CONTINUABILITY OF SOLUTIONS TO FRACTIONAL DIFFERENTIAL EQUATIONS

. This article concerns the Caputo fractional diﬀerential equation c D αa x [ n − 1] ( t ) = f ( t,x ( t )) + e ( t ) , n ≥ 2 where x [ n − 1] is the quasiderivative of x of order ( n − 1) and c D αa is the Caputo derivative of the order α ∈ (0 , 1). We study the continuability and noncon-tinuability of solutions.

The Caputo derivative given by (1.2) is the special case of Caputo derivative of order α > 0, defined as where m is the smallest integer greater than or equal to α, see e.g.[4,5,7].Fractional differential equations have attract eda great attention in the last two decades because of their importance in applications in areas of physics, chemistry, aerodynamics, etc., see e.g.monographs [4,5,9] and the references therein.
There are a lot of papers devoted to the study of asymptotic behavior of solutions of fractal differential equations, see e.g.[6,7,8,9,10,12].But results of forced fractional differential equations are relatively scarece.Equation (1.1) is studied in [7] (when n = 2 or n = 3 and a 2 ≡ 1) where sufficient conditions for boundedness of all non-oscillatory solutions are given.
A function ).We will suppose that x is nonextendable to the right, i.e., if b < ∞, then x cannot be defined at t = b.Solution x is said to be continuable if b = ∞, otherwise it is said to be noncontinuable.A continuable solution x is said to be proper if it is nontrivial in any neighbourhood of ∞.
In this article we study problem (1.1) with where Let (1.1), (1.4) have a solution x.We investigate whether or not, x is continuable.When α = 1, then (1.1) is the ordinary differential equation (t ≥ a) x [n] (t) = f t, x(t) + e(t) It is known that (1.5) can have noncontinuable solutions, see [2,8].A special case of (1.5) is the equation where Then, by [1,Lemma 4], equation (1.6) has no proper solution.
Some papers only study proper solutions of (1.1) because of their great importance.In this article, we study only the part corresponding to the continuability of solutions to (1.1).However, the methods used here can be applied for other types of Caputo differential equations.
Notation.We denote If i, j ∈ {0, 1, . . .}, i < j and c k ∈ R for i ≤ k ≤ j, then we put i k=j c k = 0.
Lemma 2.2.(i) Let x be a solution of (1.1).Then it is the solution of the nonlinear Volterra type integral equation (t ≥ a) (2.1) is the solution of (1.1) if, and only if it is the solution of (2.1).
(ii) Let a solution x of (1.1) be defined on then it is noncontinuable.If (H5) holds and x is noncontinuable then (2.2) holds and lim t→b− Let (H5) hold and x be a solution of (2.1).Then 2) holds then x is clearly noncontinuable.Let (H5) hold and let x be a noncontinuable solution of (1.1) defined on [a, b), b < ∞.We prove (2.2).So, suppose, on the contrary, that This contradicts the noncontinuability of x proves statement (2.2).
Because of Lemma 2.2(i), we will investigate (2.1) instead of (1.1) without mention it.The proofs of the main results are based on the following lemmas.
The following two lemmas are well known.

Continuable solutions
The first theorem gives a sufficient condition for all solutions of (1.1) be continuable.It is a generalization of well known theorem by Winter and Osgood [8] for differential equations.Theorem 3.1.Suppose (H4) and Then every solution of (1.1) is continuable.
Proof.Suppose, on the contrary, that x is a noncontinuable solution of (1.1) defined on [a, b).Then according to Lemma 2.2(ii), b < ∞ and where M 1 and M 2 are given by (2.8) and M = max a≤s≤b M 2 (s).From this, (3.2) and Lemma 2.6 (with This contradicts (3.1) and proves that x is continuable.
Proof.Let x be a solution of ( Then with respect to (3.7), Now, according to Lemma 2.5 (with t 0 = a, F = M 2 , condition (2.12) follows from (3.5)) we have The contradiction with (3.6) proves that x is continuable.
The following two theorems give us sets of initial conditions for which the solutions are noncontinuable.
Proof.Let β = 1.Consider the auxiliary differential equations for t ∈ [a, t 1 ), y [n] (t) = y [n−1] (t) , r 0 = min a≤t≤t1 r(t) > 0. This equation can be transformed into with Let x be a solution of (1.1) with the initial conditions We denote by I the intervals where both functions y and x are defined.We prove that x [i] (t) > y [i] (t) , t ∈ I , i = 0, . . ., n − 1 .
Let one of the following two assumptions hold: for t ≥ a and Then any solution x of (1.1) satisfying the initial conditions and its solution with the initial conditions This equation is equivalent to the system   If β = −1, the proof is similar.
(ii) The proof is similar, we use Lemma 2.7(iii) instead of Lemma 2.7(ii).
s) −α u (s) ds , Γ(x) = ∞ 0 s x−1 e −s ds , x > 0 (1.2) ) Let [a, b] ⊂ [a, ∞), and AC[a, b] the set of all functions defined on [a, b] that are absolutely continuous on [a, b].Let [a, b) ⊂ [a, ∞).Then we denote by AC loc [a, b) the set of all functions defined on [a, b) that are absolutely continuous on every compact subinterval of [a, b).
Note, by(3.17)and(3.18), condition (2.15) is valid.Now, Lemma 2.7(ii) implies the solutions of (3.21) and (3.23) and of (3.19) and (3.20) are noncontinuable.The rest of the proof is similar as the one of Theorem 3.3; only (3.17) has to be replaced by