Abstract and Applied Analysis
Volume 2012, Article ID 831468, 14 pages
http://dx.doi.org/10.1155/2012/831468
Research Article

## Ground-State Solutions for a Class of N-Laplacian Equation with Critical Growth

College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China

Received 18 May 2012; Accepted 22 July 2012

Copyright © 2012 Guoqing Zhang and Jing Sun. 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 investigate the existence of ground-state solutions for a class of N-Laplacian equation with critical growth in . Our proof is based on a suitable Trudinger-Moser inequality, Pohozaev-Pucci-Serrin identity manifold, and mountain pass lemma.

#### 1. Introduction

Consider the following -Laplacian equation: where . is the -Laplacian, the nonlinear term has critical growth.

The interest in these problems lies in that fact that the order of the Laplacian is the same as the dimension of the underlying space. The classical case of this problem that , and the the problem (1.1) reduces to has been treated by Atkinson and Peletier [1] and Berestycki and Lions [2]. They obtained the existence of ground-state solution which the nonlinear term is subcritical growth. Alves et al. [3] extend their results to the critical growth. As , do Ó and Medeiros [4] consider the following -Laplacian equation problem: where has a subcritical growth and obtain a mountain pass characterization of the ground-state solution for the problem (1.3). In the present paper, we will improve and complement some of the results cited above.

Assume the function is continuous and satisfies the following conditions: ; There exist constants such that , where ;There exist and such that , for every .

Remark 1.1. Condition implies that has a critical growth with critical exponent .
Consider the energy functional where . By a ground-state solution, we mean a solution such such that for every nontrivial solution of the problem (1.1). Let denote the best constant of Sobolev embeddings: for , that is, for all .

Now we state our main theorem in this paper.

Theorem 1.2. If satisfies , and , with then the problem (1.1) possesses a nontrivial ground-state solution.

In this paper, we complement some results [4] from subcritical case to the critical case. Furthermore, the ground-state solution to the problem (1.1) is obtained without assuming that the function is increasing for (see [5]), and the so-called Ambrosetti-Rabinowitz condition: there exists , such that for all ,

The paper is organized as follows. Section 2 contains some technical results which allows us to give a variational approach for our results. In Section 3, we prove our main results.

#### 2. The Variational Framework

For , denotes the Lebesgue spaces with the norm , denotes the Sobolev spaces with the norm . As , we have the following version of Trudinger-Moser inequality.

Lemma 2.1 (see [6]). If , and , then Moreover, if and , then there exists a constant , which depends only on , , and , such that where and is the measure of the unit sphere in .

In the sequel, since we seek positive solutions, and assume that for . Consider the following minimization problem: where , . Since the problem (1.1) is an autonomous problem, under the Schwarz symmetric process, we can minimize the problem (2.3) on the space , the subspace of formed by radially symmetric functions. Indeed, let be the Schwarz symmetrization of , we have Hence, we can minimize the problem (2.3) on the space (see [7]). Now, we defined the following notations where .

We recall that Pohozaev-Pucci-Serrin identity shows that any solutions of the problem (1.1) should satisfies the Pohozaev-Pucci-Serrin identity:

Then, as , we have .

Hence, we have the Pohozaev identity manifold:

So, we have where .

In what follows, we will show that is attained, and afterwards we prove that thereby proving that the problem (1.1) has a ground-state solution.

#### 3. The Proof of Theorem 1.2

In this section, we prove that is attained, the equality (2.9) is satisfied. Hence the proof of Theorem 1.2 is obtained.

In the following, we consider the following minimax value:

Now, we show a sufficient condition, on a sequence to get a convergence like in .

Lemma 3.1. Assume that satisfies and, and let be a sequence in such that , then we have where in .

Proof. Without loss of generality, we assume that there exist such that Let is the Schwarz symmetrization of , then we have
From , we obtain that for , there exists , such that so, we have
From , we obtain There two estimates yield
On one hand, from Lemma 2.1, we obtain that there exists a constant , which depends only on , , and such that When and . Hence, we have where are positive constants, the continuous imbedding , and .
Then, by Dominated convergence theorem, we obtain
On the other hand, where is the Schwarz symmetrization of . Notice that the estimate for all , together with the Radial Lemma [4] leads to Thus, given , there exists such that Which implies that
Using the estimate we get Hence, we obtain that

Lemma 3.2. The numbers and satisfy the inequality .

Proof. For each , since we only consider the nontrivial solutions of the problem (1.1), we divide them into two cases to consider.
Case 1. Let , we define the function by By , we obtain that there exists such that , and Hence we obtain that for small enough. On the other hand, by , we obtain that for large enough. In this way, there exists such that , That is, . Hence
Case 2. Let , since for all , we obtain As a consequence, Combining Cases 1 and 2, we obtain that .

Lemma 3.3. The number defined by (2.8) is positive, that is, .

Proof . Clearly, . Assume by contradiction that and let be a minimizing sequence in to , that is, For each , set satisfying Similarly, we have We choose , so . Then we get
In what follows, we study in the space . Firstly, we assume that there exists such that in .
On one hand, since , then , for all , when , we have and we also know that , so . From Lemma 3.1, we have Note that so we have Hence, we have
It implies that . On the other hand, since in , and the space is a reflexible Banach space, we have Since we get From which it follows that , we have an absurd. Hence, we have

Lemma 3.4. If , then .

Proof. From , we have , for all . Now we choose such that Hence, we have Let , then is continuous function, we have By a simple calculation, when , we have Hence, we have

Lemma 3.5. The number is attained, that is, there exists such that and .

Proof. Let be a minimizing sequence in for , that is, Arguing as in Lemma 3.3, we assume that . From (3.43), Lemmas 3.3 and 3.4, we obtain From Lemma 3.1, where .
By (3.43) and (3.45), we have It implies that From (3.48) and (3.49), we have If , from (3.50), we have . Consider the function defined in Lemma 3.2 relative to the function: We concludes that for small enough. On the other hand, . In this way, we obtain that there is such that , that is, Hence, from (3.48), However, from (3.52), we have . Hence, we obtain Which is contradictory with (3.53).
Thus, we obtain It implies and From (3.48) and (3.56), we obtain that with .
We obtain that is attained.

Proof of Theorem 1.2. From Lemma 3.5, there is such that we will prove that .
By Lagrange multipliers, there exists , such that for every .
Define the rescaled function , which is a nontrivial solution of (1.1) with Thus, we have So, we have
For each , one has from [4]. We obtain that there exists such that , that is, satisfied that and then Hence for every , we obtain that From (3.62) and (3.64), we obtain that .
On the other hand, for every nontrivial solution of the problem (1.1), there exists a path such that and . Consequently, .
In conclusion, we obtain Hence, the function is a ground-state solution of the problem (1.1).

#### Acknowledgment

This paper was supported by Shanghai Natural Science Foundation under Project: 11ZR1424500.

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