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Solvability of fractional boundary value problem with p-Laplacian operator at resonance
Advances in Difference Equations volume 2013, Article number: 295 (2013)
In this paper, a class of multi-point boundary value problems for nonlinear fractional differential equations at resonance with p-Laplacian operator is considered. By using the extension of Mawhin’s continuation theorem due to Ge, the existence of solutions is obtained, which enriches previous results.
In the recent years, fractional differential equations played an important role in many fields such as physics, electrical circuits, biology, control theory, etc. (see [1–7]). Thus, many scholars have paid more attention to fractional differential equations and gained some achievements (see [8–22]). For example, Wang  considered a class of fractional multi-point boundary value problems at resonance by Mawhin’s continuation theorem (see ):
where , , , , , , is the standard fractional derivative, satisfies the Carathéodory condition.
But Mawhin’s continuation theorem is not suitable for quasi-linear operators. In , Ge and Ren had extended Mawhin’s continuation theorem, which was used to deal with more general abstract operator equations. In , Pang et al. considered a higher order nonlinear differential equation with a p-Laplacian operator at resonance:
where , , and are continuous, is an integer. is a nondecreasing function with , the integral in the second part of (1.2) is meant in the Riemann-Stieltjes sense.
However, there are few articles which consider the fractional multi-point boundary value problem at resonance with p-Laplacian operator and , because p-Laplacian operator is a nonlinear operator, and it is hard to construct suitable continuous projectors. In this paper, we will improve and generalize some known results.
Motivated by the work above, our article is to investigate the multi-point boundary value problem at resonance for a class of Riemanne-Liouville fractional differential equations with p-Laplacian operator and by constructing suitable continuous projectors and using the extension of Mawhin’s continuation theorem:
where , , , , , , , , , , , , , , , , is invertible and its inverse operator is , is Riemann-Liouville standard fractional derivative, is continuous.
In order to investigate the problem, we need to suppose that the following conditions hold:
The rest of this article is organized as follows: In Section 2, we give some notations, definitions and lemmas. In Section 3, based on the extension of Mawhin’s continuation theorem due to Ge, we establish a theorem on existence of solutions for BVP (1.3).
For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory that can be found in the recent literature (see [1, 3, 14, 21, 24, 25]).
Let X and Y be two Banach spaces with norms and , respectively. A continuous operator
is said to be quasi-linear if
is a closed subset of Y,
is linearly homeomorphic to , .
Let and be the complement space of in X, then . On the other hand, suppose that is a subspace of Y, and is the complement space of in Y, so that . Let be a projector and a semi-projector, and an open and bounded set with origin . θ is the origin of a linear space.
Suppose that , is a continuous operator. Denote by N. Let . is said to be M-compact in if there is an with = and an operator continuous and compact such that for ,
and is the zero operator,
Lemma 2.1 (Ge-Mawhin’s continuation theorem )
Let and be two Banach spaces, and an open and bounded nonempty set. Suppose that is a quasi-linear operator , is M-compact in . In addition, if
where is a homeomorphism with and , then the equation has at least one solution in .
Definition 2.1 The Riemann-Liouville fractional integral of order of a function u is given by
provided the right-hand side integral is pointwise almost everywhere defined on .
Definition 2.2 The Riemann-Liouville fractional derivative of order of a function u is given by
provided the right-hand side integral is pointwise everywhere defined on , where .
Definition 2.3 Let X be a Banach space, and is a subspace. A mapping is a semi-projector if Q satisfies
, , .
Lemma 2.2 Assume that with a fractional derivative of order that belongs to . Then
for some , , where .
Lemma 2.3 Assume that , , then
Lemma 2.4 Assume that , then:
If , , , we have that
In this paper, we take with the norm , where , and with the norm . By means of the linear functional analysis theory, it is easy to prove that X and Y are Banach spaces, so we omit it.
Define the operator by
Based on the definition of domM, it is easy to find that such as , .
Define the operator , ,
Then BVP (1.3) is equivalent to the operator equation , where .
3 Main result
In this section, a theorem on existence of solutions for BVP (1.3) will be given.
Define operators , as follows:
Let us make some assumptions, which will be used in the sequel.
(H1) There exist nonnegative functions such that for all , ,
(H2) There exists a constant such that for , if for all , then
(H3) There exists a constant such that for , if for all , then
Theorem 3.1 Let be continuous and condition (H1)-(H3) hold, then BVP (1.3) has at least one solution, provided that
In order to prove Theorem 3.1, we need to prove some lemmas below.
Lemma 3.1 The operator is quasi-linear.
Proof Suppose that , by , we have
Based on , one has
which together with yields that
It is clear that . So, KerM is linearly homeomorphic to .
If , then there exists a function such that . Based on Lemmas 2.2 and 2.3, we have
which together with , and yields that , .
On the other hand, suppose that and satisfies (3.3), and let , then and , so and is a closed subset of Y. Thus, M is a quasi-linear operator. □
Lemma 3.2 Let be an open and bounded set, then is M-compact in .
Proof Define the continuous projector by
Define the continuous projector , by
Obviously, and . Thus, we have = . For any , we have
Similarly, we can get
Hence, the map Q is idempotent. Similarly, we can get , for all , . Thus, Q is a semi-projector. For any , we can get that and , conversely, if , we can obtain that , that is to say, . Thus, . Let be an open and bounded set with , for each , we can get . Thus, . Take any in the type , since , we can get . So, (2.1) holds. It is easy to verify (2.2).
By the continuity of f, it is easy to get that is continuous on . Moreover, for all , there exists a constant such that , so, we can easily obtain that , , and are uniformly bounded. By Arzela-Ascoli theorem, we just need to prove that is equicontinuous.
For , , , , , we have
Since is uniformly continuous on , so, , and are equicontinuous. Similarly, we can get that is equicontinuous. Considering that is uniformly continuous on , we have that is also equicontinuous. So, we can obtain that is compact.
For each , we have . Thus,
which together with yields that
It is easy to verify that is the zero operator. So, (2.3) holds. Besides, for any ,
which implies (2.4). So, is M-compact in . □
Lemma 3.3 Suppose that (H1), (H2) hold, then the set
Proof By Lemma 2.2, for each , we have
Combined with , we get . Thus,
By simple calculation, we get
Take any , then and . It follows from (H2) and (H3) that there exist such that , . Thus,
So, we get
where , .
Based on , we have
From (H1) and , we have
which together with , we can get
In view of (3.1), we can obtain that there exists a constant such that
Thus, we have
So, is bounded. □
Lemma 3.4 Suppose that (H2) holds, then the set
Proof For each , we have that , and . It follows from (H2) and (H3) that there exists an such that , , which implies that and . So, is bounded. □
Define the isomorphism by , , . In fact, for each , suppose that , we have
where , by the condition , there exists a unique solution for (3.4), which is , . Now, we will prove that there exists such that . Based on , we choose , where , , , , , , which together with , , , , we have and . So, we have . Thus, we can obtain that
where , which together with , we have . So, we have
Thus, there exists such that . So is well defined.
Lemma 3.5 Suppose that the first part of (H3) holds, then the set
Proof For each , we can get that , . By the definition of the set , we can obtain that
If , then is bounded because of the first part of (H2) and (H3). If , we get , obviously, is bounded. If , by the first part of (H2) and (3.5), we can obtain that , by the first part of (H3) and (3.6), we have . So, is bounded. □
Remark 3.1 If the second part of (H3) holds, then the set
Proof of Theorem 3.1 Assume that Ω is a bounded open set of X with . By Lemmas 3.1 and 3.2, we can obtain that is quasi-linear, and is M-compact on . By the definition of Ω , we have
Thus, by the homotopic property of degree, we can get
So Lemma 2.1 is satisfied, and has at least one solution in . Namely, BVP (1.3) have at least one solution in the space X. □
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The authors are grateful to those who gave useful suggestions about the original manuscript. This research is supported by the National Natural Science Foundation of China (No. 11271364) and the Fundamental Research Funds for the Central Universities (2010LKSX09).
The authors declare that they have no competing interests.
The authors contributed equally in this article. All authors read and approved the final manuscript.