ON GENERALIZED SINGULAR FRACTIONAL DIFFERENTIAL INITIAL VALUE PROBLEMS

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. The main object of the present article is the study of an abstract singular fractional differential equation subject to some initial conditions. Under appropriate assumptions on the initial conditions we obtained the explicit solution of the given singular fractional differential equation after decoupling it into two separate equations of different nature. Finally, two concrete examples of singular fractional differential initial value problems are given at the end of this work.


INTRODUCTION
Systems of differential algebraic equations (DAEs) have been used to model a large variety of areas in science and technology such as in multi-body mechanics, chemical engineering, control theory as well as incompressible fluids, see [3,4,10,13]. Thus, the theory of differential algebraic equations has known a remarkable development in the last four decades, and despite the fact that the DAEs are considered somehow as a generalization of Ordinary Differential [2], M. Plekhanova [12], and the references therein.
Let R + := [0, ∞) be the set of nonnegative real numbers, α a positive non integer number and let N = [α] + 1, where [α] is the integral part of α. We are interested in solving explicitly the following singular fractional differential initial value problem with respect to Caputo's fractional derivative in the unknown vector function x (t) : R + → X, namely subject to the initial conditions (2) x (k) (0) = v k , k = 0, 1, . . . , N − 1, where E,A ∈ B (X), so that ker E = {0} (and possibly ker A = {0}), and D α 0 + denotes the (left sided) Caputo's fractional derivative of order α > 0 initiated at 0, v 0 , v 1 , ..., v N are known vectors in X and f is a given absolutely continuous function defined on R + . Unlike the projection operators approach used in the works [6,12], we shall express the solution to problem (1)- (2) in terms of Mittag-Leffler functions and Drazin inverses [8,16] of the operators A and E, when AE = EA. In particular, if these operators are non singular, we obtain the explicit solution of a regular fractional initial value problem. The technique used in our investigation consists in decoupling the operator E into the sum of two operators, one of them is nilpotent, so that the given problem (1)-(2) is equivalent to a certain couple of manageable subproblems. In order to investigate general singular fractional initial value problems when the operators A and E don't necessarily commute we introduce a new notion of regularity that allows solving this type of problems.
The present article is organized as follows: we first start by stating some basic definitions and properties from Fractional Calculus as well as the notion of Drazin inverse of non invertible linear operators in a Banach space. Next, we investigate a certain singular fractional differential initial value problem as we establish the existence and uniqueness of the solution; moreover, we derive the explicit representation of the solution besides some illustrating examples. Finally, we finish our investigation with some concluding remarks.

PRELIMINARIES
Let (X, . ) be a complex Banach space. We denote by B (X) the Banach space of linear bounded operators from X into itself endowed with the norm A op = sup { Ax : x = 1}, for every A ∈ B (X). First of all we give some background regarding Fractional Calculus chiefly the notion of (left-sided) fractional Riemann-Liouville integral of order α > 0 of a function f : R + → X as well as its (left-sided) fractional derivative of order α in the sense of Caputo [5,9,11].
First, we recall the definition of vector-valued absolutely continuous functions over the nonnegative real line R + and taking their values in X; we have Definition 1. A function f : R + → X is said to be absolutely continuous, if for any compact interval J ⊂ R + and, for any ε > 0, there exists a positive real number δ > 0, such that for any finite set of mutually disjoint intervals [a k , b k ] ⊂ J, k = 1, 2, . . . , n, such that Remark 1. We notice that a function f : R + → X is absolutely continuous if and only if there are ϕ ∈ L 1 (R + ; X) and a constant vector c ∈ X such that f (x) = c+ x 0 ϕ(t)dt, x ≥ 0, and from which we get f (x) = ϕ (x), a.e. x ≥ 0. Notation 1. The vector space of all absolutely continuous functions on R + taking their values in X is denoted by AC (R + ; X). Moreover, we shall use the following generalization: If n ∈ N * := {1, 2, 3, ...}, then AC n R + ; X = f : R + → X : f ∈ C n−1 R + ; X and f (n−1) ∈ AC R + ; X .
Let α be a positive real constant, we define the left-sided Riemann-Liouville fractional integral of order α of an integrable function f : R + → X as follows where Γ(α) is the Gamma function, given by Γ (α) = ∞ 0 e −t t α−1 dt, which is a generalization of the factorial of an integer number. Let N = α, if α is an integer number, and N = [α] + 1, if α is not. Next, we define the left-sided Caputo's fractional derivative of order α of f : if α is not an integer.
We point out that the Caputo's fractional derivative of order α of f : R + → X is well defined whenever f ∈ AC N (R + ; X). If we define the convolution product ϕ * ψ of two functions ϕ and ψ by We have these two useful relations: We shall denote in the remaining of the present article the left-sided Caputo's fractional de- We recall the following definition, Definition 2. The index of an operator E ∈ B (X), denoted indE, is the least nonnegative integer m such that ker E m = ker E m+1 and R (E m ) = R E m+1 . In particular, if E is invertible, Moreover, if indE = m < ∞, and R (E m ) is closed, then the unique operator E D ∈ B (X) is called the Drazin inverse of E.
To prove the other relation, let z ∈ R (LA), there is x ∈ X : z = LAx = A (Lx). It follows that z ∈ R (A). Conversely, if z ∈ R (A), then there exists x ∈ X: z = Ax = LA L −1 x , and so, Suppose that indA = m, then m is the least integer number for which we have ker A m = ker A m+1 and R (A m ) = R A m+1 . Since L m and L m+1 are bijective we can apply the previous assertion of this Proposition to {L m , A m } and L m+1 , A m+1 to get We conclude that We need the following decomposition's Theorem of a bounded linear operator, see for instance [16]: Moreover, we have Here are some other interesting properties regarding the Drazin inverse of linear bounded operators, Proof. It suffices to follow the steps of the Proof of Lemma 2.21 [10] which is still valid for bounded linear operators.

Remark 2.
We note that if A, B ∈ B (X) ( not necessarily commuting), then AB is Drazin invertible if and only if BA is Drazin invertible, see [8]. If it is the case, then Furthermore, if AB = BA, then Let us state without proof some facts about the application of Laplace transform to Caputo's fractional derivative initiated at the origin. We have The inverse Laplace transform is formally given by where the integral is carried out along the line c + iy, −∞ < y < +∞, with c > a.
Let α > 0 and N = [α] + 1, if α is non integer and N = α, if α is integer, then the Laplace transform of the Caputo's fractional derivative D α 0 + g of a vector-valued function g ∈ C N (R + ; X) such that g (N) ∈ L 1 (0, T ; X), for every T > 0 and for some constants M > 0 and a > 0, is given by

MAIN RESULTS
Let us first state and solve explicitly some fractional differential equation with a nilpotent operator coefficient. The obtained solution is unique and there is no initial value imposed. We have Lemma 2. Let B, N, L ∈ B (X) such that B is invertible, N a nilpotent operator of index (of nilpotency) m ∈ N * so that BLN = LNB. Then, for any function f : R + → X such that the fractional differential equation has a unique solution given by Proof. Applying B −1 to the both sides of the first equation of (14) we find It is worth to notice that the assumption Expressing equation (16) in term of Q we obtain (17) Qξ Next, applying the operators Q k , k = 1, 2, . . . , m − 1, to equation (17) we get respectively . . .
So that, the unique solution to the fractional differential equation (14) is given by which completes the Lemma's proof.
Our next step is to establish an equivalence between the fractional differential equation (1) and a couple of appropriate fractional differential equations. We have We assume the Drazin inverse E D exists and EA = AE. Then, equation (1) is equivalent to the fractional differential system Moreover, the function y (t) = E D Ex (t) is a solution to the first equation of (18), if and only if, it satisfies the regular fractional differential equation Proof. It is worth to notice that we have E D E 2 = E D E, and so, applying the operator E D E to both sides of equation (1) we obtain at once the following fractional differential equation which is a solution to the equation (18) 1 .
Likewise, noticing that I − E D E 2 = I − E D E , and applying the operator I − E D E to both sides of equation (1) we get Therefore, z (t) satisfies equation (18) 2 . Conversely, if (y (t) , z (t)) satisfies the system (18), then, thanks to the linearity of the fractional derivative, the function x (t) = y (t) + z (t) satisfies To establish the last assertion we notice that y (t) = E D Ex (t) is already a solution to the first equation of (18), and we have Conversely, multiplying (19) by C we obtain Applying the operators A D and E D to the latter equation we get respectively

Let us now state and prove another important result which is
Hence, for any x ∈ X, we have It follows that Bx = 0, for every x ∈ X, and accordingly (21) holds.

b)⇒a): Suppose that (21) holds. Let x ∈ ker E D ∩ ker A D , then
It follows that

Remark 3. i) It is not hard to check the following inclusion by using the property E D = E D E D E and A D = A D A D A,
ii) If E, A ∈ B (X) commute, E D , A D exist, with indE = m < ∞ and indA = k < ∞, and then the relation (21) holds. Indeed, applying respectively E m and A k to the operator B we Reasoning as above we conclude that (21) holds.
Before tackling the general singular fractional differential equation we would like to investigate the homogeneous one, we have is given by is the Mittag-Leffler function of two parameters α, β > 0.
Proof. Define y (t) = E D Ex (t), then Next, applying Laplace transform to the latter equation, we obtain by virtue of the linearity of L , Setting Y (p) = (L y)(p), we infer op , then we get It follows that To simplify the solution's expression we shall put throughout Therefore, We notice that for any constant vectors b 0 , b 1 ,. . . , b N−1 ∈ X the function satisfies the following Hence y (t) is a solution to the homogeneous equation associated with (18) 1 .
Let us now obtain the closed form of the general solution to the equation (22).
Consider the following homogeneous equation associated with (18) 2 (23) Applying N m−1 to both sides of (23) we infer It follows that A D AN m−1 z (t) = 0, and thanks to the assumption (20) and Proposition 3 we by assuming of course that m − 1 > 0. Hence, N m−1 z (t) = 0, and continuing in this manner we arrive at the final result Nz (t) = 0, which in turn implies that Finally, since I − E D E y (t) = E D Ez (t) = 0, then I − E D E z (t) = z (t). It follows by virtue of Proposition 3 that Therefore, the unique solution to the differential equation (23) is the null one. Accordingly, the general solution to the singular fractional differential equation (23) is We are now in the position to establish the existence and uniqueness of the solution to the singular fractional differential initial value problem (1)-(2). We have Theorem 3. Let E, A ∈ B (X) with ker E = {0} so that AE = EA and indE = m. We assume that E and A possess bounded Drazin inverses E D and A D and both satisfy condition (20). Let f ∈ C N (R + ; X) so that T α,α * f is integrable, the composite Caputo's fractional derivative for j = 0, 1, . . . , N − 1, for some constant vectors b j , j = 0, 1, . . . , N −1, then the unique solution x (t) to problem (1)- (2) has the closed form Proof. Applying the Laplace transform L to the equation (19), we obtain by virtue of the linearity of L , Setting Y (p) = (L y)(p) and F (p) = L ( f )(p), the latter equation becomes Assuming that |p| > E D A 1/α op we obtain It follows that Therefore, for some constant vectors b j , j = 0, 1, . . . , N − 1,.
Next, to solve explicitly the fractional differential equation (18) 2 we apply the operator A D to both sides of the equation to get by virtue of Proposition 3, Next, applying Lemma 2, for B = I and L = A D , one gets the unique solution of the latter equation which is Since we have N = E I − E D E and I − E D E i = I − E D E , for i = 1, 2, . . . , m − 1, then Summing up the solutions of the above subproblems y (t) and z (t) we obtain the unique solution to the singular fractional differential initial value problem (1)-(2), that is Let us now check the given initial values. Using the derivation rule regarding integrals depending upon a certain real parameter, we get, for j = 1, . . . , N − 1, the following so that, letting t → 0 + , we obtain Regarding the uniqueness of the solution (under assumption (24)), it suffices to cope with the homogeneous problem whose solution is identically zero, and accordingly the uniqueness follows.
Remark 4. We point out that if f ≡ 0, then the compatibility assumption (24) reduces merely to v j = E D Ev j , for j = 0, 1, . . . , N − 1. Moreover, if E is nonsingular, then E D E = I, and once again, assumption (24) becomes v j = E D Ev j , for j = 0, 1, . . . , N − 1. Whence, we obtain as a unique solution in such a case as expected the function

ILLUSTRATING EXAMPLES
In order to illustrate the obtained results we consider the following examples: Example 1. Consider the following singular fractional differential initial value problem in R 4 : and f (t) = t 2 , t, 0, −t T . We notice that E and A are singular matrices whose Drazin inverses are Hence, the explicit representation of the solution is given by Therefore, the closed form of the solution to the given problem is Our second example deals this time with a singular fractional differential initial value problem in an infinite dimensional space, namely the Banach space

CONCLUSION
Combining the theory of Fractional Calculus and the theory of singular differential equations we have been able to establish the existence and uniqueness of the singular fractional differential initial value problem (1)

ACKNOWLEDGMENT
The first author would like to thank Prof. P. Kunkel for his valuable comments and constant help during her internship at Leipzig University.

CONFLICT OF INTERESTS
The author(s) declare that there is no conflict of interests.