0, 1). Then the fractional CFD of order α and n is defined by B(α) –α CF α+n CF α n (n+1) D x(t)= D D x(t) = exp (t – s) x (s) ds. 1– α 1– α Also, the fractional DCF of order α and n is defined by n t B(α) d –α CF (n) α n CF α D x(t)= D D x(t) = exp (t – s) x (s) ds. 1– α dt 1– α Here, D is the ordinary derivative. Lemma 2.1 Let n be a natural number and α ∈ (0, 1). Then –α (n) CF α CF α+n D x(t)= D x(t)+ exp t σ (α, n,0), 1– α B(α) –α n–i (i) where σ (α, n, t)= ( ) x (t). i=1 1–α 1–α Proof For each k ≥ 1, we have –α –α (k) (k–1) (k–1) exp (t – s) x (s) ds = x (t)– exp t x (0) 1– α 1– α –α –α (k–1) + exp (t – s) x (s) ds. 1– α 1– α Now by using repetition of the last relation, we get B(α) –α CF α+n CF α n (n+1) D x(t)= D D x(t) = exp (t – s) x (s) ds 1– α 1– α B(α) –α –α = exp (t – s) x (s) ds 1– α 1– α 1– α n–i B(α) –α (i) + x (t) 1– α 1– α i=1 n–i B(α) –α –α (i) – exp t x (0) 1– α 1– α 1– α i=1 B(α) –α –α = exp (t – s) x (s) ds 1– α 1– α 1– α –α + σ (α, n, t)– exp t σ (α, n,0) 1– α –α –α CF α = D x(t)+ σ (α, n, t)– exp t σ (α, n,0). 1– α 1– α Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 3 of 15 Also, we have n n–i –α B(α) –α (n) CF α n α CF α (i) D x(t)= D D x(t) = D x(t)+ x (t) 1– α 1– α 1– α i=1 B(α) –α –α = exp (t – s) x (s) ds 1– α 1– α 1– α n–i B(α) –α (i) + x (t) 1– α 1– α i=1 –α CF α = D x(t)+ σ (α, n, t). 1– α –α CF α (n) CF α+n Hence ( D ) x(t)= D x(t)+ exp( t)σ (α, n,0). 1–α CF α+n CF α (n) (k) By using Lemma 2.1,weconcludethat D x(t)=( D ) x(t)whenever x (0) = 0 for 0 ≤ k ≤ n. Lemma 2.2 Let n be a natural number, α ∈ (0, 1) and y ∈ H (0, 1). Then the solution of the CF α+n problem D x(t)= y(t) is given by (n) 1– α α x (0) x (0) n n+1  2 n x(t)= J y(t)+ J y(t)+ x(0) + tx (0) + t + ··· + t . B(α) B(α) 2! n! CF α+n CF α (n) (n) Proof By using Lemma 1.1 for the equation D x(t)= D x (t)= y(t), we get x (t)= (n) 1–α α x (0) + y(t)+ y(s) ds. By using an integration, we obtain B(α) B(α) 0 t t s 1– α α (n–1) (n–1) (n) x (t)= x (0) + tx (0) + y(s) ds + y(r) dr ds. B(α) B(α) 0 0 0 By repeating this method, we deduce that (n–2) (n–2) (n–1) (n) x (t)= x (0) + tx (0) + x (0) t s t s r 1– α α + y(r) dr ds + y(k) dk dr ds. B(α) B(α) 0 0 0 0 0 By continuing the process, we conclude that (n) 1– α α x (0) x (0) n n+1  2 n x(t)= J y(t)+ J y(t)+ x(0) + tx (0) + t + ··· + t . B(α) B(α) 2! n! On the other hand, by using some calculation, one can find that the given map x(t)is a CF α+n solution for the problem D x(t)= y(t).  Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 4 of 15 Lemma 2.3 Let n be a natural number, α ∈ (0, 1) and y ∈ H (0, 1). Then the solution of the CF α (n) problem ( D ) x(t)= y(t) is given by 1– α α n n+1 x(t)= J y(t)+ J y(t)+ x(0) + tx (0) + ··· B(α) B(α) n–i (n) n x (0) t –α n (i) + t – x (0). n! n! 1– α i=1 –α CF α (n) CF α+n Proof By using Lemma 2.2 for ( D ) x(t)= D x(t)+ exp( t)σ (α, n,0), we get 1–α 1– α –α x(t)= J y(t)– exp t σ (α, n,0) B(α) 1– α α –α n+1 + J y(t)– exp t σ (α, n,0) B(α) 1– α (n) x (0) x (0) 2 n + x(0) + tx (0) + t + ··· + t 2! n! or equivalently 1– α 1– α –α α n n n+1 x(t)= J y(t)– σ (α, n,0)J exp t + J y(t) B(α) B(α) 1– α B(α) (n) α –α x (0) x (0) n+1  2 n – σ (α, n,0)J exp t + x(0) + tx (0) + t + ··· + t B(α) 1– α 2! n! 1– α α 1– α –α n n+1 n = J y(t)+ J y(t)– σ (α, n,0)J exp t B(α) B(α) B(α) 1– α (n) α –α x (0) x (0) 1  2 n + J exp t + x(0) + tx (0) + t + ··· + t B(α) 1– α 2! n! 1– α α 1– α –α n n+1 n = J y(t)+ J y(t)– σ (α, n,0)J exp t B(α) B(α) B(α) 1– α (n) 1– α –α x (0) x (0) 2 n + 1– exp t + x(0) + tx (0) + t + ··· + t B(α) 1– α 2! n! (n) 1– α α x (0) x (0) n n+1  2 n = J y(t)+ J y(t)+ x(0) + tx (0) + t + ··· + t B(α) B(α) 2! n! n–i t –α (i) – x (0). n! 1– α i=1 Lemma 2.4 Let α ∈ (0, 1), 2 < q =2 + α <3 and y ∈ H (0, 1). The fractional differential CF q equation D x(t)= y(t) with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0 –(1–α)t αt has the unique solution of the form x(t)= G(t, s)y(s) ds, where G(t, s)= – 0 2B(α) 2B(α) (1–α)t 1–α α 2 αt whenever 0< t ≤ s <1 and G(t, s)= (t – s)+ (t – s) – – (t – s) whenever B(α) 2B(α) 2B(α) 2B(α) 0< s ≤ t <1. 1–α 2 α 3 Proof By using Lemma 2.2,weget x(t)= J y(t)+ J y(t)+ tx (0). Hence, we obtain B(α) B(α) 1–α 1 α 2 x (t)= J y(t)+ J y(t)+ x (0). By using the boundary conditions x (1) + x (0) = 0 and B(α) B(α) 1–α α 1–α α (1–α)t 1 2  2 3 1 x (1) = J y(1) + J y(1) + x (0), we have x(t)= J y(t)+ J y(t)– J y(1) – B(α) B(α) B(α) B(α) 2B(α) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 5 of 15 t t 1 (1–α)t αt 2 1–α α 2 J y(1). Thus, x(t)= y(s)(t – s) ds + y(s)(t – s) ds – y(s) ds – 2B(α) B(α) 0 2B(α) 0 2B(α) 0 1 1 αt CF q y(s)(t – s) ds = G(t, s)y(s) ds.Notethat D x(t)=0 if and only if x(t)=0. This 2B(α) 0 0 implies that the given map x(t) is a unique solution. 1–α α –αt –(1–α)t 3 Note that |G(t, s)|≤| | + | | + | | + | | < ,for t ∈ [0, 1]. Let μ, μ , μ , 1 2 B(α) 2B(α) 2B(α) 2B(α) 2B(α) k , k ∈ C [0, 1], m , m , h and g be bounded continuous functions on I := [0, 1] with M = 1 2 1 2 1 sup |μ(t)| < ∞, M = sup |μ (t)| < ∞, M = sup |μ (t)| < ∞, M = sup |k (t)| < 2 1 3 2 4 1 t∈I t∈I t∈I t∈I ∞, M = sup |k (t)| < ∞, M = sup |m (t)| < ∞, M = sup |m (t)| < ∞, M = 5 2 6 1 7 2 8 t∈I t∈I t∈I sup |h(t)| < ∞, M = sup |g(t)| < ∞, N = sup |μ (t)| < ∞, N = sup |μ (t)| < ∞, 9 1 2 t∈I t∈I t∈I t∈I 1 N = sup |μ (t)| < ∞, N = sup |K (t)| < ∞ and N = sup |K (t)| < ∞.Let α ∈ (0, 1) 3 4 5 t∈I t∈I t∈I 2 1 2 and 2 < q =2 + α < 3. Now, we investigate the CFD fractional integro-differential problem CF q   CF β CF β 1 2 D x(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t)+ k (t) D x(t) 1 2 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds,(1) 1 2 with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0, where 1 < β <2< β <3 1 2 and 1 < γ <2< ν <3. Theorem 2.5 Let ξ , ξ , ξ , ξ and ξ be nonnegative real numbers, f : [0, 1] × R → R an 1 2 3 4 5 integrable function such that f (t, x, y, z, v, w)– f t, x , y , z , v , w ≤ ξ x – x + ξ y – y + ξ z – z + ξ v – v + ξ v – v , 1 2 3 4 5 for all real numbers x, y, z, v, w, x , y , z , v , w ∈ R and t ∈ I. If < , then the problem (1) has a unique solution, where := max{ }, = [M + M + M + 1 2 3 4 1 1 2 3 2B(α) M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 3 3+4α 4 1 5 2 8 9 + + ξ + ξ M + ξ M + ξ + ξ ], = [ ][M + M + 1 2 6 3 7 4 5 2 1 2 2–β 3–β 2–γ 3–ν 2 2B(α) 1 2 M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 4 1 5 2 8 9 1+α M + + + ξ + ξ M + ξ M + ξ + ξ ], = [M + 3 1 2 6 3 7 4 5 3 1 2–β 3–β 2–γ 3–ν B(α) 1 2 M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 4 1 5 8 9 M + M + + + ξ + ξ M + ξ M + ξ + ξ ] and 2 3 1 2 6 3 7 4 5 4 2–β 3–β 2–γ 3–ν 1 2 α M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 4 1 5 2 8 9 [M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ]+ 1 2 3 1 2 6 3 7 4 5 B(α) 2–β 3–β 2–γ 3–ν 1 2 |1–β |M |2–β |M M +N 1–α N +M 1 4 4 4 2 5 5 5 [N +M +N +M +N +M +B(β –1)[ + ]+B(β –2)[ + ]+ 1 1 2 2 3 3 1 2 2 2 B(α) 2–β 3–β (2–β ) (3–β ) 1 1 2 2 M B(γ –1) M B(ν–1) 8 9 ξ + ξ M + ξ M + ξ + ξ ]. 1 2 6 3 7 4 5 2–łγ 3–ν Proof Consider the Banach space C [0, 1] equipped with the norm x = max |x(t)| + t∈I 3 3 max |x (t)| + max |x (t)| + max |x (t)|.Define themap F : C [0, 1] → C [0, 1] by t∈I t∈I t∈I R R Fx(t)= G(t, s)R(s) ds t t 1– α α = R(s)(t – s) ds + R(s)(t – s) ds B(α) 2B(α) 0 0 1 1 (1 – α)t αt – R(s) ds – R(s)(t – s) ds, 2B(α) 2B(α) 0 0 Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 6 of 15 where CF β CF β 1 2 (Rx)(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t)+ k (t) D x(t) 1 2 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 and R x (t)= μ(t)x (t)+ μ (t)x(t)+ μ (t)x (t)+ μ x (t) CF β + μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t) 2 1 1– β B(β –1) 1 1 CF β + k (t) D x(t)+ x (t) 2– β 2– β 1 1 2– β B(β –2) 2 2 CF β   CF β 2 2 + k (t) D x(t)+ x (t) + k (t) D x(t) 3– β 3– β 2 2 CF γ CF ν + f t, x(t), m (t)x (t), m (t)x (t), h(t) D x(t), g(t) D x(t) . 1 2 By using Lemma 2.4, x is a solution for the problem (1)ifand only if x is a fixed point of 0 0 the operator F.Notethat (Rx)(t)–(Ry)(t) CF β CF β 1 2 ≤ μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t)+ k (t) D x(t) 1 2 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 CF β CF β 1 2 – μ(t)y(t)+ μ (t)y (t)+ μ (t)y (t)+ k (t) D y(t)+ k (t) D y(t) 1 2 1 2 CF γ CF ν + f s, y(s), m (s)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) ds 1 2 ≤ μ(t) x(t)– y(t) + μ (t) x (t)– y (t) + μ (t) x (t)– y (t) 1 2 CF β CF β 1 2 + k (t) D x(t)– y(t) + k (t) D x(t)– y(t) 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 CF γ CF ν – f s, y(s), m (s)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) ds 1 2 ≤ M x – y + M x – y + M x – y 1 2 3 M B(β –1) M B(β –2) 4 1 5 2 + x – y + x – y 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + 1 2 3 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 7 of 15 and R x(t)– R y(t) ≤ μ (t) x(t)– y(t) + μ(t)+ μ (t) x (t)– y (t) + μ (t)+ μ (t) x (t)– y (t) + μ (t) x (t)– y (t) 1 2 1– β |k (t)B(β –1)| 1 1 1 CF β + k (t)+ k (t) D x(t)– y(t) + x (t)– y (t) 2– β 2– β 1 1 2– β |k (t)B(β –2)| 2 2 CF β 2 + k (t)+ k (t) D x(t)– y(t) + x (t)– y (t) 3– β 3– β 2 2 CF γ CF ν + f t, x(t), m (t)x (t), m (t)x (t), h(t) D x(t), g(t) D x(t) 1 2 CF γ CF ν – f t, y(t), m (t)y (t), m (t)y (t), h(t) D y(t), g(t) D y(t) . 1 2 Hence, we get R x(t)– R y(t) |1– β |M N + M 1 4 4 4 ≤ N + M + N + M + N + M + B(β –1) + 1 1 2 2 3 3 1 (2 – β ) 2– β 1 1 |2– β |M M + N 2 5 5 5 + B(β –2) + + ξ + ξ M + ξ M 2 1 2 6 3 7 (3 – β ) 3– β 2 2 M B(γ –1) M B(ν –1) 8 9 + ξ + ξ x – y. 4 5 2– łγ 3– ν On the other hand, we have 3 M B(β –1) M B(β –2) 4 1 5 2 Fx(t)– Fy(t) ≤ M + M + M + + 1 2 3 2B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν x – y and t t 1– α α α F x(t)– F y(t)= + t Rx(s)– Ry(s) ds – s Rx(s)– Ry(s) ds B(α) B(α) B(α) 0 0 1 1 (1 – α) αt – Rx(s)– Ry(s) ds – Rx(s)– Ry(s) ds 2B(α) B(α) 0 0 + s Rx(s)– Ry(s) ds 2B(α) 3 3+4α M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + 1 2 3 2 2B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν x – y 2 Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 8 of 15 and so |F x(t)– F y(t)|≤ x – y.Also, we have α 1– α F x(t)– F y(t) ≤ Rx(s)– Ry(s) ds + Rx(t)– Ry(t) B(α) B(α) + Rx(s)– Ry(s) ds B(α) 1+ α M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + 1 2 3 B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ 1 2 6 3 7 4 5 2– γ 3– ν x – y and F x(t)– F y(t) α 1– α = Rx(t)– Ry(t) + R x(t)– R y(t) B(α) B(α) α M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + + ξ + ξ M + ξ M 1 2 3 1 2 6 3 7 B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 1– α 8 9 + ξ + ξ + N + M + N + M + N + M 4 5 1 1 2 2 3 3 2– γ 3– ν B(α) |1– β |M N + M |2– β |M M + N 1 4 4 4 2 5 5 5 + B(β –1) + + B(β –2) + 1 2 2 2 (2 – β ) 2– β (3 – β ) 3– β 1 1 2 2 M B(γ –1) M B(ν –1) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– łγ 3– ν x – y. Thus, Fx – Fy≤ x – y for all x, y ∈ C [0, 1]. Put ϕ(t)=2t and φ(t)= t for all t.Now by using Theorem 1.2, F has a unique fixed point which is the unique solution for the problem (1). Lemma 2.6 Let α ∈ (0, 1) and y ∈ H (0, 1). Then the fractional differential equation (2) CF α D x(t)= y(t) with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0 has –(1–α)t –αt the unique solution x(t)= G(t, s)y(s) ds, where G(t, s)= + (t – s) when- 0 (2–α)B(α) (2–α)B(α) 1–α α (1–α)t αt ever 0< t ≤ s <1 and G(t, s)= (t – s)+ (t – s) – – (t – s) whenever B(α) 2B(α) B(α)(2–α) B(α)(2–α) 0< s ≤ t <1. 1–α α α 2 3 Proof By using Lemma 2.3,weget x(t)= J y(t)+ J y(t)+ x (0)t + x (0)t.Hence, B(α) B(α) 1–α 1–α 1 α 2 1 x (t)= J y(t)+ J y(t)+ x (0). By using the boundary conditions x (1) + x (0) = 0 B(α) B(α) 1–α 1–α 1 α 2 1  1–α 2 α 3 and x (1) = J y(1) + J y(1) + x (0), we obtain x(t)= J y(t)+ J y(t)– B(α) B(α) 1–α B(α) B(α) t t (1–α)t αt 1–α α 1 2 2 J y(1) – J y(1). Thus, x(t)= y(s)(t – s) ds + y(s)(t – s) ds – (2–α)B(α) (2–α)B(α) B(α) 0 2B(α) 0 1 1 1 (1–α)t αt CF α (2) y(s) ds – y(s)(t – s) ds = G(t, s)y(s) ds.Notethat ( D ) x(t)=0 (2–α)B(α) 0 (2–α)B(α) 0 0 if and only if x(t) = 0. This implies that the given map x(t) is a unique solution.  Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 9 of 15 –(1–α)t 1–α α –αt 2 Note that |G(t, s)|≤| |+| |+| |+| | < ,for t ∈ [0, 1]. Let α, β , β , γ , ν ∈ 1 2 B(α) 2B(α) B(α) B(α) B(α) (0, 1). Now, we investigate the DCF fractional integro-differential problem (2) CF α D x(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t) 1 2 (1) (2) CF β CF β 1 2 + λ k (t) D x(t)+ λ k (t) D x(t) 1 1 2 2 + f s, x(s), m (s)x (s), m (s)x (s), 1 2 (1) (2) CF γ CF ν h(s) D x(s), g(s) D x(s) ds (2) with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0. Theorem 2.7 Let ξ , ξ , ξ , ξ , and ξ be nonnegative real numbers, f : [0, 1] × R → R an 1 2 3 4 5 integrable function such that f (t, x, y, z, v, w)– f t, x , y , z , v , w ≤ ξ x – x + ξ y – y + ξ z – z + ξ v – v + ξ v – v 1 2 3 4 5 for all real numbers x, y, z, v, w, x , y , z , v , w and t ∈ I. If < , then the problem (2) has a unique solution, where := max{ }, = [M + M + M + 1 2 3 4 1 1 2 3 (2–α)B(α) (β –β +1)M B(β ) M B(β ) M B(γ ) (ν –ν+1)M B(ν) 2 5 2 3+α 4 1 2 8 9 + + ξ + ξ M + ξ M + ξ + ξ ], = [M + 1 2 6 3 7 4 5 2 1 2 3 2 3 B(α)(2–α) (1–β ) (1–β ) (1–γ ) (1–ν) 1 2 (β –β +1)M B(β ) M B(β ) 2 5 2 M B(γ ) (ν –ν+1)M B(ν) 4 1 2 8 9 M + M + + + ξ + ξ M + ξ M + ξ + ξ ], 2 3 2 3 1 2 6 3 7 4 2 5 3 (1–β ) (1–β ) (1–γ ) (1–ν) 1 2 (β –β +1)M B(β ) M B(β ) 2 5 2 M B(γ ) 2+α 4 1 2 8 = [M + M + M + + + ξ + ξ M + ξ M + ξ + 3 1 2 3 2 3 1 2 6 3 7 4 2 B(α)(2–α) (1–β ) (1–β ) (1–γ ) 1 2 (β –β +1)M B(β ) (ν –ν+1)M B(ν) M B(β ) 2 2 α 5 9 4 1 2 ξ ] and = [M +M +M + + + ξ + ξ M + ξ M + 5 4 1 2 3 1 2 6 3 7 3 2 3 B(α) (1–ν) (1–β ) (1–β ) 1 2 (β –β +1) M B(γ ) (ν –ν+1)M B(ν) B(β )N 1–α 1 8 9 1 1 4 ξ + ξ ]+ [N +M +N +M +N +M +M B(β ) + + 4 5 1 1 2 2 3 3 4 1 2 3 3 2 B(α) (1–γ ) (1–ν) (1–β ) (1–β ) 1 1 2 2 (2β –2β +1) (β –β +1) 2 2 ξ M B(γ ) (ν –ν+1)M B(ν) 2 2 4 8 9 M B(β ) + N B(β ) + ξ + ξ M + ξ M + + ]. 5 2 4 5 2 3 1 2 6 3 7 2 3 (1–β ) (1–β ) (1–łγ ) (1–ν) 2 2 Proof Consider the Banach space C [0, 1] equipped with the norm x = max |x(t)| + t∈I 3 3 max |x (t)| + max |x (t)| + max |x (t)|.Define themap F : C [0, 1] → C [0, 1] by t∈I t∈I t∈I R R Fx(t)= G(t, s)R(s) ds t t 1– α α = R(s)(t – s) ds + R(s)(t – s) ds B(α) 2B(α) 0 0 1 1 (1 – α)t αt – R(s) ds – R(s)(t – s) ds, B(α)(2 – α) B(α)(2 – α) 0 0 where (Rx)(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t) 1 2 (1) (2) CF β CF β 1 2 + λ k (t) D x(t)+ λ k (t) D x(t) 1 1 2 2 (1) (2) CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 0 Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 10 of 15 and R x (t)= μ(t)x (t)+ μ (t)x(t)+ μ (t)x (t)+ μ x (t) (1) CF β + μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t) 2 1 –β β 1 1 1 CF β + k (t)B(β ) D x(t)– x (t)+ x (t) 1 1 1– β (1 – β ) 1– β 1 1 1 –β β CF β + k (t)B(β ) D x(t)+ x (t) 2 2 1– β (1 – β ) 2 2 β x (t) (2) CF β – x (t)+ + k (t) D x(t) (1 – β ) 1– β 2 2 (1) (2) CF γ CF ν + f t, x(t), m (t)x (t), m (t)x (t), h(t) D x(t), g(t) D x(t) . 1 2 By using Lemma 2.6, x is a solution for the problem (2)ifand only if x is a fixed point of 0 0 the operator F.Notethat (Rx)(t)–(Ry)(t) (1) (2) CF β CF β 1 2 ≤ μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ λ k (t) D x(t)+ k (t) D x(t) 1 2 1 1 2 (1) (2) CF γ CF ν + f s, x(s), m (t)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 (1) (2) CF β CF β 1 2 – μ(t)y(t)+ μ (t)y (t)+ μ (t)y (t)+ k (t) D y(t)+ k (t) D y(t) 1 2 1 2 (1) (2) CF γ CF ν + f s, y(s), m (t)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) ds 1 2 ≤ μ(t) x(t)– y(t) + μ (t) x (t)– y (t) + μ (t) x (t)– y (t) 1 2 (1) (2) CF β CF β 1 2 + k (t) D x(t)– y(t) + k (t) D x(t)– y(t) 1 2 (1) (2) CF γ CF ν + |f s, x(s), m (t)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 (1) (2) CF γ CF ν – f s, y(s), m (t)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) | ds 1 2 ≤ M x – y + M x – y + M x – y 1 2 3 M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 + x – y + x – y 2 3 (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 ≤ M + M + M + + 1 2 3 2 3 (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 11 of 15 and R x(t)– R y(t) (β – β +1) B(β )N 1 1 4 ≤ N + M + N + M + N + M + M B(β ) + 1 1 2 2 3 3 4 1 3 2 (1 – β ) (1 – β ) 1 1 2 2 (2β –2β +1) (β – β +1) 2 2 2 2 + M B(β ) + N B(β ) + ξ + ξ M + ξ M 5 2 5 2 1 2 6 3 7 4 3 (1 – β ) (1 – β ) 2 2 ξ M B(γ ) (ν – ν +1)M B(ν) 4 8 9 + + x – y. 2 3 (1 – łγ ) (1 – ν) Thus, 2 M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 Fx(t)– Fy(t) ≤ M + M + M + + 1 2 3 2 3 (2 – α)B(α) (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) x – y and 3+ α M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 F x(t)– F y(t) ≤ M + M + M + + 1 2 3 2 3 B(α)(2 – α) (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) x – y. Also, we have 2+ α M B(β ) 4 1 F x(t)– F y(t) ≤ M + M + M + 1 2 3 B(α)(2 – α) (1 – β ) (β – β +1)M B(β ) 2 5 2 + + ξ + ξ M + ξ M 1 2 6 3 7 (1 – β ) M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ 4 5 2 3 (1 – γ ) (1 – ν) x – y and F x(t)– F y(t) α 1– α ≤ Rx(t)– Ry(t) + R x(t)– R y(t) B(α) B(α) α M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 ≤ M + M + M + + 1 2 3 2 3 B(α) (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 12 of 15 1– α (β – β +1) + N + M + N + M + N + M + M B(β ) 1 1 2 2 3 3 4 1 B(α) (1 – β ) 2 2 B(β )N (2β –2β +1) (β – β +1) 1 4 2 2 2 2 + + M B(β ) + N B(β ) 5 2 5 2 2 4 3 (1 – β ) (1 – β ) (1 – β ) 1 2 2 ξ M B(γ ) (ν – ν +1)M B(ν) 4 8 9 + ξ + ξ M + ξ M + + x – y 1 2 6 3 7 2 3 (1 – łγ ) (1 – ν) x – y. Hence, Fx – Fy≤ x – y for all x, y ∈ C [0, 1]. Put ϕ(t)=2t and φ(t)= t for all t. By using Theorem 1.2, F has a unique fixed point, which is the desired solution for the problem. Here, we provide three examples to illustrate our main results. Consider the bounded 1 3t–1 1 –6t 1 3 continuous functions μ(t)= sin(t), μ (t)= , μ (t)= e , k (t)= t + 1 2 1 100 20t+162 100 300 1 1 1 Ln(t+2) 1 2t t + , k (t)= cos(t), m (t)= e , m (t)= , h(t)=0 and g(t)= for 2 1 2 100 50 800 20 t–900 1 1 all t ∈ I = [0, 1]. Note that M = sup |μ(t)| = , M = sup |μ (t)| = , M = 1 2 1 3 t∈I t∈I 100 91 1 1 1 sup |μ (t)| = , M = sup |k (t)| = , M = sup |k (t)| = , M = sup |m (t)| = 2 4 1 5 2 6 1 t∈I 6 t∈I t∈I t∈I 30 800 100e Ln(3) 2 1 e , M = sup |m (t)| = , M = sup |h(t)| =0 and M = sup |g(t)| = .Also, 7 2 8 9 t∈I t∈I t∈I 1200 900 1  506  6 N = sup |μ (t)| = , N = sup |μ (t)| = , N = sup |μ (t)| = , N = 1 2 3 4 t∈I t∈I 1 2 t∈I 2 6 (162) 100e 1  1 sup |k (t)| = and N = sup |k (t)| = . Also, consider the function B(α)=1 for t∈I 1 t∈I 2 50 800 α ∈ (0, 1). Example 2.1 Let α ∈ (0, 1). By using Lemma 2.4, the fractional differential equation CF 2+α D x (t)= t with boundary conditions x (0) = 0, x (1) + x (0) = 0 and x (0) = 0 has 1 1 1 1 1 theuniquesolution x (t). Also by using Lemma 2.6, the fractional differential equation CF α (2) ( D ) x (t)= t with boundary conditions x (0) = 0, x (1) + x (0) = 0 and x (0) = 0 has 2 2 2 2 2 1 1 1 1 4 99 the unique solution x (t). For α = , α = , α = , α = , α = and α = we compare 100 10 5 2 5 100 the solutions x (t), x (t)and X(t)= x (t)– x (t)in Fig. 1. 1 2 1 2 Example 2.2 Consider the CFD fractional integro-differential problem –6t 12 sin(t) 3t –1 e CF D x(t)= x(t)+ x (t)+ x (t) 100 20t + 162 100 t +3t +6 3 cos(t) 5 CF CF 2 2 + D x(t)+ D x(t) 300 800 Ln(t +2) 1 8 2t   CF + f s, x(s), e x (s), x (s), 0, D x(s) ds (3) 20 t – 900 with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0, where 1 < β = <2< 5 4 8 2 3 1 1 1 β = <3 and 1 < γ = <2< ν = <3. Put f (t, x, y, z, v, w)= t + x + y + z + w + 2 3 3 91 604 200 80 M M M M 4 5 8 9 2v.Notethat = [M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ]= 1 1 2 3 1 2 6 3 7 4 5 2 2–β 3–β 2–γ 3–ν 1 2 3 3+4α M M M 4 5 8 9 0.02, = [ ][M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ] = 0.46, 2 1 2 3 1 2 6 3 7 4 5 2 2 2–β 3–β 2–γ 3–ν 1 2 M M M M 4 5 8 9 =(1 + α)[M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ] = 0.32 and 3 1 2 3 1 2 6 3 7 4 5 2–β 3–β 2–γ 3–ν 1 2 M M M M 4 5 8 9 = α[M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ]+(1– α)[N + M + 4 1 2 3 1 2 6 3 7 4 5 1 1 2–β 3–β 2–γ 3–ν 1 2 |1–β |M N +M |2–β |M M +N M M 1 4 4 4 2 5 5 5 8 6 N +M +N +M + + + + + ξ + ξ M + ξ M + ξ + ξ ]= 2 2 3 3 1 2 6 3 7 4 5 2 2 2–β 3–β 2–łγ 3–ν (2–β ) 1 (3–β ) 2 1 2 0.166. By using Theorem 2.5,the problem(3) has a unique solution. Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 13 of 15 1 1 1 1 4 99 Figure 1 The solutions of the problem with α = , α = , α = , α = , α = and α = 100 10 5 2 5 100 Example 2.3 Consider the DCF fractional integro-differential problem (2) CF D x(t) –6t sin(t) 3t –1 e = x(t)+ x (t)+ x (t) 100 20t + 162 100 t +3t +6 1 cos(t) 2 (1) (2) CF CF 2 3 + D x(t)+ D x(t) 300 800 Ln(t +2) 1 1 (2) –40t   CF + f s, x(s), e x (s), x (s), 0, D x(s) ds,(4) 20 t – 900 2 1 with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0, where α = , β = , 5 2 2 1 1 2 3 1 1 1 β = , γ = and ν = .Put f (t, x, y, z, v, w)= t + x + y + z + w +2v.Notethat 2 18 3 3 5 91 604 200 80 e (β –β +1) M 2 M (ν –ν+1)M 2 4 2 8 9 = [M +M +M + +M + ξ + ξ M + ξ M + ξ + ξ ]< 1 1 2 3 2 5 3 1 2 6 3 7 4 2 5 3 2–α (1–β ) (1–β ) (1–γ ) (1–ν) 1 2 (β –β +1)M M 2 M 3+α 5 4 2 8 0.391, =[ ][M + M + M + + + ξ + ξ M + ξ M + ξ + 2 1 2 3 1 2 6 3 7 4 2 3 2 2–α (1–β ) (1–β ) (1–γ ) 1 2 (β –β +1)M (ν –ν+1)M 2+α M 2 5 9 4 2 ξ ] < 0.225, = [M + M + M + + + ξ + ξ M + ξ M + 5 3 3 1 2 3 2 3 1 2 6 3 7 (1–ν) 2–α (1–β ) (1–β ) 1 2 (β –β +1) M (ν –ν+1)M M 2 8 9 4 2 ξ + ξ ] < 0.132 and = α[M + M + M + + M + ξ + 4 2 5 3 4 1 2 3 2 5 3 1 (1–γ ) (1–ν) (1–β ) (1–β ) 1 2 (β –β +1) M (ν –ν+1)M 1 8 9 1 ξ M + ξ M + ξ + ξ ]+(1– α)[N + M + N + M + N + M + M + 2 6 3 7 4 5 1 1 2 2 3 3 4 2 3 3 (1–γ ) (1–ν) (1–β ) 2 2 (2β –2β +1) N (β –β +1) N 2 2 ξ M (ν –ν+1)M 4 2 2 4 8 9 + M + + ξ + ξ M + ξ M + + ξ ] < 0.493. By 5 1 2 6 3 7 5 2 4 3 2 3 (1–β ) (1–β ) (1–β ) (1–łγ ) (1–ν) 1 2 2 using Theorem 2.7,the problem(4) has a unique solution. Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 14 of 15 3Conclusion It is important that researchers have some methods available enabling them to review some high order fractional integro-differential equations. In this manuscript, we intro- duce two types of new fractional derivatives entitled CFD and DCF and by using those we investigate the existence of solutions for two high order fractional integro-differential equations of such a type including the new derivatives. Acknowledgements The third author was supported by Azarbaijan Shahid Madani University. Funding Not available. Abbreviations DFC, Caputo–Fabrizio derivation followed by a differentiation; CFD, Differentiation followed by Caputo–Fabrizio derivation. Availability of data and materials Data sharing not applicable to this article as no datasets were generated or analyzed during the current study. Competing interests The authors declare that they have no competing interests. Authors’ contributions The authors declare that the study was realized in collaboration with equal responsibility. All authors read and approved the final manuscript. Author details 1 2 Department of Mathematics, Istanbul Technical University, Istanbul, Turkey. Department of Mathematics, Cankaya 3 4 University, Ankara, Turkey. Institute of Space Sciences, Bucharest, Romania. Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran. Department of Medical Research, China Medical University Hospital, China Medical University, Taichung, Taiwan. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 20 February 2018 Accepted: 24 May 2018 References 1. 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Equ. 2017, 51 (2017) 7. Baleanu, D., Mousalou, A., Rezapour, Sh.: On the existence of solutions for some infinite coefficient-symmetric Caputo–Fabrizio fractional integro-differential equations. Bound. Value Probl. 2017, 147 (2017) 8. Caputo, M., Fabrizzio, M.: A new definition of fractional derivative without singular kernel. Prog. Fract. Differ. Appl. 1(2), 73–85 (2015) 9. Caputo, M., Fabrizzio, M.: Applications of new time and spatial fractional derivatives with exponential kernels. Prog. Fract. Differ. Appl. 2(1), 1–11 (2016) 10. De La Sen, M., Hedayati, V., Atani, Y.G., Rezapour, Sh.: The existence and numerical solution for a k-dimensional system of multi-term fractional integro-differential equations. Nonlinear Anal., Model. Control 22(2), 188–209 (2017) 11. Dutta, P.N., Choudary, B.S.: A generalization of contraction principle in metric spaces. Fixed Point Theory Appl. 2008, Article ID 406386 (2008) 12. Kojabad, E.A., Rezapour, Sh.: Approximate solutions of a sum-type fractional integro-differential equation by using Chebyshev and Legendre polynomials. Adv. Differ. Equ. 2017, 351 (2017) 13. Kojabad, E.A., Rezapour, Sh.: Approximate solutions of a fractional integro-differential equation by using Chebyshev and Legendre polynomials. J. Adv. Math. Stud. 11(1), 80–102 (2018) 14. Losada, J., Nieto, J.J.: Properties of a new fractional derivative without singular kernel. Prog. Fract. Differ. Appl. 1(2), 87–92 (2015) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 15 of 15 15. Rezapour, Sh., Hedayati, V.: On a Caputo fractional differential inclusion with integral boundary condition for convex-compact and nonconvex-compact valued multifunctions. Kragujev. J. Math. 41(1), 143–158 (2017) 16. Rezapour, Sh., Shabibi, M.: A singular fractional differential equation with Riemann–Liouville integral boundary condition. J. Adv. Math. Stud. 8(1), 80–88 (2015) 17. Shabibi, M., Postolache, M., Rezapour, Sh.: Positive solutions for a singular sum fractional differential system. Int. J. Anal. Appl. 13(1), 108–118 (2017) 18. Shabibi, M., Postolache, M., Rezapour, Sh., Vaezpour, S.M.: Investigation of a multi-singular pointwise defined fractional integro-differential equation. J. Math. Anal. 7(5), 61–77 (2016) 19. Shabibi, M., Rezapour, Sh., Vaezpour, S.M.: A singular fractional integro-differential equation. UPB Sci. Bull., Ser. A 79(1), 109–118 (2017) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Boundary Value Problems Springer Journals

On high order fractional integro-differential equations including the Caputo–Fabrizio derivative

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Mathematics; Difference and Functional Equations; Ordinary Differential Equations; Partial Differential Equations; Analysis; Approximations and Expansions; Mathematics, general
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Abstract

Department of Mathematics, By using the fractional Caputo–Fabrizio derivative, we introduce two types new high Cankaya University, Ankara, Turkey Institute of Space Sciences, order derivations called CFD and DCF. Also, we study the existence of solutions for Bucharest, Romania two such type high order fractional integro-differential equations. We illustrate our Full list of author information is results by providing two examples. available at the end of the article MSC: 34A08; 34A99 Keywords: Caputo–Fabrizio derivative; Fractional integro-differential equation; High order derivation 1 Introduction Fractional integro-differential equations have been studied by many researchers from dif- ferent points of view during the last decades (see for example, [5, 10]and [15–19]). In 2015, a new fractional derivation without singular kernel was introduced by Caputo and Fabrizio ([8]). Some researchers tried to use it for solving different equations (see, for example, [2, 9]and [14]). Recently, approximate solutions of some fractional differential equations have been reviewed (see, for example, [3, 4, 6, 12, 13]and [7]). Also, one is finding some new applications for fractional derivations (see, for example, [3]). In this manuscript we consider b >0, x ∈ H (0, b)and α ∈ (0, 1). The expression of the Caputo–Fabrizio fractional derivative of order α for the function x has the form B(α) –α CF α D x(t)= exp( (t –s))x (s) ds,where t ≥ 0([1, 8]and [9]). B(α)is a normalization 1–α 0 1–α constant (B(1) = B(0) = 1). The fractional integral of order α for the function x is written CF α 1–α α as ([14]) I x(t)= x(t)+ x(s) ds,whenever 0 < α <1. If n ≥ 1and α ∈ [0, 1], then B(α) B(α) 0 CF α+n CF α+n CF α n the fractional derivative D of order n + α is defined by D x := D (D x(t)) ([6] (k) CF α n and [8]). If the function x is such that x =0 for k = 1,2,3,..., n,then D (D x(t)) = n CF α D ( D )x(t)([8]). Here, D is the ordinary derivation. Lemma 1.1 ([1]and [14]) Let 0< α <1. Then the unique solution for the problem CF α 1–α α D x(t)= y(t) is given by x(t)= x(0) + y(t)+ x(s) ds. B(α) B(α) 0 Theorem 1.2 ([11]) Let (X, d) be a complete metric space and F : X → X be a mapping such that ϕ(d(Fx, Fy)) ≤ ϕ(d(x, y)) – φ(d(x, y)), for all x, y ∈ X, where ϕ, φ : [0, 1] → [0, 1] © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 2 of 15 are continuous non-decreasing maps and ϕ(t)= φ(t)=0 if and only if t =0. Then F has a unique fixed point. 2 Main result (n) 1 Let n be a natural number, α ∈ (0, 1) and x ∈ H (0, 1). Then the fractional CFD of order α and n is defined by B(α) –α CF α+n CF α n (n+1) D x(t)= D D x(t) = exp (t – s) x (s) ds. 1– α 1– α Also, the fractional DCF of order α and n is defined by n t B(α) d –α CF (n) α n CF α D x(t)= D D x(t) = exp (t – s) x (s) ds. 1– α dt 1– α Here, D is the ordinary derivative. Lemma 2.1 Let n be a natural number and α ∈ (0, 1). Then –α (n) CF α CF α+n D x(t)= D x(t)+ exp t σ (α, n,0), 1– α B(α) –α n–i (i) where σ (α, n, t)= ( ) x (t). i=1 1–α 1–α Proof For each k ≥ 1, we have –α –α (k) (k–1) (k–1) exp (t – s) x (s) ds = x (t)– exp t x (0) 1– α 1– α –α –α (k–1) + exp (t – s) x (s) ds. 1– α 1– α Now by using repetition of the last relation, we get B(α) –α CF α+n CF α n (n+1) D x(t)= D D x(t) = exp (t – s) x (s) ds 1– α 1– α B(α) –α –α = exp (t – s) x (s) ds 1– α 1– α 1– α n–i B(α) –α (i) + x (t) 1– α 1– α i=1 n–i B(α) –α –α (i) – exp t x (0) 1– α 1– α 1– α i=1 B(α) –α –α = exp (t – s) x (s) ds 1– α 1– α 1– α –α + σ (α, n, t)– exp t σ (α, n,0) 1– α –α –α CF α = D x(t)+ σ (α, n, t)– exp t σ (α, n,0). 1– α 1– α Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 3 of 15 Also, we have n n–i –α B(α) –α (n) CF α n α CF α (i) D x(t)= D D x(t) = D x(t)+ x (t) 1– α 1– α 1– α i=1 B(α) –α –α = exp (t – s) x (s) ds 1– α 1– α 1– α n–i B(α) –α (i) + x (t) 1– α 1– α i=1 –α CF α = D x(t)+ σ (α, n, t). 1– α –α CF α (n) CF α+n Hence ( D ) x(t)= D x(t)+ exp( t)σ (α, n,0). 1–α CF α+n CF α (n) (k) By using Lemma 2.1,weconcludethat D x(t)=( D ) x(t)whenever x (0) = 0 for 0 ≤ k ≤ n. Lemma 2.2 Let n be a natural number, α ∈ (0, 1) and y ∈ H (0, 1). Then the solution of the CF α+n problem D x(t)= y(t) is given by (n) 1– α α x (0) x (0) n n+1  2 n x(t)= J y(t)+ J y(t)+ x(0) + tx (0) + t + ··· + t . B(α) B(α) 2! n! CF α+n CF α (n) (n) Proof By using Lemma 1.1 for the equation D x(t)= D x (t)= y(t), we get x (t)= (n) 1–α α x (0) + y(t)+ y(s) ds. By using an integration, we obtain B(α) B(α) 0 t t s 1– α α (n–1) (n–1) (n) x (t)= x (0) + tx (0) + y(s) ds + y(r) dr ds. B(α) B(α) 0 0 0 By repeating this method, we deduce that (n–2) (n–2) (n–1) (n) x (t)= x (0) + tx (0) + x (0) t s t s r 1– α α + y(r) dr ds + y(k) dk dr ds. B(α) B(α) 0 0 0 0 0 By continuing the process, we conclude that (n) 1– α α x (0) x (0) n n+1  2 n x(t)= J y(t)+ J y(t)+ x(0) + tx (0) + t + ··· + t . B(α) B(α) 2! n! On the other hand, by using some calculation, one can find that the given map x(t)is a CF α+n solution for the problem D x(t)= y(t).  Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 4 of 15 Lemma 2.3 Let n be a natural number, α ∈ (0, 1) and y ∈ H (0, 1). Then the solution of the CF α (n) problem ( D ) x(t)= y(t) is given by 1– α α n n+1 x(t)= J y(t)+ J y(t)+ x(0) + tx (0) + ··· B(α) B(α) n–i (n) n x (0) t –α n (i) + t – x (0). n! n! 1– α i=1 –α CF α (n) CF α+n Proof By using Lemma 2.2 for ( D ) x(t)= D x(t)+ exp( t)σ (α, n,0), we get 1–α 1– α –α x(t)= J y(t)– exp t σ (α, n,0) B(α) 1– α α –α n+1 + J y(t)– exp t σ (α, n,0) B(α) 1– α (n) x (0) x (0) 2 n + x(0) + tx (0) + t + ··· + t 2! n! or equivalently 1– α 1– α –α α n n n+1 x(t)= J y(t)– σ (α, n,0)J exp t + J y(t) B(α) B(α) 1– α B(α) (n) α –α x (0) x (0) n+1  2 n – σ (α, n,0)J exp t + x(0) + tx (0) + t + ··· + t B(α) 1– α 2! n! 1– α α 1– α –α n n+1 n = J y(t)+ J y(t)– σ (α, n,0)J exp t B(α) B(α) B(α) 1– α (n) α –α x (0) x (0) 1  2 n + J exp t + x(0) + tx (0) + t + ··· + t B(α) 1– α 2! n! 1– α α 1– α –α n n+1 n = J y(t)+ J y(t)– σ (α, n,0)J exp t B(α) B(α) B(α) 1– α (n) 1– α –α x (0) x (0) 2 n + 1– exp t + x(0) + tx (0) + t + ··· + t B(α) 1– α 2! n! (n) 1– α α x (0) x (0) n n+1  2 n = J y(t)+ J y(t)+ x(0) + tx (0) + t + ··· + t B(α) B(α) 2! n! n–i t –α (i) – x (0). n! 1– α i=1 Lemma 2.4 Let α ∈ (0, 1), 2 < q =2 + α <3 and y ∈ H (0, 1). The fractional differential CF q equation D x(t)= y(t) with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0 –(1–α)t αt has the unique solution of the form x(t)= G(t, s)y(s) ds, where G(t, s)= – 0 2B(α) 2B(α) (1–α)t 1–α α 2 αt whenever 0< t ≤ s <1 and G(t, s)= (t – s)+ (t – s) – – (t – s) whenever B(α) 2B(α) 2B(α) 2B(α) 0< s ≤ t <1. 1–α 2 α 3 Proof By using Lemma 2.2,weget x(t)= J y(t)+ J y(t)+ tx (0). Hence, we obtain B(α) B(α) 1–α 1 α 2 x (t)= J y(t)+ J y(t)+ x (0). By using the boundary conditions x (1) + x (0) = 0 and B(α) B(α) 1–α α 1–α α (1–α)t 1 2  2 3 1 x (1) = J y(1) + J y(1) + x (0), we have x(t)= J y(t)+ J y(t)– J y(1) – B(α) B(α) B(α) B(α) 2B(α) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 5 of 15 t t 1 (1–α)t αt 2 1–α α 2 J y(1). Thus, x(t)= y(s)(t – s) ds + y(s)(t – s) ds – y(s) ds – 2B(α) B(α) 0 2B(α) 0 2B(α) 0 1 1 αt CF q y(s)(t – s) ds = G(t, s)y(s) ds.Notethat D x(t)=0 if and only if x(t)=0. This 2B(α) 0 0 implies that the given map x(t) is a unique solution. 1–α α –αt –(1–α)t 3 Note that |G(t, s)|≤| | + | | + | | + | | < ,for t ∈ [0, 1]. Let μ, μ , μ , 1 2 B(α) 2B(α) 2B(α) 2B(α) 2B(α) k , k ∈ C [0, 1], m , m , h and g be bounded continuous functions on I := [0, 1] with M = 1 2 1 2 1 sup |μ(t)| < ∞, M = sup |μ (t)| < ∞, M = sup |μ (t)| < ∞, M = sup |k (t)| < 2 1 3 2 4 1 t∈I t∈I t∈I t∈I ∞, M = sup |k (t)| < ∞, M = sup |m (t)| < ∞, M = sup |m (t)| < ∞, M = 5 2 6 1 7 2 8 t∈I t∈I t∈I sup |h(t)| < ∞, M = sup |g(t)| < ∞, N = sup |μ (t)| < ∞, N = sup |μ (t)| < ∞, 9 1 2 t∈I t∈I t∈I t∈I 1 N = sup |μ (t)| < ∞, N = sup |K (t)| < ∞ and N = sup |K (t)| < ∞.Let α ∈ (0, 1) 3 4 5 t∈I t∈I t∈I 2 1 2 and 2 < q =2 + α < 3. Now, we investigate the CFD fractional integro-differential problem CF q   CF β CF β 1 2 D x(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t)+ k (t) D x(t) 1 2 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds,(1) 1 2 with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0, where 1 < β <2< β <3 1 2 and 1 < γ <2< ν <3. Theorem 2.5 Let ξ , ξ , ξ , ξ and ξ be nonnegative real numbers, f : [0, 1] × R → R an 1 2 3 4 5 integrable function such that f (t, x, y, z, v, w)– f t, x , y , z , v , w ≤ ξ x – x + ξ y – y + ξ z – z + ξ v – v + ξ v – v , 1 2 3 4 5 for all real numbers x, y, z, v, w, x , y , z , v , w ∈ R and t ∈ I. If < , then the problem (1) has a unique solution, where := max{ }, = [M + M + M + 1 2 3 4 1 1 2 3 2B(α) M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 3 3+4α 4 1 5 2 8 9 + + ξ + ξ M + ξ M + ξ + ξ ], = [ ][M + M + 1 2 6 3 7 4 5 2 1 2 2–β 3–β 2–γ 3–ν 2 2B(α) 1 2 M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 4 1 5 2 8 9 1+α M + + + ξ + ξ M + ξ M + ξ + ξ ], = [M + 3 1 2 6 3 7 4 5 3 1 2–β 3–β 2–γ 3–ν B(α) 1 2 M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 4 1 5 8 9 M + M + + + ξ + ξ M + ξ M + ξ + ξ ] and 2 3 1 2 6 3 7 4 5 4 2–β 3–β 2–γ 3–ν 1 2 α M B(β –1) M B(β –2) M B(γ –1) M B(ν–2) 4 1 5 2 8 9 [M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ]+ 1 2 3 1 2 6 3 7 4 5 B(α) 2–β 3–β 2–γ 3–ν 1 2 |1–β |M |2–β |M M +N 1–α N +M 1 4 4 4 2 5 5 5 [N +M +N +M +N +M +B(β –1)[ + ]+B(β –2)[ + ]+ 1 1 2 2 3 3 1 2 2 2 B(α) 2–β 3–β (2–β ) (3–β ) 1 1 2 2 M B(γ –1) M B(ν–1) 8 9 ξ + ξ M + ξ M + ξ + ξ ]. 1 2 6 3 7 4 5 2–łγ 3–ν Proof Consider the Banach space C [0, 1] equipped with the norm x = max |x(t)| + t∈I 3 3 max |x (t)| + max |x (t)| + max |x (t)|.Define themap F : C [0, 1] → C [0, 1] by t∈I t∈I t∈I R R Fx(t)= G(t, s)R(s) ds t t 1– α α = R(s)(t – s) ds + R(s)(t – s) ds B(α) 2B(α) 0 0 1 1 (1 – α)t αt – R(s) ds – R(s)(t – s) ds, 2B(α) 2B(α) 0 0 Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 6 of 15 where CF β CF β 1 2 (Rx)(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t)+ k (t) D x(t) 1 2 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 and R x (t)= μ(t)x (t)+ μ (t)x(t)+ μ (t)x (t)+ μ x (t) CF β + μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t) 2 1 1– β B(β –1) 1 1 CF β + k (t) D x(t)+ x (t) 2– β 2– β 1 1 2– β B(β –2) 2 2 CF β   CF β 2 2 + k (t) D x(t)+ x (t) + k (t) D x(t) 3– β 3– β 2 2 CF γ CF ν + f t, x(t), m (t)x (t), m (t)x (t), h(t) D x(t), g(t) D x(t) . 1 2 By using Lemma 2.4, x is a solution for the problem (1)ifand only if x is a fixed point of 0 0 the operator F.Notethat (Rx)(t)–(Ry)(t) CF β CF β 1 2 ≤ μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t)+ k (t) D x(t) 1 2 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 CF β CF β 1 2 – μ(t)y(t)+ μ (t)y (t)+ μ (t)y (t)+ k (t) D y(t)+ k (t) D y(t) 1 2 1 2 CF γ CF ν + f s, y(s), m (s)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) ds 1 2 ≤ μ(t) x(t)– y(t) + μ (t) x (t)– y (t) + μ (t) x (t)– y (t) 1 2 CF β CF β 1 2 + k (t) D x(t)– y(t) + k (t) D x(t)– y(t) 1 2 CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 CF γ CF ν – f s, y(s), m (s)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) ds 1 2 ≤ M x – y + M x – y + M x – y 1 2 3 M B(β –1) M B(β –2) 4 1 5 2 + x – y + x – y 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + 1 2 3 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 7 of 15 and R x(t)– R y(t) ≤ μ (t) x(t)– y(t) + μ(t)+ μ (t) x (t)– y (t) + μ (t)+ μ (t) x (t)– y (t) + μ (t) x (t)– y (t) 1 2 1– β |k (t)B(β –1)| 1 1 1 CF β + k (t)+ k (t) D x(t)– y(t) + x (t)– y (t) 2– β 2– β 1 1 2– β |k (t)B(β –2)| 2 2 CF β 2 + k (t)+ k (t) D x(t)– y(t) + x (t)– y (t) 3– β 3– β 2 2 CF γ CF ν + f t, x(t), m (t)x (t), m (t)x (t), h(t) D x(t), g(t) D x(t) 1 2 CF γ CF ν – f t, y(t), m (t)y (t), m (t)y (t), h(t) D y(t), g(t) D y(t) . 1 2 Hence, we get R x(t)– R y(t) |1– β |M N + M 1 4 4 4 ≤ N + M + N + M + N + M + B(β –1) + 1 1 2 2 3 3 1 (2 – β ) 2– β 1 1 |2– β |M M + N 2 5 5 5 + B(β –2) + + ξ + ξ M + ξ M 2 1 2 6 3 7 (3 – β ) 3– β 2 2 M B(γ –1) M B(ν –1) 8 9 + ξ + ξ x – y. 4 5 2– łγ 3– ν On the other hand, we have 3 M B(β –1) M B(β –2) 4 1 5 2 Fx(t)– Fy(t) ≤ M + M + M + + 1 2 3 2B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν x – y and t t 1– α α α F x(t)– F y(t)= + t Rx(s)– Ry(s) ds – s Rx(s)– Ry(s) ds B(α) B(α) B(α) 0 0 1 1 (1 – α) αt – Rx(s)– Ry(s) ds – Rx(s)– Ry(s) ds 2B(α) B(α) 0 0 + s Rx(s)– Ry(s) ds 2B(α) 3 3+4α M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + 1 2 3 2 2B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– γ 3– ν x – y 2 Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 8 of 15 and so |F x(t)– F y(t)|≤ x – y.Also, we have α 1– α F x(t)– F y(t) ≤ Rx(s)– Ry(s) ds + Rx(t)– Ry(t) B(α) B(α) + Rx(s)– Ry(s) ds B(α) 1+ α M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + 1 2 3 B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 8 9 + ξ + ξ M + ξ M + ξ + ξ 1 2 6 3 7 4 5 2– γ 3– ν x – y and F x(t)– F y(t) α 1– α = Rx(t)– Ry(t) + R x(t)– R y(t) B(α) B(α) α M B(β –1) M B(β –2) 4 1 5 2 ≤ M + M + M + + + ξ + ξ M + ξ M 1 2 3 1 2 6 3 7 B(α) 2– β 3– β 1 2 M B(γ –1) M B(ν –2) 1– α 8 9 + ξ + ξ + N + M + N + M + N + M 4 5 1 1 2 2 3 3 2– γ 3– ν B(α) |1– β |M N + M |2– β |M M + N 1 4 4 4 2 5 5 5 + B(β –1) + + B(β –2) + 1 2 2 2 (2 – β ) 2– β (3 – β ) 3– β 1 1 2 2 M B(γ –1) M B(ν –1) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2– łγ 3– ν x – y. Thus, Fx – Fy≤ x – y for all x, y ∈ C [0, 1]. Put ϕ(t)=2t and φ(t)= t for all t.Now by using Theorem 1.2, F has a unique fixed point which is the unique solution for the problem (1). Lemma 2.6 Let α ∈ (0, 1) and y ∈ H (0, 1). Then the fractional differential equation (2) CF α D x(t)= y(t) with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0 has –(1–α)t –αt the unique solution x(t)= G(t, s)y(s) ds, where G(t, s)= + (t – s) when- 0 (2–α)B(α) (2–α)B(α) 1–α α (1–α)t αt ever 0< t ≤ s <1 and G(t, s)= (t – s)+ (t – s) – – (t – s) whenever B(α) 2B(α) B(α)(2–α) B(α)(2–α) 0< s ≤ t <1. 1–α α α 2 3 Proof By using Lemma 2.3,weget x(t)= J y(t)+ J y(t)+ x (0)t + x (0)t.Hence, B(α) B(α) 1–α 1–α 1 α 2 1 x (t)= J y(t)+ J y(t)+ x (0). By using the boundary conditions x (1) + x (0) = 0 B(α) B(α) 1–α 1–α 1 α 2 1  1–α 2 α 3 and x (1) = J y(1) + J y(1) + x (0), we obtain x(t)= J y(t)+ J y(t)– B(α) B(α) 1–α B(α) B(α) t t (1–α)t αt 1–α α 1 2 2 J y(1) – J y(1). Thus, x(t)= y(s)(t – s) ds + y(s)(t – s) ds – (2–α)B(α) (2–α)B(α) B(α) 0 2B(α) 0 1 1 1 (1–α)t αt CF α (2) y(s) ds – y(s)(t – s) ds = G(t, s)y(s) ds.Notethat ( D ) x(t)=0 (2–α)B(α) 0 (2–α)B(α) 0 0 if and only if x(t) = 0. This implies that the given map x(t) is a unique solution.  Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 9 of 15 –(1–α)t 1–α α –αt 2 Note that |G(t, s)|≤| |+| |+| |+| | < ,for t ∈ [0, 1]. Let α, β , β , γ , ν ∈ 1 2 B(α) 2B(α) B(α) B(α) B(α) (0, 1). Now, we investigate the DCF fractional integro-differential problem (2) CF α D x(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t) 1 2 (1) (2) CF β CF β 1 2 + λ k (t) D x(t)+ λ k (t) D x(t) 1 1 2 2 + f s, x(s), m (s)x (s), m (s)x (s), 1 2 (1) (2) CF γ CF ν h(s) D x(s), g(s) D x(s) ds (2) with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0. Theorem 2.7 Let ξ , ξ , ξ , ξ , and ξ be nonnegative real numbers, f : [0, 1] × R → R an 1 2 3 4 5 integrable function such that f (t, x, y, z, v, w)– f t, x , y , z , v , w ≤ ξ x – x + ξ y – y + ξ z – z + ξ v – v + ξ v – v 1 2 3 4 5 for all real numbers x, y, z, v, w, x , y , z , v , w and t ∈ I. If < , then the problem (2) has a unique solution, where := max{ }, = [M + M + M + 1 2 3 4 1 1 2 3 (2–α)B(α) (β –β +1)M B(β ) M B(β ) M B(γ ) (ν –ν+1)M B(ν) 2 5 2 3+α 4 1 2 8 9 + + ξ + ξ M + ξ M + ξ + ξ ], = [M + 1 2 6 3 7 4 5 2 1 2 3 2 3 B(α)(2–α) (1–β ) (1–β ) (1–γ ) (1–ν) 1 2 (β –β +1)M B(β ) M B(β ) 2 5 2 M B(γ ) (ν –ν+1)M B(ν) 4 1 2 8 9 M + M + + + ξ + ξ M + ξ M + ξ + ξ ], 2 3 2 3 1 2 6 3 7 4 2 5 3 (1–β ) (1–β ) (1–γ ) (1–ν) 1 2 (β –β +1)M B(β ) M B(β ) 2 5 2 M B(γ ) 2+α 4 1 2 8 = [M + M + M + + + ξ + ξ M + ξ M + ξ + 3 1 2 3 2 3 1 2 6 3 7 4 2 B(α)(2–α) (1–β ) (1–β ) (1–γ ) 1 2 (β –β +1)M B(β ) (ν –ν+1)M B(ν) M B(β ) 2 2 α 5 9 4 1 2 ξ ] and = [M +M +M + + + ξ + ξ M + ξ M + 5 4 1 2 3 1 2 6 3 7 3 2 3 B(α) (1–ν) (1–β ) (1–β ) 1 2 (β –β +1) M B(γ ) (ν –ν+1)M B(ν) B(β )N 1–α 1 8 9 1 1 4 ξ + ξ ]+ [N +M +N +M +N +M +M B(β ) + + 4 5 1 1 2 2 3 3 4 1 2 3 3 2 B(α) (1–γ ) (1–ν) (1–β ) (1–β ) 1 1 2 2 (2β –2β +1) (β –β +1) 2 2 ξ M B(γ ) (ν –ν+1)M B(ν) 2 2 4 8 9 M B(β ) + N B(β ) + ξ + ξ M + ξ M + + ]. 5 2 4 5 2 3 1 2 6 3 7 2 3 (1–β ) (1–β ) (1–łγ ) (1–ν) 2 2 Proof Consider the Banach space C [0, 1] equipped with the norm x = max |x(t)| + t∈I 3 3 max |x (t)| + max |x (t)| + max |x (t)|.Define themap F : C [0, 1] → C [0, 1] by t∈I t∈I t∈I R R Fx(t)= G(t, s)R(s) ds t t 1– α α = R(s)(t – s) ds + R(s)(t – s) ds B(α) 2B(α) 0 0 1 1 (1 – α)t αt – R(s) ds – R(s)(t – s) ds, B(α)(2 – α) B(α)(2 – α) 0 0 where (Rx)(t)= μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t) 1 2 (1) (2) CF β CF β 1 2 + λ k (t) D x(t)+ λ k (t) D x(t) 1 1 2 2 (1) (2) CF γ CF ν + f s, x(s), m (s)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 0 Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 10 of 15 and R x (t)= μ(t)x (t)+ μ (t)x(t)+ μ (t)x (t)+ μ x (t) (1) CF β + μ (t)x (t)+ μ (t)x (t)+ k (t) D x(t) 2 1 –β β 1 1 1 CF β + k (t)B(β ) D x(t)– x (t)+ x (t) 1 1 1– β (1 – β ) 1– β 1 1 1 –β β CF β + k (t)B(β ) D x(t)+ x (t) 2 2 1– β (1 – β ) 2 2 β x (t) (2) CF β – x (t)+ + k (t) D x(t) (1 – β ) 1– β 2 2 (1) (2) CF γ CF ν + f t, x(t), m (t)x (t), m (t)x (t), h(t) D x(t), g(t) D x(t) . 1 2 By using Lemma 2.6, x is a solution for the problem (2)ifand only if x is a fixed point of 0 0 the operator F.Notethat (Rx)(t)–(Ry)(t) (1) (2) CF β CF β 1 2 ≤ μ(t)x(t)+ μ (t)x (t)+ μ (t)x (t)+ λ k (t) D x(t)+ k (t) D x(t) 1 2 1 1 2 (1) (2) CF γ CF ν + f s, x(s), m (t)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 (1) (2) CF β CF β 1 2 – μ(t)y(t)+ μ (t)y (t)+ μ (t)y (t)+ k (t) D y(t)+ k (t) D y(t) 1 2 1 2 (1) (2) CF γ CF ν + f s, y(s), m (t)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) ds 1 2 ≤ μ(t) x(t)– y(t) + μ (t) x (t)– y (t) + μ (t) x (t)– y (t) 1 2 (1) (2) CF β CF β 1 2 + k (t) D x(t)– y(t) + k (t) D x(t)– y(t) 1 2 (1) (2) CF γ CF ν + |f s, x(s), m (t)x (s), m (s)x (s), h(s) D x(s), g(s) D x(s) ds 1 2 (1) (2) CF γ CF ν – f s, y(s), m (t)y (s), m (s)y (s), h(s) D y(s), g(s) D y(s) | ds 1 2 ≤ M x – y + M x – y + M x – y 1 2 3 M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 + x – y + x – y 2 3 (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 ≤ M + M + M + + 1 2 3 2 3 (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 11 of 15 and R x(t)– R y(t) (β – β +1) B(β )N 1 1 4 ≤ N + M + N + M + N + M + M B(β ) + 1 1 2 2 3 3 4 1 3 2 (1 – β ) (1 – β ) 1 1 2 2 (2β –2β +1) (β – β +1) 2 2 2 2 + M B(β ) + N B(β ) + ξ + ξ M + ξ M 5 2 5 2 1 2 6 3 7 4 3 (1 – β ) (1 – β ) 2 2 ξ M B(γ ) (ν – ν +1)M B(ν) 4 8 9 + + x – y. 2 3 (1 – łγ ) (1 – ν) Thus, 2 M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 Fx(t)– Fy(t) ≤ M + M + M + + 1 2 3 2 3 (2 – α)B(α) (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) x – y and 3+ α M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 F x(t)– F y(t) ≤ M + M + M + + 1 2 3 2 3 B(α)(2 – α) (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ x – y 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) x – y. Also, we have 2+ α M B(β ) 4 1 F x(t)– F y(t) ≤ M + M + M + 1 2 3 B(α)(2 – α) (1 – β ) (β – β +1)M B(β ) 2 5 2 + + ξ + ξ M + ξ M 1 2 6 3 7 (1 – β ) M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ 4 5 2 3 (1 – γ ) (1 – ν) x – y and F x(t)– F y(t) α 1– α ≤ Rx(t)– Ry(t) + R x(t)– R y(t) B(α) B(α) α M B(β ) (β – β +1)M B(β ) 4 1 2 5 2 ≤ M + M + M + + 1 2 3 2 3 B(α) (1 – β ) (1 – β ) 1 2 M B(γ ) (ν – ν +1)M B(ν) 8 9 + ξ + ξ M + ξ M + ξ + ξ 1 2 6 3 7 4 5 2 3 (1 – γ ) (1 – ν) Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 12 of 15 1– α (β – β +1) + N + M + N + M + N + M + M B(β ) 1 1 2 2 3 3 4 1 B(α) (1 – β ) 2 2 B(β )N (2β –2β +1) (β – β +1) 1 4 2 2 2 2 + + M B(β ) + N B(β ) 5 2 5 2 2 4 3 (1 – β ) (1 – β ) (1 – β ) 1 2 2 ξ M B(γ ) (ν – ν +1)M B(ν) 4 8 9 + ξ + ξ M + ξ M + + x – y 1 2 6 3 7 2 3 (1 – łγ ) (1 – ν) x – y. Hence, Fx – Fy≤ x – y for all x, y ∈ C [0, 1]. Put ϕ(t)=2t and φ(t)= t for all t. By using Theorem 1.2, F has a unique fixed point, which is the desired solution for the problem. Here, we provide three examples to illustrate our main results. Consider the bounded 1 3t–1 1 –6t 1 3 continuous functions μ(t)= sin(t), μ (t)= , μ (t)= e , k (t)= t + 1 2 1 100 20t+162 100 300 1 1 1 Ln(t+2) 1 2t t + , k (t)= cos(t), m (t)= e , m (t)= , h(t)=0 and g(t)= for 2 1 2 100 50 800 20 t–900 1 1 all t ∈ I = [0, 1]. Note that M = sup |μ(t)| = , M = sup |μ (t)| = , M = 1 2 1 3 t∈I t∈I 100 91 1 1 1 sup |μ (t)| = , M = sup |k (t)| = , M = sup |k (t)| = , M = sup |m (t)| = 2 4 1 5 2 6 1 t∈I 6 t∈I t∈I t∈I 30 800 100e Ln(3) 2 1 e , M = sup |m (t)| = , M = sup |h(t)| =0 and M = sup |g(t)| = .Also, 7 2 8 9 t∈I t∈I t∈I 1200 900 1  506  6 N = sup |μ (t)| = , N = sup |μ (t)| = , N = sup |μ (t)| = , N = 1 2 3 4 t∈I t∈I 1 2 t∈I 2 6 (162) 100e 1  1 sup |k (t)| = and N = sup |k (t)| = . Also, consider the function B(α)=1 for t∈I 1 t∈I 2 50 800 α ∈ (0, 1). Example 2.1 Let α ∈ (0, 1). By using Lemma 2.4, the fractional differential equation CF 2+α D x (t)= t with boundary conditions x (0) = 0, x (1) + x (0) = 0 and x (0) = 0 has 1 1 1 1 1 theuniquesolution x (t). Also by using Lemma 2.6, the fractional differential equation CF α (2) ( D ) x (t)= t with boundary conditions x (0) = 0, x (1) + x (0) = 0 and x (0) = 0 has 2 2 2 2 2 1 1 1 1 4 99 the unique solution x (t). For α = , α = , α = , α = , α = and α = we compare 100 10 5 2 5 100 the solutions x (t), x (t)and X(t)= x (t)– x (t)in Fig. 1. 1 2 1 2 Example 2.2 Consider the CFD fractional integro-differential problem –6t 12 sin(t) 3t –1 e CF D x(t)= x(t)+ x (t)+ x (t) 100 20t + 162 100 t +3t +6 3 cos(t) 5 CF CF 2 2 + D x(t)+ D x(t) 300 800 Ln(t +2) 1 8 2t   CF + f s, x(s), e x (s), x (s), 0, D x(s) ds (3) 20 t – 900 with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0, where 1 < β = <2< 5 4 8 2 3 1 1 1 β = <3 and 1 < γ = <2< ν = <3. Put f (t, x, y, z, v, w)= t + x + y + z + w + 2 3 3 91 604 200 80 M M M M 4 5 8 9 2v.Notethat = [M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ]= 1 1 2 3 1 2 6 3 7 4 5 2 2–β 3–β 2–γ 3–ν 1 2 3 3+4α M M M 4 5 8 9 0.02, = [ ][M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ] = 0.46, 2 1 2 3 1 2 6 3 7 4 5 2 2 2–β 3–β 2–γ 3–ν 1 2 M M M M 4 5 8 9 =(1 + α)[M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ] = 0.32 and 3 1 2 3 1 2 6 3 7 4 5 2–β 3–β 2–γ 3–ν 1 2 M M M M 4 5 8 9 = α[M + M + M + + + ξ + ξ M + ξ M + ξ + ξ ]+(1– α)[N + M + 4 1 2 3 1 2 6 3 7 4 5 1 1 2–β 3–β 2–γ 3–ν 1 2 |1–β |M N +M |2–β |M M +N M M 1 4 4 4 2 5 5 5 8 6 N +M +N +M + + + + + ξ + ξ M + ξ M + ξ + ξ ]= 2 2 3 3 1 2 6 3 7 4 5 2 2 2–β 3–β 2–łγ 3–ν (2–β ) 1 (3–β ) 2 1 2 0.166. By using Theorem 2.5,the problem(3) has a unique solution. Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 13 of 15 1 1 1 1 4 99 Figure 1 The solutions of the problem with α = , α = , α = , α = , α = and α = 100 10 5 2 5 100 Example 2.3 Consider the DCF fractional integro-differential problem (2) CF D x(t) –6t sin(t) 3t –1 e = x(t)+ x (t)+ x (t) 100 20t + 162 100 t +3t +6 1 cos(t) 2 (1) (2) CF CF 2 3 + D x(t)+ D x(t) 300 800 Ln(t +2) 1 1 (2) –40t   CF + f s, x(s), e x (s), x (s), 0, D x(s) ds,(4) 20 t – 900 2 1 with boundary conditions x(0) = 0, x (1) + x (0) = 0 and x (0) = 0, where α = , β = , 5 2 2 1 1 2 3 1 1 1 β = , γ = and ν = .Put f (t, x, y, z, v, w)= t + x + y + z + w +2v.Notethat 2 18 3 3 5 91 604 200 80 e (β –β +1) M 2 M (ν –ν+1)M 2 4 2 8 9 = [M +M +M + +M + ξ + ξ M + ξ M + ξ + ξ ]< 1 1 2 3 2 5 3 1 2 6 3 7 4 2 5 3 2–α (1–β ) (1–β ) (1–γ ) (1–ν) 1 2 (β –β +1)M M 2 M 3+α 5 4 2 8 0.391, =[ ][M + M + M + + + ξ + ξ M + ξ M + ξ + 2 1 2 3 1 2 6 3 7 4 2 3 2 2–α (1–β ) (1–β ) (1–γ ) 1 2 (β –β +1)M (ν –ν+1)M 2+α M 2 5 9 4 2 ξ ] < 0.225, = [M + M + M + + + ξ + ξ M + ξ M + 5 3 3 1 2 3 2 3 1 2 6 3 7 (1–ν) 2–α (1–β ) (1–β ) 1 2 (β –β +1) M (ν –ν+1)M M 2 8 9 4 2 ξ + ξ ] < 0.132 and = α[M + M + M + + M + ξ + 4 2 5 3 4 1 2 3 2 5 3 1 (1–γ ) (1–ν) (1–β ) (1–β ) 1 2 (β –β +1) M (ν –ν+1)M 1 8 9 1 ξ M + ξ M + ξ + ξ ]+(1– α)[N + M + N + M + N + M + M + 2 6 3 7 4 5 1 1 2 2 3 3 4 2 3 3 (1–γ ) (1–ν) (1–β ) 2 2 (2β –2β +1) N (β –β +1) N 2 2 ξ M (ν –ν+1)M 4 2 2 4 8 9 + M + + ξ + ξ M + ξ M + + ξ ] < 0.493. By 5 1 2 6 3 7 5 2 4 3 2 3 (1–β ) (1–β ) (1–β ) (1–łγ ) (1–ν) 1 2 2 using Theorem 2.7,the problem(4) has a unique solution. Aydogan et al. Boundary Value Problems (2018) 2018:90 Page 14 of 15 3Conclusion It is important that researchers have some methods available enabling them to review some high order fractional integro-differential equations. In this manuscript, we intro- duce two types of new fractional derivatives entitled CFD and DCF and by using those we investigate the existence of solutions for two high order fractional integro-differential equations of such a type including the new derivatives. Acknowledgements The third author was supported by Azarbaijan Shahid Madani University. Funding Not available. Abbreviations DFC, Caputo–Fabrizio derivation followed by a differentiation; CFD, Differentiation followed by Caputo–Fabrizio derivation. Availability of data and materials Data sharing not applicable to this article as no datasets were generated or analyzed during the current study. Competing interests The authors declare that they have no competing interests. Authors’ contributions The authors declare that the study was realized in collaboration with equal responsibility. All authors read and approved the final manuscript. Author details 1 2 Department of Mathematics, Istanbul Technical University, Istanbul, Turkey. Department of Mathematics, Cankaya 3 4 University, Ankara, Turkey. Institute of Space Sciences, Bucharest, Romania. Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran. Department of Medical Research, China Medical University Hospital, China Medical University, Taichung, Taiwan. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 20 February 2018 Accepted: 24 May 2018 References 1. 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Boundary Value ProblemsSpringer Journals

Published: Jun 1, 2018

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