Existence and uniqueness of solutions for a mixed p-Laplace boundary value problem involving fractional derivatives

In this article, the existence and uniqueness of solutions for a multi-point fractional boundary value problem involving two different left and right fractional derivatives with p-Laplace operator is studied. A novel approach is used to acquire the desired results, and the core of the method is Banach contraction mapping principle. Finally, an example is given to verify the results.

In [11], a mixed fractional p-Laplace boundary value problem was studied by Liu et al.
where ϕ p (t) = |t| p-2 · t, p > 1, 1 < α, β ≤ 2, r 1 , r 2 ≥ 0, D α 0+ is Riemann-Liouville fractional derivative, c D In [2], Bai investigated the uniqueness and existence of solutions of the following fractional-order differential equation: The uniqueness result of a solution and the existence of specific solutions to problem were showed by applying Guo-Krasnoselskii's fixed point theorem.
In [4], Dang et al. proposed a fresh approach to gain the existence and uniqueness of solutions of a fourth-order two-point nonlinear differential equation In [3], Bai et al. considered a class of fourthorder nonlinear differential equation with p-Laplace operator, and the boundary conditions change from two points to multiple points compared to the above problem. In both papers, the main results were given by applying the Banach contraction mapping principle.
As far as we know, nobody used the method which was put forward by Dang [6] and Bai [3] to prove the existence and uniqueness of solutions of a nonlinear multi-point fractional-order p-Laplace boundary value problem, which has at least two different kinds of fractional derivatives. Inspired by the above-mentioned articles, the following mixed boundary value problem is studied in this work: The rest of this work is organized as follows. In Sect. 2, some related definitions and necessary lemmas of fractional calculus theory are presented, which will be applied in the main results of this article. In Sect. 3, the existence and uniqueness of solutions of the mixed fractional-order p-Laplace differential equation are proved by applying the method which was put forward by Dang and Bai. In Sect. 4, a particular example is constructed to verify the main conclusions of the paper.

Preliminaries
This section introduces some related definitions and necessary lemmas of fractional calculus theory. Definition 2.1 For given γ > 0 and the function y : (0, ∞) → R, define the left and right Riemann-Liouville fractional integrals respectively as follows: Definition 2.2 For given γ > 0, γ ∈ (n, n + 1) and the function y : (0, ∞) → R, define the left Riemann-Liouville fractional derivative and the right Caputo fractional derivative respectively as follows: Let E = C[0, 1], whose norm · is the maximum norm. Given φ ∈ C[0, 1] and the constants r 1 , r 2 ∈ R, discuss the following fractional-order mixed p-Laplace boundary value problem:  (1η) . and Using the condition k(1) = 0 yields c 1 = 0. Since k(0) = r 2 k(η), then By calculation, we can get Similarly, the solution of boundary value problem (2.4) is given by Consequently, boundary value problem (2.1) is equivalent to (2.2).

Lemma 2.2 For given φ(s)
Proof Since 1 p + 1 q = 1 and ϕ p is increasing, then The proof is completed.

Main results
This section studies the existence and uniqueness of solutions for mixed fractional-order p-Laplace boundary value problem (1.1) by applying the Banach contraction mapping principle. Given number M > 0, denote (H2) |g(x, y 2 , w 2 )g(x, y 1 , w 1 )| ≤ Q 1 |y 2y 1 | + Q 2 |w 2w 1 | for (x, y i , w i ) ∈ D M , i = 1, 2; (H3) L 1 := (q -1)M q-2 M q-1 2 (Q 1 M 1 + Q 2 ) < 1. Then the mixed boundary value problem (1.1) has a unique solution satisfying the estimation Proof First of all, define an operator A : From the continuity of G 1 (x, τ ), G 2 (τ , s), and g(x, y, w), it is not hard to see that the operator A is a continuous operator. According to Lemma 2.1, it is easy to get the following conclusion that if the mixed boundary value problem (1.1) has a solution y(x), then Consequently, for any x ∈ [0, 1], there is (x, y(x), w(x)) ∈ D M . So, from (H1), we can conclude that  = g x, y 2 (x), w 2 (x)g x, y 1 (x), w 1 (x) The proof is completed. where g(x, y, w) = -2y 2 w + 3y -2w + 4 sin(πx). We choose p = 2, that is, q = 2. By a simple computation, we obtain that M 1 = 1 8 , M 2 = 1 3 . Then a suitable number M > 0 is chosen to satisfy all the conditions of Theorem 3.1, and |y| ≤ M 24 , |w| ≤ M 3 .