Abstract

In this paper, we investigate the existence of infinitely many solutions to a fractional -Kirchhoff-type problem satisfying superlinearity with homogeneous Dirichlet boundary conditions as follows: where is a nonlocal integrodifferential operator with a singular kernel . We only consider the non-Ambrosetti-Rabinowitz condition to prove our results by using the symmetric mountain pass theorem.

1. Introduction

In recent years, the problems with fractional and nonlocal operator have attracted a lot of attention. These types of operators arise in many different contexts. We know that there are population dynamics, stratified materials, minimal surface, water waves, continuum mechanics, and so on. As far as we know, we are able to learn more about their association through referring to [16].

The problem we are going to deal with also involves fractional and nonlocal operator. Here, we will study the -Kirchhoff-type problem as follows: is an open bounded smooth domain in with Lipschitz boundary . ; with . is a real parameter. : is a Carathéodory function and is usually called nonlocal operator. It is defined as follows: for all , where The function is measurable. It has the following properties:

As the singular kernel satisfies , we call it a typical model. Hence, the fractional -Laplace operator may be defined as follows: for . Problem (1) also becomes

Usually, we write the Kirchhoff function as . Clearly, in problem (5). When , problem (5) becomes the original problem with the following fractional Laplacian form:

It is the nonlocality that is a typical characteristic of problem (6). In other words, the value of at any point relies not only on but actually also on the whole space. We know that the Dirichlet boundary condition was applied to problem in [5]. In [7], through the use of the mountain pass theorem, Servadei and Valdinoci obtained the existence of nontrivial weak solutions of problem (6). In [8], Pucci and Saldi studied the Kirchhoff-type eigenvalue problem in whole space. They proved the existence and multiplicity of nontrivial solutions. We also refer to [9] for related problems.

On the other hand, the Kirchhoff function is transformable. So far, a variety of forms of function are taken into account in many references on studying Kirchhoff-type problems; see [1018]. In addition, we notice that more attention has been focused on -Kirchhoff-type problems.

In [19], with the help of the Fountain Theorem, they studied the existence of infinitely many solutions for a fractional -Kirchhoff equation. In [20], the authors showed the existence and multiplicity of solutions to a degenerate fractional -Kirchhoff problem. However, we perceive that the condition was used widely in these papers about -Laplacian problems. We refer the interested readers to [12, 2128]. The condition is usually called (AR) condition for short. It is described as follows:And .

It was introduced for the first time by Ambrosetti and Rabinowitz in [29]. Since then, the (AR) condition has been used far and wide in more and more works involving superlinear elliptic boundary. We know that the importance of (AR) condition is to guarantee the boundedness of familiar (PS) sequences for the energy functional associated with the problem. The nonlinearity function satisfies superlinear growth under the (AR) condition.

Through (7), we can getwhere , for two constants .

However, there are still lots of functions that dissatisfy the (AR) condition, even though they are superlinear at infinity. We notice another form given by

We find that the nonlinearity is also superlinear at infinity under condition (9). Obviously, the functionalsatisfies condition (9) and dissatisfies condition (8). So it does not satisfy (7).

Motivated by the above works and [20, 24, 25, 30], we study the existence of infinitely many solutions of problem (1) without (AR) condition. Our results are extension of some problems studied by N. Van Thin in [30].

Now, we give some assumptions on the function : .There exist and such that for a.e. and all , where .   for all uniformly for .   uniformly for .There exists such that the function is decreasing if and increasing if for all There exist and satisfying such that for all and , where

Definition 1. We claim that a function is a weak solution of problem (1), if

See Section 2 for a detailed description for of .

Theorem 2. Let be a function satisfying (3). Let conditions hold. Then, for any , problem (1) has infinitely many nontrivial solutions in with unbounded energy.

Corollary 3. Let be a function satisfying (3). Let conditions hold. If condition replaces , then the conclusion of Theorem 2 holds.

Remark. Originally, Jeanjean put forward a condition that was similar to in [31]. It is easy to see that condition is equivalent to when . Actually, condition is weaker than condition . We can find that there are some functions satisfying but dissatisfying . For example,

This paper consists of the following parts. In Section 2, we give the definition and some properties for the space and some preliminary results. Section 3 verifies compactness conditions. In Section 4, we prove Theorem 2 and Corollary 3.

2. Preliminary Results

Firstly, we recall the functional space and and some lemmas, which will be used in next section for problem (1). We appoint where and . The space X is a linear of Lebesgue measurable functions from to such that the restriction to of any function in belongs to andX is endowed with the following norm:In addition, is endowed with the following norm:and is known as the Hilbert space defined by the following scalar product (see [12], Lemma 7).

We denote the usual fractional Sobolev space by , which is endowed with norm (the so-called ) as follows:

We observe that the norms (14) and (17) are not the same when , since is contained strictly in . It makes the space different from the usual classical fractional Sobolev space. Therefore, from the point of view of the variational method, the classical fractional Sobolev space is insufficient for dealing with our problem.

We recall that the space is nonempty due to (see [12], Lemma 2.1). The following conclusion is correct if a general kernel satisfies (3): Particularly, the following characterization holds when :For more details about space and , we refer to [5, 32].

Considering future works, we recall the following eigenvalue problem:It has a divergent sequence of positive eigenvalueswhose homologous eigenfunctions are denoted by . From Proposition 9 of [32], we know that can be chosen in such a way that this sequence provides an orthonormal basis in and an orthogonal basis in .

Firstly, we define where where .

Clearly, the energy functional associated with problem (1) is well defined.

For convenience, we write as . From Lemma 3.1 of [25], clearly we know that functional And if holds, So, we get that andfor all . In order to prove the conclusion of problem (1), we need some lemmas.

Lemma 4 (see [12]). Assume that (3) holds. We have the following conclusions:
(1) For any , the embedding is compact when is a bounded domain with continuous boundary.
(2) For all , the embedding is continuous.

Definition 5. Let The functional satisfies at the level , if any sequence , withhas a strongly convergent subsequence in . is the dual space of .

Theorem 6 (symmetric mountain pass theorem [33]). Assume that X is an infinite dimensional Banach space. is a finite dimensional Banach space and For any , if , it satisfies condition, and(a) is even and .(b)There exist constants such that , where(c)For any finite dimensional subspace , there exists such that on . Then possesses an unbounded sequence of critical values characterized by a minimax argument.

3. Compactness Conditions

In this section, we are going to give some lemmas about the compactness of functional and prove them.

Lemma 7. Let hold. Any bounded sequence , which satisfies as , possesses a strongly convergent subsequence in

Proof. Suppose that is bounded in . From Lemma 2.4 of [12] and Theorem 1.21 of [34], we know that is reflexive. Combining with Lemma 4, we haveWe just need to prove that strongly in .
Through the Hölder inequality and , we obtainBy (27), we getWe consider the following formula with Hlder inequality: Hence, by (27), we getThen, for convenience, we define a new linear functional on as follows:for all . By means of the Hölder inequality, we have thatHence, is a continuous functional on . Hence, we obtain thatClearly, as , since in and in . Hence, by (27), (29), and (31), we haveHence, through the boundedness of in and (34), we haveNow, we recall the Simon inequalities:for all , where relying on . Then if , by (36) and (37), we have as . When , by (36), (38), and the boundedness of in , we have as . Hence, we get strongly in .

Lemma 8. Let , and hold. Then, functional satisfies the condition.

Proof. Let hold. According to the monotonicity of , we find that there exists a positive constant such thatwhere , for all and . Let be a Cecrami sequence in . We know that it satisfiesas . By means of Lemma 7, it suffices to prove the boundedness of . Suppose that is unbounded in . Then we haveBy (43) and (44), we get Hence, we get We define . Then . So is a bounded sequence in . Through Lemma 4, there exists satisfyingas . What is more, through Lemma A.1 of [35], there exists a function satisfyingWe only need to consider two cases: and . Firstly, we consider the case . Refer to [31]; for any , we have such thatBecause of the unboundedness of , for any , we select such thatwhere is large enough, say , with . By (47) and , we getSince the function is continuous, we get thatas , for any . Through , (48), and Hlder inequality, we obtain that for any . Hence, we get thatas , for any , thanks to the Dominated Convergence Theorem and (51) and (52). As a consequence of for all , by (54) and , we haveas , for any . Therefore, by (50)–(52) and (55), we get as , for any . Hence, we infer thatNow, we will show thatwhere . Because is an open bounded smooth domain with Lipschitz boundary, we deduce that there exists a positive constant , which depends on and such thatWe notice that . Through (42), (49), and (57), we get that there exists such that for any . Then we haveThrough (41), (59), and , we haveAnd . It contradicts with (57). Hence, the sequence is bounded in .
Now we consider the case . We define . Clearly, is Lebesgue measureable. Through (44), (47), and , we haveas . Through (47), (63), and , it holds true that, as . Through the Fatou Lemma and (64), we haveNext, we discusses the case in . By , we getTherefore, there exist constants such thatfor all and . Moreover, by means of the continuity of in , we getfor all . Hence, we obtain thatfor any , thanks to (67) and (68). So we getBy (42) and (44), we get as . Consider (63), (65), (70), and the variational characterization of defined as follows:We haveWe find that (73) is a contradictory result. Thus, the sequence is bounded in . This ends the proof of Lemma 8.

4. Proof of Theorem 2 and Corollary 3

We know that is a Hilbert space. Hence, we divide into two parts. Let = , where = span. Now we defineObviously, is a finite-dimensional space.

Lemma 9. Let . We have

Proof. Clearly , so that , as . For every , there exists such that and . Through the definition of , we can get in . According to Lemma 4, the embedding implies that in . Therefore, we get , as . This implies that the proof is complete.

Proof of Theorem 2. Obviously, all norms are equivalent in . Hence, there exist two constants such thatAccording to Theorem 6 and Lemma 8, we just need to prove , and of Theorem 6. Let hold. Clearly, . Hence, condition of Theorem 6 is true. Then, we take into account the range of values of . According to (72), if , we can find and such that . So, for all , there exists and such that for any .
By and , for any , there exists such thatfor any . By Lemma 9, for any fixed , choose an integer such thatfor any Choose , for all . Note that ; from (77) and (78), we haveThen, condition (b) of Theorem 6 is true.
In the end, we demonstrate condition (c) of Theorem 6. In view of , there exist constants such thatfor any and . Considering condition , we havefor any and . Take , where . We obtainfor any . Then, by (76) and (82), we haveLet be large enough. When , . Hence, condition (c) of Theorem 6 is true. In conclusion, Theorem 2 is proven.

Proof of Corollary 3. Let hold. Similar to Theorem 2, we only need to prove inequality (62).The proof is completed.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

The contribution of Jiabin Zuo to this paper is the same as that of the first author. All authors read and approved the final manuscript.

Acknowledgments

The work is supported by the Fundamental Research Funds for Central Universities (2017B19714 and 2017B07414), National Key Research and Development Program of China (2018YFC1508106), Natural Science Foundation of Jiangsu Province (BK20180500), and Natural Science Foundation of Jilin Engineering Normal University (XYB201814 and XYB201812). The work is also supported by Program for Innovative Research Team of Jilin Engineering Normal University.