, 2013:233

# Existence results for a functional boundary value problem of fractional differential equations

• Yumei Zou
• Yujun Cui
Open Access
Research

## Abstract

In this paper, a functional boundary value problem of fractional differential equations is studied. Based on Mawhin’s coincidence degree theory, some existence theorems are obtained in the case of nonresonance and the cases of and at resonance.

## 1 Introduction

The subject of fractional calculus has gained considerable popularity and importance because of its frequent appearance in various fields such as physics, chemistry, and engineering. In consequence, the subject of fractional differential equations has attracted much attention. Many methods have been introduced to solve fractional differential equations, such as the popular Laplace transform method, the iteration method, the Fourier transform method and the operational method. For details, see [1, 2, 3] and the references therein. Recently, there have been some papers dealing with the basic theory for initial value problems of nonlinear fractional differential equations; for example, see [4, 5]. Also, there are some articles which deal with the existence and multiplicity of solutions for nonlinear boundary value problems of fractional order differential equations using techniques of topological degree theory. We refer the reader to [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16] for some recent results at nonresonance and to [17, 18, 19, 20, 21, 22, 23, 24, 25, 26] at resonance.

In [18], by making use of the coincidence degree theory of Mawhin, Zhang and Bai discussed the existence results for the following nonlinear nonlocal problem at resonance under the case :
Recently, Jiang [26] studied the existence of a solution for the following fractional differential equation at resonance under the case :
Being directly inspired by [18, 20, 26], we intend in this paper to study the following functional boundary value problems (FBVP) of fractional order differential equation:
(1.1)
(1.2)

where , and are the standard Riemann-Liouville differentiation and integration, and ; are continuous linear functionals.

In this paper, we shall give some sufficient conditions to construct the existence theorems for FBVP (1.1), (1.2) at nonresonance and resonance (both cases of and ), respectively. To the best of our knowledge, the method of Mawhin’s theorem has not been developed for fractional order differential equation with functional boundary value problems at resonance. So, it is interesting and important to discuss the existence of a solution for FBVP (1.1), (1.2). Many difficulties occur when we deal with them. For example, the construction of the generalized inverse of L. So, we need to introduce some new tools and methods to investigate the existence of a solution for FBVP (1.1), (1.2).

The rest of this paper is organized as follows. In Section 2, we give some notations and lemmas. In Section 3, we establish the existence results of a solution for functional boundary value problem (1.1), (1.2).

## 2 Preliminaries and lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions and properties can be found in the literature. The readers who are unfamiliar with this area can consult, for example, [1, 2, 4] for details.

Definition 2.1 [1, 2]

The Riemann-Liouville fractional integral of order of a function is given by

provided that the right-hand side is pointwise defined on . Here is the Gamma function given by .

Definition 2.2 [1, 2]

The Riemann-Liouville fractional derivative of order of a continuous function is given by

where , provided that the right-hand side is pointwise defined on .

We use the classical spaces with the norm , with the norm . We also use the space defined by
and the Banach space ()

with the norm .

Lemma 2.1 [2]

Let , . Assume that with a fractional integration of order that belongs to . Then the equality

holds almost everywhere on .

Remark 2.1 If u satisfies and , then . In fact, with Lemma 2.1, one has
Combine with , there is . So,

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

Lemma 2.2 [2]

Assume , then:
1. (i)
Let . If and exist, then

2. (ii)
If , , then

is satisfied at any point on for and ;
1. (iii)

Let . Then holds on ;

2. (iv)
Note that for , , we have

Lemma 2.3 [18]

is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here ‘F is uniformly bounded and equicontinuous’ means that there exists such that for every ,
and that , , for all , , , , there hold

respectively.

Next, consider the following conditions:

(A1) .

(A2) , , .

(A3) , , .

(A4) , , .

(A5) , , .

We shall prove that: If (A1) holds, then . It is the so-called nonresonance case. If (A2) holds, then . If (A3) or (A4) holds, then . If (A5) holds, then .

In the nonresonance case, FBVP (1.1), (1.2) can be transformed into an operator equation.

Lemma 2.4 Assume that (A1) holds. Then functional boundary value problem (1.1) and (1.2) has a solution if and only if the operator , defined by

has a fixed point, where .

Proof If u is a solution to , by Lemma 2.2, we get
and
Considering the linearity of (), we have

So, u is a solution to FBVP (1.1), (1.2).

If u is a solution to (1.1), by Lemma 2.1, we can reduce (1.1) to an equivalent integral equation
(2.1)
By , there is , and
(2.2)
(2.3)
Applying and to (2.2) and (2.3), respectively, we obtain
Thus,
(2.4)
(2.5)
Substituting (2.4) and (2.5) into (2.1), we obtain

The proof is complete. □

The following definitions and lemmas are a preparation for the existence of solutions to (1.1), (1.2) at resonance.

Definition 2.3 Let Y, Z be real Banach spaces, let be a linear operator. L is said to be a Fredholm operator of index zero provided that:
1. (i)

ImL is a closed subset of Z,

2. (ii)

.

Let Y, Z be real Banach spaces and be a Fredholm operator of index zero. , are continuous projectors such that , , and . It follows that is invertible. We denote the inverse of the mapping by (generalized inverse operator of L). If Ω is an open bounded subset of Y such that , the mapping will be called L-compact on , if is bounded and is compact.

We need the following known result for the sequel (Theorem 2.4 [27]).

Theorem 2.1 Let L be a Fredholm operator of index zero, and let N be L-compact on . Assume that the following conditions are satisfied:
1. (i)

for every .

2. (ii)

for every .

3. (iii)

, where is a projector as above with .

Then the equation has at least one solution in .

Let , . Let the linear operator with
be defined by . Let the nonlinear operator be defined by
Then (1.1), (1.2) can be written as

Now, we give KerL, ImL and some necessary operators at and , respectively.

Lemma 2.5 Let L be the linear operator defined as above. If (A2) holds, then
and
Proof Let . Clearly, and . Considering (A2), and . So,

If , then . Considering and (A2), we can obtain that . It yields and .

We now show that
If , then there exists such that . Hence,
for some . It yields
Therefore
On the other hand, suppose satisfies
Let
Obviously, and . Considering (A2) and the linearity of (), we have
and
It yields

The proof is complete. □

Lemma 2.6 If , then:
1. (i)

L is a Fredholm operator of index zero and .

2. (ii)
The linear operator can be defined by

3. (iii)

, where and is the norm of a continuous linear functional .

4. (iv)

The linear operator is completely continuous.

Proof Firstly, we construct the mapping defined by
(2.6)
Noting that

we get is a well-defined projector.

Now, it is obvious that . Noting that Q is a linear projector, we have . Hence, and 1. This means L is a Fredholm mapping of index zero. Taking as
then the generalized inverse of L can be rewritten
In fact, for , we have
and
which implies that is well defined on ImL. Moreover, for , we have
and for , we know
means that . So,
That is, . Since
then
It follows that
Finally, we prove that is completely continuous. Let be a bounded set. From the above discussion, we only need to prove that is equicontinuous on . For , with , we have
and

Therefore, is equicontinuous. Thus, the operator is completely continuous. The proof is complete. □

Similar to Lemmas 2.5 and 2.6, we can obtain the following lemma.

Lemma 2.7 If (A3) holds, then and
Furthermore, if also holds, then L is a Fredholm operator of index zero and . Here, the projectors , can be defined as follows:
The generalized inverse operator of can be defined by
Also,

where .

Lemma 2.8 If (A4) holds, then and
Furthermore, if also holds, then L is a Fredholm operator of index zero and . Here, the projectors , can be defined as follows:
The generalized inverse operator of can be defined by
Also,

where and is the norm of the continuous linear functional .

Lemma 2.9 If (A5) holds, then
and
Proof Let . Clearly, and . Considering (A5), and . So,
If , then . Considering and (A5), we can obtain that
We now show that
If , then there exists such that . Hence,
for some . It yields
and
Therefore,
On the other hand, suppose satisfies
Let
Obviously, and . Considering (A5) and the linearity of (), we have
and
It yields

The proof is complete. □

Lemma 2.10 If , then L is a Fredholm operator of index zero and . Furthermore, the linear operator can be defined by
Also,
Proof Firstly, we construct the mapping defined by
Let
and
We have
(2.7)
Noting that
and
we have, for each , that

So, is a well-defined projector.

Now we will show that . If , from , we have and . Considering the definitions of and , we have
Since

so , which yields . On the other hand, if , from and the definition of Q, it is obvious that , thus . Hence, .

For , from , , , we have . And for any , from , there exist constants such that . From , we obtain
(2.8)
In view of
therefore (2.8) has a unique solution , which implies and . Since , thus L is a Fredholm map of index zero. Let be defined by
Then the generalized inverse of L can be rewritten
In fact, for , we have
and
which implies that is well defined on ImL. Moreover, for , we have
and for , we know
means that . So,
That is, . Since
then
It follows that

The proof is complete. □

## 3 Main results

From Lemma 2.4, we can obtain the existence theorem for FBVP (1.1), (1.2).

Theorem 3.1 Assume that (A1) and the following conditions hold:
Then FBVP (1.1), (1.2) has a unique solution in provided that
Proof We shall prove that has a unique solution in . For each , considering the linearity of (), we have
Then
and
So,

The above inequality implies that T is a contraction. By using Banach’s contraction principle, has a unique solution in . From Lemma 2.4, FBVP (1.1), (1.2) has a unique solution in . The proof is complete. □

From Lemmas 2.5-2.8 and Theorem 2.1, we can obtain the existence theorem for FBVP (1.1), (1.2) in the case of .

Theorem 3.2 Let be a continuous function. Assume that , (A2) and the following conditions (H1)-(H3) hold:

(H1) There exist functions such that for all , ,

(H2) There exists a constant such that for , if for all , then .

(H3) There exists a constant such that either for each ,
or for each ,
Then FBVP (1.1), (1.2) has at least one solution in provided

where is the same as in Lemma  2.6.

Proof Set
Then, for , since , so , , hence
Thus, from (H2), there exists such that
Now,
and so
(3.1)
Again, for , , then and . Thus, from Lemma 2.6, we have
(3.2)
From (3.1), (3.2), we have
By this and (H1), we have
and
Therefore, is bounded. Let
For , there is , and , thus

From (H2), we get , thus is bounded.

Next, according to the condition (H3), for any , if , then either
(3.3)
or else
(3.4)
If (3.3) holds, set
here Q is given by (2.6) and is the linear isomorphism given by , , . For ,
If , then . Otherwise, if , in view of (3.3), one has

which contradicts . Thus, is bounded.

If (3.4) holds, then define the set

here J is as above. Similar to the above argument, we can show that is bounded too.

In the following, we shall prove that all the conditions of Theorem 2.1 are satisfied. Let Ω be a bounded open subset of Y such that . By Lemma 2.6 and standard arguments, we can prove that is compact, thus N is L-compact on . Then, by the above argument, we have
1. (i)

, for every ,

2. (ii)

for $u\in$