EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 3061-3078 ISSN 1307-5543 – ejpam.com Published by New York Business Global Atomic Solution of Third Order Fractional Abstract Cauchy Problem Rami Alkhateeb1,∗, Ghaith Awwad2 1 Department of Basic Sciences, Al-Ahliyya Amman University, Amman, Jordan 2 Department of Mathematics, University of Jordan, Amman, Jordan Abstract. In this paper, we find an atomic solution of the fractional abstract Cauchy problem of order three. The fractional derivative used is the conformable derivative. The main idea of the proofs are based on theory of tensor product of Banach space. 2020 Mathematics Subject Classifications: 34A55, 26A33, 34G10 Key Words and Phrases: Fractional derivatives, abstract Cauchy problem, atom function, conformable derivative, tensor product of Banach space 1. Introduction Let X be a Banach space and I = [0, 1]. Let C(I) be the Banach space of all real valued continuous function on I under the sup-norm, and C(I,X) be the Banach space of all continuous functions defined on I with values on X. In recent years, many researchers were devoted to the problem Bu´ = Au(t) + f(t)z u(0) = x0, where u ∈ Ć (I,X) and A,B are densely defined linear operators on the codomain u. This is called the Abstract Cauchy Problem which is: If f = 0 or z = 0, then the equation is homogenous, otherwise it is called non-homogenous. Now in the non homoge- nous problem we have two cases: (i) The first case, u is unknown and f is given. In this case the problem is called a direct problem. (ii) The second case, u and f are unknown. In this case the problem is called an inverse problem, and some other condi- tions and informations should be given. If B is not invertible, then the equation is called non-degenerate. There are many different techniques to solve Abstract Cauchy Problem in case (i). Tensor product is one of such techniques. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5433 Email addresses: r.alkhateeb@ammanu.edu.jo (R. Alkhateeb), ghaithmawwad@gmail.com (G. Awwad) https://www.ejpam.com 3061 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3062 In [4], a new definition called α−conformable fractional derivative was introduced, which says that: If α ∈ (0, 1), and f : E ⊆ (0,∞) → R. For x ∈ E, let: Dαf(x) = lim ε→0 f(x+ εx1−α)− f(x) ε . (1) If the limit exists then it is called the α−conformable fractional derivative of f at x. For x = 0, Dαf(0) = lim x→0 Dαf(0) if such limit exists. The new definition satisfies: (i) Dα(af + bg) = aDα(f) + bDα(g), for all a, b ∈ R. (ii) Dα(λ) = 0, for all constant functions f(t) = λ. Further, for α ∈ (0, 1] and f, g be α−differentiable at a point t, with g(t) ̸= 0. Then (iii) Dα(fg) = fDα(g) + gDα(f). (iv) Dα ( f g ) = gDα(f)−fDα(g) g2 . We list here the fractional derivatives of certain functions, (v) Dα(tp) = ptp−α. (vi) Dα α(sin ( 1 α t α ) ) = cos ( 1 α t α ) . (vii) Dα α(cos ( 1 α t α ) ) = − sin ( 1 α t α ) . (viii) Dα(e 1 α tα) = e 1 α tα . On letting α = 1 in these derivatives, we get the corresponding ordinary derivatives. One should notice that function could be α−conformable differentiable at a point but not differentiable, for example, take f(t) = 2 √ t. Then D 1 2 (f)(t) = 1. Hence D 1 2 (f)(0) = 1. But D1(f)(0) does not exist. This is not the case for the known classical fractional derivatives. For more on fractional calculus and its applications we refer to [1], [2], and [5]. 2. Atomic solution Let X and Y be two Banach space and X∗ be the dual of X. Assume x ∈ X and y ∈ Y . Define the map x⊗ y: X∗ −→ Y , by x⊗ y(x∗) = ⟨x, x∗⟩ y for all x∗ ∈ X∗. It is well known that x⊗ y is a bounded linear operator and ∥x⊗ y∥ = ∥x∥ ∥y∥ . The operator x ⊗ y is called an atom. The set X ⊗ Y = span {x⊗ y : x ∈ X and y ∈ Y } is subspace of L (X∗, Y ) . If the sum of two atoms is an atom, then either the first components are dependent or the second are dependent. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3063 An equation of the form D2αu(t) +ADαu(t) +Bu(t) = f(t) (2) is called the fractional abstract Cauchy problem of order two, where v and f are nice functions from (0,∞) to the Banach space X, Aand B are closed linear operator on X. A solution of this equation of the form v = u ⊗ x .is called an atomic solution,where v(t) = u(t)x.In this paper, we are interested in finding an atomic solution of the third order vector valued fractional differential equations. 3. Main results In this section, we prove some nice result containing certain solution of atomic problem. Consider the equation v3α (t) +Av2α (t) +Bvα (t) = f (t) , (3) where A and B are closed operators, f (t) is given and u is the unknown equation (3) was discussed in [5] for the first order. Hence,we discuss third order. Theorem 1. The equation v3α (t) +Av2α (t) +Bvα (t) = f (t) with the initial conditions v(0) = 2x0, v α(0) = x0, and v2α(0) = x0 has an atomic solution .where A and B are closed operators on X,and f is a given atomic function, f : [0,∞] → X. Now, we are looking for atomic solution of (3). So put v (t) = u (t)x, u(t) : [0,∞] → R, x is an element in the Banach space X, and consider the case when u(0) = 1, then the initial conditions given in Theorem 1 will be as follows:( v(0) = u(0)x = 2x0 which implies that x = 2x0 vα(0) = x0, and v2α(0) = x0. ) (4) Further assume f to be an atom: f = h ⊗ z where h : [0,∞] → R,and z ∈ x. So, (3) becomes: u3α ⊗ x+ u2α ⊗Ax+ uα ⊗Bx = f (t) . This can be written as: u3α ⊗ x+ u2α ⊗Ax+ uα ⊗Bx = h⊗ z. (5) There are four cases that must be discussed when solving the equation (5) as follows: Case one: u3α ⊗ x+ u2α ⊗Ax is an atom. In this case we have two situations: (i) u3α = u2α. (ii) x = Ax. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3064 Let us take situation (1).So equation (5) becomes: u3α ⊗ (x+Ax) + uα ⊗Bx = h⊗ z, (6) where h and z are given. So we have two cases: (a) u3α = uα = h = u2α. (b) x+Ax = Bx = z. In case (a), we have three cases: (i) u3α − uα = 0. This case can be solved as in [3]: r3 − r = r(r − 1)(r + 1) = 0, which gives r1 = 0, r2 = 1, and r3 = −1. Consequently, u(t) = c1 + c2e tα α + c3e − tα α . So by (4), we have c1 + c2 + c3 = 2x0, c2 − c3 = x0, and c2 + c3 = x0, which implies that c1 = x0, c2 = x0, and c3 = 0. Hence, we have u(t) = x0 + x0e tα α . (7) (ii) uα = h. Using (7), we have h = x0e tα α . Hence for an atomic solution to exist, h must = x0e tα α . (iii) u3α = h. By using (7), we have h = x0e tα α . Since u3α = uα = h = u2α = x0e tα α , then( 6) becomes x+Ax+Bx = z. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3065 So (I +A+ b)x = z, This means z will be in the intersection of the ranges (I +A+B) . Consequently, there is an atomic solution in this situation. In case (b), equation (6) becomes u3α ⊗ (x+Ax) + uα ⊗Bx = h⊗ z. So, u3α + uα = h. This is third order homogenous linear fractional differential equation. To solve it, we follow the variation of parameters method. The homogenous part can solved as in, [3]. r3 + r = r(r2 + 1) = 0, which implies that r1 = 0, r2 = i, and r3 = −i. Hence uh(t) = c1 + c2 cos ( tα α ) + c3 sin ( tα α ) . By the assumption (4), we get c1 = 3x0, c2 = −x0, and c3 = x0. Hence uh(t) = 3x0 − x0 cos ( tα α ) + x0 sin ( tα α ) . For the particular part, the ,we use variation of parameters introduced in [2]. Thus we have up(t) = 3∑ m=1 um t∫ b hWα m(τ) Wα(τ)τ1−α dτ. (8) Where b is an arbitrary positive constant, and Wα[u1(t), u2(t), u3(t)] = ∣∣∣∣∣∣ u1(t) u2(t) u3(t) uα1 (t) uα2 (t) uα3 (t) u2α1 (t) u2α2 (t) u2α3 (t) ∣∣∣∣∣∣ ,Wα 1 = ∣∣∣∣∣∣ 0 u2(t) u3(t) 0 uα2 (t) uα3 (t) 1 u2α2 (t) u2α3 (t) ∣∣∣∣∣∣ , Wα 2 = ∣∣∣∣∣∣ u1(t) 0 u3(t) uα1 (t) 0 uα3 (t) u2α1 (t) 1 u2α3 (t) ∣∣∣∣∣∣ , and Wα 3 = ∣∣∣∣∣∣ u1(t) u2(t) 0 uα1 (t) uα2 (t) 0 u2α1 (t) u2α2 (t) h ∣∣∣∣∣∣ . R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3066 Hence, Wα = ∣∣∣∣∣∣ 3x0 −x0 cos ( tα α ) x0 sin ( tα α ) 0 x0 sin ( tα α ) x0 cos ( tα α ) 0 x0 cos ( tα α ) −x0 sin ( tα α ) ∣∣∣∣∣∣ = −3x0. Which means that: Wα 1 = ∣∣∣∣∣∣ 0 −x0 cos ( tα α ) x0 sin ( tα α ) 0 x0 sin ( tα α ) x0 cos ( tα α ) 1 x0 cos ( tα α ) −x0 sin ( tα α ) ∣∣∣∣∣∣ = −x0, Wα 2 = ∣∣∣∣∣∣ 3x0 0 x0 sin ( tα α ) 0 0 x0 cos ( tα α ) 0 1 −x0 sin ( tα α ) ∣∣∣∣∣∣ = −3x0 cos ( tα α ) , and Wα 3 = ∣∣∣∣∣∣ 3 −x0 cos ( tα α ) 0 0 x0 sin ( tα α ) 0 0 x0 cos ( tα α ) 1 ∣∣∣∣∣∣ = 3x0 sin ( tα α ) . So, we have uα1 (t) = Wα 1 Wα = 1 3 , uα2 (t) = Wα 2 Wα = −3x0 cos ( tα α ) −3x0 = cos ( tα α ) , and uα3 (t) = Wα 3 Wα = 3x0 sin ( tα α ) −3x0 = − sin ( tα α ) . Consequently, up(t) = 3x0 t∫ b h 3 dτα τα−1 −x0 cos ( tα α ) t∫ b h cos ( τα α ) dτα τα−1 −x0 sin ( tα α ) t∫ b h sin ( τα α ) dτα τα−1 . So, u(t) = uh(t) + up(t), u(t) = 3x0 − x0 cos ( tα α ) + x0 sin ( tα α ) + 3x0 t∫ b h dτα τα−1 − x0 cos ( tα α ) t∫ b h cos ( τα α ) dτα τα−1 −x0 sin ( tα α ) t∫ b h sin ( τα α ) dτα τα−1 . This completes situation (1). R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3067 Considering situation (2), x = Ax, equation (5) becomes:( u3α + u2α ) ⊗ x+ uα ⊗Bx = h⊗ z. This situation has two cases: (a) u3α + u2α = uα = h. (b) x = Bx = z. In case (a), for the existence of an atomic solution, we have three situations: (i) u3α + u2α − uα = 0. This can be solved as in [3], r3 + r2 − r = r(r2 + r − 1) = 0. Which gives r1 = 0, r2 = −1 + √ 5 2 , and r3 = −1− √ 5 2 . (9) Then u(t) = c1 + c2e r2( tα α ) + c3e r3( tα α ). (10) Now, by assumptions (4), we have c1 + c2 + c3 = 2x0, r2c2 + r3c3 = x0, and r22c2 + r23c3 = x0. So, c1 = 2x0 − c2 − c3, c2 = r3 − x0 r2r3 − r22 , and c3 = r2 − x0 r2r3 − r23 . (11) From (9) and (11), we have c1 = 0, c2 = 3 + √ 5 5− √ 5 x0 , and c3 = 3− √ 5√ 5 + 5 x0. So, the equation (10) will be u(t) = 3 + √ 5 5− √ 5 x0e −1+ √ 5 2 ( t α α ) + 3− √ 5√ 5 + 5 x0e −1− √ 5 2 ( t α α ). (12) (ii) uα = h. For an atomic solution to exist h must equal to uα. Hence, from (12), we have h = 1 + √ 5 5− √ 5 x0e −1+ √ 5 2 ( t α α ) + 1− √ 5 5 + √ 5 x0e −1− √ 5 2 ( t α α ). R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3068 (iii) u3α + u2α = h. So, h = c2(r 3 2 + r22)e r2( tα α ) + c3(r 3 3 + r23)e r3( tα α ) h = 1 + √ 5 5− √ 5 x0e −1+ √ 5 2 ( t α α ) + 1− √ 5 5 + √ 5 x0e −1− √ 5 2 ( t α α ). (13) Hence, the equations (12) and (13) are equal to h. So, there is an atomic solution in this case. In case (b), equation (5) becomes( u3α + u2α ) ⊗ x+ uα ⊗Bx = h⊗ z. So, u3α + u2α + uα = h. (14) This is third order homogenous linear fractional differential equation, to solve it we follow the variation of parameters method. The homogenous part can solved as in [3]. r3 + r2 + r = r(r2 + r + 1) = 0, which gives r1 = 0, r2 = −1+i √ 3 2 , and r3 = −1−i √ 3 2 . Hence, uh(t) = c1 + e− tα 2α ( c2 cos (√ 3tα 2α ) + c3 sin (√ 3tα 2α )) . By assumption (4), we have c1 = 4x0, c2 = −2x0, and c3 = 0. So, uh(t) = 4x0 − 2x0e − tα 2α cos (√ 3tα 2α ) . Since c3 = 0, there is no particular part, and for an atomic solution to exist, h must equal zero. This completes situation (2), and hence, Case one is completed. Case two: (u3α ⊗ x+ uα ⊗Bx) is an atom. In this case we have two situations: R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3069 (i) u3α = uα. (ii) x = Bx. Considering situation (1), equation (5) becomes: u3α ⊗ (x+Bx) + u2α ⊗Ax = h⊗ z. (15) So we have two cases: (a) u3α(t) = u2α(t) = h = uα. (b) x+Bx = Ax = z. In case (a) we have three cases: (i) u3α − u2α = 0. This case can be solved as in [3]: r3 − r2 = r2(r − 1) = 0, which gives r1 = 0, r2 = 0, and r3 = 1. Consequently, u(t) = c1 + c2 tα α + c3e tα α . Hence, by the assumption (4), we have c1 = x0 , c2 = 0, c3 = x0. Hence, u(t) = x0 + x0e tα α . (16) (ii) u2α = h. From (16), we get h = x0e tα α . So for an atomic solution to exist h must = x0e tα α . (iii) u3α = h. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3070 From (16), we get h = x0e tα α . Since u3α = u2α = uα = h = x0e tα α in (ii) and (iii). Consequently, x+Bx+Ax = z. So (I+B+A)x = z which mean z will be in the intersection of the ranges of (I +B +A) . Hence, there is an atomic solution in situation (1). Now, in case (b) x + Bx = Ax = z. Hence, x + Bx = Ax, x + Bx = z, and Ax = z. So, equation (5) becomes u3α ⊗ (x+Bx) + u2α ⊗Ax = h⊗ z. (17) Substitute the equation (17) in the equation (15), we get( u3α + u2α ) ⊗Ax = h⊗ z. Hence, u3α + u2α = h. (18) This is third order homogenous linear fractional differential equation, to solve it we follow the variation of parameters method. The homogenous part can solved as in [3] as follows: r3 + r2 = 0, which gives r1 = 0, r2 = 0, and r3 = −1. Hence, uh(t) = c1 + c2 tα α + c3e −( t α α ). (19) So by the assumption (4), we have c1 = x0, c2 = 2x0, c3 = x0. Hence, (19) becomes uh(t) = x0 + 2x0 tα α + x0e −( t α α ). For the particular part we use variation of parameters introduced in [2]. Thus, by using (8), the Wronskian will given by: R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3071 Wα = ∣∣∣∣∣∣∣ x0 2x0 tα α x0e −( t α α ) 0 2 −x0e −( t α α ) 0 0 x0e −( t α α ) ∣∣∣∣∣∣∣ = 2x0e −( t α α ). So, we have Wα 1 = ∣∣∣∣∣∣∣ 0 2x0 tα α x0e −( t α α ) 0 2x0 −x0e −( t α α ) 1 0 x0e −( t α α ) ∣∣∣∣∣∣∣ = −2x0e −( t α α ) ( tα α + 1 ) , Wα 2 = ∣∣∣∣∣∣∣ x0 0 x0e −( t α α ) 0 0 −x0e −( t α α ) 0 1 x0e −( t α α ) ∣∣∣∣∣∣∣ = x0e −( t α α ), and Wα 3 = ∣∣∣∣∣∣ x0 2x0 xtα α 0 0 2x0 0 0 0 1 ∣∣∣∣∣∣ = 2. So, uα1 (t) = Wα 1 Wα = −2x0e −( t α α ) ( tα α + 1 ) 2x0e −( t α α ) = − ( tα α + 1 ) , uα2 (t) = Wα 2 Wα = e−( t α α ) 2e−( t α α ) = 1 2 , and uα3 (t) = Wα 3 Wα = 2x0 2x0e −( t α α ) = e( tα α ). Consequently, up = −x0 t∫ b h( tα α + 1) dtα tα−1 + 2x0 tα α t∫ b h 2 dtα tα−1 + x0e −( t α α ) t∫ b he tα α dtα tα−1 . Hence, u(t) = uh + up, u(t) = x0 + 2x0 tα α + x0e −( t α α ) − t∫ b h( tα α + 1) dtα tα−1 + tα α t∫ b h dtα tα−1 + e−( t α α ) t∫ b he tα α dtα tα−1 . R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3072 Now, we will take situation (2), so equation (5) becomes:( u3α(t) + uα(t) ) ⊗ x+ u2α(t)⊗Ax = h⊗ z. For this case we have two cases: (a) u3α + uα = u2α = h = uα. (b) x = Ax = z. In case (a), for the existence of an atomic solution, we have five situations. (i) u3α − u2α + uα = 0. So we have from [3] r3 − r2 + r = r(r2 − r + 1) = 0, which gives r1 = 0, r2 = 1+i √ 3 2 , and r3 = 1−i √ 3 2 . Then, u(t) = c1 + e 1 2 ( tα α )( c2 cos (√ 3tα 2α ) + c3 sin (√ 3tα 2α )) . By the assumption (4), we have c1 = 2x0, c2 = 0, and c3 = 2√ 3 x0. Hence, u(t) = 2x0 + 2√ 3 x0e 1 2 ( tα α ) sin (√ 3tα 2α ) . (20) (ii) u3α + uα = uα. u3α = 0, from [3],we have r3 = 0. Consequently, r1 = r2 = r3 = 0. So, u(t) = c1 + c2 ( tα α ) + c3 ( tα α )2 . R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3073 So, by assumption (10), we have c1 = 2x0, c2 = x0, c3 = x0 2 . Hence, u(t) = 2x0 + x0 ( tα α ) + x0 2 ( tα α )2 . (iii) u2α = h. So,from (20) we have, h = x0e 1 2 ( tα α )( cos (√ 3tα 2α ) − x0√ 3 sin (√ 3tα 2α )) . (iv) uα = h. So, from (20) we have, h = x0e 1 2 ( tα α )( cos (√ 3tα 2α ) + x0√ 3 sin (√ 3tα 2α )) . (v) u3α + uα = h. So, from (20), we have h = x0e 1 2 ( tα α ) cos (√ 3tα 2α ) . Since we don’t have same solution from (i), (ii), (iii), (iv), and (v), there is no an atomic solution in this case. This completes situation (2), and hence, Case two is completed. Case three: (u2α ⊗Ax+ uα ⊗Bx) is an atom. This has two situations: (1) u2α = uα. (2) A x = Bx. Let us take situation (1), so equation (5) becomes: u3α ⊗ x+ u2α ⊗ (Ax+Bx) = h⊗ z. So, we have two cases: (a) u3α(t) = u2α(t) = h = uα. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3074 (b) x = Ax+Bx = z. In case (a), we have four situations: (i) u3α − u2α = 0. So, we can solve it as in [3] r3 − r2 = r2(r − 1) = 0, which gives r1 = 0, r2 = 0, and r3 = 1. So, u(t) = c1 + c2 tα α + c3e ( t α α ). (21) By assumption (4), we have c1 = x0, c2 = 0, and c3 = x0. So, (21) becomes u(t) = x0 + x0e ( t α α ). (22) (ii) u2α = h. Using (21), we get h = x0e ( t α α ). (iii) u3α = uα. r3 = r. So, we have r1 = 0, r2 = 1, and r3 = −1. Hence, u(t) = c1 + c2e ( t α α ) + c3e −( t α α ). By assumption (4), we have c1 = x0.c2 = x0, and c3 = 0. Consequently, u(t) = x0 + x0e ( t α α ). (23) (iv) u3α = h. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3075 Using (21), we get h = x0e ( t α α ). Since u3α = u2α = h = uα = x0e tα α , (6) becomes x = Ax+Bx = z. So, (A+B)x = z, or (I)x = z. Which means that z will be at the range of intersection of (A+B) and I. Consequently, there is atomic solution in this case. In case (b), equation (5) becomes u3α ⊗ x+ u2α ⊗ (A+B)x = h⊗ z. So, u3α + u2α = h. This is third order homogenous linear fractional differential equation, to solve it we follow the variation of parameters method. The homogenous and particular parts can be found similarly as (18) in the case (b) in situation (1) in Case two, and u(t) = x0 + 2x0 tα α + x0e −( t α α ) − t∫ b h( tα α + 1) dtα tα−1 + tα α t∫ b h dtα tα−1 + e−( t α α ) t∫ b he tα α dtα tα−1 . Now, we will take situation (2), so equation (5) becomes: u3α ⊗ x+ (u2α + uα)⊗Ax = h⊗ z. So, we have two cases: (a) u3α(t) = u2α(t) + uα(t) = h. (b) x = Ax = z. In case (a), for an atomic solution to exist we must have three situations (i) u3α(t)− u2α(t)− uα(t) = 0. R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3076 So, we can solve it as in [3] r3 − r2 − r = r(r2 − r − 1) = 0, which gives r1 = 0, r2 = 1+ √ 5 2 , and r3 = 1− √ 5 2 . So, u(t) = c1 + c2e r2( tα α ) + c3e r3( tα α ). By assumption (4), we have c1 = 2x0, c2 = x0√ 5 , and c3 = −x0√ 5 . Hence, u(t) = 2x0 + x0√ 5 e −1+ √ 5 2 ( t α α ) − x0√ 5 e −1− √ 5 2 ( t α α ). (24) (ii) u2α(t) + uα(t) = h. So from (24), we have h = ( 1 + 2√ 5 ) x0e −1+ √ 5 2 ( t α α ) + ( 1− 2√ 5 ) x0e −1− √ 5 2 ( t α α ). So, for an atomic solution to exist hmust equal ( 1 + 2√ 5 ) x0e −1+ √ 5 2 ( t α α )+ ( 1− 2√ 5 ) x0e −1− √ 5 2 ( t α α ). (iii) u3α(t) = h. So from (24), we have h = c2r 3 2e r2t α α + c3r 3 3e r3t α α . Hence, h = ( 2√ 5 + 1)x0e 1+ √ 5 2 tα α + (1− 2√ 5 )x0e 1− √ 5 2 tα α . Since u2α(t) + uα(t) = u3α(t) = h in (ii) and (iii), there is an atomic solution . This completes situation (2), and hence, Case three is completed. Case four: (u3α ⊗ x+ u2α ⊗Ax+ uα ⊗Bx) is an atom. This has two situations: (1) u3α = u2α = uα = h. (2) x = Ax = Bx = z. Considering situation (1), equation (5) becomes: u3α ⊗ (x+Ax+Bx) = h⊗ z. So, we have seven cases: R. Alkhateeb, G. Awwad / Eur. J. Pure Appl. Math, 17 (4) (2024), 3061-3078 3077 (a) u3α = u2α. We can solve it as in [3] r3 − r2 = r2(r − 1) = 0. Hence, r1 = 0, r2 = 0, and r3 = 1. Consequently, u(t) = c1 + c2( tα α ) + c3e tα α . By assumption (4), we have c1 = x0, c2 = 0, and c3 = x0. Hence, u(t) = x0 + x0e tα α . (b) u3α = uα. We can solve it as in [3] r3 − r = r(r2 − 1) = r(r − 1)(r + 1) = 0. Hence, r1 = 0, r2 = 1, and r3 = −1. Consequently, u(t) = c1 + c2e tα α + c3e − tα α . By assumption (4), we have c1 = x0, c2 = x0, and c3 = 0. Hence, u(t) = x0 + x0e tα α . (c) u2α = uα. We can solve it as in [3] (r2 − r) = r(r − 1) = 0. Hence, r1 = 0 and r2 = 1. Consequently, u(t) = c1 + c2e tα α By assumption (4), we have c1 = x0 and c2 = x0. Hence, u(t) = x0 + x0e tα α . REFERENCES 3078 (d) u3α = h. Consequently, h = x0e tα α . So, for an atomic solution to exist h must equal x0e tα α . (e) u2α = h = x0e tα α . (f) uα = h = x0e tα α . Hence, (e) and (f) give the same result, there is an atomic solution in this case and h = x0e tα α . In situation (2), equation (5) will be e tα α ⊗ x+ e tα α ⊗Ax+ e tα α Bx = e tα α ⊗ z. So, (I +A+B)x = z. Hence, z is the image of x under (I +A+B). This completes situation (2), and hence, Case four is completed. Acknowledgements The authors are grateful to the reviewers for their careful reading and valuable sug- gestions. References [1] T Abdeljawad. On conformable fractional calculus. J. Comput. Appl. Math., 279:57– 66, 2015. [2] M Alhorani, M Abuhammad, and R Khalil. Variation of parameters for local fractional nonhomogeneous linear-differential equation. J. Math. Computer Sci., 16:140–146, 2016. [3] M Alhorani, R Khalil, and I Aldarawi. Fractional cauchy euler differential equation. J. Comput. Anal. Appl., 28:226–233, 2020. [4] R Khalil, M Alhorani, A Yousef, and M Sababheh. A new definition of fractional derivative. J. Comput. Appl. Math., 264:65–70, 2014. [5] L Rabhi, M Alhorani, and R Khalil. Inhomogeneous conformable abstract cauchy problem. Open Mathematics, 19:690–705, 2021.