Positive Solutions for a Hadamard Fractional p-Laplacian Three-Point Boundary Value Problem

This article is to study a three-point boundary value problem of Hadamard fractional p-Laplacian differential equation. When our nonlinearity grows (p − 1)-superlinearly and (p− 1)-sublinearly, the existence of positive solutions is obtained via fixed point index. Moreover, using an increasing operator fixed-point theorem, the uniqueness of positive solutions and uniform convergence sequences are also established.


Preliminaries
In this paper, we only provide the definition for the Hadamard fractional derivative; for more details about Hadamard fractional calculus, see the book [73]. In what follows, we calculate the Green's functions associated with (1). We let ϕ p (D β u(t)) = −v(t) for t ∈ [1, e]. Then, from (1) we obtain Lemma 1. The boundary value problem (10) takes the form where Proof. We use ideas in Lemma 2 of [59]. For some c i ∈ R(i = 1, 2), we have Substituting e into the above equation, and using u(e) = 0, we obtain Then, This completes the proof.
Lemma 2. The boundary value problem (12) is equivalent to the integral equation where Proof. We follow the ideas in Lemma 1. For some c i ∈ R(i = 1, 2, 3), we have Then, u(1) = u (1) = 0 implies c 2 = c 3 = 0. Consequently, we have Substituting e, ξ into the above equation, and using u(e) = au(ξ), we obtain Solving this equation, we have As a result, we obtain This completes the proof.
Note v(t) = e 1 G α (t, s) f (s, u(s)) ds s , t ∈ [1, e], and we have that (1) is equivalent to the Hammerstein type integral equation Then, (E, · ) is a real Banach space and P a cone on E. From (15), we define an operator A : E → E as follows: Note that our functions G α , G β , f are continuous, so the operator A is a completely continuous operator. Moreover, if there is a u ∈ E is a fixed point of A, then from Lemmas 1-2, we have that u is a solution for (1). Therefore, in what follows, we turn to study the existence of fixed points of the operator A.
For convenience, we define three positive constants . Then, we have the following two integral inequalities This is a direct result from Lemma 4(I2), so we omit the details.
Lemma 6 (see [74] (Lemma 2.6)). Let θ > 0 and ϕ ∈ P. Then, Lemma 7 (see [75]). Let E be a real Banach space and P a cone on E. Suppose that Ω ⊂ E is a bounded open set and that A : Ω ∩ P → P is a continuous compact operator. If there exists a ω 0 ∈ P\{0} such that then i(A, Ω ∩ P, P) = 0, where i denotes the fixed point index on P.
Lemma 8 (see [75]). Let E be a real Banach space and P a cone on E. Suppose that Ω ⊂ E is a bounded open set with 0 ∈ Ω and that A : Ω ∩ P → P is a continuous compact operator.
Lemma 9 (see [75]). Let E be a partially ordered Banach space, and x 0 , y 0 ∈ E with x 0 ≤ y 0 , D = [x 0 , y 0 ]. Suppose that A : D → E satisfies the following conditions: (i) A is an increasing operator; (ii) x 0 ≤ Ax 0 , y 0 ≥ Ay 0 , i.e., x 0 and y 0 is a subsolution and a supersolution of A; (iii) A is a completely continuous operator. Then, A has the smallest fixed point x * and the largest fixed point y * in [x 0 , y 0 ], respectively. Moreover, x * = lim n→∞ A n x 0 and y * = lim n→∞ A n y 0 .

Positive Solutions for (1)
For convenience, let First, we list assumptions for our nonlinearity f : Let Then, we have the following lemma.
Proof. Let S 1 = {u ∈ P : u = Au + λψ, ∀λ ≥ 0}, where ψ ∈ P 0 is a fixed element. We prove that S 1 is bounded in P. If u ∈ S 1 , then, from Lemma 10, we have u ∈ P 0 , and u(t) ≥ (Au)(t) for t ∈ [1, e]. Now, we consider two cases. Case 1. Let p ≥ 2. Then, we have 1 p−1 ∈ (0, 1]. From (H2), we have Consequently, from (19) and Lemma 6, we obtain Multiplying by µ(t) on both sides of (21) and integrating over [1, e], we obtain Solving this inequality, we have Note that, for u ∈ P 0 , we get , by (H2), (19) and Lemma 6 we have Multiplying by µ(t) on both sides of (22) and integrating over [1, e], we conclude that Solving this inequality, we have Noting that u ∈ P 0 , we have The above two cases imply that S 1 is bounded in P. Then, we can choose such that u = Au + λψ, for u ∈ ∂B R 1 ∩ P, ∀λ ≥ 0.
Combining the above two cases, we have that (24) holds. Then, from Lemma 8, we obtain i(A, B r 1 ∩ P, P) = 1.
Note that we can also take R 1 > r 1 such that (23) is still true. Thus, from (23) and (27), we have and hence A has at least one fixed point in (B R 1 \B r 1 ) ∩ P, i.e., (1) has at least one positive solution. This completes the proof. Proof. We can use similar methods as in Theorem 1 to provide the proof. We first prove that where ψ ∈ P is a given element, and r 2 is defined in (H4). Otherwise, there exist u ∈ ∂B r 2 ∩ P and λ ≥ 0 such that u = Au + λ ψ, and thus u(t) ≥ (Au)(t), for t ∈ [1, e]. Now, we consider two cases.
As a result, we have that (28) holds, and Lemma 7 implies that i(A, B r 2 ∩ P, P) = 0.

Case 2.
Let p ∈ (1, 2]. Then, we have 1 p−1 ≥ 1. Using (26) and (H5), we obtain Multiplying by µ(t) on both sides of the preceding inequalities and integrating over [1, e], we find Solving this inequality, we have Noting that u ∈ P 0 , we have Combining the above two cases, we have proved that S 2 is bounded in P. Then, we can choose R 2 > r 2 and Then, from Lemma 8, we have i(A, B R 2 ∩ P, P) = 1.
Thus, from (29) and (31), we have and hence A has at least one fixed point in (B R 2 \B r 2 ) ∩ P, i.e., (1) has at least one positive solution. This completes the proof.
In what follows, we consider the uniqueness of positive solutions for (1) with the boundary conditions D β u(1) = D β u(e) = 0, u(1) = u (1) = u(e) = 0. This problem is equivalent to the Hammerstein type integral equation where G β (t, s) = G 1β (t, s) for t, s ∈ [1, e]. Note that here we still use the operator A as in (16).
Lemma 11. Let w 0 (t) = Proof. We first calculate w 0 . From (14), we have Using (17) and (18)  Proof. Note that (H7) implies that A is an increasing operator, and 0 isn't a fixed point for A. Next, we shall prove that A has a subsolution and a supersolution. Let From Lemma 11, there exist a ρ > 0, b ρ > 0 such that Take and Aξ 1 ≥ ξ 1 , i.e., ξ 1 is a subsolution of A.
As a result, from Lemma 9, A has the smallest fixed point u * and the largest fixed point u * in [ξ 1 , ξ 2 ], respectively. Moreover, u * = lim n→∞ A n ξ 1 and u * = lim n→∞ A n ξ 2 .

Conclusions
In this paper we investigate the existence and uniqueness of positive solutions for the Hadamard fractional p-Laplacian three-point boundary value problem (1). We first establish some relations from the corresponding problem without the p-Laplacian operator, and use some (p − 1)−superlinearly and (p − 1)−sublinearly conditions for the nonlinearity to obtain positive solutions to problem (1). After, using an increasing operator fixed-point theorem, we obtain the unique solution to problem (1), and establish uniform converged sequences for this solution.