Hilfer-Hadamard fractional differential equations; Existence and Attractivity

This work deals with a class of Hilfer-Hadamard differential equations. Existence and stability of solutions are presented. We use an appropriate fixed point theorem.

Definition 2.1. [22]. Let (c, e) (0 ≤ c < e ≤ ∞) and τ > 0. The Hadamard left-sided fractional integral H I τ c + j of order τ > 0 is defined by When τ = 0, we set When τ = 0, we set Example 2.3. For each τ > 0 and λ ∈ R, we have In particular, when τ = 0 we have In particular, when τ = 0 we have Corollary 2.10. [21] Let σ ∈ C ϱ,log (I). Then the problem admits the following unique solution Lemma 2.11. Let χ : (c, e] × R → R be a function such that χ(·, i(·)) ∈ BC ϱ for any i ∈ BC ϱ . Then the problem (1) is equivalent to the integral equation Let ∅ ̸ = H ⊂ BC * and let T : H → H. Let the equation Definition 2.12. Solutions of equation (4) are locally attractive if there exists a ball B (i 0 , δ) in the space BC * such that, for any solutions w = w(t) and Θ = Θ(t) of equations (4) that belong to (4) is uniformly locally attractive.
Lemma 2.13. [14] Let P ⊂ BC * . Then P is relatively compact in BC * if the following conditions are satisfied: (a) P is uniformly bounded in BC * ; (b) the functions belonging to P are almost equicontinuous in R + , i.e., equicontinuous on every compact set in R + (c) the functions from P are equiconvergent, i.e., given ς > 0, there exists M (ς) > 0 such that for any t ≥ M (ς) and i ∈ P.
Theorem 2.14. (Schauder Fixed-Point Theorem [15]). Let X be a Banach space, let D be a nonempty bounded convex and closed subset of X, and let L : D → D be a compact and continuous map. Then L has at least one fixed point in D. We will give the following hypotheses:

Theorem 3.2. If (H 1 ) and (H 2 ) hold, then (1) has at least one solution which is uniformly locally attractive.
Proof. Define the operator L by We can prove that the operator L maps BC ϱ into BC ϱ . Indeed; the map L(i) is continuous on [c, +∞), and for any i ∈ BC ϱ and, for each t ∈ [c, +∞), we have Therefore, L(i) ∈ BC ϱ , which proves that the operator L (BC ϱ ) ⊂ BC ϱ . Equation (6) implies that L maps Step 1. L is continuous.
Let {i n } n∈N be a sequence converging to i in B R * . Then, Case 1. If t ∈ [c, T ], T > 0, then, since i n → i as n → ∞ and from the continuity of χ, we get Case 2. If t ∈ (T, ∞), T > 0, then (7) implies that Step 2. L (B R * ) is uniformly bounded and equicontinuous.
Since L (B R * ) ⊂ B R * and B R * is bounded, then L (B R * ) is uniformly bounded. Next let t 1 , t 2 ∈ [c, T ], t 1 < t 2 , and let i ∈ B R * . This yields Then, we get Thus, we obtain As t 1 → t 2 , the right-hand side of the inequality tends to zero.
Let t ∈ [c, +∞) and let i ∈ B R * . We have ( log t c Since As a consequence of Steps 1 − 3, we conclude that L : B R * → B R * is compact and continuous. Applying Schauder's fixed point theorem, we get that L has a fixed point i, which is a solution of problem (1) on [c, +∞).
Step 4. Assume that i 0 is solution of (1).
So, we conclude that L is a continuous function such that Moreover, if i is a solution of problem (1), then Therefore, By (9)  Hence, solutions of (1) are uniformly locally attractive.