On a fractional Caputo–Hadamard problem with boundary value conditions via different orders of the Hadamard fractional operators

We investigate the existence of solutions for a Caputo–Hadamard fractional integro-differential equation with boundary value conditions involving the Hadamard fractional operators via different orders. By using the Krasnoselskii’s fixed point theorem, the Leray–Schauder nonlinear alternative, and the Banach contraction principle, we prove our main results. Also, we provide three examples to illustrate our main results.


Introduction
In recent decades, it has become clear to researchers that studying different types of fractional differential equations is of particular importance. This is a tool to complete our modeling information.
In fact, some practical instances done in the framework of the concepts and notions of the fractional calculus show us the power of this branch of mathematics in the modeling of different natural phenomena. In the meantime, fractional differential equations and inclusions of different types play an important role to reach desired practical goals. More precisely, in recent years, some researches invoked these fractional equations to model some processes and patterns via newly defined fractional operators (see, for example, [1][2][3][4]). The techniques used in these initial value problems are based on the analytical and the existence methods. In the following, some researchers designed new fractional models and investigated them via numerical techniques (see, for example, [5][6][7][8][9][10][11][12][13][14]). Therefore, the fractional calculus has been created a powerful tool for researchers to achieve more exact findings in other applied sciences. Also for further study, notice that a lot of works about different types of fractional integro-differential equations have been published (see, for example, ), q-difference equations (see, for example, [46][47][48]), integro-differential equations involving the Caputo-Fabrizio or the Caputo-Hadamard derivatives (see, for example, [49][50][51]), hybrid equations (see, for example, [52]), approximate solutions of different fractional equations (see, for example, [53,54]), and modern models (see, for example, [55]).
It should also be noted that boundary value conditions given in this paper are general and cover many different special cases. This new type of the modeling is an abstract idea and can include various existing natural processes in the future studies. Therefore, the main purpose of this manuscript is to focus on the existence results and provide some necessary conditions for the analytical investigation and so the practical aspects of boundary value problem (1)-(2) is not our main desire here. To reach our main aim, we apply three different fixed point theorems to establish the existence and uniqueness results. These analytical results guarantee the convergence of the numerical methods to desired solution with the least error, and so this can be a reliable criterion for modeling real processes.
The rest of the paper is arranged by follows. In the next section, we recall some basic notions and definitions which are necessary in the sequel. In Sect. 3, our main existence results are presented by three different analytical techniques such as the Krasnoselskii's fixed point theorem, the Leray-Schauder nonlinear alternative, and the Banach contraction principle. In Sect. 4, we examine the validity of our theoretical findings by providing three illustrative examples. In Sect. 5, the conclusion is stated.

Preliminaries
In this section, we recall some important and basic definitions on the fractional operators.
provided that the right-hand side integral exists.
Note that the semigroup property is satisfied by the Hadamard fractional integral as follows: H I a + H I a + w(t) = H I + a + w(t) for , ∈ R + . Also, we have for , ≥ 0 and t > a [58,59]. It is clear that H I a + 1 = 1 Γ ( +1) (ln t a ) for all t > a by putting = 0 [59].
provided that the right-hand side integral exists.
whenever the right-hand side integral exists.
Assume that w ∈ AC n θ ([a, b], R) and n -1 < ≤ n. It has been proved that the solution of the Caputo-Hadamard fractional differential equation ( CH D a + w)(t) = 0 is in the form w(t) = n-1 k=0 c k (ln t a ) k , and we have for all t > a [58,59]. We need the following results.
Lemma 4 (Krasnoselskii's, [60]) Let M be a closed, bounded, convex, and nonempty subset of a Banach space E. Consider two operators Υ 1 and Υ 2 from M into E such that

Lemma 5 ([61]
) Let E be a Banach space, C a closed, convex subset of E, U an open subset of C, and 0 ∈ U . Suppose that Υ : U → C is a continuous and compact map (that is, Υ (U ) is a relatively compact subset of C). Then Υ has a fixed point in U or there is a w ∈ ∂U (the boundary of U in C) and λ ∈ (0, 1) with w = λΥ (w).

Main results
Here, we are ready to prove our main results. We first characterize the structure of the solutions of the problem (1)- (2). Consider the Banach space E = {w : w(t) ∈ C( [1, e], R)} with the norm w E = sup t∈ [1,e] |w(t)|. We first provide our key lemma.

Lemma 7 Let φ(t) ∈ E. Then w 0 is a solution for the Caputo-Hadamard problem
if and only if w 0 is a solution for the fractional integral equation Proof Let w 0 be a solution for the Caputo-Hadamard problem (3). Then, we have By using the Hadamard fractional integral of order , we obtain where b 0 , b 1 , b 2 , and b 3 are some real constants. Hence, Now by using the boundary value conditions and properties of the Hadamard and Caputo-Hadamard fractional operators, we get and By using two first boundary conditions, we obtain b 0 = b 1 = 0. By using two other boundary conditions, we obtain (5), we see that w 0 is a solution for the integral equation. For the converse part, by using some direct calculations, one can see that w 0 is a solution for the Caputo-Hadamard problem (3) whenever w 0 is a solution for the integral equation (4). This completes the proof. Now, consider the operator Υ : E → E defined by
Proof Let σ := sup t∈ [1,e] |σ (t)| and O := sup t∈ [1,e] |ψ(t, 0)|. Consider the operator Υ : E → E and the set V r := {w ∈ E : w E ≤ r} which is a closed, convex, and bounded nonempty subset of Banach space E, where r ≥ σ K * 2 +OK * 1 1-(K * 0 +LK * 1 ) and K * 1 and K * 2 are given by (7). Note that each fixed point of Υ is a solution for the Caputo-Hadamard problem (1)- (2). Let t ∈ [1, e] be given. Then, we have Thus, we have for all w 1 , w 2 ∈ V r . Hence, Υ 1 w 1 + Υ 2 w 2 ≤ r and so Υ 1 w 1 + Υ 2 w 2 ∈ V r for all w 1 , w 2 ∈ V r . Now let {w n } n≥1 be a sequence in V r with w n → w and t ∈ [1, e]. Then, we have Since ϕ is continuous, Υ 2 w n -Υ 2 w → 0, and so the operator Υ 2 is continuous on the open ball V r . Now, we show that Υ 2 is uniformly bounded. Let w ∈ V r and t ∈ [1, e]. Then, we get which implies that Υ 2 w ≤ K * 2 σ . This shows that Υ 2 is uniformly bounded. Here, we prove that Υ 2 is equicontinuous. Let t 1 , t 2 ∈ [1, e] with t 1 < t 2 . We show that Υ 2 maps bounded sets into equicontinuous sets. For each w ∈ V r , we have Note that the right-hand side is independent of w ∈ V r and converges to zero as t 1 → t 2 . This means that Υ 2 is equicontinuous. Consequently, the operator Υ 2 is relatively compact on V r and, by using the Arzela-Ascoli theorem, we conclude that Υ 2 is completely continuous. Hence, Υ 2 is compact on the open ball V r . Now, we show that Υ 1 is a contraction. Let w 1 , w 2 ∈ V r and t ∈ [1, e]. Then, we have Since K * 0 + LK * 1 < 1, Υ 1 is a contraction. Note that Υ = Υ 1 + Υ 2 . Now, by using Lemma 4, the operator Υ has a fixed point which is a solution for the Caputo-Hadamard boundary value problem (1)- (2).
Here, we are going to investigate the existence of solutions for the Caputo-Hadamard problem (1)-(2) by considering different conditions.  (7). Then the Caputo-Hadamard problem (1)-(2) has at least one solution.
Proof We first show that the operator Υ maps bounded sets of E into bounded sets. Let > 0, B = {w ∈ E : w ≤ } and t ∈ [1, e]. Then, we have Hence, Υ w ≤ K * 0 + K * 1 θ 1 ξ 1 ( ) + K * 2 θ 2 ξ 1 ( ). Now, we prove that Υ maps bounded sets into equicontinuous sets of E. Let t 1 , t 2 ∈ [1, e] with t 1 < t 2 and w ∈ B . Then, we get Note that the right-hand side tends to zero independently of w ∈ B as t 2 → t 1 . By using the Arzela-Ascoli theorem, we deduce that Υ : E → E is completely continuous. Here, we prove that the set of all solutions of the equation w = λ(Υ w) is bounded for each λ ∈ [0, 1]. Let λ ∈ [0, 1], w be such that w = λ(Υ w) and t ∈ [1, e]. Then by using computations used in the first step, we obtain w ≤ K * 0 w + K * 1 θ 1 ξ 1 ( w ) + K * 2 θ 2 ξ 2 ( w ). Thus, we conclude that By using the assumption (N 4 ), we can choose a number Ξ > 0 such that w = Ξ and Note that the operator Υ : U → E is continuous and completely continuous and also we can not find w ∈ ∂U such that w = λ(Υ w) holds for some λ ∈ (0, 1). Now, by using Lemma 5, the operator Υ has a fixed point in U which is a solution for the Caputo-Hadamard fractional integro-differential boundary value problem (1)- (2). Now by using the Banach contraction principle, we review the Caputo-Hadamard problem (1)-(2) under some different conditions. Theorem 10 Suppose that ψ : [1, e] × R → R is a continuous function satisfying assumption (N 1 ). Assume that the function ϕ : [1, e] × R → R satisfies the following condition: there is a positive constantL such that for each |ϕ(t, w 1 )ϕ(t, w 2 )| ≤L|w 1w 2 | for all w 1 , w 2 ∈ R and t ∈ [1, e].
Next example illustrates Theorem 9.
Next example illustrates Theorem 10.

Conclusions
It is known that we should increase our ability for studying of different types of fractional integro-differential equations. In this case, we could create modern software in the future by using advanced modelings of distinct phenomena. In this way, we should try to review different types of fractional integro-differential equations. In this work, we study the existence of solutions for a Caputo-Hadamard fractional integro-differential equation with boundary value conditions involving the Hadamard fractional operators via different orders. Also, we provide three examples to illustrate our main results.