The existence of solutions for p-Laplacian boundary value problem at resonance on the half-line 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:p-Laplacian; resonance; half-line; multi-point boundary value problem; continuation theorem
A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Resonance problems can be expressed as an abstract equation , where L is a noninvertible operator. When L is linear, Mawhin’s continuation theorem  is an effective tool in finding solutions for these problems, see [2-10] and references cited therein. But it does not work when L is nonlinear, for instance, p-Laplacian operator. In order to solve this problem, Ge and Ren  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 p-Laplacian:
where , , . is nonlinear when .
To the best of our knowledge, there are few papers that study the p-Laplacian boundary value problem at resonance on the half-line. In this paper, we investigate the existence of solutions for the boundary value problem
where , , .
In order to obtain our main results, we always suppose that the following conditions hold.
(H1) , , .
(H2) is continuous, , and for any , there exists a nonnegative function such that
For convenience, we introduce some notations and a theorem. For more details, see .
Let X and Y be two Banach spaces with the norms , , respectively. A continuous operator is said to be quasi-linear if
(i) is a closed subset of Y,
(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 .
Suppose that , is a continuous operator. Denote by N. Let . is said to be M-compact 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 ,
LetXandYbe two Banach spaces with the norms , , respectively, and an open and bounded nonempty set. Suppose that
is a quasi-linear operator and , M-compact. In addition, if the following conditions hold:
(C1) , , ,
then the abstract equation has 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:
Then the boundary value problem (1.1) is equivalent to .
It is clear that KerM is linearly homeomorphic to ℝ, and is closed. So, M is a quasi-linear operator.
Define , as
where , . We can easily obtain that , are projectors. Set , .
Define an operator :
where , . By (H1) and (H2), we get that is continuous.
is compact if and are both equicontinuous on any compact intervals of and equiconvergent at infinity.
Lemma 3.2 is compact.
Proof Let be nonempty and bounded. There exists a constant such that , . It follows from (H2) that there exists a nonnegative function such that
For any , , , , we have
Since are equicontinuous on , we get that are equicontinuous on .
For , , , we have
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 .
For , since
and is uniformly continuous on , for any , there exists a constant such that if , then
there exists a constant such that if , then
For , by (3.2), (3.3) and (3.4), we have
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 3.3Let be nonempty, open and bounded. Then isM-compact in .
Proof By (H2), 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,
For , we have
These, together with Lemma 3.2, mean that is M-compact in . The proof is completed. □
In order to obtain our main results, we need the following additional conditions.
(H3) There exist nonnegative functions , , with and such that
(H4) There exists a constant such that if , then one of the following inequalities holds:
Lemma 3.4Assume that (H3) and (H4) hold. The set
is bounded inX.
Proof If , then , i.e., . By (H4), there exists such that . It follows from that
Considering (H3), we have
Since , we get
By (3.5), (3.6) and (H3), we get
This, together with (3.6), means that is bounded. The proof is completed. □
Lemma 3.5Assume that (H4) holds. The set
is bounded inX.
Proof means that , and , i.e.,
By (H4), we get that . So, is bounded. The proof is completed. □
Theorem 3.1Suppose that (H1)-(H4) 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
For , we have . Then
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. □
Let us consider the following boundary value problem at resonance
where , , .
Corresponding to problem (1.1), we have , .
Take , , , . By simple calculation, we can get that conditions (H1)-(H4) hold. By Theorem 3.1, we obtain that problem (4.1) has at least one solution.
The author declares that she has no competing interests.
All results belong to WJ.
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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