EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 1, 2022, 106-125 ISSN 1307-5543 – ejpam.com Published by New York Business Global Atomic Solution of Fractional Abstract Cauchy Problem of High Order in Banach Spaces Fatema Bekraoui1, Mohammed Al Horani2, Roshdi Khalil3,∗ 1 Department of Mathematics, School of Science, The University of Jordan, Amman, Jordan Abstract. The Abstract Cauchy problem, which is a vector valued differential equation is an important equation in many branches of science. Authors usually study and discuss first and second order abstract Cauchy problem. In this paper, we try to find atomic solutions of the fractional abstract Cauchy problem with order 3α, where α ∈ (0, 1). It turned out that there are so many cases to consider in order to determine atomic solution. 2020 Mathematics Subject Classifications: 26A33 Key Words and Phrases: Tensor product of Banach spaces, atomic solution, conformable derivative, abstract Cauchy problem. 1. Introduction Let X be a Banach space and I = [0, 1]. Let C(I) be the Banach space of all real valued continuous functions on I under the sup-norm, and C(I,X) be the Banach space of all continuous functions defined on I with values in X. A classical and important differential equation that appears in physics and other branches of applied sciences is the so called Abstract Cauchy problem of the first order. One form of such equation is Bu ′ = Au(t) + f(t)z u(0) = x0 . Here u is a continuously differentiable function in C(I,X) and A,B are densely defined linear operators on the codomain of u. If f = 0 or z = 0, then the equation is homogeneous otherwise it is called non- homogeneous. Now in the non-homogeneous problem we have two cases. The first type, if u is unknown and f is given and this is called the direct problem. The second type, u ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i1.4226 Email addresses: fat9180303@ju.edu.jo (F. Bekraoui), horani@ju.edu.jo (M. Al Horani), roshdi@ju.edu.jo (R. Khalil) http://www.ejpam.com 106 © 2022 EJPAM All rights reserved. F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 107 and f are unknowns and it is called the inverse problem. If B is not invertible, then the equation is called degenerate otherwise it is called non- degenerate. It was Hille, in 1952 who introduced the Abstract Cauchy Problem. All the work since then was trying to solve the abstract Cauchy problem of the first order or second order. Most of the analysis of the Abstract Cauchy problem was using theory of semigroups of operators. We refer to [1], [2]-[4] , for more on the Abstract Cauchy problem. In this paper, we will use tensor product technique to find what is called atomic solution for the fractional Abstract Cauchy Problem, of the third order. Indeed, we will study the fractional abstract Cauchy problem of the form:{ u(3α)(t) +Au(2α)(t) +Bu(α)(t) + Cu(t) = f(t) u(0) = x0, u(α)(0) = x1, u(2α)(0) = x2. (1) where, A,B,C are closed operators with domain in range of u, and f is a given vector valued function with range in X. Here u(α) denotes the α−conformable derivative of u. Here is the definition of the conformable derivative given in [5]: For g : [0;∞) → R and 0 < α ≤ 1, the conformable fractional derivative of g of order α is defined by Dα(g)(t) = lim ε→0 g(t+ εt1−α)− g(t) ε We often write u(α) for Dαu For all t > 0, if g is α differentiable on (0; b) where b > 0 and limt→0+ g(α)(t) exists, then one can define g(α)(0) = limt→0+ g(α)(t). The α fractional integral of a function f starting from a ≥ 0 is: Iaα(f(t)) = Ia(t1−αf(t)) = ∫ t a f(s) s1−α ds For more on Conformable fractional derivative we refer to [6]-[19]. Now, we need some basic facts from theory of tensor product of Banach spaces. Let X and Y be Banach spaces, X∗ denote the dual of X. For x ∈ X and y ∈ Y define the map x⊗ y : X∗ → Y as: x⊗ y(x∗) = ⟨x, x∗⟩y, for all x∗ ∈ X∗. Clearly, x ⊗ y is a bounded linear operator and ∥ x ⊗ y ∥=∥ x ∥∥ y ∥, [ 18]. Such an operator x ⊗ y is called an atom. The set X ⊗ Y = span{x ⊗ y : x ∈ X and y ∈ Y } is a subspace of L(X∗, Y ). The following lemma, see [20]-[22], is needed in our paper. Theorem 1. Let x1 ⊗ y1 and x2 ⊗ y2 be two nonzero atoms in X ⊗ Y such that x1 ⊗ y1 + x2 ⊗ y2 = x3 ⊗ y3 F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 108 Then either x1, x2 or y1, y2 are linearly dependent. One can easily prove that: Lemma 1. If x1 ⊗ y1 = x2 ⊗ y2 , then x1, x2 are dependent and y1, y2 are dependent too. 2. Main Result Now we are interested in finding atomic solution of problem (1).That is a solution of the form v ⊗ x, with v(t) ∈ R, and x ∈ X. Also, we assume f = g ⊗ z. Substitute in (1) to get: v(3α)(t)⊗ x+ v(2α)(t)⊗Ax+ v(α)(t)⊗Bx+ v(t)⊗ Cx = g(t)⊗ z . (2) We assume the conditions v(0) = 1, v(α)(0) = 1, v(2α)(0) = 1. There are many cases to consider. Once again, the general form of third-order fractional Abstract Cauchy problem is{ u(3α)(t) +Au(2α)(t) +Bu(α)(t) + Cu(t) = h(t) u(0) = x0, u(α)(0) = x1, u(2α)(0) = x2. Where u is 3αdifferentiable function from I to X and A,B, and C are closed linear operator on X. Notice, we use u(3α) to denote DαDαDαu We are interested in finding an atomic solution of equation (1), where the right hand side of the equation is an atom. Now let the atomic solution we are looking for be u(t) = v(t)x = v ⊗ x. In this case we let h(t) = f(t)z = f ⊗ z, f is a scalar valued function. Here v, x are unknowns and f, z are given Let us rewrite (2) in the form v(3α)(t)⊗ x+ v(2α)(t)⊗Ax+ v(α)(t)⊗Bx+ v(t)⊗ Cx = f(t)⊗ z , (∗) v(0) = v(α)(0) = v(2α)(0) = 1 . (∗∗) There are many cases to consider. A. The first Case (a) v(3α) = v(2α) = v(α) = v. or (b) x = Ax = Bx = Cx F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 109 Let us consider the case where v(3α) = v(2α) = v(α) = v . (3) Now we are looking for v that satisfies (1). From (3) we have the following situations: 1. v(3α) = v(2α), 2. v(2α) = v(α), 3. v(α) = v, 4. v(3α) = v(α), 5. v(3α) = v, 6. v(2α) = v, Situation 1. v(3α) = v(2α) By the results in [12], the associated characteristic equation is r2(r − 1) = 0. Hence r = 0, 0, 1 Thus v(t) = c1 + c2 1 α tα + c3 exp 1 α tα From (∗∗), we obtain  c1 + c2 + c3 = 1 c2 + c3 = 1 c3 = 1 . Thus c1 = 0, c2 = 0, c3 = 1. Hence v(t) = exp 1 α tα . (4) Situation 4. v(3α) = v(α). Again using the results in [12], the associated characteristic equation is r(r2 − 1) = 0. Hence r = 0, 1,−1. So v(t) = c1 + c2 exp ( − 1 α tα ) + c3 exp 1 α tα Using the conditions in (∗∗) we get c1 + c2 + c3 = 1 , −c2 + c3 = 1 , c2 + c3 = 1 . Thus, c1 = 0, c2 = 0, and c3 = 1. So v(t) = exp 1 α tα F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 110 Situation 5. v(3α) = v Another use of the result [12], the associated characteristic equation is r3 − 1 = 0, so (r − 1)(r2 + r + 1) = 0. Hence r1 = 1,r2 = −1 + i √ 3 2 and r3 = −1− i √ 3 2 . So v(t) = exp ( − 1 2α tα ) ( c1 cos √ 3tα 2α + c2 sin √ 3tα 2α ) + c3 exp 1 α tα. Using (∗∗) to get  c1 + c3 = 1 −1 2 c1 + c2 + c3 = 1 −3 4c1 − c2 + c3 = 1 , from which we get c1 = 0, c2 = 0, and c3 = 1. Thus v(t) = exp 1 α tα . If we do the other situations we get the same solution: v(t) = exp 1 α t α. Thus if there is an atomic solution for this case, then v must equal to exp 1 α t α. But from Lemma 1, we must have f(t) = exp 1 α t α in order to get an atomic solution. Now, substitute v(t) = exp 1 α t α in (∗ ), we get exp 1 α tα [x+Ax+Bx+ Cx] = f(t)⊗ z . (∗ ∗ ∗) It remains to find x. From equation (∗ ∗ ∗), we have two atoms are equal. Thus the first coordinates are equal and the second coordinates are equal by Lemma 1. Thus we get x+Ax+Bx+ Cx = z . This is (I +A+B + C)x = z . (5) Hence, for the atomic solution to exist we must have x to satisfy (5), noting that z is given. So the image of x under I +A+B + C must be z. Let x = Ax = Bx = Cx = z . (6) So, x = z and it is an eigenvector (a fixed point ) for A,B, and C. Now substitute in equation (∗) to get [ v(3α)(t) + v(2α)(t) + v(α)(t) + v(t) ] ⊗ x = f(t)⊗ z . (7) F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 111 From equation (7), since x = z, we get v(3α) + v(2α) + v(α) + v = f . (8) This is a linear fractional non-homogenous differential equation. Thus using result in [12], we find the homogenous solution vh and a particular solution vp, and so, the general solution will be vg = vh + vp. Now for vh, the associated characteristic equation is (r + 1) ( r2 + 1 ) = 0, which has the roots −1,±i. Then vh(t) = c1 exp ( − 1 α tα ) + c2 cos 1 α tα + c3 sin 1 α tα . For the particular solution, we use variation of parameters introduced in [12]. Let v1 = exp ( − 1 α t α ) , v2 = cos 1 α t α and v3 = sin 1 α t α. So vp(t) = 3∑ m=1 vm(t) ∫ t a f(s)Wα m(s) Wα(s)s1−α ds = exp ( − 1 α tα )∫ t a f(s)Wα 1 (s) Wα(s)s1−α ds+ cos 1 α tα ∫ t a f(s)Wα 2 (s) Wα(s)s1−α ds +sin 1 α tα ∫ t a f(s)Wα 3 (s) Wα(s)s1−α ds where Wα = ∣∣∣∣∣∣∣ v1 v2 v3 v (α) 1 v (α) 2 v (α) 3 v (2α) 1 v (2α) 2 v (2α) 3 ∣∣∣∣∣∣∣ and Wα m is the determinant obtained from Wα by replacing mth column by the column (0, 0, 1)T , m = 1, 2, 3 . Thus vp(t) = exp ( − 1 α tα )∫ t a f(s) 2 exp ( − 1 αs α ) s1−α ds− cos 1 α tα ∫ t a f(s) (cos s+ sin s) 2s1−α ds +sin 1 α tα ∫ t a f(s) (cos s− sin s) 2s1−α ds = exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α ))− cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) +sin 1 α tαIaα ( f(t) (cos s− sin t) 2 ) Hence v(t) = vh(t) + vp(t) = c1 exp ( − 1 α tα ) + c2 cos 1 α tα + c3 sin 1 α tα + exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α )) F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 112 − cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) + sin 1 α tαIaα ( f(t) (cos t− sin t) 2 ) v(α) = −c1 exp ( − 1 α tα ) − c2 sin 1 α tα + c3 cos 1 α tα +exp ( − 1 α tα )[ −Iaα ( f(t) 2 exp ( − 1 α t α ))+ f(t) 2 exp ( − 1 α t α )] +sin 1 α tα [ Iaα ( f(t) (cos t+ sin t) 2 ) + f(t) (cos t− sin t) 2 ] +cos 1 α tα [ Iaα ( f(t) (cos t− sin t) 2 ) − f(t) (cos t+ sin t) 2 ] u(2α) = c1 exp ( − 1 α tα ) − c2 cos 1 α tα − c3 sin 1 α tα +exp ( − 1 α tα )[ Iaα ( f(t) 2 exp ( − 1 α t α ))+ f (α)(t) exp ( − 1 α t α ) + f(t) exp ( − 1 α t α ) 2 exp ( − 2 α t α ) ] − f(t) + cos 1 α tα [ Iaα ( f(t) (cos t+ sin t) 2 ) + f(t) (cos t− sin t) 2 − f (α)(t) (cos t+ sin t) 2 ] − sin 1 α tα [ Iaα ( f(t) (cos t− sin t) 2 ) − f(t) (cos t+ sin t) 2 − f (α)(t) (cos t− sin t) 2 ] But v(0) = 1, v(α)(0) = 1 and v(2α)(0) = 1. Thus c1 + c2 = 1 c3 − c1 = 1 c1 − c2 + Iaα (f(0)) = 1 ⇐⇒  c1 = 1− 1 2I a α (f(0)) c2 = 1 2I a α (f(0)) c3 = 2− 1 2I a α (f(0)) Hence v(t) = ( 1− 1 2 Iaα (f(0)) ) exp ( − 1 α tα ) + 1 2 Iaα (f(0)) cos 1 α tα + ( 2− 1 2 Iaα (f(0)) ) sin 1 α tα + exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α ))− cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) +sin 1 α tαIaα ( f(t) (cos t− sin t) 2 ) Hence, the atomic solution that satisfies the conditions is F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 113 [( 1− 1 2 Iaα (f(0)) ) exp ( − 1 α tα ) + 1 2 Iaα (f(0)) cos 1 α tα + ( 2− 1 2 Iaα (f(0)) ) sin 1 α tα ] ⊗ z + [ exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α ))− cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) +sin 1 α tαIaα ( f(t) (cos t− sin t) 2 )] ⊗ z . All other situations are handled in the same way. B. The second Case.  (i) v(3α)(t)⊗ x+ v(2α)(t)⊗Ax = v1(t)⊗ y1 and (ii) v(α)(t)⊗Bx+ v(t)⊗ Cx = v2(t)⊗ y2 , This has the following situations: Which gives four situations. Situation 1.  v(3α) = v(2α) = v1(t) and v(α) = v = v2(t) Now for v(3α) = v(2α) , the associated characteristic equation is r3 − r2 = 0. Hence using [12], we get v(t) = c1 + c2 tα α + c3 exp 1 α tα Using conditions in (∗∗) we get v(t) = exp 1 α tα For v(α) = v, the associated characteristic equation is r− 1 = 0. Hence using [6], we get v(t) = c exp 1 α tα Using conditions in (∗∗) we get v(t) = exp 1 α tα F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 114 Similarly for v(3α) = v So  x+Ax = y1 and Bx+ Cx = y2 Hence  (I +A)x = y1 and (B + C)x = y2 Now we go to our equation exp 1 α tα ⊗ [x+Ax+Bx+ Cx] = f(t)⊗ z So, f must be equal to exp 1 α t α for the atomic solution to exist and the image of x under [I +A+B + C]x = z. Situation 2.  v(3α) = v(2α) = v1(t) and Bx = Cx = y2 Now for v(3α) = v(2α) = v1(t), then v(t) = exp 1 α tα Substitute in the main equation we get exp 1 α tα ⊗ [x+Ax] + 2 exp 1 α tα ⊗Bx = f(t)⊗ z exp 1 α tα ⊗ [x+Ax+ 2Bx] = f(t)⊗ z Then for the atomic solution to exist, f = exp 1 α t α, and [I +A+ 2B]x = z. Situation 3.  x = Ax = y1 and v(α) = v = v2 Now, v(α) = v gives v(t) = exp 1 α tα So exp 1 α tα ⊗ [x+Ax+Bx+ Cx] = f(t)⊗ z exp 1 α tα ⊗ [2Ax+Bx+ Cx] = f(t)⊗ z F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 115 Hence f must equal exp 1 α t α and [2A+B + C]x = z. Situation 4.  x = Ax = y1 and Bx = Cx = y2 So [ v(3α) + v(2α) ] ⊗ x+ [ v(α) + v ] ⊗Bx = f ⊗ z. Hence we have two subcases: 1.v(3α) + v(2α) = v(α) + v or 2. Bx = x. If Bx = x, then [ v(3α) + v(2α) + v(α) + v ] ⊗ x = f ⊗ z. Hence x = z. Now we solve v(3α) + v(2α) + v(α) + v = f. This is a linear fractional differential equation of order 3α. Using [12], we get the general solution vg = vh+vp, the sum of the homogenous solution and the particular solution. For homogenous solution vh, the associated characteristic equation is r3+r2+r+1 = 0, which has the roots i,−i and −1. Then vh(t) = c1 exp ( − 1 α tα ) + c2 cos 1 α tα + c3 sin 1 α tα For the particular solution, we use variation of parameters introduced in [12]. Let u1 = exp ( − 1 α t α ) , u2 = cos 1 α t α and u3 = sin 1 α t α So vp(t) = 3∑ m=1 um(t) ∫ t a f(s)Wα m(s) Wα(s)s1−α ds = exp ( − 1 α tα )∫ t a f(s)Wα 1 (s) Wα(s)s1−α ds+ cos 1 α tα ∫ t a f(s)Wα 2 (s) Wα(s)s1−α ds+ sin 1 α tα ∫ t a f(s)Wα 3 (s) Wα(s)s1−α ds where Wα = ∣∣∣∣∣∣∣ u1 u2 u3 u (α) 1 u (α) 2 u (α) 3 u (2α) 1 u (2α) 2 u (2α) 3 ∣∣∣∣∣∣∣ and Wα m is the determinant obtained from Wα by replacing mth column by the column (0, 0, 1), m = 1, 2, 3.Thus vp(t) = exp ( − 1 α tα )∫ t a f(s) 2 exp ( − 1 αs α ) s1−α ds− cos 1 α tα ∫ t a f(s) (cos s+ sin s) 2s1−α ds F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 116 +sin 1 α tα ∫ t a f(s) (cos s− sin s) 2s1−α ds = exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α ))− cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) +sin 1 α tαIaα ( f(t) (cos s− sin t) 2 ) Hence v(t) = vh(t) + vp(t) = c1 exp ( − 1 α tα ) + c2 cos 1 α tα + c3 sin 1 α tα +exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α )) − cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) +sin 1 α tαIaα ( f(t) (cos t− sin t) 2 ) v(α)(t) = −c1 exp ( − 1 α tα ) − c2 sin 1 α tα + c3 cos 1 α tα +exp ( − 1 α tα )[ −Iaα ( f(t) 2 exp ( − 1 α t α ))+ f(t) 2 exp ( − 1 α t α )] +sin 1 α tα [ Iaα ( f(t) (cos t+ sin t) 2 ) + f(t) (cos t− sin t) 2 ] +cos 1 α tα [ Iaα ( f(t) (cos t− sin t) 2 ) − f(t) (cos t+ sin t) 2 ] v(2α)(t) = c1 exp ( − 1 α tα ) − c2 cos 1 α tα − c3 sin 1 α tα +exp ( − 1 α tα ) Iaα ( f(t) 2 exp(− 1 α tα) ) + f (α)(t) exp(− 1 α tα)+f(t) exp(− 1 α tα) 2 exp(− 2 α tα)  − f(t) + cos 1 α tα  Iaα ( f(t)(cos t+sin t) 2 ) +f(t)(cos t−sin t) 2 − f (α)(t)(cos t+sin t) 2  F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 117 − sin 1 α tα  Iaα ( f(t)(cos t−sin t) 2 ) −f(t)(cos t+sin t) 2 − f (α)(t)(cos t−sin t) 2  Using conditions in (∗∗) we get c1 + c2 = 1 c3 − c1 = 1 c1 − c2 + Iaα (f(0)) = 1 ⇔  c1 = 1− 1 2I a α (f(0)) c2 = 1 2I a α (f(0)) c3 = 2− 1 2I a α (f(0)) Hence v(t) = ( 1− 1 2 Iaα (f(0)) ) exp ( − 1 α tα ) + 1 2 Iaα (f(0)) cos 1 α tα + ( 2− 1 2 Iaα (f(0)) ) sin 1 α tα + exp ( − 1 α tα ) Iaα ( f(t) 2 exp ( − 1 α t α )) − cos 1 α tαIaα ( f(t) (cos t+ sin t) 2 ) +sin 1 α tαIaα ( f(t) (cos t− sin t) 2 ) and the atomic solution that satisfies the conditions is u(t) = v(t)⊗ z. Now, if v(3α) + v(2α) = v(α) + v = f, then (I +B)x = z We want to solve v(3α) + v(2α) = f and v(α) + v = f For v(3α) + v(2α) = f Using [12], we get the general solution vg = vh + vp, the sum of the homogenous solution and the particular solution. For homogenous solution vh, the associated characteristic equation is r2 (r + 1) = 0, which has the roots 0, 0 and −1. vh(t) = c1 + c2 tα α + c3e −tα α For the particular solution, we use variation of parameters introduced in [12]. Let u1 = 1, u2 = 1 α t α and u3 = e −tα α vp(t) = 3∑ m=1 um(t) ∫ t a f(s)Wα m(s) Wα(s)s1−α ds F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 118 = ∫ t a f(s)Wα 1 (s) Wα(s)s1−α ds+ 1 α tα ∫ t a f(s)Wα 2 (s) Wα(s)s1−α ds+ e −tα α ∫ t a f(s)Wα 3 (s) Wα(s)s1−α ds = − ∫ t a f(s) ( sα α + 1 ) s1−α ds+ tα α ∫ t a f(s) s1−α ds+ e −tα α ∫ t a f(s) e −sα α s1−α ds = −Iaα ( f(t) ( tα α + 1 )) + tα α Iaα (f(t)) + e −tα α Iaα ( f(t) e −tα α ) Hence v(t) = vh(t) + vp(t) = c1 + c2 tα α + c3e −tα α − Iaα ( f(t) ( tα α + 1 )) + tα α Iaα (f(t)) + e −tα α Iaα ( f(t) e −tα α ) For v(α) + v = f By result in [12] the solution of equation (7) is given by v(t) = e− ∫ t 0 xα−1dx + ∫ t 0 ( e− ∫ t s xα−1dxf(s)sα−1 ) ds = e− 1 α tα + ∫ t 0 ( e− 1 α (tα−sα)f(s)sα−1 ) ds But the two solution must be equal, then c3 + Iaα ( f(t) e −tα α ) = 1 and ∫ t 0 ( e− 1 α (tα−sα)f(s)sα−1 ) ds = c1 + (c2 + Iaα (f(t))) tα α − Iaα ( f(t) ( tα α + 1 )) Then v(t) = e− 1 α tα + ∫ t 0 ( e− 1 α (tα−sα)f(s)sα−1 ) ds C. The Third Case. x = Ax and Bx = Cx (a) In this case [ v(3α)(t) + v(2α)(t) ] ⊗ x+ [ v(α)(t) + v(t) ] ⊗Bx = f(t)⊗ z If x = Bx = z , (9) then v(3α)(t) + v(2α)(t) + v(α)(t) + v(t) = f(t) . F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 119 And this can be solved as before in case1. v(3α)(t) + v(2α)(t) = v(α)(t) + v(t) = f(t) . (b) Now v(3α)(t) + v(2α)(t) = v(α)(t) + v(t), the associated characteristic equation is r3 + r2 − r − 1 = 0 which has the roots −1,−1 and 1. Then v(t) = c1e 1 α tα + c2 1 α tαe− 1 α tα + c3e − 1 α tα . v(0) = v(α)(0) = v(2α)(0) = 1 gives c1 + c3 = 1 c1 + c2 − c3 = 1 c1 − 2c2 + c3 = 1 ⇔  c1 = 1 c2 = 0 c3 = 0 Hence v(t) = e 1 α tα . This forces f(t) = 2e 1 α tα , if not then there is no atomic solution. Now in case of (9 ) and from (a) we get x = Ax = Bx = Cx, so x is an eigenvector for A,B,C and x = z. In case (b), we get 2e 1 α tα + 2e 1 α tα = 2e 1 α tαz So x+Bx = z. Hence (I +B)x = z. D. The Forth Case (i) v(3α)(t)⊗ x+ v(t)⊗ Cx = v1(t)⊗ y1 and (ii) v(2α)(t)⊗ Ax+ v(α)(t)⊗Bx = v2(t)⊗ y2 Which gives Situation (1)  v(3α) = v = v1(t) and v(2α) = v(α) = v2(t) For v(3α) = v = v1(t), the associated characteristic equation is r3 − 1 = 0. Which has the roots 1, −1± √ 3i 2 . Hence using [12], we get v(t) = c1 exp 1 α tα + exp ( − 1 2α tα )( c2 cos √ 3 2α tα + c3 sin √ 3 2α tα ) . F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 120 Using conditions (∗∗) to get v(t) = exp 1 α tα . Now for v(2α) = v(α), the associated characteristic equation is r2 − r = 0. Hence using [12], we get v(t) = c1 + c2 exp 1 α tα . By the conditions (∗∗) we get v(t) = exp 1 α tα . Substitute in the main equation exp 1 α tα ⊗ [x+ Ax+Bx+ Cx] = f(t)⊗ z . So, f must be equal to exp 1 α t α for the atomic solution to exist and the image of x under [I + A+B + C]x = z. Situation (2)  v(3α) = v = v1(t) and Ax = Bx = y2 Now for v(3α) = v, we previously found v(t) = exp 1 α tα . Substitute in the main equation we get exp 1 α tα ⊗ x+ exp 1 α tα ⊗ Ax+ exp 1 α tα ⊗Bx+ exp 1 α tα ⊗ Cx = f(t)⊗ z But Ax = Bx, so exp 1 α tα ⊗ (x+ 2Ax+ Cx) = f(t)⊗ z . Then for the atomic solution to exist, f must be equal to exp 1 α t α, and [I + 2A+ C]x = z. Situation (3)  x = Cx = y1 and v(2α) = v(α) = v2 F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 121 Now for v(2α) = v(α), we already have v(t) = exp 1 α tα . Substitute in the main equation we get exp 1 α tα ⊗ x+ exp 1 α tα ⊗Ax+ exp 1 α tα ⊗Bx+ exp 1 α tα ⊗ Cx = f(t)⊗ z Since x = Cx, then exp 1 α tα ⊗ (2x+Ax+Bx) = f(t)⊗ z . Then for the atomic solution to exist, f must be equal to exp 1 α t α, and [2I +A+B]x = z. Situation (4)  x = Cx = y1 and Ax = Bx = y2 So [ v(3α) + v ] ⊗ x+ [ v(2α) + v(α) ] ⊗Ax = f ⊗ z. Then we have two cases v(3α) + v = v(2α) + v(α) = f or x = Ax = z If x = Ax, then [ v(3α) + v(2α) + v(α) + v ] ⊗ x = f ⊗ z . Then v(3α) + v(2α) + v(α) + v = f which already solved in Case 2. If v(3α) + v = v(2α) + v(α) = f , then (I + A)x = z. Now for v(3α) + v = f, the general solution is given by vg = vh + vp, the sum of the homogenous solution and the particular solution. For the homogenous solution vh, the associated characteristic equation is r3 + 1 = 0, which has the roots −1, 1± √ 3i 2 . Then vh(t) = c1 exp ( − 1 α tα ) + exp ( 1 2α tα )( c2 cos √ 3 2α tα + c3 sin √ 3 2α tα ) . For the particular solution, we use variation of parameters. Let u1 = exp ( − 1 α tα ) , u2 = exp ( 1 2α tα ) cos √ 3 2α tα and u3 = exp ( 1 2α tα ) sin √ 3 2α tα. vp(t) = 3∑ m=1 um(t) ∫ t a f(s)Wα m(s) Wα(s)s1−α ds F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 122 = exp ( − 1 α tα )∫ t a f(s)Wα 1 (s) Wα(s)s1−α ds +exp ( 1 2α tα ) cos (√ 3 2α tα )∫ t a f(s)Wα 2 (s) Wα(s)s1−α ds +exp ( 1 2α tα ) sin (√ 3 2α tα )∫ t a f(s)Wα 3 (s) Wα(s)s1−α ds = exp ( − 1 α tα )∫ t a f(s) 3s1−α ds − exp ( 1 2α tα ) cos (√ 3 2α tα ) × ∫ t a f(s) exp ( − 1 2α sα ) [ cos (√ 3 2α sα ) + sin (√ 3 2α sα )] 3 √ 3s1−α ds +exp ( 1 2α tα ) sin (√ 3 2α tα ) × ∫ t a f(s) exp ( − 1 2α sα ) [ cos (√ 3 2α sα ) − √ 3 sin (√ 3 2α sα )] 3 √ 3s1−α ds = exp ( − 1 α tα ) Iaα ( 1 3 f(t) ) − exp ( 1 2α tα ) cos (√ 3 2α tα ) × Iaα ( 1 3 f(t) exp ( − 1 2α tα )[ cos (√ 3 2α tα ) + sin (√ 3 2α tα )]) +exp ( 1 2α tα ) sin (√ 3 2α tα ) Iaα ( 1 3 f(t) exp ( − 1 2α tα ) ×[ cos (√ 3 2α tα ) − √ 3 sin (√ 3 2α tα )]) Hence v(t) = vh + vp = c1 exp ( − 1 α tα ) + exp ( 1 2α tα )( c2 cos √ 3 2α tα + c3 sin √ 3 2α tα ) +exp ( − 1 α tα ) Iaα ( 1 3 f(t) ) − exp ( 1 2α tα ) cos (√ 3 2α tα ) × F. Bekraoui, M. Al Horani, R. Khalil / Eur. J. Pure Appl. Math, 15 (1) (2022), 106-125 123 Iaα ( 1 3 f(t) exp ( − 1 2α tα )[ cos (√ 3 2α tα ) + sin (√ 3 2α tα )]) +exp ( 1 2α tα ) sin (√ 3 2α tα ) × Iaα ( 1 3 f(t) exp ( − 1 2α tα )[ cos (√ 3 2α tα ) − √ 3 sin (√ 3 2α tα )]) . For v(2α)+ v(α) = f, the associated characteristic equation is r2+ r = 0, which has the roots 0 and −1. Then vh(t) = c4 + c5 exp ( − 1 α tα ) . For the particular solution, we use variation of parameters. Let u1 = 1, u2 = exp ( − 1 α t α ) So vp(t) = ∫ t a f(s)Wα 1 (s) Wα(s)s1−α ds+ exp ( − 1 α tα )∫ t a f(s)Wα 2 (s) Wα(s)s1−α ds = ∫ t a f(s) s1−α ds− exp ( − 1 α tα )∫ t a f(s) exp ( 1 α sα ) (s)s1−α ds = Iaα (f(t))− exp ( − 1 α tα ) Iaα ( f(t) exp ( 1 α tα )) Hence v(t) = vh + vp = c4 + c5 exp ( − 1 α tα ) + Iaα (f(t))− exp ( − 1 α tα ) Iaα ( f(t) exp ( 1 α tα )) By the condition (∗∗) we get{ c4 = 2− Iaα (f(0)) c5 = −1 + Iaα (f(0)) . But the two solutions must be equal, then{ c1 = c5 c2 = c3 = 0 . 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