Abstract
By generalizing the extension of the continuous theorem of Ge and Ren and constructing suitable Banach spaces and operators, we investigate the existence of solutions for pLaplacian boundary value problems at resonance. An example is given to illustrate our results.
MSC: 34B15.
Keywords:
continuous theorem; resonance; pLaplacian; boundary value problem1 Introduction
In this paper, we will study the boundary value problem
and
A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a nontrivial solution. Mawhin’s continuous theorem [1] is an effective tool to solve this kind of problems when the differential operator is linear, see [210] and references cited therein. But it does not work for nonlinear cases such as boundary value problems with a pLaplacian, which attracted the attention of mathematicians in recent years [1115]. Ge and Ren extended Mawhin’s continuous theorem [15] and many authors used their results to solve boundary value problems with a pLaplacian, see [16,17]. In this new theorem, two projectors P and Q must be constructed. But it is difficult to give the projector Q in many boundary value problems with a pLaplacian. In this paper, we generalize the extension of the continuous theorem and show that the pLaplacian problem is solvable when Q is not a projector. And we will use this new theorem to discuss problems (1.1) and (1.2), respectively.
In this paper, we will always suppose that
2 Preliminaries
Definition 2.1[15]
Let X and Y be two Banach spaces with 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 . Let be a projector and an open and bounded set with the origin .
Definition 2.2 Suppose , is a continuous and bounded operator. Denote by N. Let . is said to be Mquasicompact in if there exists a vector subspace of Y satisfying and two operators Q, R with , , being continuous, bounded, and satisfying , continuous and compact such that for ,
(c) is the zero operator and ,
Theorem 2.1LetXandYbe two Banach spaces with the norms, , respectively, and letbe an open and bounded nonempty set. Suppose
is a quasilinear operator and that, isMquasicompact. In addition, if the following conditions hold:
then the abstract equationhas at least one solution in, where, is a homeomorphism with.
Proof The proof is similar to the one of Lemma 2.1 and Theorem 2.1 in [15]. □
We can easily get the following inequalities.
3 The existence of a solution for problem (1.1)
Let with norm , with norm , where . We know that and are Banach spaces.
Define operators , as follows:
Lemma 3.1For, there is only one constantsuch thatwithand thatis continuous.
Obviously, is continuous and strictly decreasing in ℝ. Take , . It is easy to see that , . Thus, there exists a unique constant such that , i.e. there is only one constant such that with .
For , assume , . By , and being strictly increasing, we obtain, if , then
This is a contradiction. On the other hand, if , then
This is a contradiction, too. So, we have , i.e.. So, is continuous. The proof is completed. □
It is clear that is a solution if and only if it satisfies , where . For convenience, let .
Lemma 3.2Mis a quasilinear operator.
Proof It is easy to see that .
For , if , then c satisfies (3.1). On the other hand, if , , take
By a simple calculation, we get and . Thus
By the continuity of T, we find that is closed. So, M is quasilinear. The proof is completed. □
Proof The proof is simple. Therefore, we omit it. □
Take a projector and an operator as follows:
By the continuity and boundedness of T, we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that , .
where . By (H_{2}) and the ArzelaAsscoli theorem, we can easily see that is continuous and compact, where is an open bounded set.
Lemma 3.4Assume thatis an open bounded set. ThenisMquasicompact in .
Proof It is clear that , , and . For ,
Since and , we obtain . Thus, .
i.e. Definition 2.2(c) holds. For , we have
So, Definition 2.2(d) holds. Therefore, is Mquasicompact in . The proof is completed. □
Theorem 3.1Assume that the following conditions hold.
(H_{3}) There exists a nonnegative constantKsuch that one of (1) and (2) holds:
(H_{4}) There exist nonnegative functionssuch that
Then boundary value problem (1.1) has at least one solution.
In order to prove Theorem 3.1, we show two lemmas.
Lemma 3.5Suppose (H_{3}) and (H_{4}) hold. Then the set
is bounded inX.
Proof For , we have , i.e.. By (H_{3}), there exists a constant such that . Since , , we have
It follows from , (H_{4}), and (3.2) that
thus
thus
These, together with (3.2), mean that is bounded in X. □
Lemma 3.6Assume (H_{3}) holds. Then
Proof For , we have and . By (H_{3}), we get . So, is bounded. The proof is completed. □
Proof of Theorem 3.1 Let , where r is large enough such that and .
By Lemmas 3.5 and 3.6, we know , and , .
Let , , , where is a homeomorphism with ,
If , . If , by , we get . For , we now prove that . Otherwise, if , then
Since , we have . Thus, . So, we have . A contradiction with the definition of ρ. So, , , .
By the homotopy of degree, we get
By Theorem 2.1, we can see that has at least one solution in . The proof is completed. □
Example Let us consider the following boundary value problem at resonance:
Corresponding to problem (1.1), we have , , , , , .
Take . By a simple calculation, we find that the conditions (H_{1})(H_{4}) hold. By Theorem 3.1, we obtain the result that problem (3.4) has at least one solution.
4 The existence of a solution for problem (1.2)
Let with norm , with norm , where . We know that and are Banach spaces.
Define operators , as follows:
Lemma 4.1For, there is only one constantsuch thatwith. Andare continuous, .
The proof is similar to Lemma 3.1.
It is clear that is a solution if and only if it satisfies , where . For convenience, let .
Lemma 4.2Mis a quasilinear operator.
For , if , then , satisfy (4.1). On the other hand, if , , , take
By simple calculation, we get and . Thus
By the continuity of , , we see that is closed. So, M is quasilinear. The proof is completed. □
Take a projector and an operator as follows:
By the continuity and boundedness of , , we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that , .
where . By (H_{2}) and the ArzelaAsscoli theorem, we can easily see that is continuous and compact, where is an open bounded set.
Lemma 4.3Assume thatis an open bounded set. ThenisMquasicompact in .
Proof It is clear that , , and . For ,
Since and , we obtain . Thus, .
i.e. Definition 2.2(c) holds. For , we have
Thus, Definition 2.2(d) holds. Therefore, is Mquasicompact in . The proof is completed. □
Theorem 4.1Assume that the following conditions hold:
(H_{5}) There exists a nonnegative constantLsuch that if, then either
or
(H_{6}) There exist nonnegative constants, such that one of (1) and (2) holds:
(1)
and
(2)
and
(H_{7}) There exist nonnegative functionssuch that
Then boundary value problem (1.2) has at least one solution.
In order to prove Theorem 4.1, we show two lemmas.
Lemma 4.4Suppose (H_{5})(H_{7}) hold. Then the set
is bounded inX.
Proof For , we have , i.e., . By (H_{5}) and (H_{6}), there exist constants such that , . Since , , then
It follows from , (H_{7}), and (4.2) that
thus
thus
These, together with (4.2), mean that is bounded in X. □
Lemma 4.5Assume (H_{6}) holds. Then
Proof For , we have and . By (H_{6}), we see that there exists a constant such that , . So, is bounded. The proof is completed. □
Proof of Theorem 4.1 Let , where r is large enough such that and .
By Lemmas 4.4 and 4.5, we know , and , .
Let , , , where is a homeomorphism with ,
Take the function is the same as the one in Proof of Theorem 3.1.
If , . If , by , we get . For , we now prove that . Otherwise, if , then
Since , we have either or . If , then . So, we have . This is a contradiction with the definition of ρ. If , then . Thus and . By , we get . This is a contradiction with the definition of ρ, too. So, , , .
By the homotopy of degree, we get
By Theorem 2.1, we find that (1.2) has at least one solution in . The proof is completed. □
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 improvement of the original manuscript.
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