Solvability of fractional boundary value problem with p-Laplacian operator at resonance
© Shen et al.; licensee Springer. 2013
Received: 1 January 2013
Accepted: 2 September 2013
Published: 7 November 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.
where , , , , , , is the standard fractional derivative, satisfies the Carathéodory condition.
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.
where , , , , , , , , , , , , , , , , is invertible and its inverse operator is , is Riemann-Liouville standard fractional derivative, is continuous.
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]).
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.
Lemma 2.1 (Ge-Mawhin’s continuation theorem )
where is a homeomorphism with and , then the equation has at least one solution in .
provided the right-hand side integral is pointwise almost everywhere defined on .
provided the right-hand side integral is pointwise everywhere defined on , where .
, , .
for some , , where .
- (i)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.
Based on the definition of domM, it is easy to find that such as , .
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.
Let us make some assumptions, which will be used in the sequel.
In order to prove Theorem 3.1, we need to prove some lemmas below.
It is clear that . So, KerM is linearly homeomorphic to .
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 .
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.
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.
which implies (2.4). So, is M-compact in . □
where , .
So, is bounded. □
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. □
Thus, there exists such that . So is well defined.
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. □
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. □
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).
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