Abstract
The existence of solutions for pLaplacian boundary value problem at resonance on the halfline is investigated. Our analysis relies on constructing the suitable Banach space, defining appropriate operators and using the extension of Mawhin’s continuation theorem. An example is given to illustrate our main result.
MSC: 70K30, 34B10, 34B15.
Keywords:
pLaplacian; resonance; halfline; multipoint boundary value problem; continuation theorem1 Introduction
A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a nontrivial solution. Resonance problems can be expressed as an abstract equation , where L is a noninvertible operator. When L is linear, Mawhin’s continuation theorem [1] is an effective tool in finding solutions for these problems, see [210] and references cited therein. But it does not work when L is nonlinear, for instance, pLaplacian operator. In order to solve this problem, Ge and Ren [11] proved a continuation theorem for the abstract equation when L is a noninvertible nonlinear operator and used it to study the existence of solutions for the boundary value problems with a pLaplacian:
where , , . is nonlinear when .
As far as the boundary value problems on unbounded domain are concerned, there are many excellent results, see [1215] and references cited therein.
To the best of our knowledge, there are few papers that study the pLaplacian boundary value problem at resonance on the halfline. In this paper, we investigate the existence of solutions for the boundary value problem
In order to obtain our main results, we always suppose that the following conditions hold.
(H_{2}) is continuous, , and for any , there exists a nonnegative function such that
2 Preliminaries
For convenience, we introduce some notations and a theorem. For more details, see [11].
Definition 2.1[11]
Let X and Y be two Banach spaces with the norms , , respectively. A continuous operator is said to be quasilinear if
(ii) is linearly homeomorphic to , , where domM denote the domain of the operator M.
Let and be the complement space of in X, then . On the other hand, suppose that is a subspace of Y, and that is the complement of in Y, i.e., . Let and be two projectors and an open and bounded set with the origin .
Definition 2.2[11]
Suppose that , is a continuous operator. Denote by N. Let . is said to be Mcompact in if there exist a vector subspace of Y satisfying and an operator being continuous and compact such that for ,
(c) is the zero operator and ,
Theorem 2.1[11]
LetXandYbe two Banach spaces with the norms, , respectively, andan open and bounded nonempty set. Suppose that
is a quasilinear operator and, Mcompact. In addition, if the following conditions hold:
then the abstract equationhas at least one solution in, where, is a homeomorphism with.
3 Main result
Let with norm , where . with norm . Then and are Banach spaces.
Define operators and as follows:
where
Then the boundary value problem (1.1) is equivalent to .
Obviously,
It is clear that KerM is linearly homeomorphic to ℝ, and is closed. So, M is a quasilinear operator.
where , . We can easily obtain that , are projectors. Set , .
where , . By (H_{1}) and (H_{2}), we get that is continuous.
Lemma 3.1[15]
is compact ifandare both equicontinuous on any compact intervals ofand equiconvergent at infinity.
Proof Let be nonempty and bounded. There exists a constant such that , . It follows from (H_{2}) that there exists a nonnegative function such that
Since are equicontinuous on , we get that are equicontinuous on .
Let
Then
It follows from the absolute continuity of integral that are equicontinuous on . Since is uniformly continuous on , by (3.1), we can obtain that are equicontinuous on .
and is uniformly continuous on , for any , there exists a constant such that if , then
Since
there exists a constant such that if , then
For , by (3.2), (3.3) and (3.4), we have
and
By Lemma 3.1, we get that is compact. The proof is completed. □
In the spaces X and Y, the origin . In the following sections, we denote the origin by 0.
Lemma 3.3Letbe nonempty, open and bounded. ThenisMcompact in.
Proof By (H_{2}), we know that is continuous. Obviously, . For , since is a zero operator, we get . For , . So, we have . It is clear that
and , . means that and , thus,
These, together with Lemma 3.2, mean that is Mcompact in . The proof is completed. □
In order to obtain our main results, we need the following additional conditions.
(H_{3}) There exist nonnegative functions , , with and such that
(H_{4}) There exists a constant such that if , then one of the following inequalities holds:
Lemma 3.4Assume that (H_{3}) and (H_{4}) hold. The set
is bounded inX.
Proof If , then , i.e., . By (H_{4}), there exists such that . It follows from that
Considering (H_{3}), we have
Thus,
By (3.5), (3.6) and (H_{3}), we get
So,
This, together with (3.6), means that is bounded. The proof is completed. □
Lemma 3.5Assume that (H_{4}) holds. The set
is bounded inX.
Proof means that , and , i.e.,
By (H_{4}), we get that . So, is bounded. The proof is completed. □
Theorem 3.1Suppose that (H_{1})(H_{4}) hold. Then problem (1.1) has at least one solution.
Proof Let , where . It follows from the definition of and that , , and , .
Define a homeomorphism as . If for , take the homotopy
Obviously, , . For , , if , we have
A contradiction with , . If , , take
and the contradiction follows analogously. So, we obtain , , .
By the homotopy of degree, we get that
By Theorem 2.1, we can get that has at least one solution in . The proof is completed. □
4 Example
Competing interests
The author declares that she has no competing interests.
Author’s contributions
All results belong to WJ.
Acknowledgements
This work is supported by the National Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108). The author is grateful to anonymous referees for their constructive comments and suggestions, which led to the improvement of the original manuscript.
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