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Xinguang Zhang 1 and Lishan Liu 2 and Benchawan Wiwatanapataphee 3 and Yonghong Wu 4
Recommended by Shaoyong Lai
1, School of Mathematical and Informational Sciences, Yantai University, Yantai, Shandong 264005, China
2, School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China
3, Department of Mathematics, Faculty of Science, Mahidol University, Bangkok 10400, Thailand
4, Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia
Received 2 January 2012; Accepted 15 March 2012
1. Introduction
In this paper, we discuss the existence of positive solutions for the following eigenvalue problem of a class fractional differential equation with derivatives [figure omitted; refer to PDF] where λ is a parameter, 1<α...4;2 , α-β>1 , 0<β...4;γ<1 , 0<ξ1 <ξ2 <...<ξp-2 <1 , aj ∈[0,+∞) with c=∑j=1p-2ajξjα-γ-1 <1 , and Dt is the standard Riemann-Liouville derivative. f:(0,1)×(0,+∞)×(0,+∞)[arrow right][0,+∞) is continuous, and f(t,u,v) may be singular at u=0, v=0 , and t=0,1 .
As fractional order derivatives and integrals have been widely used in mathematics, analytical chemistry, neuron modeling, and biological sciences [1-6], fractional differential equations have attracted great research interest in recent years [7-17]. Recently, ur Rehman and Khan [8] investigated the fractional order multipoint boundary value problem: [figure omitted; refer to PDF] where 1<α...4;2 , 0<β<1 , 0<ξi <1 , ζi ∈[0,+∞) with ∑i=1m-2ζiξiα-β-1 <1 . The Schauder fixed point theorem and the contraction mapping principle are used to establish the existence and uniqueness of nontrivial solutions for the BVP (1.2) provided that the nonlinear function f:[0,1]×...×... is continuous and satisfies certain growth conditions. But up to now, multipoint boundary value problems for fractional differential equations like the BVP (1.1) have seldom been considered when f(t,u,v) has singularity at t=0 and (or) 1 and also at u=0, v=0 . We will discuss the problem in this paper.
The rest of the paper is organized as follows. In Section 2, we give some definitions and several lemmas. Suitable upper and lower solutions of the modified problems for the BVP (1.1) and some sufficient conditions for the existence of positive solutions are established in Section 3.
2. Preliminaries and Lemmas
For the convenience of the reader, we present here some definitions about fractional calculus.
Definition 2.1 (See [1, 6]).
Let α>0 with α∈... . Suppose that x:[a,∞)[arrow right]... . Then the α th Riemann-Liouville fractional integral is defined by [figure omitted; refer to PDF] whenever the right-hand side is defined. Similarly, for α∈... with α>0 , we define the α th Riemann-Liouville fractional derivative by [figure omitted; refer to PDF] where n∈... is the unique positive integer satisfying n-1...4;α<n and t>a .
Remark 2.2.
If x,y:(0,+∞)[arrow right]... with order α>0 , then [figure omitted; refer to PDF]
Lemma 2.3 (See [6]).
One has the following.
(1) If x∈L1 (0,1), ν>σ>0 , then [figure omitted; refer to PDF]
(2) If ν>0, σ>0 , then [figure omitted; refer to PDF]
Lemma 2.4 (See [6]).
Let α>0 . Assume that x∈C(0,1)∩L1 (0,1) . Then [figure omitted; refer to PDF] where ci ∈... (i=1,2,...,n) , and n is the smallest integer greater than or equal to α .
Let [figure omitted; refer to PDF]
and for t,s∈[0,1] , we have [figure omitted; refer to PDF]
Lemma 2.5.
Let h∈C(0,1) ; If 1<α-β...4;2 , then the unique solution of the linear problem [figure omitted; refer to PDF] is given by [figure omitted; refer to PDF] where [figure omitted; refer to PDF] is the Green function of the boundary value problem (2.9).
Proof.
Applying Lemma 2.4, we reduce (2.9) to an equivalent equation: [figure omitted; refer to PDF] From (2.12) and noting that y(0)=0 , we have c2 =0 . Consequently the general solution of (2.9) is [figure omitted; refer to PDF] Using (2.13) and Lemma 2.3, we have [figure omitted; refer to PDF] Thus, [figure omitted; refer to PDF] and for j=1,2,...,p-2 , [figure omitted; refer to PDF] Using Dt γ-β y(1)=∑j=1p-2ajDt γ-β y(ξj ) , (2.15), and (2.16), we obtain [figure omitted; refer to PDF] So the unique solution of the problem (2.9) is [figure omitted; refer to PDF] The proof is completed.
Lemma 2.6.
The function K(t,s) has the following properties.
(1) K(t,s) > 0, for t, s∈(0,1)
(2) tα-β-1 ...(s) ...4; K(t,s) ...4; M(1-s)α-γ-1 , for t,s∈[0,1] ,
where [figure omitted; refer to PDF]
Proof.
It is obvious that (1) holds.
From (2.11), we obtain [figure omitted; refer to PDF] From (2.8), we have [figure omitted; refer to PDF] The proof is completed.
Consider the modified problem of the BVP (1.1): [figure omitted; refer to PDF]
Lemma 2.7.
Let x(t)=Iβ y(t) and y(t)∈C[0,1] ; then problem (1.1) is turned into (2.22). Moreover, if y∈C([0,1],[0,+∞)) is a solution of problem (2.22), then the function x(t)=Iβ y(t) is a positive solution of the problem (1.1).
Proof.
Substituting x(t)=Iβ y(t) into (1.1) and using Definition 2.1 and Lemmas 2.3 and 2.4, we obtain [figure omitted; refer to PDF] Consequently, Dt β x(0)=y(0)=0 . It follows from Dt γ x(t)=dn /dtnIn-γ x(t)=(dn /dtn )In-γIβ y(t)= (dn /dtn )In-γ+β y(t)=Dt γ-β y(t) that Dt γ-β y(1)=∑j=1p-2ajDt γ-β y(ξj ) . Using x(t)=Iβ y(t) , y∈C[0,1] , we transform (1.1) into (2.22).
Now, let y∈C([0,1],[0,+∞)) be a solution for problem (2.22). Using Lemma 2.3, (2.22), and (2.23), one has [figure omitted; refer to PDF] Noting [figure omitted; refer to PDF] we have [figure omitted; refer to PDF] It follows from the monotonicity and property of Iβ that [figure omitted; refer to PDF] Consequently, x(t)=Iβ y(t) is a positive solution of the problem (1.1).
Definition 2.8.
A continuous function ψ(t) is called a lower solution of the BVP (2.22), if it satisfies [figure omitted; refer to PDF]
Definition 2.9.
A continuous function [varphi](t) is called an upper solution of the BVP (2.22), if it satisfies [figure omitted; refer to PDF]
By Lemmas 2.5 and 2.6, we have the maximal principle.
Lemma 2.10.
If 1<α-β...4;2 and y∈C([0,1],R) satisfies [figure omitted; refer to PDF] and -Dt α-β y(t)...5;0 for any t∈(0,1) , then y(t)...5;0 , for t∈[0,1] .
Set [figure omitted; refer to PDF]
To end this section, we present here two assumptions to be used throughout the rest of the paper.
(B1) : f∈C((0,1)×(0,∞)×(0,∞),[0,+∞)) is decreasing in u and v , and for any (u,v)∈(0,∞)×(0,∞) ,
[figure omitted; refer to PDF]
uniformly on t∈(0,1) .
(B2) : For any μ,ν>0 , f(t,μ,ν)...2;0 , and
[figure omitted; refer to PDF]
3. Main Results
The main result is summarized in the following theorem.
Theorem 3.1.
Provided that (B1) and (B2) hold, then there is a constant λ* >0 such that for any λ∈(λ* ,+∞) , the problem (1.1) has at least one positive solution x(t) , which satisfies x(t)...5;[Lagrangian (script capital L)](t) , t∈[0,1] .
Proof.
Let E=C[0,1] ; we denote a set P and an operator Tλ in E as follows: [figure omitted; refer to PDF] [figure omitted; refer to PDF] Clearly, P is a nonempty set since ...A2;(t)∈P . We claim that Tλ is well defined and Tλ (P)⊂P .
In fact, for any ρ∈P , by the definition of P , there exists one positive number lρ such that ρ(t)...5;lρ ...A2;(t) for any t∈[0,1] . It follows from Lemma 2.6 and (B2) that [figure omitted; refer to PDF]
Setting B=max t∈[0,1] ρ(t)>0 , from (B2) , we have f(t,B/Γ(β+1),B)...2;0 . By the continuity of f(t,u,v) on (0,1)×(0,∞)×(0,∞) , we have ∫01 ...(s)f(s,B/Γ(β+1),B)ds>0 . On the other hand, [figure omitted; refer to PDF] From (3.3), one has [figure omitted; refer to PDF] It follows from Lemma 2.6 and (3.3) that [figure omitted; refer to PDF] where [figure omitted; refer to PDF] Using (3.3) and (3.6), we know that Tλ is well defined and Tλ (P)⊂P .
Next we will focus on the upper and lower solutions of problem (2.22). From (B1) and (3.2), we know that the operator Tλ is decreasing in y . Using [figure omitted; refer to PDF] and letting [figure omitted; refer to PDF] we have [figure omitted; refer to PDF]
On the other hand, letting b(t)=∫01 K(t,s)f(s,[Lagrangian (script capital L)](s),...A2;(s))ds , since f(t,u,v) is decreasing with respect to u and v , for any λ>λ1 , we have [figure omitted; refer to PDF] From (3.2), (3.3), and (B1) , for all (u,v)∈(0,∞)×(0,∞) , we have [figure omitted; refer to PDF] uniformly on t∈(0,1) . Thus there exists large enough λ* >λ1 >0 , such that, for any t∈(0,1) , [figure omitted; refer to PDF] From Lemma 2.6, one has [figure omitted; refer to PDF] Letting [figure omitted; refer to PDF] and using Lemmas 2.3 and 2.7, we obtain [figure omitted; refer to PDF] Obviously, [varphi](t),ψ(t)∈P . By (3.16), we have [figure omitted; refer to PDF] which implies that [figure omitted; refer to PDF] Consequently, it follows from (3.17)-(3.18) that [figure omitted; refer to PDF] [figure omitted; refer to PDF] From (3.16) and (3.18)-(3.20), we know that ψ(t) and [varphi](t) are upper and lower solutions of the problem (2.22), and ψ(t),[varphi](t)∈P .
Define the function F and the operator Aλ* in E by [figure omitted; refer to PDF] It follows from (B1) and (3.21) that F:(0,1)×[0,+∞)[arrow right][0,+∞) is continuous. Consider the following boundary value problem: [figure omitted; refer to PDF] Obviously, a fixed point of the operator Aλ* is a solution of the BVP (3.22). For all y∈E , it follows from Lemma 2.6, (3.21), and ψ(t)...5;...A2;(t) that [figure omitted; refer to PDF] So Aλ* is bounded. From the continuity of F(t,y) and K(t,s) , it is obviously that Aλ* :E[arrow right]E is continuous.
From the uniform continuity of K(t,s) and the Lebesgue dominated convergence theorem, we easily get that Aλ* (Ω) is equicontinuous. Thus from the Arzela-Ascoli theorem, Aλ* :E[arrow right]E is completely continuous. The Schauder fixed point theorem implies that Aλ* has at least one fixed point w such that w=Aλ* w .
Now we prove [figure omitted; refer to PDF] Let z(t)=[varphi](t)-w(t), t∈[0,1] . Since [varphi](t) is the upper solution of problem (2.22) and w is a fixed point of Aλ* , we have [figure omitted; refer to PDF]
From (3.17), (3.18), and the definition of F , we obtain [figure omitted; refer to PDF] So [figure omitted; refer to PDF] From (3.18) and (3.20), one has [figure omitted; refer to PDF] By (3.27), (3.28), and Lemma 2.10, we get z(t)...5;0 which implies that w(t)...4;[varphi](t) on [0,1] . In the same way, we have w(t)...5;ψ(t) on [0,1] . Thus we obtain [figure omitted; refer to PDF] Consequently, F(t,w(t))=f(t,Iβ w(t),w(t)), t∈[0,1] . Then w(t) is a positive solution of the problem (2.22). It thus follows from Lemma 2.7 that x(t)=Iβ w(t) is a positive solution of the problem (1.1).
Finally, by (3.29), we have [figure omitted; refer to PDF] Thus, [figure omitted; refer to PDF]
Corollary 3.2.
Suppose that condition (B1) holds, and that for any μ,ν>0 , f(t,μ,ν)...2;0 , and [figure omitted; refer to PDF] Then there exists a constant λ* >0 such that for any λ∈(λ* ,+∞) , the problem (1.1) has at least one positive solution x(t) , which satisfies x(t)...5;[Lagrangian (script capital L)](t) , t∈[0,1] .
We consider some special cases in which f(t,u,v) has no singularity at u,v=0 or t=0,1 .
We give the following assumption.
(B* 1) f∈C((0,1)×[0,∞)×[0,∞),(0,+∞)) is decreasing in u,v .
Then, f(t,u,v) is nonsingular at u=v=0 and for all u,v...5;0 , f(t,u,v)>0, t∈(0,1) , which implies that f(t,0,0)>0, t∈(0,1) . Thus [figure omitted; refer to PDF]
naturally holds; we then have the following corollary.
Corollary 3.3.
If (B* 1) holds and
(B* 2) [figure omitted; refer to PDF] then there exists a constant λ* >0 such that for any λ∈(λ* ,+∞) , the problem (1.1) has at least one positive solution x(t) , which satisfies x(t)...5;[Lagrangian (script capital L)](t) , t∈[0,1] .
Proof.
In the proof of Theorem 3.1, we replace the set P by [figure omitted; refer to PDF] and the inequalities (3.18)-(3.20) by [figure omitted; refer to PDF] Since Tλ 0,Tλ ψ(t)∈P , we have [figure omitted; refer to PDF] The rest of the proof is similar to that of Theorem 3.1.
If f(t,u,v) is nonsingular at u=0, v=0 and t=0,1 , we have the conclusion.
Corollary 3.4.
If f(t,u,v):[0,1]×[0,∞)×[0,∞)[arrow right](0,+∞) is continuous and decreasing in u and v , the problem (1.1) has at least one positive solution x(t) , which satisfies x(t)...5;[Lagrangian (script capital L)](t) , t∈[0,1] .
Example 3.5.
Consider the existence of positive solutions for the following eigenvalue problem of fractional differential equation: [figure omitted; refer to PDF]
Let [figure omitted; refer to PDF] Then f∈C((0,1)×(0,+∞)×(0,+∞),(0,+∞)) is decreasing in u and v , and for any (u,v)∈(0,∞)×(0,∞) , [figure omitted; refer to PDF] uniformly on t∈(0,1) . Thus (B1) holds.
On the other hand, for any μ,ν>0 and t∈(0,1) , [figure omitted; refer to PDF]
thus we have [figure omitted; refer to PDF]
which implies that (B2) holds. From Theorem 3.1, there is a constant λ* >0 such that for any λ∈(λ* ,+∞) the problem (3.38) has at least one positive solution x(t) and [figure omitted; refer to PDF]
Acknowledgments
This work is supported financially by the National Natural Science Foundation of China (11071141, 11126231) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017).
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Abstract
By establishing a maximal principle and constructing upper and lower solutions, the existence of positive solutions for the eigenvalue problem of a class of fractional differential equations is discussed. Some sufficient conditions for the existence of positive solutions are established.
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Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer