Sign-Changing Solutions for Kirchhoff-Type Problems with Variable Exponent

Abstract

This paper is devoted to study a class of Kirchhoff-type problems with variable exponent. By means of the perturbation technique, the method of invariant sets for the descending flow and necessary estimates and the existence of infinitely many sign-changing solutions to this problem are obtained.

1. Introduction and Main Result

In this paper, we consider the following Kirchhoff-type problems with variable exponent:where , are real numbers, is a bounded smooth domain, and is a continuous function in . We define , , and . We assume satisfies the following conditions:  , and for ;  and , where denotes the measure of and is achieved on by the function ;  there exists such that for small enough, where if and if ;  there exist such that and for . Problem (1) is a special case of the following equation: which has been studied extensively. Mathematically, (3) is a nonlocal problem as the appearance of the nonlocal term , which implies (3) is no longer a pointwise identity. This causes some mathematical difficulties which make the study of (3) particularly interesting. In the past few years, problem (3) has been studied by many authors, for example [1–17]. Many solvability conditions on the nonlinearity near zero and infinity for the problem (3) have been considered, such as the superlinear case [11] and the asymptotical linear case [14]. In addition, the authors in [18] mentioned the following growth condition on which is often assumed:  for large, where , which assures the boundedness of any Palais–Smale or Cerami sequence (see [19]). Indeed the condition may appear in different forms as follows:  there exists such that for all and , where for all (see [14]);  for all (see [16]); or  and there exists such that for any and large enough (see [11]). In the papers above, each of the conditions implies that the condition holds. Several researchers studied problem (3) with conditions weaker than . For example, a more weaker 4-superlinear growth condition is that  as uniformly in . Zhang and Perera [20], see also Mao and Zhang [11] who proved the existence of sign-changing solutions of (3) provided satisfies the condition and some extra assumptions. It is worth mentioning that the above conditions are usually for the nondegenerate case, i.e., the case . In fact, for the degenerate case , some conditions need to be strengthened (see [21]). For example, the condition should be replaced by  for large.

Indeed, if there exists such that , then the conditions and do not hold. This phenomenon does not exist for the constant exponent. Therefore, the problem we intend to study is a new phenomenon. Recently, the first author et al. in [22] studied the following semilinear elliptic equation:where , and for . They used the perturbation method (see [23–26]) to overcome the difficulty of the boundedness of Palais–Smale sequence of the Euler–Lagrange functional. Similarly, if , it is also difficult to verify the boundedness of the Palais–Smale sequence of the corresponding functional to (1). In addition, some new difficulties will be encountered due to the presence of the nonlocal term. Then, we will use the ideas of [22] and some new skills to deal with corresponding difficulties.

The main result of this paper reads as follows:

Theorem 1. Suppose that , , the conditions , , , and hold. Then, problem (1) has infinitely many sign-changing solutions.

Remark 2. Assume and , then and . Therefore, there exists domain satisfying the conditions and .

Remark 3. We use the perturbation method and Moser iteration mainly to deal with the degenerate cases. Our results include both degenerate and nondegenerate cases.
Throughout this paper, we use to denote the usual norms of . The letter stands for positive constant which may take different values at different places.

2. Solutions of the Perturbation Problem

Problem (1) has a variational structure. The corresponding functional is

Since , problem (1) is a subcritical problem and the compactness of the bounded Palais–Smale sequence of the functional is guaranteed. However, since the domain is nonempty and could be 0, the Ambrosetti–Rabinowitz condition does not hold and it is difficult to prove the boundedness of the Palais–Smale sequence. Instead, we introduce a perturbation problem.

Choose such that

Let be a smooth even function with the following properties: for , for and is monotonically decreasing on the interval . Definefor (0,1]. We will deal with the perturbation problemwhere is the characteristic function of the domain , that is, for , for , . The formal energy functional associated with (8) is defined bywhere and .

Theorem 4. Suppose that , , the conditions , , , and hold. Then, for any (0,1], there existindependent ofsuch that problem (8) has infinitely many sign-changing solutionssatisfying.

Lemma 5. ([22]). The function defined previously satisfies the following inequalities:for, whereis a positive constant.

Next, we introduce the following abstract critical point theorem (see Theorem A of [27]). Let be a Banach space, be an even -functional on . Let be a family of open convex sets of . Set

Definewhere is the genus of symmetric sets, .

Assume there exists an odd continuous map satisfying.

Given , there exists such that if , then

.

Moreover, assume

satisfies the Palais–Smale condition.

.

is nonempty.

The following abstract critical point theorem is from [27], referred as Theorem 6.

Theorem 6. Assume , , , , and hold. Then,(1).(2), as.(3)If, then.

In the following, we verify that functional satisfies all assumptions of Theorem 6.

Lemma 7. Suppose that , , the conditions and hold. Then, for any (0,1], satisfies the condition.

Proof. Let be a sequence of in . This means that there exists such thatFrom (14) and Lemma 11, we derive thatIt implies from and (14) that is bounded in . Since the functional is of subcritical growth, by a standard argument, satisfies the condition.
Given , we define by the following equation:It is easy to see that is a continuous odd mapping from into itself andDefinewhere . Obviously, and are open convex subsets of .

Lemma 8. There exists such that for , .

Proof. Assume and . By the assumption , for any , there exists such that for . Without loss of generality, we assume , and by taking as test function in equation (16), we haveWe estimateTherefore, we haveThen, we obtainChoose , , such thatThen, for , , we have and . That is, and . Similarly, we have .

Lemma 9. There exist and such that

Proof. Let , then . According to and , we can choose and such that , and for . Therefore, we haveChoose such thatThen, for , , we haveMoreover,Therefore, there exists such thatDenote , . Define , whereNow, we are in a position to prove the main result of this section.

Proof of Theorem 10. Choose such that for . Let be a family of linearly independent functions in . Denote .
By Lemma 4.1, 4.2 of [28], we have for large enough. From (17) and Lemmas 7–9, all the assumptions of Theorem 6 are fulfilled. Therefore, for are critical values of , and , as . Moreover, if , then . Hence, given an integer , for (0,1], the functional has pairs of sign-changing critical points , the corresponding critical values are . In addition, there exist independent of such thatwhere .

3. A Priori Estimate and Proof of the Main Results

In this section, we will show that solutions of perturbation problem (8) are indeed solutions of original problem (1). For this purpose, we need the following estimate:

Lemma 11. Suppose that , , the conditions and hold. If and , then there exists independent of such that .

Proof. By Lemma 11, and , we have Case 1. If , according to and (32), we obtain , then there exist independent of such that  Case 2. If , according to and (32), there exists such thatTherefore, for any , we know that there exists such thatAccording to the assumptions , we can choose , sufficiently small and such thatUsing the Young inequality, we haveNotice thatTherefore, we obtainChoose and such that . By the Hölder inequality, we deduce from (39) thatSince is a solution of problem (8), when , we haveMultiply problem (41) by and integrate, it implies from (34) and (40) thatNotice that , we can choose such thatIt follows from (42) that

Lemma 12. Suppose that , , the conditions , and hold. If is a critical point of with , then there exists a positive constant independent of such that .

Proof. Using the Sobolev embedding theorem, we haveLet and . We havefor any . Since is a solution of problem (8), we multiply problem (8) by and integrate to obtainIt implies thatOn the one hand, by the Sobolev embedding theorem, we haveOn the other hand, by the Hölder inequality and (45), we havewhere . And by the Sobolev embedding theorem, we haveAccording to (48)–(51), we obtainthen, by Lemma 11, when , we haveNotice that , we haveNow, we carry out an iteration process. Set for . By (54), we haveSince , the series and are convergent. Letting , we conclude from (45) and (55) that . The proof is complete.

Proof of Theorem 13. By Theorem 4, we know that problem (8) has a sequence of sign-changing solutions . Moreover, there exist positive constants independent of , such thatGiven a positive integer , choose , then are indeed sign-changing solutions of the original problem (1). Since is arbitrary, we claim that problem (1) has infinitely many sign-changing solutions.

Data Availability

No data were used to support the findings of this study.

Disclosure

A preprint has previously been published [29].

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Acknowledgments

This paper was supported by the National Natural Science Foundation of China (no. 11861021).