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</svg>-Laplacian Equations with Critical Growth and Indefinite Weight

Zhang, Guoqing; Yao, Ziyan

doi:https://doi.org/10.1155/2014/942092

Abstract and Applied Analysis

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Function Spaces, Compact Operators, and Their Applications

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Research Article | Open Access

Volume 2014 | Article ID 942092 | https://doi.org/10.1155/2014/942092

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

Guoqing Zhang^1,2and Ziyan Yao¹

Academic Editor: S. A. Mohiuddine

Received07 Sept 2013

Revised10 Dec 2013

Accepted20 Dec 2013

Published23 Jan 2014

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

Remark 16. The idea and proof of Lemma 15 follow as in Lemma 4.1 in [5].

3.2. Min-Max Value

In order to get a more precise information about the minimax level obtained by the Mountain Pass Theorem, we consider the following sequence of scale which is called the Moser function: Hence, we have , the support of is the ball , and

Let ; we have where From (105), we conclude that , as . Consequently, we have

Lemma 17. Suppose that (H1)–(H3) and (f1)–(f6) hold. Then there exists such that

Proof. Choose as in and such that where Suppose, by contradiction, that, for all , we get where . For each , there exists such that Thus; we have From , , we obtain Let , we have By (f6), given that , there exist and ; we have From (116) and (117), for large , we obtain Let we have Hence, the sequence is bounded. Otherwise, up to subsequences, we have , which leads to a contradiction. From (108) and (115) and it follows that From [4], we have Now, using the change of variable by straight forward computation, we have which converges to as , where Finally, let in (118); from (108) and (115), we have which implies that

Remark 18. The idea and the proof of Lemma 17 come from Lemma 3.6 in [5].

Lemma 19. There exist and with such that for all . In particular, .

Proof. See Lemma 3.3 in [10].

Corollary 20. Under the conditions (H1)–(H3) and (f1)–(f4), if , then one has

From Lemmas 13 and 19, we conclude that

Corollary 21. There exist and with compact support such that, for all ,

Lemma 22. If is a Cerami sequence for at any level with then possesses a subsequence which converges strongly to a solution of problem (1).

Proof. See Lemma 4.6 in [4].
In conclusion, we have

3.3. Multiplicity Results

In order to prove the existence of the second solution of problem (1) follows by the minimum argument and Ekeland's Variational Principle.

Proposition 23. Under the conditions (H1)–(H3) and (f1)–(f6), there exists such that, for each with , problem (1) has a solution via Mountain Pass Theorem.

Proof. See Proposition 4.1 in [5].

Proposition 24. There exists such that, for each with , (1) has a minimum type solution with , where is defined in (130).

Proof. See Proposition 5.1 in [5].

Proposition 25. If is sufficiently small, then the solutions of problem (1) obtained in Propositions 23 and 24 are distinct.

Proof. By Propositions 23 and 24, there exist sequences , in such that Suppose by contradiction that . As in the proof of Lemma 15, we have Hence, by (f2) and (f3) and Generalized Lebesgue's Dominated Convergence Theorem, we obtain that there exists such that
Claim 1. , as . Indeed, by (f2) and (f3), we have Hence, on the domain and , we have Since is bounded, and using (137), we have For , we have Since is bounded, we have Combining (139) and (141), we have Similarly, we also have Combining (136), (142), and (143), we obtain Hence, the claim is proved.
Claim 2. . Indeed, we have From [10], we have On the other hand, since , we have By the inequality , we have Hence, we have in and . It is a contradiction. The proof is complete.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are grateful to the reviewers for some valuable suggestions. This paper was supported by Shanghai Natural Science Foundation Project (no. 11ZR1424500) and Shanghai Leading Academic Discipline Project (no. XTKX2012).

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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.

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