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Abstract and Applied Analysis

Volume 2012 (2012), Article ID 568120, 31 pages

http://dx.doi.org/10.1155/2012/568120

## Existence Results for Quasilinear Elliptic Equations with Indefinite Weight

Department of Mathematics, Tokyo University of Science, Kagurazaka 1-3, Shinjuku-ku 162-8601, Japan

Received 19 January 2012; Accepted 7 March 2012

Academic Editor: Juan J. Nieto

Copyright © 2012 Mieko Tanaka. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We provide the existence of a solution for quasilinear elliptic equation in under the Neumann boundary condition. Here, we consider the condition that as and as . As a special case, our result implies that the following -Laplace equation has at least one solution: in on for every , , and with . Moreover, in the nonresonant case, that is, is not an eigenvalue of the -Laplacian with weight , we present the existence of a solution of the above -Laplace equation for every , and .

#### 1. Introduction

In this paper, we consider the existence of a solution for the following quasilinear elliptic equation:

where is a bounded domain with boundary , denotes the outward unit normal vector on , , and . We assume that is a Carathéodory function on satisfying and that is bounded on a bounded set (admitting in the nonresonant case). Here, is a map which is strictly monotone in the second variable and satisfies certain regularity conditions (see the following assumption ). The equation contains the corresponding -Laplacian problem as a special case. Although the operator is nonhomogeneous in the second variable in general, we assume that is asymptotically -homogeneous at infinity in the following sense .

Throughout this paper, we assume that the map satisfies the following assumptions and : there exist a positive function and a continuous function on such that , where for all and (i); (ii) there exists such that (iii) there exists such that (iv) there exists such that

A similar hypothesis to () is considered in the study of quasilinear elliptic problems (cf. [1, Example 2.2], [2–6]). It is easily seen that many examples as in the above references satisfy the condition (). Also, the following example satisfies our hypotheses: In particular, for , that is, stands for the usual -Laplacian , we can take in (). Conversely, in the case where holds in , by the inequalities in Remark 1.4 (ii) and (iii), we see whence .

Concerning the weight , throughout this paper, we assume that holds, where denotes the Lebesgue measure of a measurable set .

Because is asymptotically -homogeneous at infinity, the solvability of our equation is related to the following homogeneous equation (see Theorem 1.1): where is the positive function as in (). We say that is an eigenvalue of if the equation has a nontrivial solution.

There are few existence results of a solution to our equation (and also the -Laplace equation). For example, if and hold, then the standard argument guarantees the existence of a solution. For the -Laplacian as a special case of our problem, it is shown in [7] that the equation has a unique positive solution provided , and , where is the principal eigenvalue defined in Section 2.1 with . In [8], although the resonant case where or is considered under the assumptions to , its result does not cover the case of with , where (, ) is th eigenvalue of the -Laplacian with weight . For the Laplace problem under the Neumann boundary condition, we can refer to [9, 10]. Under the Dirichlet boundary condition, the existence results for the Laplace problem are well known when and is not an eigenvalue of the Laplacian (cf. [11]). Moreover, under the Dirichlet (or blow-up) boundary condition, many authors study various equations involving the -Laplace (Laplace) operator with (indefinite) weight. For example, we refer to [12] for boundary blow-up problems with Laplacian, [13] for periodic reaction-diffusion problems and [14, 15] for singular quasilinear elliptic problems.

Recently, the present author shows the existence of a solution for our problem in the case where is between the principal eigenvalue and the second eigenvalue in [6] (for ). In addition, a similar situation is treated in [5]. However, existence results are not seen in the case when is greater than the second eigenvalue for our problem. Therefore, the first purpose of this paper is to present an existence result of a solution in the nonresonant case where is not an eigenvalue of . Then, it studied the existence of at least one solution in the resonant case under assumptions that cover the case with and .

For the proof of our result, it is necessary to study the weighted eigenvalue problem . Thus, in Section 2, we introduce two sequences and of an eigenvalue of defined by Ljusternik-Schnirelman theory or Drábek-Robinson's method (cf. [16]), respectively. Then, we show several properties of above eigenvalues. In Section 3, we give the proof in the nonresonant case by using . In Sections 4 and 5, we handle the resonant case.

##### 1.1. Statements of Our Existence Results

First, we state the existence result of a solution in the nonresonant case.

Theorem 1.1. *Assume that is not an eigenvalue of . Then, has at least one solution.*

To state our existence result in the resonant case, we introduce some conditions. Set where is the function as in (). there exist and such that there exist and such that there exist and such that there exist and such that

Theorem 1.2. *Assume one of the following conditions: *(i)* and () or () hold; *(ii)*, and one of (), (), () and () hold; *(iii)*, and () or () hold; ** Then, has at least one solution.*

In the special case where and for , we easily see that () or () holds with provided or , respectively. Therefore, the following result is proved according to Theorem 1.2.

Corollary 1.3. *Let , and . Then, the following equation has at least one solution:
*

##### 1.2. Properties of the Map

In what follows, the norm on is given by , where denotes the norm of for (). Setting , then we can easily see that for every .

*Remark 1.4. *It is easily seen that the following assertions hold under condition (): (i)for all , is maximal monotone and strictly monotone in ; (ii) for every ; (iii) for every ; (iv) is convex in for all and satisfies the following inequalities:
for every , where and are the positive constants in ().

The following result is proved in [3]. It plays an important role for our poof.

Proposition 1.5 (see [3, Proposition 1]). *Let be the map defined by
**
for , . Then, has the property, that is, any sequence weakly convergent to with strongly converges to .*

#### 2. The Weighted Eigenvalue Problems

##### 2.1. Preliminaries

The following lemmas can be easily shown by way of contradiction because is equivalent to (note that is positive). Here, we omit the proofs (refer to [7]).

Lemma 2.1. *Assume . Then, there exists a constant such that for every with .*

Lemma 2.2. *Assume that and . Then, there exists a constant such that
**
for every . *

Lemma 2.3. *Assume that in . Then, for every there existed such that
**
for every .*

First, we recall the following principle eigenvalue : Because of , we have the following result as the same argument as in the case of the -Laplacian.

Proposition 2.4 (see [7, Proposition 2.2]). *The following assertions hold: *(i)If holds, then ; (ii)If holds, then is a simple eigenvalue and it admits a positive eigenfunction. In addition, the open interval contains no eigenvalues of .

Lemma 2.5. *Assume . Then, one has for every and with .*

*Proof. *We choose a minimizer for because Proposition 2.4 guarantees the existence of it. Then, for every , we have
by the definition of . By applying the same argument to a minimizer for , we obtain for with .

##### 2.2. Other Eigenvalues

Here, we introduce two unbounded sequences and as follows: where denotes the Krasnoselskii genus of (see [17, Definition 5.1] for the definition) and denotes the usual unit sphere in . We see that is defined by Ljusternik-Schnirelman theory and it is known that the definition of is introduced by Drábek and Robinson ([16]) under the -Laplace Dirichlet problem with .

*Remark 2.6. *The following assertions can be shown easily: (i); (ii) and for every ; (iii) for every ; (iv) for every ; (v) and for every , see [18] for the proof of (ii).

Define a functional on by for . Because is a regular value of , it is well known that the norm of the derivative at of the restriction of to is defined as follows: where denotes the tangent space of at , that is, . Here, we recall the definition of the Palais-Smale condition for .

*Definition 2.7. * is said to satisfy the bounded Palais-Smale condition if any bounded sequence such that has a convergent subsequence. Moreover, we say that satisfies the Palais-Smale condition at level if any sequence such that and as has a convergent subsequence. In addition, we say that satisfies the Palais-Smale condition if satisfies the Palais-Smale condition for every .

The following result can be proved by the same argument as in [19, Proposition 3.3] (which treats the case of the -Laplacian, i.e., ) because of . Here, we omit the proof.

Lemma 2.8. *The following assertions hold: *(i)* satisfies the bounded Palais-Smale condition; *(ii)* satisfies the Palais-Smale condition provided . *

Proposition 2.9. * and are eigenvalues of such that
*

*Proof. *In the case of , since satisfies the Palais-Smale condition, we can apply the first deformation lemma on manifold (refer to [20]). Thus, by the standard argument, we can prove that and are critical values of . This means that and are eigenvalues of by the Lagrange multiplier rule. In addition, we can easily show by the standard argument via the first deformation lemma on manifold (refer to [21, Proposition 3.14.7], [22] or [17] in the case of a Banach space). Hence, holds because of for every .

In the case of , by the same argument as in [18], our conclusion can be proved. For readers' convenience, we give a sketch of the proof. For , we define and . Moreover, we set minimax values and of by
Because any Palais-Smale sequence of is bounded, it is easily shown that satisfies the Palais-Smale condition (refer to [19, Proposition 3.3]) Hence, it can be proved that and are critical values of . Furthermore, it follows from the argument as in [18, Lemma 3.5] that and as . Therefore, by noting that is -homogeneous, we can obtain a solution with for in , on , where or . Because of , it follows from the standard argument that has a subsequence strongly convergent to a solution for
where . Thus, and are eigenvalues of . To prove , by considering a function for , we have (refer to Proposition 2.10). Because we can apply our fist assertion to (note ), we obtain .

Proposition 2.10. *Let if and if . Then, the following assertions hold: *(i)if in , then ; (ii)if in , then ; (iii)if and in , then . * Moreover, the same conclusion holds for .*

*Proof. *We only treat because we can give the proof for similarly.(i) Let in . Fix an arbitrary . Then, by the definition of , there exists a such that . Set for (note ), then holds. Therefore, by the definition of , we have
because of for every . Since is arbitrary, we obtain .(ii) Let in and fix an arbitrary . By the definition of , there exists a such that . Since is compact and , we set . Then, due to Hölder's inequality and in , there exists an such that
for every and . Therefore, by a similar argument to (i), we obtain
for sufficiently large . Hence, follows. Since is arbitrary, our conclusion is proved.(iii) Let in and . We fix an arbitrary . Due to our assertion (ii), there exists an such that . For every , by the definition of , we can take satisfying .

Here, we will prove

If satisfies , then we obtain
by Lemma 2.2 and Hölder's inequality (note and ), where is a constant (independent of and ) obtained by the continuity of into . Therefore, if we take an satisfying for every , then we obtain
for every provided and . Similarly, in the case where changes sign, for every satisfying , we have
Hence, by taking a sufficiently large , we get the inequality
for every with and . In the case of in , by using Lemma 2.3 instead of Lemma 2.2, we have a similar inequality
for every provided (some sufficiently large ). Consequently, our claim follows from (2.15), (2.17), and (2.18).

Let us return to the proof of (iii). Because
holds by (2.13), and the continuity of into , we see the inequality
for every and (some sufficiently large ). By considering , we obtain
Because of , we get for sufficiently large , and hence our conclusion holds.

Finally, we recall the second eigenvalue of obtained by the mountain pass theorem. where .

Since holds, the following result can be shown by the same argument as in [19] (although they handle the asymmetry case, it is sufficient to consider the case of in this paper). See [19, Theorem 3.2] for the proof.

Theorem 2.11. * is an eigenvalue of which satisfies . Moreover, there is no eigenvalues of between and . *

Now, we have the following result.

Proposition 2.12. *
holds, where is a minimax value defined by (2.22).*

*Proof. *First, we prove the inequality . Because is an eigenvalue (note that the following equation is homogeneous), we can choose a solution with for
Note that is a sign-changing function because any eigenfunction associated with any eigenvalue greater than the principal eigenvalue changes sign (refer to [18, Proposition 4.3]). Thus, we have
by taking as test function (recall that ). Hence, we may assume that by the normalization. Set . Then, because is homeomorphic to , there exists such that . Since the value of is equal to on , we obtain
by the definition of and .

Next, we will prove the inequality by dividing into two cases: and .

Case of : by way of contradiction, we assume that . Then, follows from Theorem 2.11. Note that satisfies the Palais-Smale condition in this case (see Lemma 2.8), and hence we can apply the first deformation lemma to . Therefore, by the standard argument (cf. [22], [17, Lemma 5.6]), we see that , where . This means that is an infinite set, that is, the following equation has infinite many solutions:
due to the Lagrange multiplier's rule. This contradicts to the fact described as in Proposition 2.4 that is simple. As a result, we have shown that holds in the case of (note ).

Case of : According to Proposition 2.10 (i) for , we have for every since we can apply the first result to . Because we prove by the same argument as in [6, Lemma 2.9] (for the case ), our conclusion is proved by taking in the inequality .

#### 3. Proof of Theorem 1.1

We define a functional on as follows: for ((1.15) or (1.9) for the definition of , , and ). It is easily seen that is well defined and class of on by (1.1), (1.16) and the continuity of .

*Remark 3.1. *Let be a critical point of , namely, satisfies the equality
for every . Then, by the Moser iteration process (refer to Theorem C in [4]). Therefore, () follows from the regularity result in [23]. Furthermore, due to [24, Theorem 3], satisfies in the distribution sense and the boundary condition
for every (see [24] for the definition of ). Since and for every , satisfies the Neumann boundary condition, that is, for every .

##### 3.1. The Palais-Smale Condition in the Nonresonant Case

First, we recall the definition of the Palais-Smale condition.

*Definition 3.2. *A functional on a Banach space is said to satisfy the Palais-Smale condition at if a Palais-Smale sequence at level , namely,
has a convergent subsequence. We say that satisfies the Palais-Smale condition if satisfies the Palais-Smale condition at any . Moreover, we say that satisfies the bounded Palais-Smale condition if any bounded sequence such that is bounded and as has a convergent subsequence.

Concerning the Palais-Smale condition, we state the following result developed from [6, Proposition 7].

Proposition 3.3. *If is not an eigenvalue of , then satisfies the Palais-Smale condition.*

*Proof. *Let be a Palais-Smale sequence of , namely,
for some . It is sufficient to prove only the boundedness of because the operator described in Proposition 1.5 has the property.

To prove the boundedness of , it suffices to show that is bounded because of the inequality (obtained by (1.1)) and the following inequality:
where we use Remark 1.4 (iii) in the last inequality. By way of contradiction, we may assume that as by choosing a subsequence if necessary. Set . Then, since the inequality (3.6) guarantees that is bounded in , we may suppose, by choosing a subsequence, that in and in for some .

Here, we will prove that
where . Fix an arbitrary . It follows from (1.1) that there exists a such that
Then, we obtain
Since we are assuming that as , there exists such that for every
holds. This shows that because is arbitrary.

Here, we recall the following result proved in [6]:
for every . Thus, by considering
we see that strongly converges to in (note that -Laplacian has the property). Therefore, by taking a limit in for any and by noting (3.7) and (3.11), we know that is a nontrivial solution (note ) of
This means that is an eigenvalue of . This is a contradiction. Hence, is bounded.

##### 3.2. Key Lemmas

To show the linking lemma, we define .

Lemma 3.4. *Let be odd and . Then, for every with , where is the set introduced in (3.14) and is the upper hemisphere in with boundary .*

*Proof. *Fix any such that . If satisfies , then holds. So, we may assume that for every . Define as follows:
By the definition of , there exists such that . Since is odd and is even, we may suppose . So, this yields the inequality , whence holds.

Lemma 3.5. *Let . Then, there exists such that
**
where is defined by (2.5).*

*Proof. *Choose such that . By the definition of , there exists such that
Due to the compactness of , we put . By the property of the function as in () and Young's inequality, for every there exist constants and such that
for every and . Moreover, the hypothesis (1.1) ensures that for every there exist constants satisfying
for every and a.e. . Hence, we have
for every , , and since , (3.17), (3.18) and (3.19), where and . By taking and satisfying , we show that as . Thus, our conclusion follows because is symmetric.

##### 3.3. The Case

Lemma 3.6. *Let and . Then, is bounded from below, coercive and weakly lower semicontinuous (w.l.s.c.) on .*

*Proof. * is w.l.s.c. on because is convex and continuous on (cf. [25, Theorem 1.2]). Thus, is also w.l.s.c. on since the inclusion from to is compact.

Choose such that , where . By an easy estimation, (3.18) and (3.19) as in Lemma 3.5, we have
for every and .

Let satisfy . Then, the following inequality follows from Lemma 2.2:
where is a positive constant independent of with and .

For every such that , we obtain
by the definition of , Lemma 2.1 and , where is a constant obtained by Lemma 2.1.

Consequently, if we choose a satisfying , then we obtain positive constants and (independent of ) such that