Multiplicity Results to a Conformable Fractional Differential Equations Involving Integral Boundary Condition

In this article, by using topological degree theory couple with the method of lower and upper solutions, we study the existence of at least three solutions to Riemann-Stieltjes integral initial value problem of the typeDαx(t) = f(t, x), t ∈ [0, 1], x(0) = ∫ 0 x(t)dA(t), whereDαx(t) is the standard conformable fractional derivative of order α, 0 < α ≤ 1, and f ∈ C([0, 1] ×R,R). Simultaneously, the fixed point theorem for set-valued increasing operator is applied when considering the given problem.


Introduction
In recent years, fractional differential equations have exerted tremendous influence on some mathematical models of research processes and phenomena in many fields such as electrochemistry, heat conduction, underground water flow, and porous media.A growing number of papers deal with the existence or multiplicity of solutions of initial value problem and boundary value problem for fractional differential equations [1][2][3][4][5][6][7][8][9][10][11].Recently, the authors [12] give an interesting fractional derivative called the "conformal fractional derivative", which depends on the limit definition of the function derivative.Moreover, readers can find in [13] the properties of conformable fractional derivatives that are similar to ordinary differential ones.Other related work on conformable fractional differential equation can be found in [12][13][14][15][16][17] and the references therein.
Inspired by the above work, we consider the existence and multiple solutions of the following fractional differential equation involving integral boundary condition: where  ∈ ([0,1] × R, R), ∫ 1 0 ()() denotes the Riemann-Stieltjes integral with positive Stieltjes measure.  () is the standard conformable fractional derivative of order 0 <  ≤ 1 of  at  > 0, defined by If   () exists on (0, 1), then In order to study the existence and multiplicity results of the problem (4), this paper is arranged as follows: after this introduction, in Section 2, we briefly show some necessary definitions and lemmas that are used to prove our main results.In Section 3, we shall define a modified bounded function to discuss the existence of solutions for a conformable fractional differential equation with initial value condition.Moreover, in this part we employ two fixed point theorems, which are Schauder's fixed point theorem and the fixed point theorem for set-valued increasing operator, respectively.In addition, using two pairs of upper and lower solutions and the property of degree theory, we deduce that problem (4) has at least three solutions.
en we have the following: (1) if   () ≥ 0 for all  ∈ (, ), then  is increasing on [, ]; (2) if   () ≤ 0 for all  ∈ (, ), then  is decreasing on The following lemma is a direct consequence of the application of the definition of conformable fractional derivative and Lemma 7. (1) If  has an extreme value at  0 ∈ (, ), then   ( 0 ) = 0; (2) if  has a maximum (minimum) value at  0 = , then (3) if  has a maximum (minimum) value at  0 = , then Theorem 9 ([36, 37] (fixed point theorem for set-valued increasing operator)).Let  be a partially ordered set,  be a nonempty closed set of , and  :  → 2  be a set-valued increasing operator.Assume that (i) any totally ordered subset of  is a relatively compact, (ii) for all  ∈ ,  is a compact set in , (iii) there exist  0 ∈  and  ∈  0 such that  0 ≤ .
en  has a fixed point in ; that is, there exists  * ∈  such that  * ∈  * .
Throughout the paper, we list some hypotheses.
is lower and upper solution of problem (4) with V() ≤ ().

Main Results
Based on the above preparations and the assumptions mentioned, we first consider the following initial value problem (IVP) where  ∈ R. We suppose that ( 1 ) holds.Applying Lemma 2, it is easy to prove that ( 12) is equivalent to the following integral equation Define integral operator  :  →  by Then,  is a solution of ( 12) if and only if  ∈  is a solution of the operator equation ( − ) = 0, that is, a fixed point of operator .
Secondly, we consider the modified initial value problem The above initial value problem ( 16) is equivalent to the following integral equation: For any  ∈ , we have Choose  1 > (1 + )/, and let Obviously, Ω is a bounded convex subset in .From the above argument, the operator T : Ω → Ω defined as follows is well defined, It is easy to verify that T : Ω → Ω is compact.By Schauder's fixed point theorem, T has a fixed point  in Ω.Subsequently,  is the solution of problem (16).Also, by the definition of  V and Definition 5, we have and Hence V and  are lower and upper solutions of (16).
Our final job is to apply the method in [38] to illustrate that if (16) has upper and lower solutions, hence any solution  of differential equation ( 16) must satisfy V() ≤ () ≤ (),  ∈ [0, 1] and then is a solution of (12).
For case (i), by the definition of  V and Definition 5, we can get which contradicts with (24).This implies that ℎ() has no positive local maximum on [0, 1].
Above we have studied the solution of ( 4), and then we present the main results of this paper.
Proof.Similar to Theorem 10, we define the following modified function Obviously  V 1  2 is continuous and bounded on [0, 1] × R, so there exists  > 0 such that where  2 = ∫ 1 0   ().We consider the modified boundary value problem It is clear that the solution () of the above problem satisfies V 1 () ≤ () ≤  2 () and then is a solution of problem (4).
By Lemma 2, it is easy to know that ( 36) is equivalent to the following integral equation: Integrating (37) with respect to () form 0 to 1, Combining with (37) and (38), we reduce (36) to the following equivalent integral equation: Now, we prove the existence of at least three solutions of the problem (36).Note that Then,  is a solution of (36) if and only if  ∈  is a solution of the equation ( −  1 ) = 0, that is, a fixed point of  1 .For  ∈ Ω 1 , we get So  1 (Ω 1 ) ⊂ Ω 1 and  1 : Ω 1 → Ω 1 is compact.By the topological degree theory, we have Let and It follows from the fact V 2 ≰  1 that then