Abstract
In this paper, we are concerned with the system of the Schrödinger-Maxwell equations where are constants, and . Under appropriate assumptions on and , we prove the existence of positive solutions in the case via the truncation technique. Moreover, suppose that may change sign, we also obtain the multiplicity of solutions for the case .
1. Introduction and Main Results
Consider the following system of Schrödinger-Maxwell equations: where are constants, and . Such a system is also called the Schrödinger-Poission equation which is obtained while looking for the existence of standing waves for nonlinear Schrödinger equations interacting with an unknown electrostatic field. For more details on the physical aspects, we refer the reader to [1] and the references therein.
On the potential , we make the following assumptions:
() and is bounded below.
() there exists a constant such that the set is nonempty and , where denote the Lebesgue measure in .
() is nonempty and has smooth boundary and .
In recent years, system (1) has been widely studied under various conditions on and . The greatest part of the literature focuses on the study of the system for and being constants or radially symmetric functions. We refer the reader to [2–9].
When the potential is neither a constant nor radially symmetric, in [10, 11], the existence of ground state solutions is proved for . In [12–14], the existence of nontrival solution is obtained via the variational techniques in a standard way under the following condition:
() , where is a constant. Moreover, for any , , where denote the Lebesgue measure in .
It is worth mentioning that conditions () were first introduced by Bartsch and Wang [15] to guarantee the compact embedding of the functional space. If replacing by more general assumptions and , the compactness of the embedding fails and this situation becomes more complicated. Recently, [16, 17] considered this case. The authors studied the following problem where is a parameter, the potential may change sign and is either the superlinear or sublinear in as . Very recently, Liu and Mosconi [18] considered the following system with a coercive sign-changing potential and a 3-sublinear nonlinearity:
By using a linking theorem, the authors obtained the existence of nontrivial solutions. Nextly, Gu, Jin, and Zhang [19] investigated the existence of sign-changing solutions for system (3). By using the method of invariant sets of descending flow, the multiple radial sign-changing solutions are obtained in the subquadratic case as small. For more results about the Schrödinger-Poisson systems, we refer the reader to [20–23] and the reference therein.
Here, we should point out that for the power-type nonlinearity , in order to get the boundedness of a (PS) sequence, the methods heavily rely on the restriction . Meanwile, the condition cannot guarantee the compactness of the embedding of into the Lebesgue spaces , . This prevents from using the variational techniques in a standard way. Motivated by the works mentioned above and [24–28], in the present paper, we are mostly interested in sign-changing potentials and consider system (1) with more general potential , , and the range of . Our main results are as follows:
Theorem 1. Suppose that , ()- , , , and hold. Then, system (1) possesses at least a nontrivial solution for small and large.
Remark 2. It is known that it is difficult to get the boundedness of a (PS) sequence when dealing with the case . To overcome the difficulty, motivated by [24, 25], we use the truncation technique to obtain a bounded Cerami sequence for small. In this case, the conditions and of Theorem 1.3 in [23] cannot be used. Moreover, in the process of proving the convergence of a bounded Cerami sequence, we use the observation that the condition makes the less strong influence of the nonlocal term (The conclusions remain valid if ). In this sense, Theorem 1 can be viewed as an improvement of Theorem 1.3 in Zhao et al. [23].
Theorem 3. Suppose that (), , , , and hold. Then, system (1) possesses infinitely many distinct pairs of nontrivial solutions whenever is sufficiently large.
Remark 4. In Theorem 3, is allowed to be sign-changing for . We obtain the multiplicity of solutions for (1).
2. Preliminaries
Let is the usual Sobolev space with the standard inner product and norm
In our problem, we work in the space defined by with the inner product and the norm where , . Obviously, it follows from that the embedding is continuous.
As in [26], let and denote the orthogonal complement of in by . Consider the eigenvalue problem
In view of , the quadratic form is weakly continuous. We have the following proposition.
Proposition 5 ([26], Lemma 8). Suppose , , and . Then, for each fixed , (i) as (ii) is a nonincreasing continuous function of , where is sequence of positive eigenvalues of problem satisfying as and the corresponding eigenfunctions .Let Then, Moreover dim for every fixed .
In the sequel, we denote the usual -norm by , and stands for different positive constants. For any , denotes the open ball of radius centered at . Since the continuity of the following embedding there are constants and such that
It is well known that system (1) is the Euler-Lagrange equation of the functional defined by
Evidently, the action functional belongs to and its critical points are the solutions of system (1). It is easy to know that exhibits a strong indefiniteness, namely, it is unbounded both from below and from above on infinitely dimensional subspaces. This indefiniteness can be removed using the reduction method described in [29], by which we are led to study a one variable functional that does not present such a strongly indefiniteness.
Actually, considering for all , the linear functional defined in by
If , the Hölder inequality and the Sobolev inequality imply while for , we have
Hence, by the Lax-Milgram theorem, there exists a unique such that
Moreover, we can write an integral expression for in the form:
So, we can consider the functional defined by . Then,
It follows from (16), (17), (18), and the Sobolev inequality that
Thus, is a well-defined functional with derivative given by
By the proposition 2.3 in [12], we know that is a critical point of if and only if is a critical point of and .
To complete the proof of our theorem, we need the following results.
Theorem 6 (see [30]). Let be a real Banach space with its dual space , and suppose that satisfies for some and with . Let be characterized by where is the set of continuous paths joining 0 and . Then, there exists a sequence such that
Theorem 7 see ([31], Theorem 9.12). Let be an infinite dimensional Banach space and let be even, satisfy (PS), and . If , where is finite dimensional and satisfies.
there are constants such that , and
for each finite dimensional subspace , there is an such that on ,
then, possesses an unbounded sequence of critical values.
Lemma 8 ([17], Lemma 2.3). The function possess the following properties: (1)The mapping maps bounded sets of into bounded sets of ;(2)If in , then in ;(3)The mapping is continuous.
3. Proof Of Theorem 1
In this section, we give the proof of Theorem 2 by using Theorem 6. Let be a cut-off function satisfying , if , if , and for each . For every we consider the truncated functional defined by
It is easy to see that is of . Moreover, for each , we have
Lemma 9. Suppose that and , and , hold. Then, (i)there exists a ,with (ii)there exists independent of and such that , where ,
Proof. For any , by (19), we have
Let and for some . Let be such that . Set , then for . Hence, it is easy to see that
Since , we have that as . Taking for large enough, we have (iii)since there exists a constant (independent of , and ) such that .
Lemma 10. Suppose that and hold. Then, there exists such that .
Proof. For any , we have
Since , we conclude that there exists such that for all with
From Lemmas 9 and 10 and Theorem 6, we thus deduce that there exist a Cerami sequence such that
Lemma 11. Let , , , and be satisfied. Then, there exists such that for each , if is sequence satisfying (32), then has a convergent subsequence in for small enough.
Proof. Let be a Cerami sequence satisfying (32). Let , we show that . We first prove that . Suppose by contradiction that there exist a subsequence of , still denoted by , such that , we obtain which is a contradiction by Lemma 9. Suppose that there exists no subsequence of which is uniformly bounded by . Then, we deduce that . We handle the case of (The conclusions remain valid if ). By (22), we obtain This is contradiction by choosing sufficiently small. Since , passing to a subsequence if necessary, we can assume that there exists such that , in . In view of the Sobolev embedding theorems and Lemma 8, in implies. Furthermore, for any , we have In fact, if , we just need to show that Note that The first limit on the right is 0 by the fact in and so is the second limit because in and . While for , on the one hand, since , using (35) (c), we have as . On the other hand, since , for any there exists such that Moreover, in view of the Sobolev embedding theorem, implies that Hence, for large , we obtain Consequently, From above inequality and (39), one has Then, implies that Let . It follows from , that Moreover, let . Then, . By the Sobolev inequalities and Hölder inequality, one has We know Letting be so large that the term in the brackets above is positive when , we get in . Since and , it follows that in . This completes the proof.
Proof of Theorem 2. Note that is also a Cerami sequence of satisfying , the conclusion follows from Lemmas 9, 10, and 11 and Theorem 6.
4. Proof Of Theorem 3
In this section, while is sign-changing, we study the existence of solutions of (1) for the case and give the proof of Theorem 3. Without loss of generality, we assume that .
Lemma 13. Let , and be satisfied. Then, there exist such that for all with .
Proof. By proposition 5, for each fixed , there exists a positive integer such that for and for . Thus, for any , we have for all , where . Since , the conclusion follows by choosing sufficiently small.
Lemma 14. Let , and be satisfied. Then, for any finite dimensional subspace , there is a large such that on .
Proof. Since all norms are equivalent in a finite dimensional space, there is a constant such that By (22), there is a constant such that Hence, for all , Since , there is a large such that on .
Lemma 15. Let , and be satisfied. Then, there exists such that, for each , satisfies the condition.
Proof. Let be a sequence, that is and . If is unbounded in , up to a subsequence, we can assume that
as , after passing to a subsequence. Set , we can assume that in and a.e..
If , since is weakly continuous, we have
a contradiction. If , then the set has the positive Lebesgue measure. For , one has as , Fatou’s lemma shows that as . Thus, by (22), we obtain
This is a contradiction. This implies is bounded in . Going if necessary to a subsequence, we can assume that in . The following proof is similar to the proof of Lemma 11. We omit details of this.
Proof of Theorem 3. Obviously, . Furthermore, is even. The conclusion follows from Lemmas 13, 14, and 15, and Theorem 7.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgment
This work is supported partially by the Foundation of Baoshan University (BYPY202016).