The Existence of Solutions for a Nonlinear Fractional Multi-Point Boundary Value Problem at Resonance

Xiaoling Han and Ting Wang Department of Mathematics, Northwest Normal University, Lanzhou 730070, China Correspondence should be addressed to Xiaoling Han, hanxiaoling@nwnu.edu.cn Received 16 May 2011; Accepted 16 June 2011 Academic Editor: Nikolai Leonenko Copyright q 2011 X. Han and T. Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We discuss the existence of solution for a multipoint boundary value problem of fractional differential equation. An existence result is obtained with the use of the coincidence degree theory.


Introduction
In this paper, we study the multipoint boundary value problem D α 0 u t f t, u t , D α−1 0 u t , D α−2 0 u t e t , 0 < t < 1, 1.1 β j ξ j 0, n j 1 International Journal of Differential Equations f : 0, 1 × R 3 → R satisfying the Carathéodory conditions, e ∈ L 1 0, 1 .D α 0 and I α 0 are the standard Riemann-Liouville derivative and integral, respectively.We assume, in addition, that where Γ is the Gamma function.Due to condition 1.3 , the fractional differential operator in 1.1 , 1.2 is not invertible.
Fractional differential equation can describe many phenomena in various fields of science and engineering.Many methods have been introduced for solving fractional differential equations, such as the popular Laplace transform method, the iteration method.For details, see 1, 2 and the references therein.
Recently, there are some papers dealing with the solvability of nonlinear boundary value problems of fractional differential equation, by use of techniques of nonlinear analysis fixed-point theorems, Leray-Schauder theory, etc. , see, for example, 3-6 .But there are few papers that consider the fractional-order boundary problems at resonance.Very recently 7 , Y. H. Zhang and Z. B. Bai considered the existence of solutions for the fractional ordinary differential equation subject to the following boundary value conditions: where n > 2 is a natural number, n − 1 < α ≤ n is a real number, f : 0, 1 × R n → R is continuous, and e ∈ L 1 0, 1 , σ ∈ 0, ∞ , and η ∈ 0, 1 are given constants such that ση α−1 1. D α 0 and I α 0 are the standard Riemann-Liouville derivative and integral, respectively.By the conditions, the kernel of the linear operator is one dimensional.
Motivated by the above work and recent studies on fractional differential equations 8-18 , in this paper, we consider the existence of solutions for multipoint boundary value problem 1.1 , 1.2 at resonance.Note that under condition 1.3 , the kernel of the linear operator in 1.1 , 1.2 is two dimensional.Our method is based upon the coincidence degree theory of Mawhin  The rest of this paper is organized as follows.In Section 2, we give some notation and Lemmas.In Section 3, we establish an existence theorem of a solution for the problem 1.1 , 1.2 .

Background Materials and Preliminaries
For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory, and these definitions can be found in the recent literature 1, 2 .
Definition 2.1.The fractional integral of order α > 0 of a function y : 0, ∞ → R is given by provided the right side is pointwise defined on 0, ∞ .And we let I 0 0 y t y t for every continuous y : 0, ∞ → R.
Definition 2.2.The fractional derivative of order α > 0 of a function y : 0, ∞ → R is given by where n α 1, provided the right side is pointwise defined on 0, ∞ .
We use the classical space C 0, 1 with the norm x ∞ max t∈ 0,1 |x t |.Given μ > 0 and N μ 1, one can define a linear space where x ∈ C 0, 1 and c i ∈ R, i 1, 2, . . ., N − 1.By means of the linear function analysis theory, one can prove that with the norm Lemma 2.4 see 7 .F ⊂ C μ 0, 1 is a sequentially compact set if and only if F is uniformly bounded and equicontinuous.Here, uniformly bounded means that there exists M > 0 such that for every and equicontinuous means that for all ε > 0, ∃δ > 0 such that

2.6
Let Z L 1 0, 1 with the norm g 1 Definition 2.6.We say that the map f : 0, 1 × R → R satisfies the Carathéodory conditions with respect to L 1 0, 1 if the following conditions are satisfied: i for each z ∈ R, the mapping t → f t, z is Lebesgue measurable, ii for almost every t ∈ 0, 1 , the mapping z → f t, z is continuous on R, iii for each r > 0, there exists ρ r ∈ L 1 0, 1 , R such that, for a.e., t ∈ 0, 1 and every |z| ≤ r, we have |f t, z | ≤ ρ r t .
Define L to be the linear operator from dom

2.7
We define N : Y → Z by setting 2.10 then D α 0 u t g t a.e., t ∈ 0, 1 and, if hold.Then u t satisfies the boundary conditions 1.2 , that is, u ∈ dom L , and we have g ∈ Z | g satisfies 2.11 ⊆ Im L .

2.12
Let u ∈ dom L , then for D α 0 u ∈ Im L , we have which, due to the boundary value condition 1.2 , implies that D α 0 u satisfies 2.11 .In fact, from I 3−α 0 u 0 0, we have c 3 0, from u 1 Hence, Im L ⊆ g ∈ Z | g satisfies 2.11 .

2.16
International Journal of Differential Equations Therefore, Im L g ∈ Z | g satisfies 2.11 .

2.17
Consider the continuous linear mapping Q 1 : Z → Z and Q 2 : Z → Z defined by β j ξ j 0 ξ j − s g s ds.

2.18
Using the above definitions, we construct the following auxiliary maps R 1 , R 2 : Z → Z:

2.19
Since the condition 1.4 holds, the mapping Q : Z → Z defined by is well defined.Recall 1.4 and note that

2.21
International Journal of Differential Equations 7 and similarly we can derive that

2.22
So, for g ∈ Z, it follows from the four relations above that

2.23
that is, the map Q is idempotent.In fact, Q is a continuous linear projector.Note that g ∈ Im L implies Qg 0. Conversely, if Qg 0, then we must have R 1 g R 2 g 0; since the condition 1.4 holds, this can only be the case if and assume that g s as α−1 bs α−2 is not identically zero on 0, 1 , then, since g ∈ Im L , from 2.11 and the condition 1.4 , we derive a b 0, which is a contradiction.Hence, Im L ∩ Im Q {0}; thus, Z Im L ⊕ Im Q .Now, dim Ker L 2 co dim Im L , and so L is a Fredholm operator of index zero.
Let P : Y → Y be defined by

2.24
Note that P is a continuous linear projector and

2.25
It is clear that Y Ker L ⊕ Ker P .Note that the projectors P and Q are exact.Define K P : Im L → dom L ∩ Ker P by

2.26
International Journal of Differential Equations Hence, we have

2.28
In fact, if g ∈ Im L , then LK P g D α 0 I α 0 g g.Also, if u ∈ dom L ∩ Ker P , then from boundary value condition 1.2 and the fact that u ∈ dom L ∩ Ker P , we have c 1 c 2 c 3 0. Thus,

2.31
With arguments similar to those of 7 , we obtain the following Lemma.Lemma 2.8.

The Main Results
Assume that the following conditions on the function f t, x, y, z are satisfied: H1 there exists a constant A > 0, such that for u ∈ dom L \ Ker L satisfying International Journal of Differential Equations 9 H2 there exist functions a, b, c, d, r ∈ L 1 0, 1 and a constant θ ∈ 0, 1 such that for all x, y, z ∈ R 3 and a.e., t ∈ 0, 1 , one of the following inequalities is satisfied: H3 there exists a constant B > 0 such that for every a, b ∈ R satisfying a 2 b 2 > B, then either then for u ∈ Ω 1 , Lu λNu; thus, λ / 0, Nu ∈ Im L Ker Q , and hence QNu t 0 for all t ∈ 0, 1 .By the definition of Q, we have

3.7
International Journal of Differential Equations so

3.8
Now by 3.8 , we have 3.9 Note that I − P u ∈ Im K P dom L ∩ Ker P for u ∈ Ω 1 , then, by 2.28 and 2.30 ,

3.10
Using 3.9 and 3.10 , we obtain where

3.12
If the first condition of H2 is satisfied, then we have Note that θ ∈ 0, 1 and a 1 b 1 c 1 < 1/τ, so there exists M 1 > 0 such that D α−1 0 u ∞ ≤ M 1 for all u ∈ Ω 1 .The inequalities 3.14 and 3.15 show that there exist M , that is, Ω 1 is bounded given the first condition of H2 .If the other conditions of H2 hold, by using an argument similar to the above, we can prove that Ω 1 is also bounded. Let We define the isomorphism J : Im Q → Ker L by If the first part of H3 is satisfied, let

3.20
If λ 1, then a b 0, and if a 2 b 2 > B, then by H3 , λ a 2 b 2 1 − λ aR 1 N at α−1 bt α−2 bR 2 N at α−1 bt α−2 < 0, 3.21 which, in either case, obtain a contradiction.If the other part of H3 is satisfied, then we take and, again, obtain a contradiction.Thus, in either case,
In the following, we will prove that all the conditions of Theorem 1.1 are satisfied.Set Ω to be a bounded open set of Y such that U 3 i 1 Ω i ⊂ Ω. by Lemma 2.8, the operator K P I − Q N : Ω → Y is compact; thus, N is L-compact on Ω, then by the above argument, we have i Lu / λNx, for every u, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1 , ii Nu / ∈ Im L , for every u ∈ Ker L ∩ ∂Ω.
Finally, we will prove that iii of Theorem 1.1 is satisfied.Let H u, λ ±Iu 1 − λ JQNu, where I is the identity operator in the Banach space Y .According to the above argument, we know that H u, λ / 0, ∀u ∈ ∂Ω ∩ Ker L , and Y is a Banach space.Definition 2.5.By a solution of the boundary value problem 1.1 , 1.2 , we understand a function u ∈ C α−1 0, 1 such that D α−1 0 u is absolutely continuous on 0, 1 and satisfies 1.1 , 1.2 .

Theorem 3 . 2 .
H3 stand for the images of u t at α−1 bt α−2 under the maps R 1 N and R 2 N, respectively.If (H1)-(H3) hold, then boundary value problem 1.1 -1.2 has at least one solution provided that
18. Now, we will briefly recall some notation and abstract existence result.Im Q .It follows that L| dom L ∩Ker P : dom L ∩ Ker P → Im L is invertible.We denote the inverse of the map by International Journal of Differential Equations 3K P .If Ω is an open-bounded subset of Y such that dom L ∩Ω / ∅, the map N : Y → Z will be called L-compact on Ω if QN Ω is bounded and K P I − Q N : Ω → Y is compact.The theorem that we used is Theorem 2.4 of 18 .
Let Y, Z be real Banach spaces, let L : dom L ⊂ Y → Z be a Fredholm map of index zero, and let P : Y → Y, Q : Z → Z be continuous projectors such that Im P Ker P , Ker Q Im L , and Y Ker L ⊕ Ker P , Z Im L ⊕