Abstract

In the present paper, we consider the following Hamiltonian elliptic system : , A new existence result of nontrivial solutions is obtained for the system via variational methods for strongly indefinite problems, which generalizes some known results in the literatures.

1. Introduction and Main Results

The goal of this paper is to study the existence of solutions for the nonperiodic elliptic systems in Hamiltonian form where , , , and . Such a system arises when one is looking for stationary solutions to certain systems of optimal control (Lions [1]) or systems of diffusion equations (Itô [2] and Nagasawa [3]).

In recent years, the systems like or similar to in the whole space were studied by a number of authors. Most of these works focused on the case . An usual way to overcome this difficulty is to consider the corresponding functional in the space of radially symmetric functions. In this way, De Figueiredo and Yang [4] considered the system where , , and They proved that system (1) has a radial solution pair under the assumptions that and are superlinear in and radially symmetric with respect to , and with . This result was later generalized by Sirakov [5] to the system In [6], Zhao et al. considered periodic asymptotically linear elliptic systems where the potential is periodic and has a positive bound from below and and are periodic in , asymptotically linear in as . By using critical point theory of strongly indefinite functionals, they obtained the existence of a positive ground state solution as well as infinitely many geometrically distinct solution for systems (3) under the assumptions that and are odd in . For other results, we refer readers to [717].

Without assumption of periodicity or radially symmetric about nonlinearities, the problem is quite different in nature and there has not been much work done up to now. In a recent paper [18], Wang et al. considered the following nonperiodic elliptic systems in Hamiltonian form: and obtained the following theorem.

Theorem A (see [18, Theorem ]). Suppose that the following conditions are satisfied: .There is such that if .  , where with as uniformly in , and there exist , such that , where .  There is with for , and as uniformly in bounded set of , and for all , where and .Then system (4) has one solution.

In the present paper, we are interested in the existence of solutions for Hamiltonian-elliptic systems involving gradient terms and nonperiodic superquadratic nonlinearities. The class of problems treated here has several difficulties. First, the problem is set on ; a main difficulty when dealing with this problem is the lack of compactness of the Sobolev embedding theorem. Second, the variational functional is strongly indefinite. Therefore, the classic critical point theorem cannot be applied directly. Third, the nonlinearities are nonperiodic in variable and superquadratic at infinity; the method in [6] cannot be applied to obtain the existence of solutions. Finally, the appearance of the gradient terms in the systems also brings us some difficulties; in this case, the variational framework in [18] cannot work any longer. Inspired by recent works of Zhao and Ding [19], we are going to investigate the existence of solutions for the Hamiltonian elliptic systems . By using the critical point theory of strongly indefinite functional which was developed recently by Bartsch and Ding [10, 20] and the reduction methods which was developed in [21, 22], we obtain an existence result of problem , which generalizes Theorem A.

Our fundamental assumptions are as follows: is 1-periodic in for and ., and . is 1-periodic in for .   , where with as uniformly in , and there exist , such that , where . as uniformly in . whenever , where . for some and all with large enough.There is with for , and as uniformly in bounded set of , for all and whenever .

Now we can state our main result.

Theorem 1. Let , , , and be satisfied. Then system has at least one nontrivial solution.

Remark 2. Theorem 1 extends and improves Theorem A. First, we only need to assume that the potential is periodic and has a positive bound from below. Second, the conditions and can be obtained by and . In fact, by (), we know that Consequently, by the conditions and (5), it is easy to see that and hold. Furthermore, similar to the proof of Lemma  2.2(i) in [23], the condition can be obtained by and (5). Indeed, since , we can obtain that . For some , . If , then there exists such that Choose so large that whenever . Then, by , (6), and (7), we obtain It follows that Third, the condition is weaker than the condition and the condition is weaker than the condition . Finally, summing up the above discussion, Theorem A is the special case of Theorem 1 corresponding to .
Throughout this paper, we always assume that denote any positive constant and may be different in different places. For , we define , where .

2. Variational Setting

In this section, we will establish variational framework for the system . For the convenience of notation, let denote the usual -norm and be the usual -inner product. Let and be two Banach spaces with norms and ; we always choose the equivalent norm on the product space . In particular, if and are two Hilbert spaces with inner products and , we choose the inner product on the product space . In order to continue the discussion, we need the following notations. Set and is a matrix operator. Let denote the Schrödinger operator. Denote and Then can be rewritten as

Denote by and the spectrum and the essential spectrum of the operator , respectively. Set ; then we have the following lemmas.

Lemma 3 (see [19, Lemma ]). Suppose that and are satisfied. Then the operator is a self-adjoint operator on with domain .

Lemma 4 (see [19, Lemma ]). Let , , and be satisfied. Then(1) = ; that is, has only essential spectrum;(2) and is symmetric with respect to origin;(3).It follows from Lemmas 3 and 4 that the space possesses the orthogonal decomposition such that is negative definite (resp., positive definite) in (resp. ). Let denote the absolute value of and be the square root of . Let be the Hilbert space with the inner product and norm . possesses an induced decomposition which are orthogonal with respect to the inner products and (the above results can be found in [19]).

Lemma 5 (see [19, Lemma ]). and are equivalent norms. Therefore, embeds continuously into for any and compactly into for , and there exists constant such that On we define the following functional: where . It follows from that, for any , there is such that for all . Thus, Lemma 5 implies that is well defined on . Lemma 4 implies that is strongly indefinite; such type functional appeared extensively when one considers differential equations via critical point theory; see, for example, [2427] and the references therein. Our hypotheses imply that (see Lemma  3.10 in [27]) and a standard argument shows that the critical points of are solutions of .

3. The Abstract Critical Point Theorem

In order to study the critical points of , we now recall a abstract critical point theorem developed recently in [10, 20]. Let be a Banach space with direct sum and corresponding projections onto . We assume that the Banach space is separable and reflexive. Let be a dense subset; for each there is a seminorm on defined by Denote by and the topology induced by seminorm family and the weak-topology on , respectively. denotes the weak-topology on . Now, some notations and definitions are needed.

For a functional , we write , , and .

Suppose() for any , is -closed, and is continuous;()for any , there exists such that for all ;()there exists such that , where .

Theorem 6 (see [10] or [20]). Let be satisfied and suppose there are and with such that where . Then, has a -sequence with . Moreover, if satisfies the -condition for all then has a critical point with .

Lemma 7 (see [10] or [20]). Let Suppose(1) is bounded from below;(2) is sequentially lower semicontinuous; that is, in implies ;(3) is sequentially continuous;(4) is and is sequentially continuous.

Then satisfies .

4. The Limit Equation

In this section, we study the following limit equation related to , where is given in , , and . By virtue of , we have firstly the following lemma.

Lemma 8. and possess the following properties.(i) as and for some .(ii) as .(iii) whenever , where .(iv) whenever large enough.(v) as .

Proof. (i) It is clear by and .
(ii) By , for any , there is such that whenever . Hence for all . Observe that it follows from that as . Letting we get for all .
(iii) Since as , it follows from that (iv) By , for large enough Letting , we obtain (v) It follows from (ii)–(iv) that as .
Now, we set and define the functional By Lemma 8, for any , there is such that It follows from (30) and (31) that is well defined and its critical points are solutions of .

Lemma 9. possesses the following properties:(1) is weakly sequentially lower semicontinuous and is weak sequentially continuous.(2)There is such that , where .

Proof. (1) Suppose in . Going if necessary to a subsequence, we can assume in for and a.e. in . Hence a.e. in . Thus Next it is sufficient to show that is weak sequentially continuous. Indeed, by (30) and in , , it follows from Theorem  A.2 in [27] that Furthermore, for each fixed , one has that, for any , there exists such that Hence, for large , it follows from (30), (33), (34), and Hölder inequality that Therefore,(2) For any , it follows from (31) that The conclusion follows because .
Now, we choose a number such that . From Lemma 8(ii), there is such that whenever . Let be the spectrum family of the operator . It follows from Lemma 4 that is a infinite dimension subspace of and

We have the following result.

Lemma 10. For any finite dimensional subspace of ,

Proof. If not, then there are and with such that for all . Denote , passing to a subsequence if necessary; we can assume that , , and . Then which yields that We claim that . Indeed, if not then it follows from (41) that . Thus , which contradicts with . By (38), we get Hence, there exists such that where . Note that Hence Now the desired conclusion follows from this contradiction.

As a consequence, we have the following.

Lemma 11. Let be given by Lemma 9. Then, letting with , there is such that , where .

Lemma 12. Let be any -sequence for . That is, is such that as . Then it is bounded and . Moreover, there is a subsequence still denoted by satisfying and , as .

Proof. Let be such that Then, for large , one has which implies . If is unbounded in , up to a subsequence if necessary, we can assume that . Set . Then and for each . Note that Hence, one has On the other hand, for and , set Then, by Lemma 8, we have for all and as . For large , one has and Consequently, for large and , whenever one has Since implies For any , we choose . Using the Hölder inequality we have as uniformly in and as . Therefore, as . Let be given. It follows from Lemma 8 that there is such that for all . Consequently, we have for all . By Lemma 8, we define and . By (55), we can take so large that for all . For fixed , it follows from (57) that there is such that Now the combination of (58)–(60) implies that for This contradicts with (50). Hence, is bounded. Passing to a subsequence if necessary, we can assume that in , in for , and a.e. in . Since is an autonomous system, by and Lemma 8, we know that all of the conditions of Lemma  5.7 in [20] are satisfied. Hence we have and , as .

Lemma 13. Let be the critical set of . Then and is attained.

Proof. Set and . Then . For any and , using the fact that one has This yields , and hence . satisfies . By virtue of Lemma 7 and conclusion (1) of Lemma 9, it follows that holds. From conclusion  (2) of Lemma 9, we know that holds. Thus, combining with Lemma 11, we know that all of the conditions of Theorem 6 hold. Then there is sequence for with . By Lemma 12, sequence is bounded in . For the concentration functions , there are only two cases needed to be considered: vanishing and nonvanishing. If vanishing occurs, by the vanishing lemma Hence which implies that . Hence as . This contradicts with as . So nonvanishing occurs; that is, there exist , and such that Choose such that Thus . Setting , by the invariance under translation of , is a -sequence of and . From we see that , and hence is a nontrivial critical point of . Therefore If , one has which implies . If , let be such that as . Then is -sequence. By Lemma 12, we can assume that . Then By Lemma 8(i), we see that, for any , there is such that It follows from Lemma 8(iv) that there exists constant such that By the continuity of , there exists such that Note that for some . Hence, there exists such that for all Denote ; we obtainBy Hölder inequality (), we have Hence , a contradiction. Hence . Finally, we show that there is with . Let be such that . Then, as before, is bounded, and one may assume Now that is, . This completes the proof.

Definition 14 (see Ackermann [28]). A mapping from Banach space to another Banach space is called BL-split, if for every weakly convergent sequence with it holds that in .
In what follows, we use the idea of [21, 28]. For fixed , we introduce the functional by Hence one has for all , which implies that is strictly concave. Moreover which implies that as . Now, it follows from Lemma 9 that is weakly sequentially upper semicontinuous. Hence, there is a unique strict maximum point for , which is also the only critical point of on and satisfies for all , , . Now, we define the reduced functional by We have the following lemma.

Lemma 15. is a critical point of if and only if is a critical point of . Moreover, the following conclusions hold:(1) and .(2) is a bounded map.(3), , are all BL-splits.(4) in if in .(5) is -invariant, i.e., for all .

Proof. It follows from that and Hence, combining with , , and , we know that all of the conditions of Lemma  2.6 in [19] hold. So, by Lemma  2.6 in [19], the desired conclusions can be obtained.
By Lemma 15, the critical points of and are in one-to-one correspondence via the injective map from into . Consequently, let Then In particular

Lemma 16. The sequence is a bounded -sequence of if and only if is a bounded -sequence of .

Proof. Let be a sequence of ; that is, and . Since , we have By (81) and Lemma 15, for all , we have and Hence, one has Similarly Consequently Hence That is, is sequence of . The inverse is obvious. By Lemma 15, the boundedness of yields the equivalence between the boundedness of and . This completes the proof.

Next, we discuss the mountain pass geometry of the reduce functional . One has the following Lemma.

Lemma 17. possesses the mountain pass geometry:(1)There is such that , where .(2)There is some such that .

Proof. By (31) and (80), for any , we have Hence, for small , conclusion (1) holds.
Similar to the proof of Lemma 10, we can obtain that Hence conclusion (2) holds.

Remark 18. Lemma 17 implies that is an isolated critical point of . Therefore there is a such that for all , where .

Remark 19. If we set then, by Lemma  4.7 in [18], we have that,. for each , there is a unique such that and .

Lemma 20. Let be such that and set . Then

Proof. For any , by (80), we obtain Since , we have . Hence, combining with Remark 19, we obtain . Consequently, one has

5. Proof of the Main Result

In this section we give the proof of Theorem 1. Let be the set of critical points for . First, we study the linking structure for the functional . Similar to the proof of Lemma 9, we have the following three lemmas.

Lemma 21. is nonnegative and weakly sequentially lower semicontinuous.

Lemma 22. is weakly sequentially continuous.

Lemma 23. There exists such that , where .

Lemma 24. There is such that for any with and

Proof. For any , by , we have Thus, the conclusion follows easily from Lemma 10.
In particular, set with , , and , where is given in Lemma 24. Then we have the following lemma.

Lemma 25. .

Proof. By Lemmas 23 and 24, we have . Since is weakly upper semicontinuous on , there is some with such that . From and Lemma 20, we obtain

In what follows, we discuss sequences of . Firstly we have the following.

Lemma 26. Any -sequence for is bounded.

Proof. The proof is similar to Lemma 12. We omit the details here.

Lemma 27 (see [23, TheoremA.2]). Let be an open set in and be a function such that for some and . Suppose , is a bounded sequence in , and and in for all . Then, passing to a subsequence, there exists a sequence such that in and where and is such that for , for , is a sequence of constants with as , the space with the norm and the space with the norm

Lemma 28. Let be -sequence of . Then, passing to a subsequence if necessary, we can assume that in . Furthermore, either (i) or(ii) and there exist a positive integer , and sequences , , such that, after extraction of a subsequence of where and is a cut-off function such that for , for .

Proof. First, Lemma 26 implies that any sequence of is bounded; hence it is a bounded sequence. Passing to a subsequence if necessary, we can assume that with , a.e. in and in . Let be a cut-off function such that for , for and define ; then in . Indeed, by Lemma 5, implies that for any there is a corresponding such that Hence, using the Lebesgue dominated convergence theorem, we obtain Now we set . Then, by , one has . Moreover, we claim that is a bounded -sequence for with . In fact, by , for any , there is a such that uniformly in bounded set of . On the one hand, by (19), taking , , and in Lemma 27, we know that On the other hand, by (18), taking , and . It follows from Lemma 27 that where Hence, for each with , one has where and the constant is independent of . ConsequentlyFor any , observe that Hence, it follows from (110) and (114) that and then as . Consequently, by (19), (31), and (117), for large , we obtain Consequently In what follows, we claim that Indeed, since in for , by (18), (30), and Theorem  A.2 in [27], we have Hence, for any with , it follows from (109) and (121) that Hence, we obtain In addition, we have and if and only if in . In fact, if , then Hence, using (74) and Hölder inequality (), we obtain This implies that in . The inverse is obvious. Now, assume that conclusion (i) is false. Then For the concentration functions , there are only two cases needed to be considered: vanishing and nonvanishing. If vanishing occurs, by the vanishing lemma and (30)-(31), one has Hence This is a contradiction. Thus, nonvanishing occurs; that is, there is a sequence and constants , such that for large . We may choose and such that, passing to a subsequence, for all . Note that , , and . We know that is bounded -sequence of . Hence, by Lemma 12, passing to a subsequence if necessary, we can assume that ,as , where . Note that Hence Since and , we obtain that and . There are now two possibilities to consider: or .
Set . If , repeat the above arguments but replace and by and , respectively. We obtain that if and only if . Consequently, we obtain that . Hence the lemma holds with , (where is chosen to ensure that ) and .
If , then we argue again as above with and replaced by and , respectively, and we obtain with . After at most steps, we obtain the desired conclusion.

As a straight consequence of Lemma 28, we have the following.

Lemma 29. satisfies condition for all .

We are now in a position to complete the proof of Theorem 1.

Lemma 30. satisfies .

Proof. For any and , using the definition of and , one has This yields , and hence . satisfies .

Proof of Theorem 1. By Lemma 30, satisfies . It follows from Lemmas 7, 21, and 22 that holds. By Lemma 23, we know that holds. Lemma 24 shows that the linking condition of Theorem 6 holds. These, together with Lemma 25, yield -sequence with for . By virtue of Lemma 29, we can assume that as . Furthermore, we have and . This implies that . The proof is completed.

6. Summary

The purpose of this paper is to investigate the existence of nontrivial solutions for a class of Hamiltonian elliptic system in in the case that the nonlinearity may not satisfy the standard Ambrosetti-Rabinowitz condition . We obtain rather general conditions for the existence of nontrivial solutions, which extend and improve some recent results in the literature. In this paper, since we have no assumption of periodic about nonlinearity, the problem is more difficult and interesting. We will point out that if the following condition is satisfied:(HP),then for any , we have , , and . Hence, combining with Lemmas 7, 21, 22, 23, and 30, similar to the proof of Lemmas 11 and 13, we have the following.

Theorem 31. and is attained; that is, the problem has a ground state solution.

Conflicts of Interest

The authors declare that they have no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work is supported partially by the National Natural Science Foundation of China (11426037), the Applied Basic Research Programs of Yunnan Province (2012FD060), Outstanding Young Teachers Training Program of Yunnan Province, and Youth Academic and Technical Leaders of Baoshan Municipality.