Solvability for p-Laplacian boundary value problem at resonance on the half-line
© Jiang; licensee Springer 2013
Received: 5 April 2013
Accepted: 23 August 2013
Published: 11 September 2013
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.
where , , . is nonlinear when .
where , , .
In order to obtain our main results, we always suppose that the following conditions hold.
(H1) , , .
For convenience, we introduce some notations and a theorem. For more details, see .
Definition 2.1 
is a closed subset of Y,
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 
is the zero operator and ,
Theorem 2.1 
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.
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.
where , . We can easily obtain that , are projectors. Set , .
where , . By (H1) and (H2), we get that is continuous.
Lemma 3.1 
is compact if and are both equicontinuous on any compact intervals of and equiconvergent at infinity.
Lemma 3.2 is compact.
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 .
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 3.3 Let be nonempty, open and bounded. Then is M-compact in .
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.
is bounded in X.
This, together with (3.6), means that is bounded. The proof is completed. □
is bounded in X.
By (H4), we get that . So, is bounded. The proof is completed. □
Theorem 3.1 Suppose 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 , .
and the contradiction follows analogously. So, we obtain , , .
By Theorem 2.1, we can get that has at least one solution in . The proof is completed. □
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.
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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