The Existence of Positive Solutions for Fractional Differential Equations with Integral and Disturbance Parameter in Boundary Conditions

and Applied Analysis 3 where G (t, s)

At the same time, while using the methods of the differential equations to solve the actual problems, it is inevitable that there always exists disturbance which will have great influence on the existence of the solutions.In paper [35,36], the authors studied nonlinear nonlocal boundary value problem with nonhomogeneous boundary conditions   () +  (,  () ,   ()) = 0,  ∈ (0, 1) ,  (0) − where they discussed the impact of disturbance parameters  1 ,  2 on the existence of the solution and obtained some meaningful conclusions.And then, the authors further studied the 2nth order nonlinear nonlocal boundary value problem with nonhomogeneous boundary conditions; see [37].
The purpose of this paper is to study the impact of the disturbance parameter  on the existence of positive solutions and obtain sufficient conditions for the boundary value problem (1) to have at least one positive solution, at least two solutions, and no solutions.Under certain conditions, we obtain that there exists a constant  * > 0, which separates R + into two disjoint subintervals  = [0,  * ] and  = ( * , +∞) such that the boundary value problem (1) has at least two positive solutions for each  ∈ (0,  * ), at least one positive solution for  = 0 and  =  * , and no positive solutions for  ∈ .The main tools we applied are the upper and lower solutions method, fixed point index theory, and the Schauder fixed point theorem.
This paper is organized as follows.In Section 2, we introduce the basic definitions and the basic properties of the integral kernel.In Section 3, we study the comparison principles and the basic lemmas.In Section 4, we consider the existence and nonexistence of the positive solutions of the boundary value problem ( 1), and we study the impact of the disturbance parameter  on the existence of positive solutions.

Preliminaries
In this section, we give some basic definitions and lemmas which play an important role in our research.
Lemma 4 (See [2]).Suppose  > 0 and  ∈   ().Then where  is the smallest integer greater than or equal to .Throughout this paper, we assume the following conditions hold.
Similarly, we can obtain the following lemma.
For convenience, we denote () = (, ()).Similar to the proof of Lemma 6, we can obtain Substituting  0 and  1 into (25), we can obtain We multiply by the function  on both sides of ( 27), integrate from 0 to 1, and get That is, On the other hand, if  ∈ () is the solution of the integral equation ( 24), we have It is easy to see that   ∈ ().
We can easily verify that  satisfies the boundary value problem (1).

Comparison Principle and the Existence of Solutions
Definition 12. Let  ∈  2 () and     ∈ ().One says that  is a lower solution of the boundary value problem (1), if where 6

Abstract and Applied Analysis
Let  ∈  2 () and     ∈ ().We say that  is an upper solution of the boundary value problem (1), if where where () is defined in (H2).
We can easily obtain the following lemma from the definition of  * , where  * is defined by Lemma 7.   (67 Then  is a solution of (66) if and only if  is a fixed point of the operator T. It is easy to see T :  →  0 .
Next we can prove that T is completely continuous.
Since  2 (, ) is continuous on  × , we have  2 which is uniformly continuous on  × .It implies that for any  > 0, there exists  1 > 0, when  1 ,  (72) Thus, we have proved T is equicontinuous.By Arzela-Ascoli theorem, we know that T() is relatively compact.
We can easily show that T is continuous since  is continuous.Hence, T is completely continuous.
Since  is bounded and T is completely continuous, we can get T has at least one fixed point  by Schauder fixed point theorem, that is, there exists a solution  of the boundary value problem (66).
Then  =  +  * is a solution of the boundary value problem (65).
We can prove that if each solution  of the boundary value problem (65) satisfies () ≤ () ≤ () for  ∈ , then  is a solution of the boundary value problem (1).
We can easily verify that  and  are a lower solution and an upper solution of the boundary value problem (1) and  ≤ .
By Theorem 17, we have that the boundary value problem (1) has a positive solution  and  * ≤  ≤ .
(2) If there exists a constant  0 >  such that the boundary value problem (77) has a positive solution, by (1), we can show that for each  with 0 ≤  ≤  0 , the boundary value problem (1) has a positive solution.So, the boundary value problem (1) has a positive solution, for  = , which is a contradiction.
Therefore, if there exists a constant  > 0 such that the boundary value problem (77) does not have positive solutions, then for each  > , the boundary value problem (1) does not have positive solutions.
which is a contradiction.
(1) If 0 ≤  0 < , then the boundary value problem (1) has at least one positive solution when  ≥ 0 and  is small enough. ( That is () ⊂ .By the Schauder fixed point theorem, we can get that  has at least one fixed point on .In view of Lemma 15,  =  +  * is a positive solution of the boundary value problem (1).
(2) The boundary value problem (1) has at least two positive solutions for each  ∈ (0,  * ), has at least one positive solution for  = 0 and  =  * , and does not have positive solutions for each  ∈ .
Proof.We have the following four steps to prove the conclusions of Theorem 22.
It follows that Λ is a bounded set from  ∞ > / 2 and Theorem 20 (2).
Step 4. We prove the boundary value problem (1) has at least two positive solutions when  ∈ (0,  * ).
Let  be a solution of the boundary value problem (1) with  replaced by .By Theorem 18, the boundary value problem (1) has a positive solution  1 with  1 ≤ .
Similarly, let  be a solution of the boundary value problem (1) with  replaced by  and  ≤  1 .Hence,  ≤  1 ≤ .
Let  =  and  = .We can easily verify  and  are a lower solution and an upper solution of the boundary value problem (1), respectively, and  < .