Positive Solution for the Nonlinear Hadamard Type Fractional Differential Equation with p-Laplacian

We study the following nonlinear fractional differential equation involving the p-Laplacian operator Dβ(φp(D α u(t))) = f(t, u(t)), 1 < t < e, u(1) = u(1) = u(e) = 0, Du(1) = Du(e) = 0, where the continuous function f : [1, e] × [0, +∞) → [0, +∞), 2 < α ≤ 3, 1 < β ≤ 2. D denotes the standard Hadamard fractional derivative of the order α, the constant p > 1, and the p-Laplacian operator φp(s) = |s| p−2 s. We show some results about the existence and the uniqueness of the positive solution by using fixed point theorems and the properties of Green’s function and the p-Laplacian operator.

By the use of the fixed point theorem on cones, Chai in [20] obtained the existence and multiplicity of positive solutions for a class of boundary value problem of fractional differential equation with -Laplacian operator: where 0 <  ≤ 1, 2 <  ≤ 3, 0 ≤  < 1, 0 ≤  ≤ 1,  > 0 is a parameter, and   0+ ,   0+ are the standard Caputo fractional derivatives.By the properties of Green function and Schauder fixed point theorem, several existence and nonexistence results and the uniqueness of positive solutions are acquired.
Lu et al. in [24] considered the following fractional boundary value problem with -Laplacian operator: where 2 <  ≤ 3, 1 <  ≤ 2, and   0+ ,   0+ are the standard Riemann-Liouville fractional derivatives.By the properties of Green's function, the Guo-Krasnosel'skii fixed point theorem, the Leggett-Williams fixed point theorem, and the upper and lower solutions method, some new results on the existence of positive solutions are gained.
Definition 1 (see [1,Page 111]).The th Hadamard fractional order derivative of a function  : [1, +∞)  → R is defined by where  > 0,  = [] + 1, and [] denotes the largest integer which is less than or equal to .Correspondingly, the th Hadamard fractional order integral of  : [1, +∞)  → R is defined by where Γ is the gamma function.
To the best of our knowledge, there are few contributions to the Hadamard type with -Laplacian operator; we fill the gap in this paper.In fact, we will discuss the existence and the uniqueness of the positive solutions of (5).The structure of this paper goes on as follows.In Section 2, we will introduce some basic lemmas that will be used.In Section 3, we first give some existence results including Theorems 10 and 11, Corollary 12, and Theorem 13.Then, we will prove Theorems 14 and 15 which reveal the uniqueness of the solution.In Section 4, we give two examples to illustrate our results.

Preliminary Results
In this section, we will first recall the following preliminary facts that will be used in our main results.
Next, we give several lemmas which will be applied in the proofs of our main results.Lemma 6.Let () be the solution of the problem (5); then it can be described as below: where Proof.Putting () = (, ()), we have   (  (  ())) = ().By Lemma 2 and the fact that The boundary value hypotheses give   (1) =   () = 0.So we can get that Therefore, Notice the fact that  −1  =   , 1/ + 1/ = 1; we have Putting This, combined with the fact that (1) =   (1) =   () = 0, yields Thus, where This completes the proof of Lemma 6.
Let (28) Evidently, the solutions of boundary value problem (5) are the corresponding fixed points of the operator .
Lemma 8. Suppose that  :  →  is an operator as above; then  :  →  is completely continuous.
Next, the continuity of (, ) implies that, for any  > 0, there exists a constant  such that, for any That is, Ω is equicontinuity.By the means of Arzela-Ascoli theorem [26], we have that  :  →  is completely continuous.This completes the proof of Lemma 8.
In the final part of this section, we list the following basic properties of the -Laplacian operator.

Proofs of the Main Results
In this section, first, we consider the existence of the solutions of problem (5).
Proof.The proof of Corollary 12 is similar to the one of Theorem 11.So we omit the detail.
Theorem 13.Suppose that (, ) : [1, ]×[0, +∞) is a positive continuous function and there exists a constant  > 0 such that Then the boundary value problem (5) has at least one positive solution. Proof.
From Lemma 8, we know  :  →  is completely continuous.Assume that there exist  ∈ ,  ∈ (0, 1) such that  = .Then we have Thus, By (43), we can imply that ‖‖ < , which means that  ∉ .That is to say, there is no  ∈  such that  =  for some  ∈ (0, 1).Therefore, by Lemma 5, we conclude that the problem ( 5) has at least one positive solution.This completes the proof of Theorem 13.Now we turn to the uniqueness of solution for boundary value problem (5). where By the third hypothesis, 0 <  1 < 1, which implies that () = V().And this completes the proof of Theorem 14.
By using the same way, we can prove the last one of our main uniqueness results.

Examples
In this section we give several examples to illustrate our main results.
Example 16.Consider the boundary value problem: Then the boundary value problem has a unique positive solution.

3 .
If  is a nonempty closed, bounded, and convex subset of a Banach space  and  :  →  is completely continuous, then  has a fixed point in .Let  be a Banach space, let  ⊆  be a cone, and let Ω 1 , Ω 2 be two bounded open balls of  centered at the origin with Ω 1 ⊂ Ω 2 .Suppose that  :  ∩ (Ω 2 \ Ω 1 ) →  is a completely continuous operator such that either