In this study, we consider the following sublinear Schrödinger equations where satisfies some sublinear growth conditions with respect to u and is not required to be integrable with respect to x. Moreover, V is assumed to be coercive to guarantee the compactness of the embedding from working space to for all . We show that the abovementioned problem admits at least one solution by using the linking theorem, and there are infinitely many solutions when is odd in u by using the variant fountain theorem.

1. Introduction and Main Results

In this study, we consider the following nonlinear Schrödinger equations:where . The existence and multiplicity of solutions for (1) have attracted much attention of mathematicians. As we know, the existence of solutions for problem (1) is important in finding the solutions for

With the development of the variational methods in past decades, many mathematicians used this tool to show the existence and multiplicity of solutions for differential equations with superlinear, sublinear, asymptotic linear, and mixed nonlinearities (see [125]). In this study, we mainly consider the sublinear case. Many mathematicians have considered this topic with being integrable in x. For example, in 2012, Zhang and Wang [23] assumed V to be coercive and satisfies the following: , where is a constant and is a positive continuous function such that .In addition, they obtained infinitely many nontrivial solutions for (1). By using the dual fountain theorem, Bahrouni [1] showed (1) admitted infinitely many solutions when , where and a, b satisfy the following conditions: and there exists such that for any . and .Later, Chen and Tang [6] relaxed assumption , and they obtained the following theorem.

Theorem 1. (see [6])Suppose the following conditions hold and . For any , there exist , , and such that There exist an open set and , such that where , .

Then, problem (1) possesses at least one nontrivial solution. Moreover, if for all , then problem (1) possesses infinitely many nontrivial solutions.

In Theorem 1, V is not required to be coercive, and the authors took advantage of the decay of to show the convergence of the asymptotic critical points sequence. Recently, Zhang et al. [24] introduced some different sublinear conditions of f to obtain infinitely many nontrivial solutions for (1). They assumed the weight functions to be integrable in with and .

After then, Hou et al. [7] considered the fractional Schrödinger-type equations with the similar conditions to and . In the abovementioned studies, the results depend on the integrability of the weight functions. There are still some mathematicians considered the case without any integrable weight functions. In 2009, Benrhouma et al. [3] considered the following problem:where and . They used the minimizing method on a Nehari manifold and concentration-compactness principle due to Lions [10, 11] to show the existence of unique positive solution for (2). Then, Benrhouma [2] improved this result by assuming the following: and ., a.e. .

In 2013, Bao and Han [5] considered (2), where and V is a sign-changing function while 0 lies in the spectrum of and obtained infinitely many solutions for this resonant problem. In a recent study, Isernia [8] generalizes the results in [2] to a class of sublinear fractional Schrödinger equations. Motivated by the abovementioned studies, we introduce some new sublinear growth conditions without weight functions to obtain the existence and multiplicity of solutions for (1). First of all, we recall the following coercive condition which is proposed by Wang and Han [14] to show the existence of infinitely many small solutions for a class of Kirchhoff equations with local sublinear nonlinearities by using the symmetric mountain pass theorem. There are constants and such that for any

Remark 1. Condition is introduced by Wang and Han in [14] to obtain a compactness embedding theorem. In addition, the authors assumed that V possesses the global positive lower bound. Although we do not assume that V is positive in the whole space in this study, the compactness embedding theorem still holds in the same way.

Now, we state our main results.

Theorem 2. Suppose that satisfies and f satisfies the following: There exist and such that for all and there exists such that uniformly in x. Letting , there exist , , and such that There exists such thatThen, there exists at least one nontrivial solution for problem (1).

Theorem 3. Suppose that the conditions of Theorem 2 hold and. for any and .Then, there exist infinitely many solutions for problem (1).

Remark 2. In Theorem 2, is sublinear at infinity and can be a sign-changing function. LetThen, satisfies the conditions of Theorem 2, but the conditions in [18, 20, 2325].

2. Preliminaries

First, we show the decomposition of the space. Similar to [15], we denote by the self-adjoint extension of the operator with the domain . Let be the spectral family of and be the square root of . Define . commutes with , and is the polar decomposition of . Let be the domain of . Define the norm on E aswhere and in the sequel, denotes the usual inner product. Then, E is a Hilbert space. The following compact embedding theorem is obtained with the condition .

Lemma 1. Suppose that holds. Then, E is compact embedded into for any .

Proof. The proof of this lemma is similar to the proof of Lemma 1 in [14].

Remark 3. From Lemma 1, for any , there exists a constant such thatIt follows from Lemma 1 that the spectrum consists of eigenvalues numbered in , and a corresponding system of eigenfunctions forms an orthogonal basis in . Let , , and . Set , , and . Then, withwhere and .
The following lemma can easily be obtained by . We omit its proof here.

Lemma 2. Suppose that holds, then for any , there exists such thatLet be the corresponding functional defined byIt is known that the critical points of I are the solutions for problem (1). The following lemma shows that I is differentiable in E.

Lemma 3. Suppose the conditions of Theorem 2 hold, then andwhich yields that

Proof. By (12), it is easy to see that I is well defined. Subsequently, we show (16) holds. For any u, , , and , it follows from Lemma 2 and (12) thatIt follows from the definition of I, (18), the mean value theorem, and the Lebesgue’s dominated convergence theorem that there exists such thatThen, we obtain (16). Subsequently, we show is continuous. LetAssume that in E as . By Lemma 1, we obtain thatwhich implies that for any and with , there exists such thatfor n large enough. Since is continuous in t, we getThen, for n large enough, we havewhich yields that as . Hence, is weak continuous. Therefore, ψ is compact and is continuous. The proof of this lemma is finished.

3. Proof of Theorem 1

In order to apply the link theorem by Rabinowitz (see [6], Theorem 5.29) to obtain the critical points of I, we assume in the following proof. Obviously, I and J share the same critical points.

Lemma 4. Suppose the conditions of Theorem 1 hold, then J satisfies the condition.

Proof. Let be a sequence, then there exists a constant such thatFirst, we show that is bounded in E. Arguing in an indirect way, we assume that as . It follows from (25), and thatLetthen, we haveSince is a finite-dimensional subspace of E, there exists such thatMoreover, there exist , such thatwhich implies thatChoosing , we can deduce from (26) and (12) thatfor some , , , where . By (25), we obtainThen, it follows from , (32), and (33) thatfor some , , which implies that as . Together with (31), we obtain as , which is a contradiction. Hence, is bounded in E. Then, there exists such that , which implies that uniformly in for any . Similar to (4), we deduce thatwhich shows that as since is of finite dimensions. Hence, J satisfies the condition.

Lemma 5. Suppose the conditions of Theorem 1 hold, then for any , , there are and such that on and , where .

Proof. From , for any with and , there exists such thatSince , for , one hasfor some . Then, we can easily find the desired ρ and δ.

Lemma 6. Suppose the conditions of Theorem 1 hold, then as .

Proof. For any , from , there exists such thatwhich, together with Lemma 2, implies thatWe finish the proof of this lemma.

Proof of Theorem 2. If , we can deduce from Lemma 4–6 that J possesses a minimax-type critical point, which is a solution for (1). If , let . Similar to the proof of Lemma 5, there exists small and such thatIn addition, by Lemma 6, we can choose large enough such thatLet . Since S and link, by the linking theorem (see [6], Theorem 5.29), J has a critical point with .

Proof of Theorem 3. In order to obtain infinitely many solutions for problem (1), we need the variant fountain theorem by Zou in [25]. Let E be a Banach space with the norm and with for any . Set and . Consider the following functional defined by

Lemma 7. (see [25])Suppose that the functional defined above satisfies the following: maps bounded sets to bounded sets uniformly for . Furthermore,for all;; as on any finite dimensional subspace of E; There exists such thatThen, there exist , such that , as . In particular, if has a convergent subsequence for every k, then has infinitely many nontrivial critical points satisfying as .
Subsequently, we show I possesses infinitely many critical points when is even in t, which are obtained by the variant fountain theorem. Define for any . Let E be decomposed as follows . Set and . In the following proof, we show that I possesses geometrical structure of the variant fountain theorem. In order to make use of the variant fountain theorem, we setThen, we have the following lemmas.

Lemma 8. There exists a sequence as such that

Proof. Setfor any . Since for , we can see that . So as for some for any . For every , there exists such that and . Furthermore, for any , we can see that , which implies thatThen, we have as in E. It follows from Lemma 1 that in . Hence, we conclude that for any . For any , by the definitions of and Lemma 2, we obtainLet . Obviously, we have as . When k large enough, we can conclude thatMoreover, it follows from (48) thatTherefore, when k large enough, we can deduce thatSince , as , we haveThen, we obtain our conclusion.

Lemma 9. There exists a sequence for all such that

Proof. For any , . It follows from thatfor some and any . LetThen, we obtain thatThe proof is completed.
It is easy to see condition is satisfied in Lemma 4 since is even in t. Similar to Lemma 4 in [13], it can easily be deduced from that as on any finite dimensional subspace of E. Then, the condition in Lemma 7 is satisfied. From Lemma 8 and Lemma 9, we can see that all the conditions of Lemma 1 are satisfied. Then, there exist , such thatSimilar to Lemma 4, there exists such that in E as for any . Then, we can deduce that possesses infinitely many nontrivial critical points. Therefore, (1) has infinitely many nontrivial solutions.

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this article.


This work was supported by the National Natural Science Foundation of China (no. 11801472), the China Scholarship Council (no. 201708515186), the Youth Science and Technology Innovation Team of Southwest Petroleum University for Nonlinear Systems (no. 2017CXTD02), and the Science and Technology Innovation Team of Education Department of Sichuan for Dynamical System and its Applications (no. 18TD0013).