Abstract and Applied Analysis

Volume 2014 (2014), Article ID 349304, 8 pages

http://dx.doi.org/10.1155/2014/349304

## Infinitely Many Nontrivial Solutions of Resonant Cooperative Elliptic Systems with Superlinear Terms

^{1}School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan 455000, China^{2}School of Mathematical Sciences and LPMC, Nankai University, Tianjin 300071, China

Received 4 March 2014; Accepted 11 May 2014; Published 9 June 2014

Academic Editor: Adrian Petrusel

Copyright © 2014 Guanwei Chen and Shiwang Ma. 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 study a class of resonant cooperative elliptic systems and replace the Ambrosetti-Rabinowitz superlinear condition with general superlinear conditions. We obtain ground state solutions and infinitely many nontrivial solutions of this system by a generalized Nehari manifold method developed recently by Szulkin and Weth.

#### 1. Introduction and Main Results

We consider the following cooperative elliptic system:
where is a bounded smooth domain in and . The nonlinearities are the gradient of some function; that is, there exists a function such that . For system (1), we are interested in the* resonant* case; that is,
where , denotes the spectrum of the matrix , and and denotes the eigenvalues of the Laplacian on with zero boundary condition.

If , and on , then (1) reduces to the following single elliptic equation: The authors [1, 2] have considered the strongly resonant single elliptic equation (3) with odd nonlinearities and obtained a finite number of solutions. Li and Zou [3] investigated (3) by using the Morse theory.

The system (1) has been studied by many authors under asymptotically linear or sublinear assumptions on nonlinearities; see [4–10]. In [4], the variational structure was established and several existence results were obtained by minimax techniques under a condition which was called nonquadraticity at infinity. Ma [5] established the existence of infinitely many solutions for (1) with odd nonlinearities by the minimax techniques. Ma [6] and Zou et al. [9] established the existence and multiplicity of solutions for (1) via the computations of the critical groups and the Morse theory. By using a penalization technique and the Morse theory, Pomponio [7] established the existence and multiplicity of solutions of (1). However, very little is known about the existence of infinitely many solutions for resonant single elliptic equation and elliptic systems (both cooperative and noncooperative). Zou [8] considered (1) and, by using the methods used in [1], obtained infinitely many solutions under the oddness and boundedness assumptions on the nonlinearities. Zou [10] proved that (1) has infinitely many solutions under the oddness assumption and some growth assumptions near . Recently, the author [11] also obtained the existence of* ground state* solutions for (1) in the* nonresonant case* (i.e., ) by a variant weak linking theorem. For related topics, including noncooperative elliptic systems, we refer the readers to [12–18] and references cited therein.

Let and denote, respectively, the usual norm and the inner product in . Assume that for all , ; for some and , where if and if as uniformly in ; as uniformly in ; for all , ; and , and , ; and , .

In this paper, we obtain ground state solutions of (1), that is, nontrivial solutions corresponding to the least energy of the action functional of (1). Moreover, if is even in , we obtain infinitely many solutions of (1). Our main results based on a generalized Nehari manifold method developed in [2]. Now, our main results read as follows.

Theorem 1. *If hold, then (1) admits a ground state solution.*

Theorem 2. *If and , for all , hold, then (1) has infinitely many solutions such that .*

Considering the following superquadratic condition there exists a constant such that which is now known as Ambrosetti-Rabinowitz superlinear condition. As we all know, our proofs will be more easier if we use the assumption (4). But we replace condition (4) with more general superquadratic conditions. As is shown in next examples, our assumptions are reasonable and there are cases in which Ambrosetti-Rabinowitz condition (4) is not satisfied.

*Example 3. *Let
where and is a continuous function. Clearly, satisfies .

*Example 4. *Let
where is a continuous function, , if , and if . Note that
It is not hard to check that satisfies but does not satisfy condition (4).

The rest of the present paper is organized as follows. In Section 2, we firstly establish the variational framework of (1), and then we give some preliminary lemmas, which are useful in the proofs of our main results. In Section 3, we give the detailed proofs of our main results.

#### 2. Variational Framework and Preliminaries

Here and in what follows, we use to denote the norm of . Let where is the usual Sobolev space with the norm generated by the inner product , for . For and in , the induced inner product and norm on are given, respectively, by

Let and ; then , , , and . For any , let be the subspaces of , where the quadratic form is positive definite, negative definite, and zero, respectively. Let Set where denotes the identity from to . We introduce an operator Then is a bounded self-adjoint operator from to and with . The space splits as , where and are invariant under is negative, and is positive definite. More precisely, there exists a positive constant such that Here and in what follows, for any , we always denote by , and the vectors in with , and . Note that and are finite. Furthermore, implies that .

For problem (1), we consider the following functional: where . We define an equivalent inner product and the corresponding norm on given, respectively, by where . Thus, can be rewritten as Under our assumptions, and the derivative is given by where . By the discussion of [4], the (weak) solutions of system (1) are the critical points of the functional .

We let By definition, it is easy to see that contains all nontrivial critical points of . We define for the subspace, and the convex subset, of , where .

We assume are satisfied from now on. Obviously, and imply that for each there is such that

Lemma 5. *Let , where , , and ; then
*

*Proof. *We fix and . For , we let , so . Let
We need to show that whenever (i.e., ). We first consider the case . Then and by . We may therefore assume from now on. It is not hard to check that and imply that
Suppose that attains its maximum on at some point ; then
*Case **1.* If , since , it follows from and that
*Case 2.* If , it follows from (25) and that ; it follows from the fact , , and that
Therefore, whenever .

*Lemma 6. Let ; then for any , . Hence is the unique global maximum of .*

*Proof. *Let and . Since , we have
which together with Lemma 5 and the facts and implies that
The proof is finished.

*In what follows, we let
*

*Lemma 7. One has the following facts.There exists such that , where .Consider for every .*

*Proof. *() For , we have and as by (21) and ; hence the second inequality follows if is sufficiently small. Since for every there is such that , the first inequality is a consequence of Lemma 6.

() For , we have
thus . The proof is finished.

*Lemma 8. If is a compact subset, then there exists such that on for every .*

*Proof. *Without loss of generality, we may assume that for every . Suppose by contradiction that there exist and () such that for all and as . Passing to a subsequence, we assume . Let ; then
Thus .

Therefore, after passing to a subsequence. We may assume that in and a.e. in after passing to a subsequence. If , then . Hence as ; it follows from and Fatou’s lemma that
which contradicts with (32). If , then it follows from that and . Thus . Therefore, (33) still holds. We also get a contradiction. The proof is finished.

*Lemma 9. For each , the set consists of precisely one point which is the unique global maximum of .*

*Proof. *By Lemma 6, it suffices to show that . Since , we may assume that . By Lemma 8, there exists such that on . By Lemma 7(), for small , and since on . It is easy to see that is weakly upper semicontinuous on ; therefore, for some . This is a critical point of , so for all . Consequently, , as required.

*Lemma 10. is coercive on ; that is, as , .*

*Proof. *Arguing by contradiction, suppose that there exists a sequence such that and for some . Let . Then in and a.e. in after passing to a subsequence. By Lemma 7(), . By Sobolev compact embedding theorem, we get
If , then it follows from (21), (34), and that as for every . Since for , Lemma 6 implies that
This yields a contradiction if . Hence , which implies that . Therefore, ; it follows again from and Fatou’s lemma that ; therefore,
as , a contradiction. The proof is finished.

*Lemma 11. The map , (see Lemma 9) is continuous.*

*Proof. *Let . By a standard argument, the continuity of in is reduced to the following assertion:

To prove the above assertion, we let be a sequence with . Without loss of generality, we may assume that for all , so that
By Lemma 8, there exists such that
Hence is bounded by Lemma 10. Note that and are finite. Passing to a subsequence, we may assume that
where by Lemma 7(). Therefore, we have
Note that . By Lemma 9, we have
which together with (40), (41), and Fatou’s lemma implies that
On the other hand, Lemma 9 implies that . Therefore, Lemma 9 implies that . It follows from (41) that . The proof is finished.

*We now consider the functional
which is continuous by Lemma 11.*

*Lemma 12. Consider and for , .*

*Proof. *We put , so we have . Let . Choose such that for and let . We may write with . Then , and the function , is continuous by Lemma 11. By Lemma 9 and the mean value theorem, we have
with some . By a similar reasoning, we also have
with some . Combining (45) and (46), we conclude that the directional derivative exists and is given by
Hence is linear (and continuous) in and depends continuously on . So the assertion follows by [19, Proposition 1.3].

*Next we consider the unit sphere in . We note that the restriction of the map to is a homeomorphism with inverse given by
We also consider the restriction of to .*

*Lemma 13. One has the following facts: and for , is a Palais-Smale sequence of if and only if is a Palais-Smale sequence of ,one has . Moreover, is a critical point of if and only if is a critical point of , and the corresponding critical values coincide.*

*Proof. *() is a direct consequence of Lemma 12.

To prove (), let be a sequence such that , and let . Since for every we have an orthogonal splitting
and for all , we find that and using (),
By Lemma 7() and Lemma 10, we have . Hence is a Palais-Smale sequence for if and only if is a Palais-Smale sequence for .

In () the proof is similar as in () but easier.

*3. Proofs of Main Results *

*3. Proofs of Main Results**We are now in a position to prove our main results.*

*Proof of Theorem 1. *From Lemma 7() we know that . Moreover, if satisfies , then is a minimizer of and therefore a critical point of , so that is a critical point of by Lemma 13. It remains to show that there exists a minimizer of . By Ekeland’s variational principle [19], there exists a sequence with and as . Put for . Then and as by Lemma 13(). By Lemma 10, is bounded and hence in after passing to a subsequence. As a result of the Sobolev compact embedding theorem, we get
If , then it follows from (21), (51), and Hölder’s inequality that
Note that as implies that
Therefore, we have as , which is contrary to Lemma 7(). So and .

*For any , assumption () implies that
Therefore, . We prove that and there is such that . By , Fatou’s lemma, and the boundedness of , we have
where the last inequality follows from the definition of . Hence and because .*

*Proof of Theorem 2. *The functional is of class on by Lemma 13. It is obviously even since is even. implies that is a critical point of . We will show that satisfies the Palais-Smale condition. Suppose that is bounded and . Then is bounded by Lemma 13; hence is also bounded by Lemma 10. Therefore, we can take a subsequence still denoted by such that
Employing Lemma 13 again, we have
By (56) and (57), we can easily get that in . Hence (see (48) for the definition of ).

Let
where denotes the usual Krasnoselskii genus (see, e.g., [20, 21]) and the infimum is taken over all closed subsets with . Since are well defined and positive for all . Now standard arguments using the deformation lemma, see, for example, [19–21], imply that all are critical values of and (that is seen as [20, Proposition 9.33]). Hence, setting , we have
By , the integrand above is nonnegative, so implies that . The proof finished.

*Conflict of Interests*

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

*Acknowledgments*

*Acknowledgments**Research is supported by the Tianyuan Fund for Mathematics of NSFC (Grant no. 11326113) and the Key Project of Natural Science Foundation of Educational Committee of Henan Province under Grant 13A110015 of China.*

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