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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 942092, 14 pages

http://dx.doi.org/10.1155/2014/942092

## Multiple Solutions for a Class of -Laplacian Equations with Critical Growth and Indefinite Weight

^{1}College of Sciences, University of Shanghai for Science and Technology, Shanghai 200093, China^{2}Department of Mathematical Sciences, University of Nevada Las Vegas, Las Vegas, NV 89154-4020, USA

Received 7 September 2013; Revised 10 December 2013; Accepted 20 December 2013; Published 23 January 2014

Academic Editor: S. A. Mohiuddine

Copyright © 2014 Guoqing Zhang and Ziyan Yao. 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

Using the suitable Trudinger-Moser inequality and the Mountain Pass Theorem, we prove the existence of multiple solutions for a class of -Laplacian equations with critical growth and indefinite weight , , , , where , is an indefinite weight, behaves like and does not satisfy the Ambrosetti-Rabinowitz condition, and .

#### 1. Introduction

In this paper, we consider the existence of multiple solutions for the -Laplacian elliptic equations with critical growth and singular potentials where , , , , , is the -Laplacian, the indefinite weight , and is the classes of rearrangement of ; satisfies the following conditions:(H1), ,(H2), or , for some ,(H3), where , is the best constant that is,

Note that if is a measurable function which satisfies (H2), there exists such that .

Recently, -Laplacian equations had been studied by many authors. Marcos do Ó [1] studied the existence of nontrivial solutions for the following -Laplacian equations with critical growth: where is bounded smooth domain in . Adimurthi and Sandeep [2] proved that the singular Trudinger-Moser inequality holds if and only if , where , , , and , and studied the corresponding critical exponent problem. For the unbounded domain, Li and Ruf [3] proved that, if we replace the -norm of in the supermum by the standard Sobolev norm, the supermum can still be finite. Adimurthi and Yang [4] obtained the following Trudinger-Moser inequality where , , , and , and studied the existence of nontrivial solution for the corresponding -Laplacian equations with critical growth. In particular, using inequality (6) and the Mountain Pass Theorem, Lam and Lu [5] studied the following nonuniformly elliptic equations of -Laplacian type of the form where , and obtained the existence and multiplicity results of problem (7).

On the other hand, some authors have studied the case for the nonlinear term which does not satisfy the Ambrosetti-Rabinowitz condition. Lam and Lu [6, 7] studied the existence of nontrivial solutions for the -Laplacian equations and systems and polyharmonic equations without Ambrosetti-Rabinowitz conditions, respectively. Miyagaki and Souto [8] discussed a class of superlinear problems for the polynomial case without Ambrosetti-Rabinowitz conditions. Motivated by a suitable Trudinger-Moser inequality, we assume the following growth conditions on the nonlinearity :(f1)the function is continuous, for some constants and for all , (f2), for all , , where (f3)there exists such that for all , ;(f4), uniformly on . We state our main result in this paper.

Theorem 1. *Suppose that (H1)–(H3) and (f1)–(f4) are satisfied and . Furthermore, assume that *(f5)*, uniformly on ,* *and there exists such that*(f6)*uniformly on compact subsets of , where and are defined in Section 3. Then there exists such that, for each , problem (1) has at least two nontrivial weak solutions.*

In this paper, as the function is an indefinite weight, we establish a singular Trudinger-Moser inequality (see Lemma 8) and investigate the eigenvalue problem corresponding to problem (1). Using the singular Trudinger-Moser inequality, the eigenvalue problem and the Mountain Pass Theorem, we prove the multiplicity result for problem (1). Furthermore, condition (f2) is used by Lam and Lu [5], and it implies that the function does not satisfy the Ambrosetti-Rabinowitz condition.

The paper is organized as follows. In Section 2, we recall some important lemmas and consider the eigenvalue problem corresponding to problem (1). Section 3 is devote to prove Theorem 1.

#### 2. Preliminary Results

##### 2.1. Key Lemmas

Now, we define the following Sobolev space and the corresponding norm,

From the Radial Lemma [9, 10], we have for all being radially symmetric, where is the surface area of the unit sphere in . is a rearrangement of if where denotes the Lebesgue measure.

Lemma 2 (see [11]). *Let satisfy and . Then there exists such that
*

*Proof. *Assume that . Since
Then, by (H2), there exists such that
Therefore, we have

*Remark 3. *In this paper, we denote as positive (possibly different) constants.

*Remark 4. *If , then satisfies (H1)–(H3).

Lemma 5. *If (H1)–(H3) are satisfied, then*(1)*the embedding is continuous, for all ;*(2)*the embedding is compact, for all .*

*Proof. *(1) From Lemma 2 and Sobolev-Poincare inequality, we obtain the conclusion.(2)Let satisfy for all , and we assume

In view of (H3), for every , there exists such that
Hence, we have
From (19), we have in and is the ball centered at with radius . This together with (21) leads to . Since is arbitrary, we have
Hence, for every , we have

Lemma 6. * is a reflexive Banach space.*

*Proof. *Suppose that , , we have
and there exists , using the following inequality
such that
Hence, is uniformly convex. We obtain that is a reflexive Banach space.

Now, we define the functional
then the functional is well defined by Lemma 5. Moreover, is the functional on and ; we have
Hence, the critical point of the functional is the weak solution of problem (1).

Lemma 7. *Let , , and ; then for some and , one has
*

*Proof. *Let ; is the Schwarz symmetrization of ; we can conclude that
Let . It is easy to obtain that is increasing with respect to . If ; then there holds
Now, we prove that there exists a uniform constant such that, for all radially decreasing symmetric functions and ,
where . In the following, assume that is radially decreasing function in and . Take sufficiently large; that is, . By the radial lemma, for all , we have and
Define the set . Assume that is nonempty; then for all and we have
Since
so we have
Thus, we obtain
From Hardy-Littlewood inequality, we have
Since , we have . Let
Applying the mean value theorem to the function , we obtain that there exists which satisfies
such that
So we have
By , we have
and ,
Obviously, , and
Let ; we obtain
Hence, we obtain that (32) holds. For , , we have
Since and , , that is, , , we have

As the proof of Lemma 7, we can obtain the following.

Lemma 8. *For , , and , , and , one has
*

##### 2.2. The Eigenvalue Problem

We consider the following eigenvalue problem:

Now, we denote the set , and define where .

Lemma 9 (see [12]). *Let , be two continuous functions in differentiable a.e., and
*

Then , a.e. in if and only if for some .

Proposition 10. *Assume that (H1)–(H3) hold; then is the lowest eigenvalue of Problem (50) and is principal.*

*Proof. *From Lemma 2, we have . Furthermore, any minimizing sequence is bounded. Up to a subsequence, there exists such that
Hence, we have
and consequently we have
From Lemma 5, we obtain that is weakly closed in . By the Lagrange Multipliers rule, is an eigenvalue of problem (50). Moreover for any , so that possesses a nonnegative eigenfunction. We conclude that the eigenvalue is principal from Harnack inequality in [13].

Proposition 11. *The eigenvalue is isolated. That is, there exists , such that there are no other eigenvalues of problem (50) in the interval .*

*Proof. *Assume by contradiction there exists a sequence of eigenvalue of problem (50) with . Let be an eigenfunction associated with . Then satisfies
We define
The coercivity of the functional implies that is a bounded sequence. Hence is bounded in . So there exists a subsequence (still denoted) and such that
and . On the other hand, we have
and . So we conclude that is an eigenfunction associated with and . Then we conclude from the convergence in measure of the sequence towards that
where denotes the negative set of , which contradicts Proposition 11.

Proposition 12. *The first eigenvalue is simple, in the sense that the eigenfunctions associated with it are merely constant multiples of each other.*

*Proof. *Let , be two eigenfunctions associated with . We assume without restriction that , ; then satisfies . Testing it with function , we get
Let , from Lemma 9, we have
The function , where , belongs to and then it is admissible for the weak formulation of , a.e., and
It follows from (62) and (63) that we have
Let ; we have . By Lemma 9, there exists such that .

#### 3. The Proof of Theorem 1

##### 3.1. Palais-Smale Sequence

Now, we check that the functional satisfies the geometric conditions of the Mountain Pass Theorem.

Lemma 13. *Suppose that (H1)–(H3) and (f1)–(f5) hold. Then there exists such that, for , there exists such that if . Furthermore, can be chosen such that , as .*

*Proof. *From (f5), for every , there exists such that implies
Moreover, using (f1), for each and , we find a constant such that
Combining (65) and (66), we have
Since the embedding is continuous, we obtain
Thus, we have
Since and and letting , we choose such that . Thus, if is sufficiently small, we find some such that if and even as .

Lemma 14. *If , there exists , with , such that .*

*Proof. *Let , with compact support . By (f4), we obtain that for , there exists a positive constant such that for every ,
Then, we have
Choose and and let
we have as . Setting with being sufficient large, we obtain the conclusion.

It is well known that the failure of the (PS) compactness condition creates some difficulties in studying the class of elliptic problems involving critical growth. In Lemma 15, instead of (PS) sequence, we analyze the compactness of Cerami sequences of the functional .

Lemma 15. *Let be a Cerami sequence of ; that is,
**
Then there exists a subsequence of (still denoted by ) and such that
**
where . Furthermore, is a nontrivial weak solution of problem (1).*

*Proof. *Let , , as ; we have
where as . Let in (76); we have
Suppose that
Set
we have . From [5], we have
However, since is the Cerami sequence at the level , we have that
Then there exists some constant such that
which implies that
Let ; then we have
So we can conclude that
Note that ; by Fatou Lemma, (80), and (85), we get a contradiction. So which means that in .

Let , such that
For any given , let , by (f1); there exists such that
Since , we have
and by (f3), , and , we have
Since in and the embedding is compact from and the Hölder inequality, we have . By Lemma 7, we have .

Let in (89) and ; we get
Note that and ; we suppose that . Since , we have
By (f2) and , we have
Moreover, we have
which is a contraction to (75); this proves that is bounded in . Thus, we have
From Lemma 5, the embedding is compact for all . If , we get
From (f1), the Trudinger-Moser inequality, and the Hölder inequality, we have . From Lemma 2.1 in [14], we have
For any fixed , set
Because is bounded, is a finite set. From Lemma 4.4 in ([4]), for any compact set , we have
Now, we prove that
It is enough to prove that for any and there holds
We take with and on . Then is a bounded sequence. Choosing and in (76), we have
Adapting an argument similar to [4], we have
Since is finite, it follows that a.e. This implies, up to a subsequence that in . Let in (76), and in ; we obtain