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

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1 Aydogan et al. Boundary Value Problems :9 R E S E A R C H Open Access On high order fractional integro-differential equations including the Caputo Fabrizio derivative elike S. Aydogan 1,DumitruBaleanu 2,3*, Asef ousalou 4 Shahram Rezapour 4,5 * Correspondence: dumitru@cankaya.edu.tr 2 Department of athematics, Cankaya University, Ankara, Turkey 3 Institute of Space Sciences, Bucharest, Romania Full list of author information is available at the end of the article Abstract By using the fractional Caputo Fabrizio derivative, we introduce two types new high order derivations called CFD DCF. Also, we study the existence of solutions for two such type high order fractional integro-differential equations. We illustrate our results by providing two examples. SC: 34A8; 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 different pointsofview during thelastdecades seeforexample, 5, In 215, a new fractional derivation without singular kernel was introduced by Caputo Fabrizio 8. Some researchers tried to use it for solving different equations see, for example, 2, 914. Recently, approximate solutions of some fractional differential equations have been reviewed see, for example, 3, 4, 6, 12, Also, one is finding some new applications for fractional derivations see, for example, 3. In this manuscript we consider b >,x H 1, b, 1. The expression of the Caputo Fabrizio fractional derivative of order for the function x has the form CF D xt= 1 exp 1 t sx s ds,wheret 1, 89. isanormalization constant B1 = B = 1. The fractional integral of order for the function x is written xs ds,whenever< <1.Ifn 1, 1, then as 14 CF I xt= 1 xt the fractional derivative CF D n of order n is defined by CF D n x := CF D D n xt 6 8. If the function x is such that x k =fork = 1,2,3,...,n, then CF D D n xt = D n D xt8. Here, D is the ordinary derivation. Lemma Let < <1.Then the unique solution for the problem xs ds. CF D xt=yt is given by xt=x 1 yt Theorem Let X, d be a complete metric space F : X X be a mapping such that ϕdfx, Fy ϕdx, y φdx, y, for all x, y X, where ϕ, φ :, 1, 1 The Authors 218. This article is distributed under the terms of the Creative Commons Attribution 4. International License which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original authors the source, provide a link to the Creative Commons license, indicate if changes were made.

2 Aydogan et al. Boundary Value Problems :9 Page 2 of 15 are continuous non-decreasing maps ϕt=φt=if only if t =.Then F has a unique fixed point. 2 ain result Let n be a natural number,, 1 x n H 1, 1. Then the fractional CFD of order n is defined by CF D n xt= CF D D n xt = 1 Also, the fractional DCF of order n is defined by D n xt=d n D xt = d n 1 dt n Here, D is the ordinary derivative. exp t s x n1 s ds. 1 Lemma 2.1 Let n be a natural number, 1. Then D n xt= CF D xtexp n 1 t σ, n,, where σ, n, t= n 1 i=1 1 n i x i t. exp t s x s ds. 1 Proof For each k 1, we have exp t s x k s ds = x k 1 t exp 1 1 t x k 1 t exp t s x k 1 s ds. 1 1 Now by using repetition of the last relation, we get CF D n xt= CF D D n xt = 1 = n exp 1 1 n n i x i t 1 1 i=1 1 exp 1 t = n 1 1 n exp exp t s 1 t s 1 x s ds n i x i 1 i=1 t s x s ds 1 σ, n, σ, n, t exp 1 t n = CF D xtσ, n, t exp 1 1 t x n1 s ds σ, n,.

3 Aydogan et al. Boundary Value Problems :9 Page 3 of 15 Also, we have D n xt=d n D xt n = CF D xt 1 1 = 1 1 n 1 i=1 n 1 n = CF D xtσ, n, t. 1 n n i x i t 1 i=1 exp t s x s ds 1 n i x i t Hence D n xt= CF D n xtexp tσ, n,. 1 By using Lemma 2.1, weconcludethat CF D n xt = D n xt wheneverx k = for k n. Lemma 2.2 Let n be a natural number,, 1 y H 1, 1. Then the solution of the problem CF D n xt=yt is given by xt= 1 Jn yt Jn1 ytx tx t 2 x 2! t n xn. Proof By using Lemma 1.1 for the equation CF D n xt= CF D x n t=yt, we get x n t= ys ds. By using an integration, we obtain x n 1 yt x n 1 t=x n 1 tx n 1 ys ds s yr dr ds. By repeating this method, we deduce that x n 2 t=x n 2 tx n 1 t2 2 xn 1 s yr dr ds s r yk dk dr ds. By continuing the process, we conclude that xt= 1 Jn yt Jn1 ytx tx t 2 x 2! t n xn. On the other h, by using some calculation, one can find that the given map xt isa solution for the problem CF D n xt=yt.

4 Aydogan et al. Boundary Value Problems :9 Page 4 of 15 Lemma 2.3 Let n be a natural number,, 1 y H 1, 1. Then the solution of the problem D n xt=yt is given by xt= 1 Jn yt t n xn Jn1 ytx tx n n i x i. 1 tn i=1 Proof By using Lemma 2.2 for D n xt= CF D n xtexp tσ, n,,weget 1 xt= 1 Jn yt exp 1 t σ, n, Jn1 yt exp 1 t σ, n, or equivalently x tx t 2 x 2! t n xn xt= 1 Jn yt 1 σ, n,jn exp 1 t Jn1 yt σ, n,jn1 exp 1 t x tx t 2 x t n xn 2! = 1 Jn yt 1 Jn1 yt σ, n,j n exp 1 t J1 exp = 1 Jn yt 1 1 exp = 1 Jn yt tn n i=1 1 1 t x tx t 2 x t n xn 2! 1 Jn1 yt σ, n,j n exp 1 t 1 t x tx t 2 x t n xn 2! Jn1 ytx tx t 2 x 2! n i x i. t n xn Lemma 2.4 Let, 1, 2 < q =2 <3 y H 1, 1. The fractional differential equation CF D q xt=yt with boundary conditions x =, x 1 x = x = has the unique solution of the form xt = 1 t Gt, sys ds, where Gt, s = t 2 2 whenever <t s <1 Gt, s= 1 t s 2 t s2 1 t 2 t t s whenever 2 <s t <1. Proof By using Lemma 2.2, wegetxt= 1 x t= 1 J1 yt x 1 = 1 J1 y1 J2 yt J3 yttx. Hence, we obtain J2 ytx. By using the boundary conditions x 1 x = J2 y1 x, we have xt= 1 J2 yt J3 yt 1 t 2 J1 y1

5 Aydogan et al. Boundary Value Problems :9 Page 5 of 15 t 2 J2 y1. Thus, xt = 1 t 2 t yst s ds 2 yst s2 ds 1 t 2 ys ds yst s ds = Gt, sys ds. NotethatCF D q xt=ifonlyifxt=.this implies that the given map xt is a unique solution. Note that Gt, s 1 2 t 2 1 t < 3 2 2,fort, 1. Let μ, μ 1, μ 2, k 1, k 2 C 1, 1, m 1, m 2, h g be bounded continuous functions on I :=, 1 with 1 = sup t I μt <, 2 = sup t I μ 1 t <, 3 = sup t I μ 2 t <, 4 = sup t I k 1 t <, 5 = sup t I k 2 t <, 6 = sup t I m 1 t <, 7 = sup t I m 2 t <, 8 = sup t I ht <, 9 = sup t I gt <, N 1 = sup t I μ t <, N 2 = sup t I μ 1 t <, N 3 = sup t I μ 2 t <, N 4 = sup t I K 1 t < N 5 = sup t I K 2 t <. Let, 1 2 < q =2 < 3. Now, we investigate the CFD fractional integro-differential problem CF D q xt=μtxtμ 1 tx tμ 2 tx tk 1 t CF D β 1 xtk 2 t CF D β 2 xt f s, xs, m 1 sx s, m 2 sx s, hs CF D γ xs, gs CF D ν xs ds, 1 with boundary conditions x =, x 1 x = x =, where 1 < β 1 <2<β 2 <3 1 < γ <2<ν <3. Theorem 2.5 Let ξ 1, ξ 2, ξ 3, ξ 4 ξ 5 be nonnegative real numbers, f :, 1 R 5 R an integrable function such that f t, x, y, z, v, w f t, x, y, z, v, w ξ 1 x x ξ2 y y ξ3 z z ξ4 v v ξ5 v v, for all real numbers x, y, z, v, w, x, y, z, v, w R t I. If < 1, then the problem 2 1 has a unique solution, where := max{ 1, 2, 3, 4 }, 1 = Bβ 1 1 5Bβ 2 2 ξ 1 ξ 2 6 ξ 3 7 ξ 8 Bγ 1 4 ξ 9 Bν 2 2 γ 5, 2 = Bβ 1 1 5Bβ 2 2 ξ 1 ξ 2 6 ξ 3 7 ξ 8 Bγ 1 4 ξ 9 Bν 2 2 γ 5, 3 = Bβ 1 1 5Bβ 2 2 ξ 1 ξ 2 6 ξ 3 7 ξ 8 Bγ 1 4 ξ 9 Bν 2 2 γ 5 4 = Bβ 1 1 5Bβ 2 2 ξ 1 ξ 2 6 ξ 3 7 ξ 8 Bγ 1 4 ξ 9 Bν 2 2 γ 5 1 N 1 1 N 2 2 N 3 3 Bβ β 1 4 N Bβ β 2 5 5N 5 2 ξ 1 ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 2 łγ ξ 5 9 Bν 1. Proof Consider the Banach space CR 3, 1 equipped with the norm x = max t I xt max t I x t max t I x t max t I x t.definethemapf : CR 3, 1 C3 R, 1 by Fxt= = 1 Gt, srs ds 1 t 2 Rst s ds 2 Rs ds t 2 Rst s 2 ds Rst s ds,

6 Aydogan et al. Boundary Value Problems :9 Page 6 of 15 where Rxt=μtxtμ 1 tx tμ 2 tx tk 1 t CF D β 1 xtk 2 t CF D β 2 xt f s, xs, m 1 sx s, m 2 sx s, hs CF D γ xs, gs CF D ν xs ds R x t=μtx tμ txtμ 1 tx tμ 1 x t μ 2 tx tμ 2 tx tk 1 tcf D β 1 xt 1 β1 k 1 t CF D β 1 xt Bβ 1 1 x t 2 β2 k 2 t CF D β 2 xt Bβ 2 2 x t k 2 3 β tcf D β 2 xt 2 f t, xt, m 1 tx t, m 2 tx t, ht CF D γ xt, gt CF D ν xt. By using Lemma 2.4, x is a solution for the problem 1ifonlyifx is a fixed point of the operator F.Notethat Rxt Ryt μtxtμ 1tx tμ 2 tx tk 1 t CF D β 1 xtk 2 t CF D β 2 xt f s, xs, m 1 sx s, m 2 sx s, hs CF D γ xs, gs CF D ν xs ds μtytμ 1 ty tμ 2 ty tk 1 t CF D β 1 ytk 2 t CF D β 2 yt f s, ys, m 1 sy s, m 2 sy s, hs CF D γ ys, gs CF D ν ys ds μt xt yt μ1 t x t y t μ2 t x t y t k1 t CF D β 1 xt yt k2 t CF D β 2 xt yt f s, xs, m1 sx s, m 2 sx s, hs CF D γ xs, gs CF D ν xs ds f s, ys, m 1 sy s, m 2 sy s, hs CF D γ ys, gs CF D ν ys ds Bβ 1 1 5Bβ Bγ 1 ξ 1 ξ 2 6 ξ 3 7 ξ 4 2 γ ξ 5 9 Bν Bβ 1 1 5Bβ 2 2 ξ 1 ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 2 γ ξ 5 9 Bν 2

7 Aydogan et al. Boundary Value Problems :9 Page 7 of 15 R xt R yt μ t xt yt μtμ 1 t x t y t μ1 tμ 2 t x t y t μ2 t x t y t 1 β 1 k 1 tk 1 2 β t CF D β 1xt yt k 1tBβ 1 1 x t y t 1 2 β 2 k 2 tk 2 3 β t CF D β 2 xt yt k 2 tbβ 2 2 x t y t 2 f t, xt, m 1 tx t, m 2 tx t, ht CF D γ xt, gt CF D ν xt f t, yt, m 1 ty t, m 2 ty t, ht CF D γ yt, gt CF D ν yt. Hence, we get R xt R yt 1 β1 4 N 1 1 N 2 2 N 3 3 Bβ β2 5 Bβ N 5 3 β 2 2 ξ 4 8 Bγ 1 2 łγ ξ 5 9 Bν 1 2 β 1 2 N 4 4 ξ 1 ξ 2 6 ξ 3 7. On the other h, we have Fxt Fyt Bβ 1 1 5Bβ Bγ 1 ξ 1 ξ 2 6 ξ 3 7 ξ 4 2 γ = 1 ξ 5 9 Bν 2 1 F xt F yt= t t Rxs Rys ds s Rxs Rys ds 1 1 t 1 Rxs Rys ds Rxs Rys ds s Rxs Rys ds Bβ 1 1 5Bβ Bγ 1 ξ 1 ξ 2 6 ξ 3 7 ξ 4 2 γ = 2 ξ 5 9 Bν 2

8 Aydogan et al. Boundary Value Problems :9 Page 8 of 15 so F xt F yt 2.Also,wehave F xt F yt 1 Rxs Rys ds 1 Rxt Ryt Rxs Rys ds Bβ 1 1 5Bβ 2 2 ξ 1 ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 2 γ = 3 ξ 5 9 Bν 2 F xt F yt = Rxt Ryt 1 R xt R yt Bβ 1 1 5Bβ Bγ 1 ξ 4 2 γ 1 β1 4 Bβ 1 1 ξ 5 9 Bν 2 2 β 1 2 N Bγ 1 ξ 1 ξ 2 6 ξ 3 7 ξ 4 2 łγ = 4. 1 ξ 1 ξ 2 6 ξ 3 7 N 1 1 N 2 2 N 3 3 Bβ 2 2 ξ 5 9 Bν 1 2 β2 5 5 N 5 3 β 2 2 Thus, Fx Fy for all x, y CR 3, 1. Put ϕt=2t φt=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, 1 y H 1, 1. Then the fractional differential equation CF D 2 xt =yt with boundary conditions x =, x 1 x = x = has the unique solution xt = 1 t Gt, sys ds, where Gt, s = 2 t t s whenever <t s <1 Gt, s= 1 t s 2 t s2 1 t 2 t 2 t s whenever 2 <s t <1. Proof By using Lemma 2.3,wegetxt= 1 J2 yt J3 ytx t 1 x t.hence, x t= 1 J1 yt J2 yt 1 1 x. By using the boundary conditions x 1 x = x 1 = 1 J1 y1 J2 y1 1 1 x, we obtain xt = 1 J2 yt J3 yt 1 t 2 J1 t y1 2 J2 y1. Thus, xt = 1 t t yst s ds 2 yst s2 ds 1 t 1 2 ys ds t 2 if only if xt =. This implies that the given map xt is a unique solution. yst s ds = Gt, sys ds. NotethatCF D 2 xt =

9 Aydogan et al. Boundary Value Problems :9 Page 9 of 15 Note that Gt, s 1 t 1 t < 2 2,fort, 1. Let, β 1, β 2, γ, ν, 1. Now, we investigate the DCF fractional integro-differential problem D 2 xt=μtxtμ1 tx tμ 2 tx t λ 1 k 1 t D β 1 1 xtλ2 k 2 t D β 2 2 xt f s, xs, m 1 sx s, m 2 sx s, hs D γ 1 xs, gs D ν 2 xs ds 2 with boundary conditions x =, x 1 x = x =. Theorem 2.7 Let ξ 1, ξ 2, ξ 3, ξ 4, ξ 5 be nonnegative real numbers, f :, 1 R 5 R an integrable function such that f t, x, y, z, v, w f t, x, y, z, v, w ξ 1 x x ξ 2 y y ξ 3 z z ξ 4 v v ξ 5 v v for all real numbers x, y, z, v, w, x, y, z, v, w t I. If < 1, then the problem has a unique solution, where := max{ 1, 2, 3, 4 }, 1 = Bβ 1 β2 1 β β 21 5 Bβ 2 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 Bγ ν 4 ξ 2 ν1 9 Bν 3 1 γ 2 5, 1 ν 3 2 = Bβ 1 β2 1 β β 21 5 Bβ 2 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 Bγ ν 4 ξ 2 ν1 9 Bν 1 γ 2 5, 1 ν = Bβ 1 β2 1 β β 21 5 Bβ 2 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 Bγ 4 1 γ 2 ν ξ 2 ν1 9 Bν 5 1 ν 3 4 = Bβ 1 β2 1 β β 21 5 Bβ 2 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 Bγ ν 4 ξ 2 ν1 9 Bν 1 γ ν 3 N 1 1 N 2 2 N Bβ 1 β2 1 β 11 Bβ 1N 4 1 β β Bβ 2 2β2 2 2β 21 N 1 β Bβ 2 β2 2 β 21 ξ 1 β ξ 2 6 ξ 3 7 ξ 4 8 Bγ ν2 ν1 9 Bν. 1 łγ 2 1 ν 3 Proof Consider the Banach space CR 3, 1 equipped with the norm x = max t I xt max t I x t max t I x t max t I x t.definethemapf : CR 3, 1 C3 R, 1 by where Fxt= = 1 Gt, srs ds 1 t 2 Rst s ds Rs ds 2 t 2 Rxt=μtxtμ 1 tx tμ 2 tx t Rst s 2 ds λ 1 k 1 t D β 1 1 xtλ2 k 2 t D β 2 2 xt Rst s ds, f s, xs, m 1 sx s, m 2 sx s, hs D γ 1 xs, gs D ν 2 xs ds

10 Aydogan et al. Boundary Value Problems :9 Page 1 of 15 R x t=μtx tμ txtμ 1 tx tμ 1 x t μ 2 tx tμ 2 tx tk 1 t D β 1 1 xt 2 β1 k 1 tbβ 1 CF D β 1 xt 1 β 1 3 β2 k 2 tbβ 2 CF D β 2 xt 1 β 2 β 2 1 β 2 2 x t x t 1 β 2 β 1 1 β 1 2 x t β2 2 1 β 2 3 x t k 2 t D β 2 2 xt 1 x t 1 β 1 f t, xt, m 1 tx t, m 2 tx t, ht D γ 1 xt, gt D ν 2 xt. By using Lemma 2.6, x is a solution for the problem 2ifonlyifx is a fixed point of the operator F.Notethat Rxt Ryt μtxtμ 1 tx tμ 2 tx tλ 1 k 1 t D β 1 1 xtk2 t D β 2 2 xt f s, xs, m 1 tx s, m 2 sx s, hs D γ 1 xs, gs CF D ν 2 xs ds μtytμ 1 ty tμ 2 ty tk 1 t D β 1 1 ytk2 t D β 2 2 yt f s, ys, m 1 ty s, m 2 sy s, hs D γ 1 ys, gs CF D ν 2 ys ds μt xt yt μ 1 t x t y t μ 2 t x t y t k 1 t D β 1 1 xt yt k 2 t D β 2 2 xt yt f s, xs, m 1 tx s, m 2 sx s, hs D γ 1 xs, gs D ν 2 xs ds f s, ys, m 1 ty s, m 2 sy s, hs D γ 1 ys, gs D ν 2 ys ds Bβ 1 1 β 1 β2 2 β Bβ β Bγ ξ 1 ξ 2 6 ξ 3 7 ξ 4 1 γ ξ 2 5 ν 2 ν 1 9 Bν 1 ν Bβ 1 1 β 1 β2 2 β Bβ β 2 3 ξ 1 ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 γ 2 ξ 5 ν 2 ν 1 9 Bν 1 ν 3

11 Aydogan et al. Boundary Value Problems :9 Page 11 of 15 R xt R yt N 1 1 N 2 2 N Bβ 1 β2 1 β 1 1 Bβ 1N 4 1 β β Bβ 2 2β2 2 2β 2 1 N 1 β Bβ 2 β2 2 β 2 1 ξ 1 β ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 łγ ν2 ν 1 9 Bν 2 1 ν 3. Thus, Fxt Fyt Bβ β 1 β2 2 β Bβ β Bγ ξ 1 ξ 2 6 ξ 3 7 ξ 4 1 γ ξ 2 5 = 1 ν 2 ν 1 9 Bν 1 ν 3 F xt F yt Bβ 1 1 β 1 β2 2 β Bβ β 2 3 ξ 1 ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 γ 2 ξ 5 ν 2 ν 1 9 Bν 1 ν 3 = 2. Also, we have F xt F yt Bβ β 1 2 F xt F yt β2 2 β Bβ 2 1 β 2 3 ξ 1 ξ 2 6 ξ Bγ ξ 4 1 γ ξ ν 2 ν 1 9 Bν ν 3 = 3 Rxt Ryt 1 R xt R yt Bβ 1 1 β 1 β2 2 β Bβ β Bγ ξ 1 ξ 2 6 ξ 3 7 ξ 4 1 γ ξ ν 2 ν 1 9 Bν ν 3

12 Aydogan et al. Boundary Value Problems :9 Page 12 of 15 1 N 1 1 N 2 2 N Bβ 1 β2 1 β β 1 3 Bβ 1N 4 1 β 1 5Bβ 2 2 2β2 2 2β 2 1 N 1 β Bβ 2 β2 2 β β 2 3 ξ 1 ξ 2 6 ξ 3 7 ξ 4 8 Bγ 1 łγ ν2 ν 1 9 Bν 2 1 ν 3 = 4. Hence, Fx Fy for all x, y CR 3, 1. Put ϕt =2t φt =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 continuous functions μt = 1 1 sint, μ 1t = 3t 1 2t162, μ 2t = 1 1 e 6t, k 1 t = 1 3 t3 1 1 t 1 5, k 2t = 1 8 cost, m 1t =e 2t, m 2 t = Lnt2 1, ht =gt = 2 t 9 for all t I =, 1. Note that 1 = sup t I μt = 1 1, 2 = sup t I μ 1 t = 1 91, 3 = sup t I μ 2 t = 1, 1e 6 4 = sup t I k 1 t = 1 3, 5 = sup t I k 2 t = 1 8, 6 = sup t I m 1 t = e 2, 7 = sup t I m 2 t = Ln3 12, 8 = sup t I ht = 9 = sup t I gt = 1 9.Also, N 1 = sup t I μ t = 1 1, N 2 = sup t I μ 56 1 t =, N = sup t I μ 2 t = 6, N 1e 6 4 = sup t I k 1 t = 1 5 N 5 = sup t I k 2 t = 1 8, 1.. Also, consider the function =1for Example 2.1 Let, 1. By using Lemma 2.4, the fractional differential equation CF D 2 x 1 t =t with boundary conditions x 1 =, x 1 1 x 1 = x 1 = has theuniquesolutionx 1 t. Also by using Lemma 2.6, the fractional differential equation D 2 x 2 t=t with boundary conditions x 2 =, x 2 1 x 2 = x 2 = has the unique solution x 2 t. For = 1 1, = 1 1, = 1 5, = 1 2, = 4 99 = we compare 5 1 the solutions x 1 t, x 2 txt=x 1 t x 2 tinfig.1. Example 2.2 Consider the CFD fractional integro-differential problem CF D 12 sint 3t 1 5 xt= xt 1 2t 162 x t e 6t 1 x t t 3 3t 6 CF D 3 2 xt cost CF D 5 2 xt 3 8 f s, xs, e 2t x s, Lnt 2 x s,, 2 1 CF D 8 3 xs ds 3 t 9 with boundary conditions x =, x 1 x = x =, where 1 < β 1 = 3 2 <2< β 2 = 5 2 <31<γ = 4 3 <2<ν = 8 3 <3.Putf t, x, y, z, v, w= 2 91 t 3 64 x 1 2 y 1 8 z 1 e 18 w 2v. Notethat 1 = ξ 1 ξ 2 6 ξ 3 7 ξ γ ξ 5 9 =.2, 2 = ξ 1 ξ 2 6 ξ 3 7 ξ γ ξ 5 9 =.46, 3 = ξ 1 ξ 2 6 ξ 3 7 ξ γ ξ 5 9 =.32 4 = ξ 1 ξ 2 6 ξ 3 7 ξ γ ξ N 1 1 N 2 2 N β N β N 5 ξ 1 ξ 2 6 ξ 3 7 ξ łγ ξ 5 6 =.166. By using Theorem 2.5,theproblem3hasa uniquesolution.

13 Aydogan et al. Boundary Value Problems :9 Page 13 of 15 Figure 1 The solutions of the problem with = 1 1, = 1 1, = 1 5, = 1 2, = = 1 Example 2.3 Consider the DCF fractional integro-differential problem D 2 5 2xt = sint 3t 1 xt 1 2t 162 x t e 6t 1 x t t 3 3t 6 CF D f s, xs, e 4t x s, 1xt cost D 2 3 2xt 8 Lnt 2 x s,, 2 1 D 1 5 2xs ds, 4 t 9 with boundary conditions x =, x 1 x = x =, where = 2 5, β 1 = 1 2, β 2 = 2 3, γ = 1 3 ν = 1 5.Putft, x, y, z, v, w= 2 91 t 3 64 x 1 2 y 1 8 z 1 w 2v.Notethat e 18 1 = β 1 β β 21 5 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 ν 4 ξ 2 ν1 9 1 γ 2 5 < 1 ν 3.391, 2 = β2 1 β β 21 5 ξ 1 β ξ 2 6 ξ 3 7 ξ γ 2 ν ξ 2 ν1 9 5 <.225, 1 ν 3 3 = β2 1 β β 21 5 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 ν 4 ξ 2 ν1 9 1 γ 2 5 < ν 3 4 = β 1 β β 21 5 ξ 1 β ξ 2 6 ξ 3 7 ξ 8 ν 4 ξ 2 ν1 9 β 1 γ N 1 ν N 2 2 N β 1 3 N 4 2β 1 β β 21 5 N 5β 2 1 β β 21 ξ 1 β ξ 2 6 ξ 3 7 ξ 4 8 ν ξ 2 ν1 9 1 łγ 2 5 <.493. By 1 ν 3 using Theorem 2.7,theproblem4 has a unique solution.

14 Aydogan et al. Boundary Value Problems :9 Page 14 of 15 3 Conclusion It is important that researchers have some methods available enabling them to review some high order fractional integro-differential equations. In this manuscript, we introduce two types of new fractional derivatives entitled CFD DCF 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 adani University. Funding Not available. Abbreviations DFC, Caputo Fabrizio derivation followed by a differentiation; CFD, Differentiation followed by Caputo Fabrizio derivation. Availability of data 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 approved the final manuscript. Author details 1 Department of athematics, Istanbul Technical University, Istanbul, Turkey. 2 Department of athematics, Cankaya University, Ankara, Turkey. 3 Institute of Space Sciences, Bucharest, Romania. 4 Department of athematics, Azarbaijan Shahid adani University, Tabriz, Iran. 5 Department of edical Research, China edical University Hospital, China edical University, Taichung, Taiwan. Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps institutional affiliations. Received: 2 February 218 Accepted: 24 ay 218 References 1. Abdeljawad, T., Baleanu, D.: On fractional derivatives with exponential kernel their discrete versions. Rep. ath. Phys. 81, Alsaedi, A., Baleanu, D., Etemad, S., Rezapour, Sh.: On coupled systems of time-fractional differential problems by using a new fractional derivative. J. Funct. Spaces 216, ArticleID Area, I., Losada, J., Nieto, J.J.: A note on the fractional logistic equation. Phys. A, Stat. ech. Appl. 444, Aydogan, S.., Baleanu, D., ousalou, A., Rezapour, Sh.: On approximate solutions for two higher-order Caputo Fabrizio fractional integro-differential equations. Adv. Differ. Equ. 217, Baleanu, D., Hedayati, V., Rezapour, Sh., Al Qurashi,..: On two fractional differential inclusions. SpringerPlus 5, Baleanu, D., ousalou, A., Rezapour, Sh.: A new method for investigating some fractional integro-differential equations involving the Caputo Fabrizio derivative. Adv. Differ. Equ. 217, Baleanu, D., ousalou, A., Rezapour, Sh.: On the existence of solutions for some infinite coefficient-symmetric Caputo Fabrizio fractional integro-differential equations. Bound. Value Probl. 217, Caputo,., Fabrizzio,.: A new definition of fractional derivative without singular kernel. Prog. Fract. Differ. Appl. 12, Caputo,., Fabrizzio,.: Applications of new time spatial fractional derivatives with exponential kernels. Prog. Fract. Differ. Appl. 21, De La Sen,., Hedayati, V., Atani, Y.G., Rezapour, Sh.: The existence numerical solution for a k-dimensional system of multi-term fractional integro-differential equations. Nonlinear Anal., odel. Control 222, Dutta, P.N., Choudary, B.S.: A generalization of contraction principle in metric spaces. Fixed Point Theory Appl. 28, Article ID Kojabad, E.A., Rezapour, Sh.: Approximate solutions of a sum-type fractional integro-differential equation by using Chebyshev Legendre polynomials. Adv. Differ. Equ. 217, Kojabad, E.A., Rezapour, Sh.: Approximate solutions of a fractional integro-differential equation by using Chebyshev Legendre polynomials. J. Adv. ath. Stud. 111, Losada, J., Nieto, J.J.: Properties of a new fractional derivative without singular kernel. Prog. Fract. Differ. Appl. 12,

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