CAPUTO-HADAMARD APPROACH APPLICATIONS: SOLVABILITY FOR AN INTEGRO-DIFFERENTIAL PROBLEM OF LANE AND EMDEN TYPE

The present paper is dealing with a new direction in the Caputo-Hadamrd approach. It is concerned with the solvability of an integro differential problem of type Lane and Emden. The studied problem involves Caputo-Hadamard derivative with new different fractional orders. The main results of existence of solutions are based on the contraction principle of Banach, however, for the existence of solutions, the use of Scheafer fixed point theorem is applied to prove the result. Three examples are discussed at the end of this work.


INTRODUCTION
In the present paper, we are concentrating on the investigation of the existence and uniqueness of solutions for the following problem of nonlinear fractional differential equation of Lane-Emden singular type: where C H D β , C H D α and C H D σ are the derivatives in the sense of Caputo-Hadamard, I ρ denotes the Hadamard integral of order ρ, with: A, B > 0, J = [1, e], the functions f , g ∈ C J × R 2 , R and h is defined over J. To the best of our knowledge, this is the first time where such problem is investigated.
The structure of our paper is as follows: In Section 2, we will recall some preliminary related to fractional calculus and Caputo-Hadamard derivatives. In Section 3, we apply the integral inequality theory combined with the fixed point theory for study the questions of existence and uniqueness of solutions for the considered problem. In Section 4, three illustrative examples are presented and discussed in details.

PRELIMINARIES ON FRACTIONAL CALCULUS
In this section, we recall the basic definitions, properties and lemmas involving Caputo-Hadamard derivatives, for more details, one can consult the references [15,16,18]. We begin this section by the following definition: provided that the right-hand side integral exists. Now, we recall the following lemmas: Then the Caputo-Hadamard fractional differential equation ,t > a > 0, and the following formula holds where c k ∈ R, k = 0, 1, 2, ..., n − 1.
Before proving our main results, we introduce the following important result. It deals with the integral representation of the above considered problem. We have: Lemma 2.4: Let us take a function G ∈ C(J, R). Therefore, the unique integral solution of the problem is given by Proof: By applying Lemma 2.3, we can reduce (2) to the following equivalent integral problem: for some real constants c 1 , c 2 , c 3 .

MAIN RESULTS
For computational convenience, we need to introduce the following notions: . Now, we define S : E → E as follows: We begin by taking into account the following hypotheses: (H 1 ) : The functions f , g are continuous over J × R 2 and h is continuous over J.
(H 2 ) :There exist nonnegative constants M i and N i , i = 1, 2, such that for all t ∈ J, x i , y i ∈ R : we put (H 3 ) : There exist nonnegative constants L 1 , L 2 , L 3 , such that for all t ∈ J, x i ∈ R, i = 1, 2 : Setting the following quantities: , , , Now, we are in a good position to present our first uniqueness of solution for (1): and (H 2 ) are satisfied and we have also: Then, (1) has a unique solution on J.
Proof: We shall proceed to prove that S is contractive. For x, y ∈ E, we can write which implies that Also, one can see that We have also Thus, we obtain the inequality: We conclude that S is contractive. As a consequence of Banach contraction principle, we deduce that S has a unique fixed point which is the exact solution of (1). Now, we shall use the following lemma to prove the second main result. If one has bounded the set then S has a fixed point in E. Proof: We need the bounded ball: B r = {x ∈ E : x E ≤ r} and we proceed as follows: Step 1: The application S is continuous on E.
The proof is trivial and then it can be omitted.
Step 2: Uniform boundedness: For all x ∈ C r and by (H 3 ) we have Hence, for any x ∈ C r , we obtain Sx E < +∞, which means in particular that the operator S is uniformly bounded on C r .
Step 3: The application S maps bounded sets into equicontinuous sets of E: Let t 1 ,t 2 ∈ J with t 1 < t 2 and let C r be the above bounded set of E, for all x ∈ C r , we have .
Similarly as before, we have .
The right hand sides of (22) and (23) tend to zero independently of (u 1 , u 2 ) as t 1 → t 2 .
As a consequence of Steps 1, 2 and 3, thanks to Ascoli-Arzela theorem, we conclude that S is completely continuous.
Step 4: The set is bounded.
As a consequence of the above Schaefer theorem, we conclude that S has a fixed point which is a solution of (1).
Thus, Theorem 3.1 implies that the problem (26) has a unique solution on [1, e] .