RANK TWO SOLUTIONS OF THE ABSTRACT CAUCHY PROBLEM

In this paper we discuss solutions of the Abstract Cauchy Problem Bu(α) (t) = Au(t)+ f (t) ..........................(E) u(0) = x0 that are of the form u = u1⊗δ 1 +u2⊗δ 2, where u and f are functions defined on [0,a] with values in the Hilbert space `2.

where A and B are densely defined linear operators on the Banach space X , u ∈ C 1 (I, X) , f ∈ C (I, X) and x 0 ∈ X.
In this paper we study the non-homogeneous α−Abstract Cauchy Problem using the tensor product technique when the non-homogeneous part of the problem is a two rank function.In other words f in Problem (E) is equal to f 1 ⊗ δ 1 + f 2 ⊗ δ 2 , where f 1 , f 1 are real-valued continuous functions on I and δ 1 , δ 1 are two orthogonal unit vectors in 2 , the Banach space consisting of the square summable sequences.We did prove the existence of a unique solution for this type of problems when u has the form u 1 ⊗ δ 1 + u 2 ⊗ δ 2 , where u 1 , u 2 are real-valued continuously α−differentiable functions on I, and with some conditions on the operators A and B.

Basic Facts On Conformable Fractional Derivatives
There are many definitions available in the literature for fractional derivatives.The main ones are the Riemann Liouville definition and the Caputo definition, see [8] .
Such definitions have many setbacks such as (i) The Riemann-Liouville derivative does not satisfy D α a (1) = 0 (D α a (1) = 0 for the Caputo derivative), if α is not a natural number.
(ii) All fractional derivatives do not satisfy the known formula of the derivative of the product of two functions: (iii) All fractional derivatives do not satisfy the known formula of the derivative of the quotient of two functions: .
(iv) All fractional derivatives do not satisfy the chain rule: (v) All fractional derivatives do not satisfy: (vi) All fractional derivatives, specially Caputo definition, assumes that the function f is differentiable.
We refer the reader to [3] for more results on Caputo and Riemann -Liouville Definitions.
Recently, the authors in [ 2 ], gave a new definition of fractional derivative which is a natural extension to the usual first derivative.So many papers since then were written, and many equations were solved using such definition.The definition goes as follows: T α is called the conformable fractional derivative of f of order α.
Further, many functions behave as in the usual derivative.Here are some formulas We refere to [ ] for more on fractional derivative.

Two Rank Solution
Let u be an α− differentiable function on [0, b] with values in the Hilbert space 2   Consider the problem where A and B are densely defined linear operators on the Banach space In this section, we study the problem Where we assume that

and we
assume A and B are densely defined closed operators on 2 .
One of our main results is the following: , where a i j = Aδ j , δ i , i, j = 1, 2.
Taking the inner product of δ 1 and δ 2 to both sides of (1), we get And u (α) Since {δ 1 , δ 2 } is an orthonormal set, we get from equations (2) and (3) : And Now, equations (4) and (5) represent a non-homogeneous system of two linear α−differential equations where The general solution of system (6) U g is the sum of the homogeneous solution and a particular solution.The details are as follows: Now, the corresponding homogeneous system of (6) is For such A we have the following cases: Case 1: A has distinct real eigenvalues (i.e λ 1 = λ 2 ).Then the general solution of system where ξ 1 and ξ 2 are the corresponding eigenvectors of λ 1 and λ 2 , respectively.
Case 2: A has equal eigenvalues (i.e λ 1 = λ 2 = λ ), then we have the following sub-cases: Case 2.1: λ has two linearly independent eigenvectors ξ 1 and ξ 2 .Then the general solution of system (7) is given by Case 2.2: λ has a single linearly independent eigenvector ξ .Then the general solution of system (7) is given by where η satisfies equation A − λ I η = ξ and I is the identity matrix.Then the general solution of system (7) is given by Now, to get a particular solution, form the matrix Ψ (t) = (e λ 1 t ξ 1 : e λ 2 t ξ 2 ) which is known as the fundamental matrix of the system.From [10] it is known that the inverse of Ψ exists and a particular solution to system (6) is given by the formula: where J α ( f ) is defined by x (1−α) dx, and the integral is the usual Rimann improper integral, and α ∈ (0, 1).Furthermore, for any of the above cases the general solution of system (6) is of the form Where U h (t) is the general solution of the corresponding homogeneous system and U p (t) is the particular solution of the system.By the initial condition u (0) = x 0 , we have Since ξ 1 and ξ 2 are linearly independent eigenvectors, then the matrix (ξ 1 : ξ 2 ) is invertible.
Multiplying (ξ 1 : ξ 2 ) −1 to both sides, we get Taking the inner product of δ 1 and δ 2 to both sides, we get So, the Problem has a unique solution.
By equating the coefficients of δ 1 , δ 2 , θ 1 and θ 2 in both sides, we have a 13 u 1 (t) = 0 and a 23 u 2 (t) = 0 So, we have the following cases: Case (i) : a 13 = 0 and a 23 = 0.This case contradicts the assumption on a 13 and a 23 .
Then equation (6) becomes By taking the inner product of δ i to both sides of (9), we have Equation ( 10) is first order linear α−differential equation, and has a general solution of the form And by the initial condition u (0) = x 0 , we have Taking the inner product of δ i to both sides, we have And hence the Problem has a unique solution.
Proof : Let D = diag (λ 1 , λ 2 ) be the matrix representation of B 2 with respect to {θ 1 , θ 2 } .Now, if λ 1 = 0 and λ 2 = 0, then B 2 is invertible and we can use Theorem 5.1, so the problem has a unique solution. Assume Hence, Since Bθ 1 = λ 1 θ 1 and Bθ 2 = 0, we have Taking the inner product of θ 1 and θ 2 with both sides of (5.11), we get where Then (15) is a first order linear α−differential equation and it has a general solution of the form where the integral is the usual Rimann improper integral, and α ∈ (0, 1).
Substituting (16) in (14) , we get From (16), (17) and the initial condition u (0) = x 0 , we can determine constant c uniquely as follows By taking the inner product of θ 1 with both sides, we get c = x 0 , θ 1 Thus, Problem (E 2 ) has a unique solution.
Now, by the assumption on A , without loss of generality, we can assume that Aθ 2 , δ 1 = 0.
Taking the inner product of δ 1 to both sides of equation (11) , we get