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Abstract and Applied Analysis
Volume 2012, Article ID 831468, 14 pages
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

Academic Editor: Norimichi Hirano

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.


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

1. Introduction

Consider the following 𝑁-Laplacian equation: Δ𝑁𝑢+|𝑢|𝑁2𝑢=𝑓(𝑢),in𝑁,𝑢>0,in𝑁,𝑢𝑊1,𝑁𝑁,(1.1) where 𝑁2. Δ𝑁𝑢=div(|𝑢|𝑁2𝑢) is the 𝑁-Laplacian, the nonlinear term 𝑓(𝑢) has critical growth.

The interest in these problems lies in that fact that the order of the Laplacian is the same as the dimension 𝑁 of the underlying space. The classical case of this problem that 𝑁=2, and the the problem (1.1) reduces to Δ𝑢+𝑢=𝑓(𝑢),in2,(1.2) 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 𝑓(𝑢) is subcritical growth. Alves et al. [3] extend their results to the critical growth. As 𝑁2, do Ó and Medeiros [4] consider the following 𝑁-Laplacian equation problem: Δ𝑁𝑢=𝑔(𝑢),in𝑁,(1.3) where 𝑔 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 𝑓 is continuous and satisfies the following conditions: (𝑔1)lim𝑠0+(𝑓(𝑠)/𝑠|𝑠|𝑁2)=0; (𝑔2) There exist constants 𝛼0,𝑏1,𝑏2>0 such that |𝑓(𝑠)|𝑏1|𝑠|𝑁1+𝑏2[exp(𝛼0|𝑠|𝑁/(𝑁1))𝑆𝑁2(𝛼0,𝑠)], where 𝑆𝑁2(𝛼0,𝑠)=𝑁2𝑘=0(𝛼𝑘0/𝑘!)|𝑠|𝑁𝑘/(𝑁1);(𝑔3)There exist 𝜆>0 and 𝑞>𝑁 such that 𝑓(𝑠)𝜆𝑠𝑞1, for every 𝑠0.

Remark 1.1. Condition (𝑔2) implies that 𝑓 has a critical growth with critical exponent 𝛼0.
Consider the energy functional 𝐼𝑊1,𝑁(𝑁)1𝐼(𝑢)=𝑁𝑁||||𝑢𝑁+|𝑢|𝑁𝑑𝑥𝑁𝐹(𝑢)𝑑𝑥,(1.4) where 𝐹(𝑠)=𝑠0𝑓(𝑡)𝑑𝑡. By a ground-state solution, we mean a solution such 𝜔𝑊1,𝑁(𝑁) such that 𝐼(𝜔)𝐼(𝑢) for every nontrivial solution 𝑢 of the problem (1.1). Let 𝐶𝑞>0 denote the best constant of Sobolev embeddings: 𝑊1,𝑁𝑁𝐿𝑞𝑁,(1.5) for 𝑞(𝑁,+), that is, 𝐶𝑞𝑁|𝑢|𝑞𝑑𝑥𝑁/𝑞𝑁||||𝑢𝑁+|𝑢|𝑁𝑑𝑥,(1.6) for all 𝑢𝑊1,𝑁(𝑁).

Now we state our main theorem in this paper.

Theorem 1.2. If 𝑓 satisfies (𝑔1),(𝑔2), and (𝑔3), with 𝜆>𝑞𝑁𝑞(𝑞𝑁)/𝑁𝐶𝑞𝑞/𝑁,(1.7) 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 𝑠𝑓(𝑠)|𝑠|𝑁1(1.8) is increasing for 𝑠>0 (see [5]), and the so-called Ambrosetti-Rabinowitz condition: there exists 𝜃>𝑁, such that for all 𝑥𝑁, 0<𝜃𝐹(𝑥,𝑢)𝑢𝑓(𝑥,𝑢).(1.9)

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 1𝑝, 𝐿𝑝(𝑁) denotes the Lebesgue spaces with the norm 𝑢𝐿𝑝(𝑁)=(𝑁|𝑢|𝑝𝑑𝑥)1/𝑝, 𝑊1,𝑝(𝑁) denotes the Sobolev spaces with the norm 𝑢𝑊1,𝑝(𝑁)=(𝑁(|𝑢|𝑝+|𝑢|𝑝)𝑑𝑥)1/𝑝. As 𝑝=𝑁, we have the following version of Trudinger-Moser inequality.

Lemma 2.1 (see [6]). If 𝑁2, 𝛼>0 and 𝑢𝑊1,𝑁(𝑁), then 𝑁exp𝛼|𝑢|𝑁/(𝑁1)𝑆𝑁2(𝛼,𝑢)𝑑𝑥<.(2.1) Moreover, if 𝑢𝑁𝐿𝑁(𝑁)1,𝑢𝐿𝑁(𝑁)𝑀< and 𝛼<𝛼𝑁, then there exists a constant 𝐶, which depends only on 𝑁, 𝑀, and 𝛼, such that 𝑁exp𝛼|𝑢|𝑁/(𝑁1)𝑆𝑁2(𝛼,𝑢)𝑑𝑥𝐶(𝑁,𝑀,𝛼),(2.2) where 𝛼𝑁=𝑁𝜔1/(𝑁1)𝑁1 and 𝜔1/(𝑁1) is the measure of the unit sphere in 𝑁.

In the sequel, since we seek positive solutions, and assume that 𝑓(𝑠)=0 for 𝑠0. Consider the following minimization problem: 1min𝑁𝑁||||𝑢𝑁𝑑𝑥𝑁𝐺(𝑢)𝑑𝑥=0,(2.3) where 𝑔(𝑠)=𝑓(𝑠)𝑠|𝑠|𝑁2, 𝐺(𝑠)=𝑠0𝑔(𝑡)𝑑𝑡=𝐹(𝑠)(1/𝑁)𝑠𝑁. Since the problem (1.1) is an autonomous problem, under the Schwarz symmetric process, we can minimize the problem (2.3) on the space 𝑊1,𝑁rad(𝑁), the subspace of 𝑊1,𝑁(𝑁) formed by radially symmetric functions. Indeed, let 𝑢 be the Schwarz symmetrization of 𝑢, we have 𝑁𝐺𝑢𝑑𝑥=𝑁𝐺(𝑢)𝑑𝑥,𝑁||𝑢||𝑁𝑑𝑥𝑁||||𝑢𝑁𝑑𝑥.(2.4) Hence, we can minimize the problem (2.3) on the space 𝑊1,𝑁rad(𝑁) (see [7]). Now, we defined the following notations 1𝑚=inf{𝐼(𝑢)𝑢isnontrivialsolutionoftheproblem(1.1)}𝐴=inf𝑁𝑁||||𝑢𝑁𝑑𝑥𝑁𝐺(𝑢)𝑑𝑥=0𝑏=inf𝛾Γmax[]𝑡0,1𝐼(𝛾(𝑡)),(2.5) where Γ={𝛾𝐶([0,1],𝑊1,𝑁rad(𝑁))𝛾(0)=0,𝐼(𝛾(1))<0}.

We recall that Pohozaev-Pucci-Serrin identity shows that any solutions 𝑢 of the problem (1.1) should satisfies the Pohozaev-Pucci-Serrin identity: (𝑁𝑝)𝑁||||𝑢𝑝𝑑𝑥=𝑁𝑝𝑁𝐺(𝑢)𝑑𝑥.(2.6)

Then, as 𝑝=𝑁, we have 𝑁𝐺(𝑢)𝑑𝑥=0.

Hence, we have the Pohozaev identity manifold: 𝒫=𝑢𝑊1,𝑁𝑁{0}(𝑁𝑝)𝑁||||𝑢𝑝𝑑𝑥=𝑁𝑝𝑁=𝐺(𝑢)𝑑𝑥𝑢𝑊1,𝑁𝑁{0}𝑁.𝐺(𝑢)𝑑𝑥=0(2.7)

So, we have 𝐴=inf𝑢𝒫1𝑁𝑁||||𝑢𝑁𝑑𝑥,𝑚=inf𝑢𝜏𝐼(𝑢),(2.8) where 𝜏={𝑢𝑊1,𝑁(𝑁){0}𝐼(𝑢)=0}.

In what follows, we will show that 𝐴 is attained, and afterwards we prove that 𝑚=𝐴=𝑏,(2.9) 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 𝐴 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: 𝑐=inf0𝑣𝑊1,𝑁𝑁max𝑡0𝐼(𝑡𝑣).(3.1)

Now, we show a sufficient condition, on a sequence {𝑣𝑛} to get a convergence like 𝐹(𝑣𝑛)𝐹(𝑣) in 𝐿1(𝑁).

Lemma 3.1. Assume that 𝑓 satisfies (𝑔1) and(𝑔2), and let {𝑣𝑛} be a sequence in 𝑊1,𝑁rad(𝑁) such that 𝑣𝑛𝑁𝐿𝑁(𝑁)1,𝑣𝑛𝐿𝑁(𝑁)𝑀<, then we have 𝑁𝐹𝑣𝑛𝑑𝑥𝑁𝐹(𝑣)𝑑𝑥,(3.2) where 𝑣𝑛𝑣 in 𝑊1,𝑁(𝑁).

Proof. Without loss of generality, we assume that there exist 𝑣𝑊1,𝑁rad(𝑁) such that 𝑣𝑛𝑣,in𝑊1,𝑁rad𝑁,𝑣𝑛𝑣,a.e.inN.(3.3) Let 𝑣 is the Schwarz symmetrization of 𝑣, then we have 𝑁𝑣𝛼exp0|𝑣|𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑣𝑑𝑥=𝑁𝑣𝛼exp0||𝑣||𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑣𝑑𝑥,𝑁|𝑣|𝑁𝑑𝑥=𝑁||𝑣||𝑁𝑑𝑥.(3.4)
From (𝑔1), we obtain that for 𝜖>0, there exists 𝛿>0, such that 𝑓(𝑠)𝜖|𝑠|𝑁1,for|𝑠|<𝛿,(3.5) so, we have 𝜖𝐹(𝑠)𝑁|𝑠|𝑁,for|𝑠|<𝛿.(3.6)
From (𝑔2), we obtain 𝐹(𝑠)𝐶1|𝑠|𝑁+𝐶2𝛼|𝑠|exp0|𝑠|𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑢,for|𝑠|𝛿.(3.7) There two estimates yield 𝐹(𝑠)𝐶3|𝑠|𝑁+𝐶2𝛼|𝑠|exp0|𝑠|𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑢,for𝑠>0.(3.8)
On one hand, from Lemma 2.1, we obtain that there exists a constant 𝐶, which depends only on 𝑁, 𝑀, and 𝛼 such that 𝑁𝛼||𝑣exp𝑛||𝑁/(𝑁1)𝐶.(3.9) When 𝑣𝑛𝑁𝐿𝑁(𝑁)1,𝑣𝑛𝐿𝑁(𝑁)𝑀< and 𝛼<𝛼𝑁. Hence, we have |𝑥|𝑟𝐹𝑣𝑛𝐶3𝑁||𝑣𝑛||𝑁𝑑𝑥+𝐶2|𝑥|𝑟||𝑣𝑛||𝛼exp0||𝑣𝑛||𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑣𝑛𝑑𝑥𝐶3𝑀𝑁+𝐶2|𝑥|𝑟||𝑣𝑛||𝛼exp0||𝑣𝑛||𝑁/(𝑁1)𝑑𝑥𝐶3𝑀𝑁+𝐶3|𝑥|𝑟||𝑣𝑛||𝜇1/𝜇|𝑥|𝑟exp𝛽𝛼0||𝑣𝑛||𝑁/(𝑁1)𝑑𝑥1/𝛽𝐶3𝑀𝑁+𝐶4𝑀|𝑥|𝑟exp𝛽𝛼0||𝑣𝑛||𝑁/(𝑁1)𝑑𝑥1/𝛽𝐶5,(3.10) where 𝐶𝑖(𝑖=2,3,4,5) are positive constants, the continuous imbedding 𝑊1,𝑁(𝑁)𝐿𝜇(𝑁), 1/𝜇+1/𝛽=1 and 𝛽𝛼0<𝛼𝑁.
Then, by Dominated convergence theorem, we obtain |𝑥|𝑟𝐹𝑣𝑛𝑑𝑥|𝑥|𝑟𝐹(𝑣)𝑑𝑥.(3.11)
On the other hand, |𝑥|>𝑟𝐹𝑣𝑛𝑑𝑥𝐶3|𝑥|>𝑟||𝑣𝑛||𝑁𝑑𝑥+𝐶2|𝑥|>𝑟||𝑣𝑛||𝛼exp0||𝑣𝑛||𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑣𝑛𝑑𝑥|𝑥|>𝑟||𝑣𝑛||𝛼exp0||𝑣𝑛||𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑣𝑛=𝑑𝑥𝑗=𝑁1𝛼𝑗0𝑗!|𝑥|>𝑟||𝑣𝑛||||𝑣𝑛||𝑁𝑗/(𝑁1)=𝑑𝑥𝑗=𝑁1𝛼𝑗0𝑗!|𝑥|>𝑟||𝑣𝑛||||𝑣𝑛||𝑁𝑗/(𝑁1)𝑑𝑥,(3.12) where 𝑣𝑛 is the Schwarz symmetrization of 𝑣𝑛. Notice that the estimate |𝑥|>𝑟1|𝑥|1+𝑁𝑗/(𝑁1)𝑑𝑥=𝜔𝑁1𝑟𝑡𝑁1𝑡1+𝑁𝑗/(𝑁1)=𝜔𝑑𝑡𝑁1𝑟𝑁𝑗/(𝑁1)𝑁+1𝑁1𝑁𝑗/(𝑁1)𝜔𝑁1𝑟,(3.13) for all 𝑗𝑁1, together with the Radial Lemma [4] leads to 𝑗=𝑁1𝛼𝑗0𝑗!|𝑥|>𝑟||𝑣𝑛||||𝑣𝑛||𝑁𝑗/(𝑁1)𝑁𝑑𝑥𝑀𝜔𝑁11/𝑁𝑗=𝑁1𝛼𝑗0𝑁𝑗!𝜔𝑁1𝑗/(𝑁1)𝑀𝑁𝑗/(𝑁1)|𝑥|>𝑟|𝑥|1𝑁𝑗/(𝑁1)𝑑𝑥𝐶(𝑁)𝑟.(3.14) Thus, given 𝛿>0, there exists 𝑟>0 such that |𝑥|>𝑟||𝑣𝑛||𝑁𝑑𝑥<𝛿,|𝑥|>𝑟𝛼exp0||𝑣𝑛||𝑁/(𝑁1)𝑆𝑁2𝛼0,𝑣𝑛𝑑𝑥<𝛿.(3.15) Which implies that |𝑥|>𝑟𝐹𝑣𝑛𝑑𝑥𝐶𝛿,|𝑥|>𝑟𝐹(𝑣)𝑑𝑥𝐶𝛿.(3.16)
Using the estimate ||||𝑁𝐹𝑣𝑛||||||||𝐹(𝑣)𝑑𝑥|𝑥|𝑟𝐹𝑣𝑛||||+||||𝐹(𝑣)𝑑𝑥|𝑥|>𝑟𝐹𝑣𝑛||||,𝐹(𝑣)𝑑𝑥(3.17) we get lim𝑛||||𝑁𝐹𝑣𝑛𝑁||||𝐹(𝑣)𝑑𝑥𝐶𝛿,(3.18) Hence, we obtain that 𝑁𝐹𝑣𝑛𝑑𝑥𝑁𝐹(𝑣)𝑑𝑥.(3.19)

Lemma 3.2. The numbers 𝐴 and 𝑐 satisfy the inequality 𝐴𝑐.

Proof. For each 𝑣𝑊1,𝑁(𝑁){0}, since we only consider the nontrivial solutions of the problem (1.1), we divide them into two cases to consider.
Case 1. Let 𝑣+=max{𝑣,0}0, we define the function by (𝑡)=𝑁𝐺(𝑡𝑣)𝑑𝑥=N𝑡𝐹(𝑡𝑣)𝑁𝑣𝑁𝑁𝑑𝑥.(3.20) By (𝑔1), we obtain that there exists 𝛿>0,0<𝑐0<1 such that |𝑠|<𝛿, and ||||𝑓(𝑠)<𝑐0|𝑠|𝑁1.(3.21) Hence (𝑡)𝑁0𝑡𝑣𝑐0|𝑠|𝑁11𝑑𝑥𝑁𝑁𝑡𝑁𝑣𝑁=𝑐𝑑𝑥0𝑁𝑁|𝑡𝑣|𝑁1𝑑𝑥𝑁𝑁|𝑡𝑣|𝑁𝑑𝑥,(3.22) we obtain that (𝑡)<0 for 𝑡 small enough. On the other hand, by (𝑔2), we obtain that (𝑡)>0 for 𝑡 large enough. In this way, there exists 𝑡0>0 such that (𝑡0)=0, That is, 𝑡0𝑣𝒫. Hence 1𝐴𝑁𝑅𝑁||𝑡0𝑣||𝑁𝑡𝑑𝑥=𝐼0𝑣max𝑡0𝐼(𝑡𝑣).(3.23)
Case 2. Let 𝑣+=max{𝑣,0}=0, since 𝑓(𝑠)=0 for all 𝑠<0, we obtain max𝑡0𝐼(𝑡𝑣)=+.(3.24) As a consequence, 𝐴𝑐.(3.25) Combining Cases 1 and 2, we obtain that 𝐴𝑐.

Lemma 3.3. The number 𝐴 defined by (2.8) is positive, that is, 𝐴>0.

Proof . Clearly, 𝐴0. Assume by contradiction that 𝐴=0 and let {𝑢𝑛} be a minimizing sequence in 𝑊1,𝑁() to 𝐴, that is, 1𝑁𝑁||𝑢𝑛||𝑁𝑑𝑥𝐴=0with𝑁𝐺𝑢𝑛𝑑𝑥=0.(3.26) For each 𝜆𝑛>0, set 𝑣𝑛(𝑥)=𝑢𝑛(𝑥/𝜆𝑛) satisfying 1𝑁𝑁||𝑣𝑛||𝑁1𝑑𝑥=𝑁𝑁||𝑣𝑛||𝑁𝑑𝑥.(3.27) Similarly, we have 𝑁𝐺𝑣𝑛𝑑𝑥=𝜆𝑁𝑛𝑁𝐺𝑢𝑛𝑑𝑥=0,𝑁||𝑣𝑛||𝑁𝑑𝑥=𝜆𝑁𝑛𝑁||𝑢𝑛||𝑁𝑑𝑥.(3.28) We choose 𝜆𝑁𝑛=1/𝑁|𝑢𝑛|𝑁𝑑𝑥, so 𝑁|𝑣𝑛|𝑁𝑑𝑥=1. Then we get 1𝑁𝑁||𝑣𝑛||𝑁𝑑𝑥𝐴=0,𝑁||𝑣𝑛||𝑁𝑑𝑥=1,𝑁𝐺𝑣𝑛𝑑𝑥=0.(3.29)
In what follows, we study in the space 𝑊1,𝑁rad(𝑁). Firstly, we assume that there exists 𝑣𝑊1,𝑁rad(𝑁) such that 𝑣𝑛𝑣 in 𝑊1,𝑁rad(𝑁).
On one hand, since (1/𝑁)𝑁|𝑣𝑛|𝑁𝐴=0, then 𝑁0>0, for all 0<𝜖<1, when 𝑛>𝑁0, we have 𝑁|𝑣𝑛|𝑁𝑑𝑥<𝜖<1 and we also know that 𝑁|𝑣𝑛|𝑁𝑑𝑥=1, so 𝑣𝑛𝐿𝑁(𝑁)𝑀<. From Lemma 3.1, we have 𝑁𝐹𝑣𝑛𝑑𝑥𝑁𝐹(𝑣)𝑑𝑥.(3.30) Note that 𝑁𝐺𝑣𝑛𝑑𝑥=𝑁𝐹𝑣𝑛1𝑁||𝑣𝑛||𝑁𝑑𝑥=0,(3.31) so we have 𝑁𝐹𝑣𝑛1𝑑𝑥=𝑁𝑁||𝑣𝑛||𝑁1𝑑𝑥=𝑁.(3.32) Hence, we have 𝑁1𝐹(𝑣)𝑑𝑥=𝑁.(3.33)
It implies that 𝑣0. On the other hand, since 𝑣𝑛𝑣 in 𝑊1,𝑁rad(𝑁), and the space 𝑊1,𝑁rad(𝑁) is a reflexible Banach space, we have lim𝑛1inf𝑁𝑁||𝑣𝑛||𝑁1𝑑𝑥𝑁𝑁||||𝑣𝑁𝑑𝑥0.(3.34) Since lim𝑛1inf𝑁𝑁||𝑣𝑛||𝑁𝑑𝑥=𝐴=0,(3.35) we get 1𝑁𝑁||||𝑣𝑁𝑑𝑥=0.(3.36) From which it follows that 𝑣=0, we have an absurd. Hence, we have 𝐴>0.(3.37)

Lemma 3.4. If 𝜆>(𝑞𝑁/𝑞)(𝑞𝑁)/𝑁𝐶𝑞𝑞/𝑁, then 𝑐<1/𝑁.

Proof. From (𝑔3), we have 𝑓(𝑠)>𝜆𝑠𝑞1, for all 𝑠0. Now we choose 𝜓W1,𝑁rad(𝑁) such that 𝜓0,𝜓𝑁𝑞=𝐶𝑞1,𝜓𝑊1,𝑁(𝑁)=1.(3.38) Hence, we have 𝑐max𝑡0𝐼(𝑡𝜓)=max𝑡01𝑁𝑁||||(𝑡𝜓)𝑁+||||𝑡𝜓𝑁𝑑𝑥𝑁𝐹(𝑡𝜓)𝑑𝑥=max𝑡0𝑡𝑁𝑁𝑁0𝑡𝜓𝑓(𝑠)𝑑𝑠𝑑𝑥max𝑡0𝑡𝑁𝑁𝜆𝑁0𝑡𝜓𝑠𝑞1𝑑𝑠𝑑𝑥=max𝑡0𝑡𝑁𝑁𝜆𝑡𝑞𝑞𝑁𝜓𝑞.𝑑𝑥(3.39) Let 𝐾(𝑡)=𝑡𝑁/𝑁(𝜆𝑡𝑞/𝑞)𝑁𝜓𝑞𝑑𝑥, then 𝐾(𝑡) is continuous function, we have 𝐾(𝑡)=𝑡𝑁1𝜆𝑡𝑞1𝑁𝜓𝑞𝑑𝑥=0.(3.40) By a simple calculation, when 𝑡0=(1/𝜆𝑁𝜓𝑞𝑑𝑥)1/(𝑞𝑁)>0, we have max𝑡>0𝑡𝐾(𝑡)=𝐾0=1𝑁1𝜆𝑁𝜓𝑞𝑑𝑥𝑁/(𝑞𝑁)𝜆𝑞𝑁𝜓𝑞1𝑑𝑥𝜆𝑁𝜓𝑞𝑑𝑥𝑞/(𝑞𝑁)=𝑞𝑁𝜆𝑁𝑞𝑁/(𝑞𝑁)𝐶𝑞(𝑞/𝑁)(𝑁/(𝑞𝑁))<𝑞𝑁𝑁𝑞𝑞𝑁𝑞((𝑞𝑁)/𝑁)(𝑁/(𝑞𝑁))𝐶𝑞(𝑞/𝑁)(𝑁/(𝑞𝑁))𝐶𝑞𝑞/(𝑞𝑁)=1𝑁.(3.41) Hence, we have 1𝑐<𝑁.(3.42)

Lemma 3.5. The number 𝐴 is attained, that is, there exists 𝑢𝑊1,𝑁rad(𝑁) such that 𝐴=𝑁|𝑢|𝑁𝑑𝑥 and 𝑁𝐺(𝑢)𝑑𝑥=0.

Proof. Let {𝑢𝑛} be a minimizing sequence in 𝑊1,𝑁rad(𝑁) for 𝐴, that is, 1𝑁𝑁||𝑢𝑛||𝑁𝑑𝑥𝐴(𝑛),𝑁𝐺𝑢𝑛𝑑𝑥=0.(3.43) Arguing as in Lemma 3.3, we assume that 𝑁|𝑢𝑛|𝑁𝑑𝑥=1. From (3.43), Lemmas 3.3 and 3.4, we obtain lim𝑛𝑁||𝑢𝑛||𝑁𝑑𝑥=𝑁𝐴𝑁𝑐<1.(3.44) From Lemma 3.1, 𝑁𝐹𝑢𝑛𝑑𝑥𝑁𝐹(𝑢)𝑑𝑥(3.45) where 𝑢𝑛𝑢in𝑊1,𝑁(𝑁),as𝑛.
By (3.43) and (3.45), we have 𝑁𝐹𝑢𝑛1𝑑𝑥=𝑁𝑁||𝑢𝑛||𝑁1𝑑𝑥=𝑁,𝑁1𝐹(𝑢)𝑑𝑥=𝑁.(3.46) It implies that 1𝑢0,(3.47)𝑁𝑁||||𝑢𝑁𝑑𝑥lim𝑛1inf𝑁𝑁||𝑢𝑛||𝑁𝑑𝑥=𝐴,(3.48)𝑁|𝑢|𝑁𝑑𝑥lim𝑛inf𝑁||𝑢𝑛||𝑁𝑑𝑥=1.(3.49) From (3.48) and (3.49), we have 𝑁𝐺(𝑢)𝑑𝑥=𝑁1𝐹(𝑢)𝑑𝑥𝑁𝑁|𝑢|𝑁1𝑑𝑥=𝑁1𝑁𝑁|𝑢|𝑁𝑑𝑥0.(3.50) If 𝑁𝐺(𝑢)𝑑𝑥0, from (3.50), we have 𝑁𝐺(𝑢)𝑑𝑥>0. Consider the function defined in Lemma 3.2 relative to the function: (𝑡)=𝑁𝐺(𝑡𝑢)𝑑𝑥.(3.51) We concludes that (𝑡)<0 for 𝑡 small enough. On the other hand, (1)=𝑁𝐺(𝑢)𝑑𝑥>0. In this way, we obtain that there is 𝑡0(0,1) such that (𝑡0)=0, that is, 𝑁𝐺𝑡0𝑢𝑑𝑥=0.(3.52) Hence, from (3.48), 10<𝑁𝑁||𝑡0𝑢||𝑁1𝑑𝑥=𝑁𝑡𝑁0𝑁||||𝑢𝑁𝑑𝑥𝑡𝑁0𝐴<𝐴.(3.53) However, from (3.52), we have 𝑡0𝑢𝒫. Hence, we obtain 1𝑁𝑁||𝑡0𝑢||𝑁𝑑𝑥𝐴.(3.54) Which is contradictory with (3.53).
Thus, we obtain 𝑁𝐺(𝑢)𝑑𝑥=0.(3.55) It implies 𝑢𝒫 and 1𝑁𝑁||||𝑢𝑁𝑑𝑥𝐴.(3.56) From (3.48) and (3.56), we obtain that 1𝑁𝑁||||𝑢𝑁𝑑𝑥=𝐴,(3.57) with 𝑁𝐺(𝑢)𝑑𝑥=0,𝑢0.
We obtain that 𝐴 is attained.

Proof of Theorem 1.2. From Lemma 3.5, there is 𝑢𝑊1,𝑁rad(𝑁){0} such that 1𝑁𝑁||||𝑢𝑁𝑑𝑥=𝐴,𝑁𝐺(𝑢)𝑑𝑥=0.(3.58) we will prove that 𝑚=𝑏=𝐴.
By Lagrange multipliers, there exists 𝜌, such that 𝑁||||𝑢𝑁2𝑢𝑣𝑑𝑥=𝜌𝑁𝑔(𝑢)𝑣𝑑𝑥,(3.59) for every 𝑣𝑊1,𝑁(𝑁).
Define the rescaled function 𝑢𝜌1/𝑁=𝑢(𝜌1/𝑁𝑥), which is a nontrivial solution of (1.1) with 𝑁||𝑢𝜌1/𝑁||𝑁𝑑𝑥=𝑁||||𝑢𝑁𝑑𝑥,𝑁𝐺𝑢𝜌1/𝑁𝑑𝑥=𝜌𝑁𝐺(𝑢)𝑑𝑥=0.(3.60) Thus, we have 𝑢𝑚𝐼𝜌1/𝑁=1𝑁𝑁||𝑢𝜌1/𝑁||𝑁𝑑𝑥𝑁𝐺𝑢𝜌1/𝑁1𝑑𝑥=𝑁𝑁||||𝑢𝑁𝑑𝑥=𝐴.(3.61) So, we have 𝑚𝐴.(3.62)
For each 𝛾Γ, one has 𝛾([0,1])𝒫 from [4]. We obtain that there exists 𝑡0[0,1] such that 𝛾(𝑡0)𝒫, that is, 𝛾(𝑡0) satisfied that 𝑁𝐺(𝛾(𝑡0))𝑑𝑥=0 and then 1𝐴𝑁𝑁||𝑡𝛾0||𝑁1𝑑𝑥𝑁𝑁𝐺𝛾𝑡0𝛾𝑡𝑑𝑥=𝐼0.(3.63) Hence 𝐴𝐼(𝛾(𝑡0))max𝑡[0,1]𝐼(𝛾(𝑡)) for every 𝛾Γ, we obtain that 𝐴𝑏.(3.64) From (3.62) and (3.64), we obtain that 𝑚𝐴𝑏.
On the other hand, for every nontrivial solution 𝜔𝑊1,𝑁(𝑁) of the problem (1.1), there exists a path 𝛾𝜔Γ such that 𝜔𝛾𝜔([0,1]) and max𝑡[0,1]𝐼(𝛾𝜔(𝑡))=𝐼(𝜔). Consequently, 𝑏𝐼(𝜔),𝑏𝑚.
In conclusion, we obtain 𝑚=𝐴=𝑏.(3.65) Hence, the function 𝑢𝜌1/𝑁 is a ground-state solution of the problem (1.1).


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


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