The Existence of Solutions for a Fractional 2 m-Point Boundary Value Problems

By using the coincidence degree theory, we consider the following 2m-point boundary value problem for fractional differential equationD 0 u t f t, u t , D α−1 0 u t , D α−2 0 u t e t , 0 < t < 1, I3−α 0 u t |t 0 0, Dα−2 0 u 1 ∑m−2 i 1 aiD α−2 0 u ξi , u 1 ∑m−2 i 1 biu ηi , where 2 < α ≤ 3, D 0 and I 0 are the standard Riemann-Liouville fractional derivative and fractional integral, respectively. A new result on the existence of solutions for above fractional boundary value problem is obtained.


Introduction
Fractional differential equations have been of great interest recently.This is because of the intensive development of the theory of fractional calculus itself as well as its applications.Apart from diverse areas of mathematics, fractional differential equations arise in a variety of different areas such as rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, and many other branches of science see 1-4 and references cited therein .The research of fractional differential equations on boundary value problems, as one of the focal topics has attained a great deal of attention from many researchers see 5-13 .However, there are few papers which consider the boundary value problem at resonance for nonlinear ordinary differential equations of fractional order.In 14 , Hu and Liu studied the following BVP of fractional equation at resonance: where 1 < α ≤ 2, D α 0 is the standard Caputo fractional derivative.

Journal of Applied Mathematics
In 15 , Zhang and Bai investigated the nonlinear nonlocal problem where 1 < α ≤ 2, they consider the case βη α−1 1, that is, the resonance case.In 16 , Bai investigated the boundary value problem at resonance D α 0 u t f t, u t , D α−1 0 u t e t , 0 < t < 1, is considered, where 1 < α ≤ 2 is a real number, D α 0 and I α 0 are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and f : 0, 1 × R 2 → R is continuous and e t ∈ L 1 0, 1 , m ≥ 2, 0 < ξ i < 1, β i ∈ R, i 1, 2, . . ., m − 2 are given constants such that m−2 i 1 β i 1.In this paper, we study the 2m-point boundary value problem satisfies Carathéodory conditions, D α 0 and I α 0 are the standard Riemann-Liouville fractional derivative and fractional integral, respectively. Setting

1.6
In this paper, we will always suppose that the following conditions hold: C1 : We say that boundary value problem 1.4 and 1.5 is at resonance, if BVP D α 0 u t 0, has u t at α−1 bt α−2 , a, b ∈ R as a nontrivial solution.The rest of this paper is organized as follows.Section 2 contains some necessary notations, definitions, and lemmas.In Section 3, we establish a theorem on existence of solutions for BVP 1.4 -1.5 under nonlinear growth restriction of f, basing on the coincidence degree theory due to Mawhin see 17 .
Now, we will briefly recall some notation and an abstract existence result.
Let Y, Z be real Banach spaces, L : dom L ⊂ Y → Z a Fredholm map of index zero,s and

Preliminaries
For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory.These definitions can be found in the recent literature 1-16, 18 .
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, ∞ , where Γ • is the Gamma function.
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, ∞ .
Definition 2.3.We say that the map f : 0, 1 × R n → R satisfies Carathéodory conditions with respect to L 1 0, 1 if the following conditions are satisfied: i for each z ∈ R n , the mapping t → f t, z is Lebesgue measurable; ii for almost every t ∈ 0, 1 , the mapping t → f t, z is continuous on R n ; 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 .
We use the classical Banach space C 0, 1 with the norm L 0, 1 with the norm Definition 2.5.For n ∈ N, we denote by AC n 0, 1 the space of functions u t which have continuous derivatives up to order n − 1 on 0, 1 such that u n−1 t is absolutely continuous: Lemma 2.6 see 15 .Given μ > 0 and N μ 1 we can define a linear space By means of the linear functional analysis theory, we can prove that with the Remark 2.7.If μ is a natural number, then C μ 0, 1 is in accordance with the classical Banach space C n 0, 1 .
Lemma 2.8 see 15 .f ⊂ C μ 0, 1 is a sequentially compact set if and only if f is uniformly bounded and equicontinuous.Here uniformly bounded means there exists M > 0, such that for every and equicontinuous means that ∀ε > 0, ∃δ > 0, such that
Definition 2.10 see 16 .Let I α 0 L 1 0, 1 , α > 0 denote the space of functions u t , represented by fractional integral of order α of a summable function: Let Z L 1 0, 1 , with the norm y

2.13
Then boundary value problem 1.4 and 1.5 can be written as Lu Nu.

Main Results
Lemma 3.1.Let L be defined by 2.12 , then

3.1
Proof.In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that I α 0 D −α 0 for α < 0. Let Lu D α 0 u, by Lemma 2.9, D α 0 u t 0 has solution

3.3
Let y ∈ Z and let

3.4
Then D α 0 u t y t a.e.t ∈ 0, 1 and, if Let u ∈ dom L. Then for D α 0 u ∈ Im L, we have where which, due to the boundary value condition 1.5 , implies that satisfies 3.5 .In fact, from

3.11
The proof is complete.
Lemma 3.2.The mapping L : domL ⊂ Y → Z is a Fredholm operator of index zero, and where define by K p : ImL → domL ∩ KerP by 14 Proof.Consider the continuous linear mapping

3.15
Using the above definitions, we construct the following auxiliary maps T 1 : Z → Z and T 2 : Z → Z:

3.16
Since the condition C2 holds, the mapping defined by Qy t T 1 y t t α−1 T 2 y t t α−2 3.17 is well defined.Recall C2 and note that

3.18
and similarly we can derive that

3.19
So, for y ∈ Z, it follows from the four relations above that Qy,

3.20
that is, the map Q is idempotent.In fact Q is a continuous linear projector.Note that y ∈ Im L implies Qy 0. Conversely, if Qy 0, so then we must have Q 1 y Q 2 y 0; since the condition C2 holds, this can only be the case if In fact Ker Q Im L, take y ∈ Z in the form y y−Qy Qy so that y − Qy ∈ Ker Q Im L, Qy ∈ Im Q, thus, Z Im L Im Q, Let y ∈ Im L ∩ Im Q and assume that y at α−1 bt α−2 is not identically zero on 0, 1 .Then, since y ∈ Im L, from 3.5 and the condition C2 , we have

3.24
Journal of Applied Mathematics but we derive a b 0, which is a contradiction.Hence, Im L ∩ Im Q {0}; thus Z Im L ⊕ Im Q.Now, dimKerL 2 codimImL and so L is a Fredholm operator of index zero.Let P : Y → Y be defined by

3.26
Note that P is a continuous linear projector and

3.27
It is clear that Y Ker P ⊕ Ker L.
Note that the projectors P and Q are exact.Define by K p : Im L → dom L ∩ Ker P by

3.28
Hence we have

3.31
In fact, if y ∈ Im L, then LK p y t D α 0 I α 0 y t y t .

3.32
Also, if u ∈ dom L ∩ Ker P , then where and from the boundary value condition 1.5 and the fact that u ∈ dom L ∩ Ker P , Pu 0,

3.35
This shows that K p L| dom L∩Ker P −1 .The proof is complete.Using 3.16 , we write

3.36
By Lemma 2.8 and a standard method, we obtain the following lemma.

Lemma 3.3 see 16 . For every given e
Assume that the following conditions on the function f t, x, y, z are satisfied.H1 There exist functions a t , b t , c t , d t , r t ∈ L 1 0, 1 , and a constant θ ∈ 0, 1 such that for all x, y, z ∈ R 3 , t ∈ 0, 1 , one of the following inequalities is satisfied: H2 There exists a constant A > 0, such that for H3 There exists a constant B > 0 such that for every a, b ∈ R satisfying a 2 b 2 > B then either Remark 3.4.T 1 N at α−1 bt α−2 and T 2 N at α−1 bt α−2 from H3 stand for the images of u t at α−1 bt α−2 under the maps T 1 N and T 2 N, respectively.

3.44
Then for x ∈ Ω 1 , Lx λNx thus λ / 0, Nx ∈ Im L Ker Q, and hence QNx t for all t ∈ 0, 1 .By the definition of Q, we have

3.45
and so

3.46
Therefore, we have

3.47
Note that I − P x ∈ dom L ∩ Ker P for all x ∈ Ω 1 .Then, by Lemma 3.2, we have 48 so, we have , A is a constant.This is for all x ∈ Ω 1 .If the first condition of H1 is satisfied, then, we have

3.50
where D r 1 e 1 n/m, and consequently, for

3.53
Therefore, for all x ∈ Ω 1 , we can prove that Ω 1 is also bounded.If 3.38 or 3.39 holds, similar to the above argument, we can prove that Ω 1 is bounded too.Lemma 3.6.Suppose H3 holds, then the set Lemma 3.7.Suppose H3 holds, then the set Proof.We define the isomorphism J : Ker L → Im Q by J at α−1 bt α−2 at α−1 bt α−2 .

3.58
If the first part of H3 is satisfied, let

3.60
If λ 1, then a b 0, and if a 2 b 2 > B, then by H3 which, in either case, is a contradiction.If the other part of H3 is satisfied, then we take and, again, obtain a contradiction.Thus, in either case Remark 3.8.Suppose the second part of H3 holds, then the set

3.65
Then by Lemma 1.1, Lx Nx has at least one solution in dom L ∩ Ω, so the boundary value problem 1.4 and 1.5 has at least one solution in the space C α−1 0, 1 .The proof is finished.
Corresponding to the problem 1.4 -1.
u t satisfies the boundary conditions 1.5 .That is, u ∈ dom L and we have y ∈ Z | y satisfies 3.4 ⊆ Im L.3.6

Theorem 3 . 9 .
If C1 -C2 and H1 -H3 hold, then the boundary value problem 1.4 -1.5 has at least one solution.Proof.Set Ω to be a bounded open set of Y such that ∪ 3 i 1 Ω ⊂ Ω.It follows from Lemmas 3.2 and 3.3 that L is a Fredholm operator of index zero, and the operator K p I − Q N : Ω → Y is 16 Journal of Applied Mathematics compact N, thus, is L-compact on Ω.By Lemmas 3.5 and 3.6, we get that the following two conditions are satisfied: i Lx / λNx for every x, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1 ;ii Nx / ∈ Im L, for every x ∈ Ker L ∩ ∂Ω.Finally, we will prove that iii of Lemma 1.1 is satisfied.Let H x, λ ±λJx 1 − λ QNx, where I is the identity operator in the Banach space Y .According to Lemma 3.7 or Remark 3.8 , we know that H x, λ / 0, for all x ∈ ∂Ω ∩ Ker L, and thus, by the homotopy property of degree, deg QN| Ker L , Ker L ∩ Ω, 0 deg H •, 0 , Ker L ∩ Ω, 0 deg H •, 1 , Ker L ∩ Ω, 0 deg ±I, Ker L ∩ Ω,

Lemma 1.1. Let L be a Fredholm operator of index zero and let N be
10It follows that L| dom L∩Ker P : dom L ∩ Ker P → Im L is invertible.We denote the inverse of the map by K 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 andK p I − Q N : Ω → Y is compact.The lemma that we used is 17, Theorem 2.4 .