Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 RESEARCH Open Access Abstract fractional integro-differential equations involving nonlocal initial conditions in a-norm Rong-Nian Wang*, Jun Liu and De-Han Chen * Correspondence: rnwang@mail ustc.edu.cn Department of Mathematics, NanChang University, NanChang, JiangXi 330031, People’s Republic of China Abstract In the present paper, we deal with the Cauchy problems of abstract fractional integro-differential equations involving nonlocal initial conditions in a-norm, where the operator A in the linear part is the generator of a compact analytic semigroup New criterions, ensuring the existence of mild solutions, are established The results are obtained by using the theory of operator families associated with the function of Wright type and the semigroup generated by A, Krasnoselkii’s fixed point theorem and Schauder’s fixed point theorem An application to a fractional partial integrodifferential equation with nonlocal initial condition is also considered Mathematics subject classification (2000) 26A33, 34G10, 34G20 Keywords: Cauchy problem of abstract fractional evolution equation, Nonlocal initial condition, Fixed point theorem, Mild solution, α-norm Introduction Let (A, D(A)) be the infinitesimal generator of a compact analytic semigroup of bounded linear operators {T(t)} t≥0 on a real Banach space (X, ||·||) and Ỵ r(A) Denote by Xa, the Banach space D(Aa) endowed with the graph norm ||u||a = ||Aau|| for u Ỵ Xa The present paper concerns the study of the Cauchy problem for abstract fractional integro-differential equation involving nonlocal initial condition ⎧c β ⎪ Dt u(t) = Au(t) + F(t, u(t), u(κ1 (t))) ⎪ ⎨ t + K(t − s)G(s, u(s), u(κ2 (s)))ds, ⎪ ⎪ ⎩ u(0) + H(u) = u0 t ∈ [0, T], (1:1) in Xa, where c Dβ, such that ||Aα T(t)||L(X) ≤ Mα −δt e tα (c) A-ais a bounded linear operator in X with D(Aa) = Im(A-a) (d) If Aα Pβ (t)x (4) For all x Ỵ X and t Ỵ (0, ∞), Cα = ≤ Cα t−αβ x , where Mα β (2−α) (1+β(1−α)) We can also prove the following criterion Lemma 2.3 The functions t → Aα Pβ (t) and t → Aα Sβ (t) are continuous in the uniform operator topology on (0, +∞) Proof Let ε > be given For every r > 0, from (2.1), we may choose δ1, δ2 > such that Mα r αβ δ1 −α β (s)s ds ≤ ε , Mα r αβ ∞ δ2 −α β (s)s ds ≤ ε (2:2) Then, we deduce, in view of the fact t ® AaT(t) that is continuous in the uniform operator topology on (0, ∞) (see [[30], Lemma 2.1]), that there exists a constant δ > such that δ2 β (s) β β Aα T t1 s − Aα T t2 s δ1 for t1, t2 ≥ r and |t1 - t2| implies that Aα Pβ (t) is continuous in the uniform operator topology for t > A similar argument enable us to give the characterization of continuity on Aα Pβ (t) This completes the proof ■ Lemma 2.4 For every t > 0, the restriction of Sβ (t)to Xa and the restriction of Pβ (t)to Xa are compact operators in Xa Proof First consider the restriction of Sβ (t) to Xa For any r > and t > 0, it is sufficient to show that the set {Sβ (t)u; u ∈ Br }is relatively compact in Xa, where Br := {u Ỵ Xa; ||u||a ≤ r} Since by Lemma 2.1, the restriction of T(t) to Xa is compact for t > in Xa, for each t > and ε Ỵ (0, t), ∞ ε tβ s uds; u ∈ Br β (s)T = T tβ ε ∞ ε β (s)T tβ s − tβ ε uds; u ∈ Br Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page of 16 is relatively compact in Xa Also, for every u Î Br, as ∞ β (s)T tβ s uds → Sβ (t)u, ε→0 ε in X a , we conclude, using the total boundedness, that the set {Sβ (t)u; u ∈ Br } is relatively compact, which implies that the restriction of Sβ (t) to X a is compact The same idea can be used to prove that the restriction of Pβ (t) to Xa is also compact ■ The following fixed point theorems play a key role in the proofs of our main results, which can be found in many books Lemma 2.5 (Krasnoselskii’s Fixed Point Theorem) Let E be a Banach space and B be a bounded closed and convex subset of E, and let F1, F2 be maps of B into E such that F1x + F2y Î B for every pair x, y Î B If F1 is a contraction and F2 is completely continuous, then the equation F1x + F2x = x has a solution on B Lemma 2.6 (Schauder Fixed Point Theorem) If B is a closed bounded and convex subset of a Banach space E and F : B ® B is completely continuous, then F has a fixed point in B Main results Based on the work in [[15], Lemma 3.1 and Definition 3.1], in this paper, we adopt the following definition of mild solution of Cauchy problem (1.1) Definition 3.1 By a mild solution of Cauchy problem (1.1), we mean a function u Ỵ C([0, T]; Xa) satisfying u(t) = Sβ (t)(u0 − H(u)) + t (t − s)β−1 Pβ (t − s)(F(s, u(s), u(κ1 (s))) s + K(s − τ )G(τ , u(τ ), u(κ2 (τ )))dτ )ds for t Ỵ [0, T] Let us first introduce our basic assumptions (H0) 1, 2 Ỵ C([0, T]; [0, T]) and K Ỵ C([0, T]; ℝ+) (H1) F, G : [0, T] ì Xa ì Xa đ X are continuous and for each positive number k Ỵ N, there exist a constant g Ỵ [0, b(1 - a)) and functions k(·) Ỵ L1/g(0, T; ℝ+), jk(·) Ỵ L∞(0, T; ℝ+) such that sup u α, v α ≤k sup u α, v α ≤k F(t, u, v) ≤ ϕk (t) and lim inf G(t, u, v) ≤ φk (t) and lim inf ϕk k→+∞ L1/γ (0,T) k k→+∞ φk L∞ (0,T) k = σ1 < ∞, = σ2 < ∞ (H2) F, G : [0, T] × Xa × Xa ® X are continuous and there exist constants LF, LK such that F(t, u1 , v1 ) − F(t, u2 , v2 ) ≤ LF ( u1 − u2 α + v1 − v2 α ), G(t, u1 , v1 ) − G(t, u2 , v2 ) ≤ LG ( u1 − u2 α + v1 − v2 α ) Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page of 16 for all (t, u1, v1), (t, u2, v2) ẻ [0, T] ì Xa ì Xa (H3) H : C([0, T]; Xa) ® Xa is Lipschitz continuous with Lipschitz constant LH (H4) H : C([0, T]; Xa) ® Xa is continuous and there is a h Î (0, T) such that for any u, w Î C([0, T]; Xa) satisfying u(t) = w(t)(t Ỵ[h, T]), H(u) = H(w) (H5) There exists a nondecreasing continuous function F : + đ + such that for all u ẻ Θk, H(u) ≤ α (k), and (k) = μ < ∞ k lim inf k→+∞ Remark 3.1 Let us note that (H4) is the case when the values of the solution u(t) for t near zero not affect H(u) We refer to [19]for a case in point T K(t)dt We are now ready to state our main results in In the sequel, we set k := this section Theorem 3.1 Let the assumptions (H0), (H1) and (H3) be satisfied Then, for u0 Ỵ Xa, the fractional Cauchy problem (1.1) has at least one mild solution provided that 1−γ (1 − α)β − γ MLH + Cα σ1 T (1−α)β−γ 1−γ + Cα σ2 kT (1−α)β < (1 − α)β (3:1) Proof Let v Ỵ C([0, T]; Xa) be fixed with |v|a ≡ From (3.1) and (H1), it is easy to see that there exists a k0 > such that M( u0 α + LH k0 + (1−α)β + 1−γ (1 − α)β − γ H(ν) α ) + Cα Cα kT (1 − α)β φ k0 T (1−α)β−γ k0 t ( u)(t) = Sβ (t)(u0 − H(u)) + s by (t − s)β−1 Pβ (t − s) F(s, u(s), u(κ1 (s))) K(s − τ )G(τ , u(τ ), u(κ2 (τ )))dτ ds u)(t) +( u)(t), t ∈ [0, T] It is easy to verify that (Γu)(·) Ỵ C([0, T]; Xa) for every u ∈ pair v, u ∈ k0 and t Ỵ [0, T], by (H1) a direct calculation yields ≤ v)(t) +( L1/γ (0,T) + ( ϕk0 ≤ k0 L∞ (0,T) Consider a mapping Γ defined on := ( 1−γ u)(t) α Sβ (t)(u0 − H(v)) t α + (t − s)β−1 Aα Pβ (t − s) L(X) k0 Moreover, for every F(s, u(s), u(κ1 (s))) s + K(s − τ ) G(τ , u(τ ), u(κ2 (τ )))dτ ds ≤ M( u0 t +Cα α + LH k0 + H(ν) α ) β(1−α)−1 (t − s) (ϕk0 (s) + k φ k0 L∞ (0,T) )ds ≤ M( u0 +Cα ≤ k0 + LH k0 + H(ν) α ) 1−γ 1−γ T (1−α)β−γ (1 − α)β − γ α ϕk0 L1/γ (0,T) Cα kT (1−α)β (1 − α)β φ k0 L∞ (0,T) Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page of 16 That is, v + u ∈ k0 for every pair v, u ∈ k0 Therefore, the fractional Cauchy problem (1.1) has a mild solution if and only if the operator equation Γ1u + Γ2u = u has a solution in k0 In what follows, we will show that Γ1 and Γ2 satisfy the conditions of Lemma 2.5 From (H3) and (3.1), we infer that Γ1 is a contraction Next, we show that Γ2 is completely continuous on k0 We first prove that Γ2 is continuous on k0 Let {un }∞ ⊂ k0 be a sequence such n=1 that un ® u as n ® ∞ in C([0, T]; Xa) Therefore, it follows from the continuity of F, G, 1 and 2 that for each t Î [0, T], F(t, un (t), un (κ1 (t))) → F(t, u(t), u(κ1 (t))) as n → ∞, G(t, un (t), un (κ1 (t))) → G(t, u(t), u(κ2 (t))) as n → ∞ Also, by (H1), we see t (t − s)β−1−αβ F(s, un (s), un (κ1 (s))) − F(s, u(s), u(κ1 (s))) ds t ≤ (t − s)β−1−αβ ϕk0 (s)ds ≤2 1−γ 1−γ (1 − α)β − γ T (1−α)β−γ ϕk0 L1/γ (0,T) , and t s (t − s)β−1−αβ K(s − τ ) G(τ , un (τ ), un (κ2 (τ ))) −G(τ , u(τ ), u(κ2 (τ ))) ≤ 2k t φ k0 L∞ (0,T) dτ ds β−1−αβ (t − s) ds ≤ 2kT (1−α)β (1 − α)β φ k0 L∞ (0,T) Hence, as ( un )(t) t ≤ Cα −( u)(t) α β−1−αβ (t − s) t +Cα F(s, un (s), un (κ1 (s))) − F(s, u(s), u(κ1 (s))) (t − s)β−1−αβ s K(s − τ ) ds G(τ , un (τ ), un (κ2 (τ ))) −G(τ , u(τ ), u(κ2 (τ ))) dτ ds, we conclude, using the Lebesgue dominated convergence theorem, that for all t Ỵ [0, T], ( un )(t) −( u)(t) α → 0, as n → ∞, which implies that | un − u|α → 0, as n → ∞ This proves that Γ2 is continuous on k0 It suffice to prove that Γ2 is compact on k0 For the sake of brevity, we write t N (t, u(t)) = F(t, u(t), u(κ1 (t))) + K(t − τ )G(τ , u(τ ), u(κ2 (τ )))dτ Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page of 16 Let t Ỵ [0, T] be fixed and ε, ε1 > be small enough For u ∈ by k0, we define the map ε,ε1 t−ε ∞ ( ε,ε1 βτ u)(t) = β (τ )T((t − s)β τ )N (s, u(s))dτ ds ε1 t−ε ∞ β = T(ε ε1 ) βτ β (τ )T((t − s)β τ − εβ ε1 )N (s, u(s))dτ ds ε1 { Therefore, from Lemma 2.1 we see that for each t Ỵ (0, T], the set ε,ε1 u)(t); u ∈ k0 } is relatively compact in Xa Then, as ( ≤ u)(t) t ε1 −( ε,ε1 u)(t) βτ (t − s) 0 t β−1 α β (τ )T((t α ∞ βτ (t − s)β−1 + t−ε ε1 t ≤ βMα − s)β τ )N (s, u(s))dτ ds β(1−α)−1 (t − s) β (τ )T((t − s)β τ )N (s, u(s))dτ ds φ k0 (ϕk0 (s) + k ε1 L∞ (0,T) )ds t (t − s)β(1−α)−1 (ϕk0 (s) + k + φ k0 ∞ L∞ (0,T) )ds t−ε 1−γ (1 − α)β − γ ≤ βMα kT (1−α)β (1 − α)β + φ k0 τ τ 1−α β (τ )dτ β (τ )dτ 1−γ ϕk0 T (1−α)β−γ ε1 L∞ (0,T) τ 1−α L1/γ (0,T) β (τ )dτ 1−γ (1 − α)β − γ βMα (2 − α) (1 + β(1 − α)) + ε1 α 1−α k φ k0 (1 − α)β → as ε, ε1 → 0+ + L∞ (0,T) ε 1−γ ϕk0 L1/γ (0,T) ε (1−α)β−γ (1−α)β in view of (2.1), we conclude, using the total boundedness, that for each t Ỵ [0, T], the set { u)(t); u ∈ k0 } is relatively compact in Xa On the other hand, for 0, there would exist uk Ỵ Θ k and t k Ỵ [0, T] such that ||(Γ w uk )(t k)|| a >k Thus, we have k< ( w uk )(tk ) α Cα + max 0≤s≤T ≤ M( u0 F(s, ν, ν) α+ H(u) α ) + +k max 0≤s≤T (1 − α)β 2kCα (LF + kLG )T (1−α)β (1 − α)β G(s, ν, ν) T (1−α)β Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page 12 of 16 Dividing on both sides by k and taking the lower limit as k ® +∞, we get 2Cα (LF + kLG )T (1−α)β , (1 − α)β 1≤ this contradicts (3.2) Also, for u, v ∈ ( w u)(t) t −( a direct calculation yields w v)(t) α β−1 (t − s) = k0, Pβ (t − s) F(s, u(s), u(κ1 (s))) − F(s, v(s), v(κ1 (s))) s + K (s − τ )(G(τ , u(τ ), u(κ2 (τ ))) − G(τ , v(τ ), v(κ2 (τ ))))dτ ds t ≤ Cα α β−1−αβ (t − s) s + ≤ Cα K(s − τ ) F(s, u(s), u(κ1 (s))) − F(s, v(s), v(κ1 (s))) G(τ , u(τ ), u(κ2 (τ ))) − G(τ , v(τ ), v(κ2 (τ ))) t (t − s)β−1−αβ LF ( u(s) − v(s) α + s +LG K(s − τ )( u(τ ) − v(τ ) α + ds dτ ds u(κ1 (s)) − v(κ1 (s)) α ) u(κ2 (τ )) − v(κ2 (τ )) α )dτ ds 2Cα T (1−α)β (LF + kLG ) |u − v|α , (1 − α)β ≤ which together with (3.2) implies that Γw is a contraction mapping on k0 Thus, by the Banach contraction mapping principle, Γw has a unique fixed point uw ∈ k0, i.e., t (t − s)β−1 Pβ (t − s)(F(s, uw (s), uw (κ1 (s))) uw = Sβ (t)(u0 − H(w)) + s K(s − τ )G(τ , uw (τ ), uw (κ2 (τ )))dτ )ds + for t Ỵ [0, T] Step Write η k0 = {u ∈ C([η, T]; Xα ); It is clear that η k0 u(t) α ≤ k0 for all t ∈ [η, T]} is a bounded closed convex subset of C([h, T]; Xa) Based on the argument in Step 1, we consider a mapping F on (F w)(t) = uw , η k0 into itself Moreover, for w1 , w2 ∈ 2Cα T (1−α)β (LF + kLG ) |uw1 − uw2 |α ≤ M (1 − α)β H(w1 ) − H(w2 ) α , that is, sup t∈[η,T] defined by t ∈ [η, T] It follows from (H5) and (3.2) that F maps from Step 1, we have 1− η k0 F w1 (t) − F w2 (t) α →0 as w1 → w2 in η k0 , η k0, Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page 13 of 16 η which yields that F is continuous Next, we prove that F has a fixed point in k0 It will suffice to prove that F is a compact operator Then, the result follows from Lemma 2.6 Let’s decompose the mapping F = F1 + F2 as (F1 w)(t) = Sβ (t)(u0 − H(w)), (F2 w)(t) = t (t − s)β−1 Pβ (t − s)(F(s, uw (s), uw (κ1 (s))) s + K(s − τ )G(τ , uw (τ ), uw (κ2 (τ )))dτ )ds Since assumption (H5) implies that the set H(w); w ∈ η k0 is bounded in Xa, it fol- lows from Lemma 2.4 that for each t Ỵ [h, T], (F1 w)(t); w ∈ η k0 is relatively com- pact in Xa Also, for h ≤ t1 ≤ t2 ≤ T, (Sβ (t2 ) − Sβ (t1 ))(u0 − H(w)) independently of w ∈ η k0 α →0 as t2 − t1 → This proves that the set (F1 w)(·); w ∈ η k0 is equicontin- uous Thus, an application of Arzela-Ascoli’s theorem yields that F1 is compact Observe that the set ⎧ ⎫ t ⎬ ⎨ η F(t, u(t), u(κ1 (t))) + K(t − τ )G(τ , u(τ ), u(κ2 (τ )))dτ ; t ∈ [0, T], w ∈ k0 ⎭ ⎩ is bounded in X Therefore, using Lemma 2.1, Lemma 2.2 and Lemma 2.3, it is not difficult to prove, similar to the argument with Γ2 in Theorem 3.1, that F2 is compact Hence, making use of Lemma 2.6 we conclude that F has a fixed point w∗ ∈ q = uw∗ Then, η k0 Put t q(t) = Sβ (t) u0 − H w∗ (t − s)β−1 Pβ (t − s) F(s, q(s), q(κ1 (s))) + s K (s − τ )G(τ , q(τ ), q(κ2 (τ )))dτ ) ds, + t ∈ [0, T] Since uw∗ = F w∗ = w∗ (t ∈ [η, T]), H(w∗ ) = H(q) and hence q is a mild solution of the fractional Cauchy problem (1.1) This completes the proof ■ Example In this section, we present an example to our abstract results, which not aim at generality but indicate how our theorem can be applied to concrete problem We consider the partial differential equation with Dirichlet boundary condition and nonlocal initial condition in the form Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page 14 of 16 ⎧ ⎪c ⎪ ∂ u(t, x) = ∂ u(t, x) + q1 (t)|u(sin(t), x)| ⎪ t ⎪ ⎪ ∂x2 + |u(sin(t), x)| ⎪ ⎪ t ⎪ ⎪ q2 (s) |u(s, x)| ⎨ + ds, ≤ t ≤ 1, ≤ x ≤ π , + |u(s, x)| + (t − s) ⎪ ⎪ u(t, 0) = u(t, π ) = 0, ≤ t ≤ 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u(0, x) = ln[eu(s,x) (|u(s, x)| + 1)]ds + u (x), ≤ x ≤ π , ⎪ ⎩ (4:1) η where the functions q1, q2 are continuous on [0, 1] and 0, t ∈ R Therefore, we have Sβ (t)u = Pβ (t)u = ∞ n=1 ∞ n=1 Eβ (−n2 tβ )(u, yn )yn , u ∈ X; Sβ (t) L(X) ≤ 1, eβ (−n2 tβ )(u, yn )yn , u ∈ X; Pβ (t) L(X) ≤ β (1 + β) for all t ≥ 0, where Eb(t) := Eb,1(t) and eb(t) := Eb, b(t) The consideration of this section also needs the following result Lemma 4.1 [31]If w ∈ D(A ), then w is absolutely continuous, w’ Ỵ X, and w = A2 w Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 Page 15 of 16 Define F(t, u(t), u(κ1 (t)))(x) = q1 (t)|u(sin(t), x)| , + |u(sin(t), x)| , κ1 (t) = sin(t), κ2 (t) = t, t2 + q2 (t)|u(t, x)| G(t, u(t), u(κ2 (t)))(x) = , + |u(t, x)| K(t) = H(u)(x) = η ln[eu(s,x) (|u(s, x)| + 1)]ds Therefore, it is not difficult to verify that F, G : [0, 1] × X × X → X and 2 H : C([0, 1]; X ) → X are continuous, 2 F(t, u(t), u(κ1 (t))) − F(t, v(t), v(κ1 (t))) ≤ μ1 A− L(X) u(κ1 (t)) − v(κ1 (t)) 1, G(t, u(t), u(κ1 (t))) − G(t, v(t), v(κ1 (t))) ≤ μ2 A− L(X) where μi := suptỴ H(u) u(t) − v(t) [0, 1]|qi(t)|, and for any u satisfying |u| ≤ k, 1 1, = A H(u)(·) = H(u) (·) ≤ 2(1 − η)k in view of Lemma 4.1 Now, we note that the problem (4.1) can be reformulated as the abstract problem (1.1) and the assumptions (H0), (H2), (H4) and (H5) hold with α= , T = 1, LF = μ , LG = μ2 , Thus, when − η + 4M (μ1 + μ2 ) < 2 (k) = 2(1 − η)k, μ = 2(1 − η) such that condition (3.2) is verified, (4.1) has at least one mild solution due to Theorem 3.2 Acknowledgements We would like to thank the referees for their valuable comments and suggestions This research was supported in part by NNSF of China (11101202), NSF of JiangXi Province of China (2009GQS0018), and Youth Foundation of JiangXi Provincial Education Department of China (GJJ10051) Authors’ contributions All authors contributed equally to the manuscript and read and approved the final manuscript Competing interests The authors declare that they have no competing interests Received: 17 December 2010 Accepted: 16 August 2011 Published: 16 August 2011 References Anh Vo, V, Leonenko, NN: Spectral analysis of fractional kinetic equations with randomdata J Stat Phys 104, 1349–1387 (2001) doi:10.1023/A:1010474332598 Eidelman, SD, Kochubei, AN: Cauchy problem for fractional diffusion equations J Differ Equ 199(2), 211–255 (2004) doi:10.1016/j.jde.2003.12.002 Metzler, R, Klafter, J: The random walk’s guide to anomalous diffusion: a fractional dynamics approach Phys Rep 339, 1–77 (2000) doi:10.1016/S0370-1573(00)00070-3 Miller, KS, Ross, B: An Introduction to the Fractional Calculus and Differential Equations Wiley, New York (1993) Wang et al Advances in Difference Equations 2011, 2011:25 http://www.advancesindifferenceequations.com/content/2011/1/25 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Podlubny, I, Petraš, I, Vinagre, BM, O’Leary, P, Dorčak, L: Analogue realizations of fractional-order controllers: fractional order calculus and its applications Nonlinear Dyn 29, 281–296 (2002) doi:10.1023/A:1016556604320 Samko, SG, Kilbas, AA, Marichev, OI: Fractional Integrals and Derivatives, Theory and Applications Gordon and Breach, Yverdon (1993) Mainardi, F: Fractional calculus: some basic problems in continuum and statistical mechanics In: Carpinteri A, Mainardi F (eds.) 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Cite this article as: Wang et al.: Abstract fractional integro-differential equations involving nonlocal initial conditions in a-norm Advances in Difference Equations 2011 2011:25 Page 16 of 16... of mild solutions in [14] for fractional differential equations with nonlocal initial conditions and in [15] for fractional neutral differential equations with nonlocal initial conditions and time... solutions for fractional differential equations with nonlocal initial conditions in a-norm using the contraction mapping principle and the Schauder’s fixed point theorem have been investigated in [16]