Research Article | Open Access

Inbo Sim, Yong Hoon Lee, "A New Solution Operator of One-Dimensional -Laplacian with a Sign-Changing Weight and Its Application", *Abstract and Applied Analysis*, vol. 2012, Article ID 243740, 15 pages, 2012. https://doi.org/10.1155/2012/243740

# A New Solution Operator of One-Dimensional -Laplacian with a Sign-Changing Weight and Its Application

**Academic Editor:**Ferhan M. Atici

#### Abstract

We establish a new solution operator for the following problem , , , where , . may be singular at the boundary or change signs or may not be in so that this solution operator can cover larger class of than previously known ones. As an application, by checking complete continuity of the solution operator, we show the existence of nontrivial solutions for -Laplacian , , , where a parameter and and and may change signs or may be beyond of .

#### 1. Introduction

The solution operator approach has been used to show the existence of solutions for wide class of nonlinear differential equations. As it is wellknown, a solution operator plays a key role when we employ nonlinear analytic methods such as degree theory, fixed point index theory, and bifurcation theory. In this paper, we are concerned with a solution operator for one-dimensional -Laplacian with a sign-changing singular weight.

Let us summarize a brief history about solution operators for one-dimensional -Laplacian with Dirichlet boundary condition where , , and .

First, when , a solution operator was introduced by Manásevich and Mawhin [1, 2]. In this case, we know that all solutions of are in so that by a solution of problem , we understand a function with absolutely continuous which satisfies . Key step of establishing a solution operator for this case is to show that, for each , the equation has a unique solution where is a continuous function. Defining problem can be equivalently written as .

Secondly, when and satisfies a solution operator was studied by Agarwal et al. [3]. We note that a function satisfying (1.3) needs not be in . In this case, we know that solutions of may not be in so that, by a solution of problem , we understand a function with absolutely continuous which satisfies . Define a function by for . It is easy to see that is continuous, strictly increasing in and . Thus, has a unique zero in , say . Then, we see that and defining problem can be equivalently written as .

Now let us consider the case that may change signs and belongs to which is defined as Defining by where , it is not hard to check that , but .

For but , existence of solution for (1.1) is not known so that Manásevich-Mawhin's approach cannot be applied directly. For sign changing, the uniqueness of zeros of function in (1.4) is not guaranteed so that Agarwal et al.’s approach cannot be applied directly either. The aim of this paper is to introduce a new solution operator for under this situation.

As an application, we will make use of newly established solution operator to show the existence of nontrivial solutions with respect to a parameter for the following type of nonlinear problem where a parameter, , , and .

This kind of problems can be classified as cases , , or , and, under additional nonlinear growth conditions, it is easy to see that bifurcation phenomena usually happen for the case of . In this paper, we will employ the global continuation theorem to get an existence result. One of the major steps for proof is to show the complete continuity of corresponding solution operator for . There have been many studies for nonnegative with and . Many authors are interested in finding positive solutions mainly employing the fixed-point index theory on a cone. Since may change signs in our case, this method is not suitable. Hai [4] investigated semilinear boundary value problems with a sign-changing weight; however, his method is also restricted to fit our quasilinear case.

The paper is organized as follow. In Section 2, we establish a solution operator for , when . In Section 3, we prove the continuity and compactness of the solution operator for problem . In Section 4, we obtain the existence of non-trivial solutions for .

#### 2. A Solution Operator

In this section, we construct a new solution operator for .

Theorem 2.1. *For each , there is a unique solution for .*

The proof follows the next two lemmas. We first show the existence of a solution for .

Lemma 2.2. *For each , there is a unique constant such that a function defined by
**
satisfies and solves . Furthermore, we have
*

*Remark 2.3. *For ,
Thus for ,
where
Similarly, we get an upper bound of on . It is clear that .

*Proof of Lemma 2.2. *Define
Then, by Lebesgue dominated convergence theorem, we see that and are continuous on . Because of the strict monotonicity of , it is easy to show that is strictly increasing on and is strictly decreasing on .

We claim that , as , and , as . Since for , for sufficiently large , we have
as . Since for , for sufficiently large , we have
as . Similarly, since for , for negatively large , we have
as . Since for , for negatively large , we have
as . Thus, by intermediate value theorem, there is a unique such that . Put
Then, and solves .

From the uniqueness of , the following is obvious.

Corollary 2.4. *Suppose that is a function defined in Lemma 2.2. Then, has the following property:
**
for all .*

The following lemma guarantees the uniqueness of solutions for .

Lemma 2.5. *Assume that, for , satisfies
**
If or on , then or on , respectively.*

*Proof. *We only prove the first case. We may use similar argument to show the second case. Let be a solution of , , and assume for . We want to show , . If this is false, then there exists with . Hence, there is an interval with such that , and . Let be the positive maximum of on . Then, and . Integrating both sides and over , , we have
Integrating both sides of the above equalities over , we get
Thus, we obtain
which is a contradiction. This completes the proof.

Let us define Then, problem can be equivalently written as .

#### 3. Application: Compactness

In this and the following sections, as an application of new solution operator, we consider a problem of one-dimensional -Laplacian with a singular weight as follows: where a parameter, , and . Note that we do not need any other assumptions on except the continuity in this section. Employing the solution operator established in Section 2, we may rewrite equivalently as where is defined by The purpose of this section is to show that is completely continuous on . For this, we need to know the properties of function .

Lemma 3.1. * sends bounded sets in into bounded sets in .*

*Proof. *Assume that and are bounded on and , respectively. For convenience, we denote . Suppose that is unbounded in . Then, there is a subsequence such that or as . Here, we consider the first case . We can prove the second case by the same argument. As in (2.2) of Lemma 2.2, we have
As in the proof of Lemma 2.2, we get the following inequality for the left-hand side
for sufficiently large , where . On other hand, as in the proof of Lemma 2.2, we get the following inequality for the right-hand side
for sufficiently large , and this is a contradiction.

Lemma 3.2. * is continuous on .*

*Proof. *Assume that in and in . Then, there exists a constant such that
Lemma 3.1 implies that is bounded in . Hence, it has a convergent subsequence denoted by . Let
as . As in (2.2) of Lemma 2.2, we have
By Lebesgue dominated convergence theorem, as , we get
By the uniqueness of , we obtain , and this proves the continuity of .

We have the following corollary using similar argument in the proof of Lemma 3.2.

Corollary 3.3. *Let satisfy as . Then, as .*

Theorem 3.4. * is continuous and compact on .*

* Proof. *First, we show the continuity of . Assume that on and in . Then, from the continuity of and Lebesgue dominated convergence theorem, we have

Next, we show that is uniformly bounded. Let be a bounded subset of , and let , then using the fact that sends bounded sets on into bounded sets on and taking , we obtain
where is a constant and the above upper bound is independent of .

Finally, we show that is equicontinuous. By Lemma 3.1, is bounded so by taking , we obtain for ,
Since , the above upper bound tends to as does. For case , , we may handle similar way. For case , (or case ), we can compute the following:
where . Thus, by Arzela-Ascoli Theorem, is compact, and the proof is complete.

#### 4. Application: Existence

In this section, we show the existence of nontrivial solutions for problem : where a parameter, , and . In addition, we assume that (F) and .

Based on the solution operator and its complete continuity established in Section 3, we show the existence of nontrivial solutions of for all . We will employ the following theorem the so-called global continuation theorem for the proof of our existence theorem.

Let be a Banach space and completely continuous with , and consider We denote as the set of solutions for (4.1), and .

Theorem 4.1 (see [5, Theorem 3.2]). *Let be a Banach space and continuous and compact and . Then, contains a pair of unbounded components in and , respectively, and .*

We now state our main existence theorem.

Theorem 4.2. *Assume that and holds . Then has non-trivial solutions for all .*

Let us take with the norm . We know that, to solve , it is enough to solve
where is defined in Section 3. And we have shown that is continuous and compact in Section 3. Since is obvious, employing Theorem 4.1, we obtain an unbounded continuum . If we provide *a priori* boundedness of solutions for , then the unbounded continuum guarantees the existence of solutions for all . Therefore, to complete the proof of Theorem 4.2, we need to give *a priori* estimate of solutions for .

Lemma 4.3. *Assume that and holds . Let be a solution for with . Then, there exists a constant which depends only on such that .*

*Proof. *Let us define a function which satisfies the following property: as . Since , for given , there exists such that
Then, since , for given above, there exists such that for all with ,
Since is continuous on , for given above, there exists a maximum value for on , that is, for all with , we have . Let us divide the interval into and as
respectively. Then, we can estimate solution as follows:
where with or can be given as
By Lemma 3.1, if for some , then there exists a constant such that . Thus, on , we have
where and
On , using Corollaries 2.4 and 3.3, and , we obtain
Therefore, we have
so that
and this completes the proof.

#### Acknowledgment

The authors express their thanks to Professor Yuki Naito for valuable suggestions. The first author was supported by the 2010 Research Fund of the University of Ulsan.

#### References

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