Research

# Solvability of boundary value problem with p-Laplacian at resonance

Weihua Jiang

Author Affiliations

College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei, 050018, P.R. China

Boundary Value Problems 2014, 2014:36  doi:10.1186/1687-2770-2014-36

 Received: 29 October 2013 Accepted: 24 January 2014 Published: 7 February 2014

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### 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 p-Laplacian boundary value problems at resonance. An example is given to illustrate our results.

MSC: 34B15.

##### Keywords:
continuous theorem; resonance; p-Laplacian; boundary value problem

### 1 Introduction

In this paper, we will study the boundary value problem

(1.1)

and

(1.2)

where , , , , .

A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Mawhin’s continuous theorem [1] is an effective tool to solve this kind of problems when the differential operator is linear, see [2-10] and references cited therein. But it does not work for nonlinear cases such as boundary value problems with a p-Laplacian, which attracted the attention of mathematicians in recent years [11-15]. Ge and Ren extended Mawhin’s continuous theorem [15] and many authors used their results to solve boundary value problems with a p-Laplacian, 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 p-Laplacian. In this paper, we generalize the extension of the continuous theorem and show that the p-Laplacian 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

(H1) are nonnegative and , where .

(H2) is continuous in .

### 2 Preliminaries

Definition 2.1[15]

Let X and Y be two Banach spaces with 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 . 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 M-quasi-compact 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 ,

(a) ,

(b) , ,

(c) is the zero operator and ,

(d) .

Theorem 2.1LetXandYbe two Banach spaces with the norms, , respectively, and letbe an open and bounded nonempty set. Suppose

is a quasi-linear operator and that, isM-quasi-compact. In addition, if the following conditions hold:

(C1) , , ,

(C2) ,

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.

Lemma 2.1For any, we have

(1) , .

(2) , .

In the following, we will always suppose that q satisfies .

### 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:

where , , c satisfying

(3.1)

Lemma 3.1For, there is only one constantsuch thatwithand thatis continuous.

Proof For , let

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 quasi-linear 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 quasi-linear. The proof is completed. □

Lemma 3.3, , , , .

Proof The proof is simple. Therefore, we omit it. □

Take a projector and an operator as follows:

where . Obviously, , and .

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 , .

Define an operator as

where . By (H2) and the Arzela-Asscoli theorem, we can easily see that is continuous and compact, where is an open bounded set.

Lemma 3.4Assume thatis an open bounded set. ThenisM-quasi-compact in .

Proof It is clear that , , and . For ,

Since and , we obtain . Thus, .

For , we get

i.e. Definition 2.2(c) holds. For , we have

So, Definition 2.2(d) holds. Therefore, is M-quasi-compact in . The proof is completed. □

Theorem 3.1Assume that the following conditions hold.

(H3) There exists a nonnegative constantKsuch that one of (1) and (2) holds:

(1) , , , ,

(2) , , , .

(H4) There exist nonnegative functionssuch that

where, if; , if.

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 (H3) and (H4) hold. Then the set

is bounded inX.

Proof For , we have , i.e.. By (H3), there exists a constant such that . Since , , we have

(3.2)

It follows from , (H4), and (3.2) that

If , by Lemma 2.1, we get

thus

where , .

If , by Lemma 2.1, we get

thus

where , .

These, together with (3.2), mean that is bounded in X. □

Lemma 3.6Assume (H3) holds. Then

is bounded inX, where.

Proof For , we have and . By (H3), 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 ,

Define a function

For , we have . Thus

If , . If , by , we get . For , we now prove that . Otherwise, if , then

(3.3)

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:

(3.4)

where .

Corresponding to problem (1.1), we have , , , , , .

Take . By a simple calculation, we find that the conditions (H1)-(H4) 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:

where , , , , satisfy

(4.1)

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 quasi-linear operator.

Proof It is easy to get .

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 quasi-linear. The proof is completed. □

Take a projector and an operator as follows:

where . Obviously, , and .

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 , .

Define an operator as

where . By (H2) and the Arzela-Asscoli theorem, we can easily see that is continuous and compact, where is an open bounded set.

Lemma 4.3Assume thatis an open bounded set. ThenisM-quasi-compact in .

Proof It is clear that , , and . For ,

Since and , we obtain . Thus, .

For , we get

i.e. Definition 2.2(c) holds. For , we have

Thus, Definition 2.2(d) holds. Therefore, is M-quasi-compact in . The proof is completed. □

Theorem 4.1Assume that the following conditions hold:

(H5) There exists a nonnegative constantLsuch that if, then either

or

(H6) There exist nonnegative constants, such that one of (1) and (2) holds:

(1)

and

(2)

and

(H7) There exist nonnegative functionssuch that

where, if; , if.

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 (H5)-(H7) hold. Then the set

is bounded inX.

Proof For , we have , i.e., . By (H5) and (H6), there exist constants such that , . Since , , then

(4.2)

It follows from , (H7), and (4.2) that

If , by Lemma 2.1, we get

thus

where , .

If , by Lemma 2.1, we get

thus

where , .

These, together with (4.2), mean that is bounded in X. □

Lemma 4.5Assume (H6) holds. Then

is bounded inX, where.

Proof For , we have and . By (H6), 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.

For , we have . Thus

If , . If , by , we get . For , we now prove that . Otherwise, if , then

(4.3)

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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