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Abstract and Applied Analysis
Volume 2014 (2014), Article ID 942092, 14 pages
Multiple Solutions for a Class of -Laplacian Equations with Critical Growth and Indefinite Weight
1College of Sciences, University of Shanghai for Science and Technology, Shanghai 200093, China
2Department 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.
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 .
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  studied the existence of nontrivial solutions for the following -Laplacian equations with critical growth: where is bounded smooth domain in . Adimurthi and Sandeep  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  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  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  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  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 , and it implies that the function does not satisfy the Ambrosetti-Rabinowitz condition.
2. Preliminary Results
2.1. Key Lemmas
Now, we define the following Sobolev space and the corresponding norm,
Lemma 2 (see ). 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
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 ). 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 .
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
we have . From , 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 , we have For any fixed , set Because is bounded, is a finite set. From Lemma 4.4 in (), 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 , we have Since is finite, it follows that a.e. This implies, up to a subsequence that in . Let in (76), and in