In this paper, we consider the following nonlinear Schrödinger-Poisson systems. Under suitable conditions on , , , and , when , we obtain two nontrivial solutions for the problem and when is odd and , we obtain infinitely many solutions for the problem.

1. Introduction

In this paper, we consider the following nonlinear Schrödinger-Poisson equations on

Such problem arises when one is looking for standing wave solutions for the nonlinear Schrödinger equationcoupled with the Poisson equation

It is well known that the Schrödinger-Poisson systems have a strong physical meaning because they appear in quantum mechanics models and in semiconductor theory (see [13]). Problem (1) is a special case of the following Schrödinger-Poisson system:

Cerami and Vaira in [4] studied the existence of positive solutions for the problem where they considered , , and . After that, many researchers have focused on the problem under various conditions (see [5, 6]).

In recent years, the Schrödinger-Poisson system has been studied widely under variant assumptions on , , and (see [710]). Because the problem is set on the whole space , it is well known that the main difficulty of this problem is the lack of compactness for Sobolev embedding, and then, it is usually difficult to prove that a minimizing sequence or a sequence is strongly convergent if we seek solutions by variational methods. In order to overcome this difficulty, most of them dealt with the situation where is a positive constant or being radially symmetric (see [1113]). When is not a constant and not radially symmetric, there have been many works by developing various variational techniques (see [4, 1417]).

In paper [18], Sun et al. considered the systemwhere , , , and are measurable functions satisfying suitable assumptions. They obtained infinitely many solutions in with negative energy. Since is allowed to be supercritical, the usual Sobolev space cannot be used for the study; to overcome this difficulty we introduce a new space which is motivated by [19].

We should also mention another recent paper [20]; Wang et al. considered a similar problemand a nontrivial solution is obtained (see [20], Theorem 2). Note that there is an inhomogeneous term on the right-hand side of the first equation. Here and the potential satisfies a coercive condition so that the working space can be compactly embedded into Lebesgue spaces.

Now, we turn to our problem (1). Since the space dimension is , the critical Sobolev exponent . For we denote

Note that is the conjugate exponent in Hölder inequality. To state our results on the problem (1), we make the following assumptions:

(V) satisfies .

(K) , for a.e. .

(g1) ; for some and , we have

(g2) There exist , , and ; such that

(h1) and ; for .

(h2) , , and for .

Remark 1. It follows from (g1) and (g2) that for and small enough.

Under these assumptions, it is clear that the zero function is the trivial solution of problem (1). Our main results on the existence of multiple nontrivial solutions are the following theorems.

Theorem 2. Suppose (V), (K), (g1), (g2), and (h1) are satisfied; then, problem (1) has two nontrivial solutions.

Theorem 3. Suppose (V), (K), (g1), (g2), and (h2) are satisfied. If is odd, then problem (1) admits infinitely many solutions.

The paper is organized as follows. In Section 2, we give some useful notions and set up the variational framework of the problem. In Section 3, we prove Theorem 2, and the proof of Theorem 3 is given in Section 4. For simplicity, throughout this paper, we denote the norm on with by .

2. Preliminary

Thanks to condition (V), the normis an equivalent norm on . Let ; then, we have a continuous embedding . Hence, there is a constant such that

For later use, we also denote by the best Sobolev constant for the continuous embedding .

For , it is well known that the Poisson equationhas a unique solution in . Note that according to [21], Theorem 2.2.1,

Consequently, in . We also know that there exists such that

See [14], Lemma 1.1.

Define a functional ,

Under our assumptions, it is easy to see that . According to Benci and his collaborators [1, 22], it is well known that if is a critical point of , then is a (weak) solution of (1).

To find critical points of , some compactness conditions, such as the well-known Palais-Smale condition ( for short), are crucial. To establish the condition for , we need the following results.

Proposition 4. Let such thatfor , , and . Then, the functional ,is of class and is compact.

Remark 5. Proposition 4 is a special case of do Ó ([23], Lemma 1). Note that since the embedding is continuous, the statement of Proposition 4 remains valid if we replace with .

Motivated by Ruiz ([13], Lemma 2.1), we also have the following result.

Lemma 6. Suppose and in . Then, in .

Proof. As in the proof of [13], Lemma 2.1, we define linear functionals ,It can be shown that and are continuous. Note thatby the isometry between and via the Riesz representation theorem; it suffices j. Let ; we choose such thatNow, let with ; using the Hölder inequality, we haveSince is bounded in andby the compactness of the Sobolev embedding, letting in (21), we deduceuniformly for . It follows that in . The proof is completed.☐

Remark 7. Even though we have this lemma in hand, we do not know how to deducefrom . However, we can still deduce the condition for our functional (see the proof of Lemma 9).

3. Proofs of Theorem 2

To prove Theorem 2, we will apply the truncated method, see e.g. [24]. For a function , we set . Then, we define the truncated functional ,

It is well known that is of class ; the critical points of are solutions of the truncated problem

Moreover, suppose is a critical point of , then . Hence, is a solution of (1).

Lemma 8. Under the assumptions of Theorem 2, the functional is coercive. As a consequence, is bounded from below.

Proof. By (g1), we haveNote that and on ; using the Hölder inequality and the factwe haveSince , we get as . Thus, we have proved Lemma 8.☐

Lemma 9. Under the assumptions of Theorem 2, the functional satisfies the condition.

Proof. Let be a sequence of . By Lemma 8, is bounded. Hence, up to a subsequence, we have in . Firstly, we haveTo apply Proposition 4, let ,By our assumptions (g1 ) and (h1 ), we can apply Proposition 4 (see also Remark 5) and deduceTherefore, since is bounded in , we haveUsing Lemma 6 and the continuous embedding , we see that in . Hence, by the boundedness of in , we deduceCombining (30)–(34), we obtainSince the integral in the final line is nonnegative, we deduce that in ; thus, we have proved Lemma 9.☐

Now we are ready to present the proof of Theorem 2.

Proof of Theorem 2. Firstly, we claim that the zero function is not a minimizer of . For this purpose, we choose a nonnegative function . Suppose ; then, for all , we have . Thus, using (g2) and noting that , we deduceBecause and , we see thatfor small enough. So is not a minimizer of .
By Lemma 8 and Lemma 9, we know that is bounded from below and satisfies the condition. It is well known that there exists a minimizer of (see e.g., [25], Corollary 2.5). By the claim above, we see that and it is a critical point of . As mentioned at the beginning of this section, and it is a nontrivial solution of (1).
In a similar manner, by considering ,we can obtain another nontrivial solution , which is nonpositive on . This completes the proof of Theorem 2.☐

4. Proofs of Theorem 3

In this section, we prove Theorem 3. Note that, in our Theorem 3, since is allowed to be supercritical, the usual space cannot be used as our framework for the study of problem (1). For this reason, motivated by [18, 19], we introduce a new space as our working space. Let be the completion of under the norm

For a nonnegative measurable function and , we define the weighted Lebesgue spaceand it is associated with the seminorm

Motivated by [18, 19], let be the completion of with respect to the norm

Then, is a Banach space.

Lemma 11. If (V) and (g1 ) are satisfied, then we have the compact embedding . Furthermore, we also have the compact embedding .

Proof. By our assumption on , using the results in [26] (see ([26], page 255)), the embedding is well defined and compact. The compactness of follows from the continuity of .

Lemma 12 ([19], Lemma 2.2). Given , there is such that, for any and for , we have

Now, let us define the variational functional corresponding to problem (1). We set as

By Lemma 11, all the integrals in (45) are well defined and converge; we know that the weak solutions of problem (1) correspond to the critical points of functional with derivative given by

Lemma 13. Under the assumptions of Theorem 3, the functional is coercive.

Proof. Because is the best Sobolev constantusing (g1 ) and Hölder inequality, we getFor large enough, (48) together with Lemma 12 give thatbecause . This implies that is coercive on .

In general, to prove the condition, the reflexivity of the space is needed. However, we do not know whether is reflexive, but we can still prove the following.

Lemma 14. Under the assumptions of Theorem 3, the functional satisfies the condition.

Proof. From Lemma 13, we can deduce that every sequence of is bounded in , and is also bounded in . Therefore, we can assume that for some , up to a subsequenceFirst, we show that . For any , sincewe haveNow, we claim thatIndeed, by (14), we see that is bounded in , hence up to a subsequence in , we haveMoreover, by the Hölder inequality, we getwhere . Therefore, (56) and (57) give thatThus, (53) holds.
Next, we verify (54) and (55). It is easy to see that the sequence is bounded in . Since a.e. in ; applying the Brezis-Lieb lemma, up to a subsequence, we haveMoreover, ; thus, we have (54). Similarly, using (g1 ) and the Lebesgue theorem, we have (55). Letting in (52), we havethat is, .
Next, we prove . By Lemma 11 and the fact that and , we haveOn the other hand, by , the Hölder inequality, and Sobolev inequality, we get , so we can see that is bounded in ; hence, we can assumeHence, by the weak lower semicontinuity of the norm , we getBy Fatou’s lemma, we haveIt follows from (63) and (64) thatFrom (61) and (65), we see thatFrom this and the weak lower semicontinuity of the norm , we see that up to a subsequence,then in .
Finally, we show that . It suffices to show thatIn fact, from in , we have in and in . Hence, by Hölder inequality and Sobolev inequality, we getUsing (61), (67), and (69) we get (68). Thus,The proof is completed.☐

Having verified the condition, we investigate the geometry of . First, we note that obviously, is even (by our assumption on ).

Letand we define

Then, is an infinitely dimensional Banach space under the norm . Therefore, from [19], Lemma 3.3, we know that the seminorm is a norm on .

Let be the class of the closed and symmetric (with respect to the origin) subsets of . For , we define the genus by

If such a minimum does not exist, we define . The main properties of the genus can be found in [27, 28]; we omit them here.

Lemma 15. Given , there is such that

Proof. Given , let be a -dimensional subspace of ; as in the proof of Theorem 2, we can choose and such that , if . (for small enough).
Denote ; is a sphere in . Then, we haveso, by the monotonicity property of genus, we haveThe proof is completed.☐


From Lemma 15, we can define a sequence of real numbers


By Lemma 13, is coercive and bounded from below. That is, , for . For , denote . Then, a standard argument (see [29]) gives the following.

Lemma 16. All the are critical values of . Furthermore, if , then .

Proof of Theorem 3. Because is even and note that by Lemma 15, , thus , for . This final Lemma 16 gives the existence of infinitely many critical points of . So problem (1) has infinitely many solutions. This completed the proof of Theorem 3.☐

Data Availability

No date were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.


This research was supported by the National Natural Science Foundation of China (No. 11501143) and Doctoral research project of Guizhou Normal University (No. GZNUD [2017]27, GZNUD [2019]14).