Research Article | Open Access

Gang Wang, Wenbin Liu, Jinyun Yang, Sinian Zhu, Ting Zheng, "The Existence of Solutions for a Fractional 2-Point Boundary Value Problems", *Journal of Applied Mathematics*, vol. 2012, Article ID 841349, 18 pages, 2012. https://doi.org/10.1155/2012/841349

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

**Academic Editor:**Yongkun Li

#### Abstract

By using the coincidence degree theory, we consider the following 2-point boundary value problem for fractional differential equation , , where and 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.

#### 1. 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 , is the standard Caputo fractional derivative.

In [15], Zhang and Bai investigated the nonlinear nonlocal problem where , they consider the case , that is, the resonance case.

In [16], Bai investigated the boundary value problem at resonance is considered, where is a real number, and are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and is continuous and are given constants such that

In this paper, we study the 2-point boundary value problem where , , satisfies Carathéodory conditions, and are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.

Setting

In this paper, we will always suppose that the following conditions hold:(): ():

We say that boundary value problem (1.4) and (1.5) is at resonance, if BVP has , 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 , basing on the coincidence degree theory due to Mawhin (see [17]).

Now, we will briefly recall some notation and an abstract existence result.

Let be real Banach spaces, a Fredholm map of index zero,s and continuous projectors such that It follows that is invertible. We denote the inverse of the map by . If is an open-bounded subset of such that , the map will be called -compact on if is bounded and is compact.

The lemma that we used is [17, Theorem 2.4].

Lemma 1.1. *Let be a Fredholm operator of index zero and let be L-compact on . Assume that the following conditions are satisfied: *(i)*, for all ; *(ii)*, for all ; *(iii)*,**where is a projection as above with and is any isomorphism. Then the equation has at least one solution in .*

#### 2. 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 of a function is given by
provided the right side is pointwise defined on , where is the Gamma function.

*Definition 2.2. *The fractional derivative of order of a function is given by
where , provided the right side is pointwise defined on .

*Definition 2.3. *We say that the map satisfies Carathéodory conditions with respect to if the following conditions are satisfied: (i)for each , the mapping is Lebesgue measurable; (ii)for almost every , the mapping is continuous on ; (iii) for each , there exists such that for a.e. and every , we have .

Lemma 2.4 (see [15]). *Assume that with a fractional derivative of order that belongs to . Then
**
for some , where is the smallest integer greater than or equal to .*

We use the classical Banach space with the norm with the norm

*Definition 2.5. *For , we denote by the space of functions which have continuous derivatives up to order on such that is absolutely continuous: = and is absolutely continuous in .

Lemma 2.6 (see [15]). *Given and we can define a linear space
**
where . By means of the linear functional analysis theory, we can prove that with the
** is a Banach space.*

*Remark 2.7. *If is a natural number, then is in accordance with the classical Banach space .

Lemma 2.8 (see [15]). * is a sequentially compact set if and only if is uniformly bounded and equicontinuous. Here uniformly bounded means there exists , such that for every **
and equicontinuous means that , , such that
*

Lemma 2.9 (see [1]). *Let . Assume that with a fractional integration of order that belongs to . Then the equality
**
holds almost everywhere on .*

*Definition 2.10 (see [16]). *Let denote the space of functions , represented by fractional integral of order of a summable function: .

Let , with the norm , defined by Lemma 2.6, with the norm , where is a Banach space.

Define to be the linear operator from to with we define by setting Then boundary value problem (1.4) and (1.5) can be written as .

#### 3. Main Results

Lemma 3.1. *Let L be defined by (2.12), then
*

* Proof. *In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that for .

Let , by Lemma 2.9, has solution
Combine with (1.5), So,
Let and let

Then a.e. and, if
hold, then satisfies the boundary conditions (1.5). That is, and we have

Let . Then for , we have
where
which, due to the boundary value condition (1.5), implies that satisfies (3.5). In fact, from we have , from , , we have
Hence,
Therefore,
The proof is complete.

Lemma 3.2. *The mapping is a Fredholm operator of index zero, and
**
where
**
define by by
**
and .*

* Proof. *Consider the continuous linear mapping and defined by
Using the above definitions, we construct the following auxiliary maps and :
Since the condition (C2) holds, the mapping defined by
is well defined.

Recall (C2) and note that
and similarly we can derive that
So, for , it follows from the four relations above that
that is, the map is idempotent. In fact is a continuous linear projector.

Note that implies . Conversely, if , so
but
then we must have ; since the condition (C2) holds, this can only be the case if , that is, . In fact , take in the form so that , thus, , Let and assume that is not identically zero on . Then, since , from (3.5) and the condition (C2), we have
So
but
we derive , which is a contradiction. Hence, ; thus .

Now, and so is a Fredholm operator of index zero.

Let be defined by
Note that is a continuous linear projector and
It is clear that .

Note that the projectors and are exact. Define by by
Hence we have
then
and thus

In fact, if , then
Also, if , then
where
and from the boundary value condition (1.5) and the fact that , , , we have , thus
This shows that . The proof is complete. Using (3.16), we write

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

Lemma 3.3 (see [16]). *For every given is completely continuous.*

Assume that the following conditions on the function are satisfied.(H1)There exist functions , and a constant such that for all , , one of the following inequalities is satisfied:(H2)There exists a constant , such that for satisfying for all , we have (H3)There exists a constant such that for every satisfying then either

or

*Remark 3.4. * and from (H3) stand for the images of under the maps and , respectively.

Lemma 3.5. *Suppose (H1)-(H2) hold, then the set
**
is bounded.*

* Proof. *Take
Then for , thus , and hence for all . By the definition of , we have . It follows from (H2) that there exists , such that .

Now
and so
Therefore, we have
Note that for all . Then, by Lemma 3.2, we have
so, we have
where is a constant. This is for all . If the first condition of (H1) is satisfied, then, we have
where , and consequently, for
so
But and , so there exists such that
Therefore, for all ,
we can prove that is also bounded.

If (3.38) or (3.39) holds, similar to the above argument, we can prove that is bounded too.

Lemma 3.6. *Suppose (H3) holds, then the set
**
is bounded.*

* Proof. *Let
for and ; thus . By (H3), , that is, is bounded.

Lemma 3.7. *Suppose (H3) holds, then the set
**
is bounded.*

*Proof. *We define the isomorphism by

If the first part of (H3) is satisfied, let
For every ,
If , then , and if , 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
for all , that is, is bounded.

*Remark 3.8. *Suppose the second part of (H3) holds, then the set
is bounded.

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 to be a bounded open set of such that . It follows from Lemmas 3.2 and 3.3 that is a Fredholm operator of index zero, and the operator is compact , thus, is L-compact on . By Lemmas 3.5 and 3.6, we get that the following two conditions are satisfied: (i) for every ; (ii), for every .

Finally, we will prove that (iii) of Lemma 1.1 is satisfied. Let , where is the identity operator in the Banach space . According to Lemma 3.7 (or Remark 3.8), we know that , for all , and thus, by the homotopy property of degree,
Then by Lemma 1.1, has at least one solution in , so the boundary value problem (1.4) and (1.5) has at least one solution in the space . The proof is finished.

#### 4. An Example

Let us consider the following boundary value problem: where Corresponding to the problem (1.4)-(1.5), we have that , , , , , , , , , and then there is Again, taking , , then