Negative Energy Solutions for a New Fractional <span class="nowrap"><svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-4.5269pt" id="M1" height="16.7875pt" version="1.1" viewBox="-0.0657574 -12.2606 31.8932 16.7875" width="31.8932pt"><g transform="matrix(.017,0,0,-0.017,0,0)"><path id="g113-113" d="M570 304C570 398 525 448 414 448C385 448 343 445 312 434L329 511L321 518C297 504 262 482 244 460L233 411C195 397 159 381 128 358L135 332C160 347 189 360 224 373L111 -147C97 -210 84 -218 17 -231L13 -257L254 -247L259 -218L233 -216C183 -212 177 -202 189 -142L218 -1C238 -10 266 -12 283 -12C351 3 429 48 483 105C543 168 570 242 570 304ZM482 289C482 161 380 33 304 33C278 33 248 51 233 69L303 396C326 400 352 403 369 403C428 403 482 380 482 289Z"/></g><g transform="matrix(.017,0,0,-0.017,10.177,0)"><path id="g113-41" d="M300 -147C201 -63 143 98 143 270S200 602 300 686L282 710C136 610 70 450 70 271V270C70 89 136 -72 282 -170L300 -147Z"/></g><g transform="matrix(.017,0,0,-0.017,16.115,0)"><path id="g113-121" d="M536 404C536 423 520 448 491 448C445 448 398 404 308 283L286 338C255 416 242 448 217 448C182 448 138 402 92 341L111 321C149 368 169 378 178 378C188 378 198 363 214 320L254 212C185 117 138 65 107 65C98 65 85 69 82 75C79 82 73 86 65 86C44 86 23 60 23 39C23 7 44 -12 71 -12C119 -12 168 33 265 177L306 61C321 17 347 -12 373 -12C413 -12 465 33 507 96L491 119C459 84 432 60 413 60C395 60 378 92 358 148L321 250C341 279 369 310 389 332C417 363 439 382 456 382C466 382 475 376 481 368C486 361 492 358 496 358C513 358 536 381 536 404Z"/></g><g transform="matrix(.017,0,0,-0.017,25.708,0)"><path id="g113-42" d="M275 270C275 450 212 609 64 710L45 686C145 604 203 442 203 270S147 -63 45 -147L64 -170C213 -68 275 89 275 270Z"/></g></svg>-</span>Kirchhoff Problem without the (AR) Condition

Bu, Weichun; An, Tianqing; Ye, Guoju; Taarabti, Said

doi:https://doi.org/10.1155/2021/8888078

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Volume 2021 | Article ID 8888078 | https://doi.org/10.1155/2021/8888078

Negative Energy Solutions for a New Fractional -Kirchhoff Problem without the (AR) Condition

Weichun Bu,¹Tianqing An,¹Guoju Ye,¹and Said Taarabti²

Academic Editor: Maria Alessandra Ragusa

Received09 Sept 2020

Accepted11 Mar 2021

Published23 Mar 2021

Abstract

In this paper, we investigate the following Kirchhoff type problem involving the fractional -Laplacian operator. , where is a bounded domain in with Lipschitz boundary, are constants, is a function defined on , , and , is the fractional -Laplacian operator, , for any , , is a given positive parameter, and is a continuous function. By using Ekeland’s variational principle and dual fountain theorem, we obtain some new existence and multiplicity of negative energy solutions for the above problem without the Ambrosetti-Rabinowitz ((AR) for short) condition.

1. Introduction and the Main Results

In this article, we investigate the existence and multiplicity of solutions for the Kirchhoff type problem involving the fractional -Laplacian operator. More precisely, we consider the following problem : where is a bounded domain in with Lipschitz boundary and are constants, and for all . is a given positive parameter, and is a continuous function. The operator is the fractional -Laplacian operator, which is defined by where stands for the Cauchy principal value. Especially, when , the operator reduced to the usual fractional -Laplace operator. In the past few decades, many people studied the fractional -Laplace operator problem, and we refer the readers to [1–9]. We point out that Behboudi et al. studied the existence of mountain pass solution for the nonlocal fractional -Laplacian problem in [10] by using the variational method. The more general problem is the following -Laplacian equation

The study for this problem and the related knowledge on fractional Sobolev spaces with variable exponent please refer to [11, 12] and the references therein.

In the recent years, the fractional calculus and the related problems are studied widely, see for example, the literature [13, 14]. The elementary knowledge of fractional derivative and integral can be found in [15, 16]. Especially, the Caputo derivative is discussed in [17]. These references will be helpful for our research later.

The following Kirchhoff equation was introduced by Kirchhoff in [18]: where are constants that have physical meaning. We call the problem () a new problem of Kirchhoff type because it contains the new Kirchhoff term: which makes the problem more interesting, meaningful, and difficult. Up to now, the study on the Kirchhoff type problems involving -Laplacian operator and the fractional -Laplacian operator is very active, see [19–23]. Many results concerning the existence and multiplicity of solutions have been appeared. Especially, [24] proved the existence of two weak solutions by using the variational methods in Orlicz-Sobolev spaces, [25] studied a class of Kirchhoff nonlocal fractional equations, and obtained the existence of three solutions. We also mention that [26] discussed a class of -Kirchhoff equations via the fountain theorem and dual fountain theorem, and [27] studied the existence of nonnegative solutions for a Kirchhoff type problem driven by a nonlocal integrodifferential operator.

It is well known that the (AR) condition plays an important role in verifying the Palais-Smale condition. However, there are a lot of functions not satisfying this condition. Hence, many people pay attention to find the new reasonable conditions instead of the (AR) condition, see [28–31] and the references therein. Motivated by these work, we use Ekeland’s variational principle and dual fountain theorem to study the existence and multiplicity of negative energy solutions for a new fractional -Kirchhoff problem without the (AR) condition. Our results generalize the related work in two ways. Firstly, we deal with the problem in the fractional framework and [32] consider only the integer framework. Secondly, compare to [33], we add a new Kirchhoff function and consider the case of variable exponents.

Throughout this paper, the nonlinearity is a Carathodory function satisfying

(f1): There exists a positive constant such that where ;

(f2): , for uniformly;

(f3): , , where ;

(f4):): There exists positive constants such that ;

(f5): , .

Definition 1. We say is a weak solution of problem , if for any , where will be introduced in Section 2.
It is well known that a weak solution for problem is a critical point of the following energy functional defined on by for all . Moreover, we have for any . Under our assumptions, is well defined in and . We say that a weak solution for problem is a negative energy solution if the energy .
In order to reduce our statements, we define the function space and will be introduced in Section 2. The main results of this paper are as follows.

Theorem 2. Assume that the function satisfies (f1)–(f4) and Then, there exists a such that for any , the problem has at least one solution with negative energy.

Theorem 3. Assume that the function satisfies (f1)–(f5) and Then, for any , problem has infinitely many solutions in with negative energy converging to .
The rest of this paper is organized as follows. In Section 2, some basic properties of the variable exponent fractional Sobolev spaces and Lebesgue spaces are given. In Section 3, it is proved that the functional satisfies the Cerami compactness condition in certain energy levels. In Section 4, by using Ekeland’s variational principle, we give the proof of Theorem 2. Finally, in Section 5, we prove Theorem 3 by using the dual fountain theorem.
Throughout this paper, for simplicity, we use letters to denote positive constants in different cases, and we will specify them whenever it is necessary.

2. Preliminary Results

In this section, we recall some preliminary results of generalized Lebesgue spaces with variable exponent and generalized fractional Sobolev spaces which will be used later. The readers can consult [34–36] and the references therein for more details. Let and be a bounded domain in with Lipschitz boundary, and be continuous functions satisfying

We assume that is symmetric; that is, for all , such that

For any , we introduce the variable exponent Lebesgue space as endowed with the so-called Luxemburg norm

Define a mapping by

Lemma 4 (See [37]). The space is separable, uniformly convex, and reflexive and its conjugate space is , where is the conjugate function of , i.e., for all , and the Hölder type inequality holds.
Note that for any function and , there exists a continuous embedding for any .

Lemma 5 (see [38]). Suppose that . Then, the following properties hold The fractional Sobolev spaces with variable exponent are defined by For , let Define the corresponding variable exponent norm and set Then, the () becomes an uniformly convex and reflexive Banach spaces (see [39]). Let denote the closure of in endowed with the norm and then () is also an uniformly convex and reflexive Banach spaces. is the dual spaces of .

Theorem 6 (see [39]). Let be a smooth bounded domain and , be continuous variable exponents with for and for , and (13) and (15) be satisfied. Assume that is a continuous function and such that for . Then, there exists a constant such that for any , Thus, the spaces is continuously embedded in for any . Furthermore, this embedding is compact. If , then there exists a constant such that

Lemma 7 (See [40]). For all , define the functional by Then, (1) is a bounded and strictly monotone operator(2) is a mapping of type , i.e., if in and then in (3) is a homeomorphism

Lemma 8 (see [41]). (1)If , then(2)If , thenWe state Ekeland’s variational principle and dual fountain theorem which will be used in the proofs of Theorems 2 and 3.

Theorem 9 (Ekeland’s variational principle, see [42]). Let be a complete metric spaces with metric , and let be a lower semicontinuous function, bounded from below and not identical to . Let be given and be such that Then, there exists such that and for each , one has The space is a separable and reflexive real Banach space, and there exists and such that and

Set , we define

Theorem 10 (Dual Fountain Theorem, see [43]). Assume that satisfies , and for every , there exist such that
(B1) ;
(B2) ;
(B3) ;
(B4) satisfies the condition for every
Then, has a sequence of negative critical values converging to 0.

3. Cerami Compactness Condition

We discuss the compactness properties of our energy functional related to the condition and condition.

Definition 11. Let , and we say that satisfies the Cerami condition at the level ( for short), if any sequence with possesses a convergent subsequence in .

Definition 12. Let , and we say that satisfies condition at the level ( for short), if any sequence , namely, , with possesses a convergent subsequence in .
Assume that is a bounded sequence in . By Theorem 6, there exists such that

Lemma 13. Let such that (38) holds. Then, passing to a subsequence, the following properties hold

Proof. (i)By Hölder’s inequality (Lemma 4),Theorem 6, and (38), we haveand thus (ii)By virtue of conditions (f1) and (f2), for any , there exists such thatcombining with Hölder’s inequality, Theorem 6 and (38),and it follows that which implies that The following lemma about the condition will play an important role in the proof of our main results.

Lemma 14. Let the function satisfy (f1), (f3), and (f4), then the energy functional satisfies the condition, where precisely .

Proof. Step 1. We prove that is bounded in . Let be a sequence, which implies that where as
We claim that is a bounded sequence. Suppose to the contrary that Denote then with . Up to subsequences, for some , we get in in There are only two cases need to be discussed and
First, let us consider the case . In view of (f4), (8), (9), (45), (46), and Theorem 6, we obtain which implies , and this is a contradiction according to the conditions and (see (11)).
In the case of , setting , then from (46) implies that for , we have as . In view of (f3), we see that so, using Fatou’s Lemma, we have From (f1) and (f3), there exist , such that and as , and then we have , hence by (46), which implies that Using (8), (45), (46), (49), (51), and Theorem 6, we obtain and this is a contradiction according to the condition (see (11)). Then, the sequence is bounded in .
Step 2. We prove that has a convergent subsequence in . Since , is bounded in , we have therefore, we obtain or and we deduce from Lemma 13 that Since is bounded in , passing to a subsequence, we may assume (i)If , then strongly converges to in (ii)If , thenis not true. Because is bounded in , so, is bounded. (iii)If , then . We define a function byand then and it follows that Since in , which implies that according to the Hölder’s inequality, we get Hence, we obtain With a slight modification of the above proof, we can also prove the following conclusion; so, we have omitted the details. By Hölder’s inequality, (38) and (42), we deduce that Combining with (65) and (66), we get While since and it follows that as , i.e., by virtue of (61), we obtain by the fundamental lemma of the variational method (see [43]). It follows that . So, Hence, for , we have This is a contradiction with ; then, is not true and similarly to (ii), and we obtain that is bounded. So, it can be concluded from the above discussion (ii) and (iii) that Thus, invoking the condition (see (2) of Lemma 7), we deduce that the converges to in .
Therefore, the energy functional satisfies the condition.

Lemma 15. Let the function satisfy (f1), (f3), and (f4), and then the energy functional satisfies the condition, where precisely .

Proof. Consider a sequence such that , , in as By the same method used in the proof of Lemma 14, we can prove that has a strongly convergent subsequence in . The details are omitted.

Remark 16. Since the and conditions hold for the energy , we discuss the negative energy solutions for the problem .

4. Proof of Theorem 2

We firstly prove two lemmas.

Lemma 17. Let assumptions (f1), (f2), and (f3) hold, and then there are constants and such that , where and

Proof. For any and , by (f1) and (f2), there exists such that Let be such that and combining with (74), we deduce that By Theorem 6, Lemma 5, and Lemma 8, there exist positive constants such that Hence, we obtain Choose and take Note that (see (11)), and we can conclude that there exists a constant such that Therefore, taking , we have whenever and . This completes the proof.

Lemma 18. Let assumptions (f1), (f2,) and (f3) hold, and then where is given by Lemma 17 and .

Proof. By (f2) and (f3), for any , there exists such that Let , combining with (82), we have Since (see (11)), for , it is small enough. Hence, we obtain that where is given by Lemma 17 and .
Now, we give the proof of Theorem 2.

Proof. By Lemma 17 and Lemma 18, we know that whenever . Set So, by Theorem 9 (Ekeland’s variational principle), there exists such that for all . In that way, by (86)–(88), we have so that .
Define the functional by Like that, (88) implies that for all with , and thus is a strict local minimum of . Furthermore, for all , where , taking small enough, we have which implies that Thus, In the above inequality, replacing with , we have Hence, Passing to the limit in (87) and (95), we conclude that there is a sequence such that To sum up, by Lemma 14, there exists a strongly convergent sequence , still denoted by , such that in . Consequently, is a solution of , with . The proof of Theorem 2 is completed.

5. Proof of Theorem 3

In order to prove Theorem 3, we need the following result.

Lemma 19 (see [44], Lemma 18). Assume that , for any and denote then
We give the proof of Theorem 3.

Proof. The function satisfies hypotheses (f1)–(f5),and it is obvious that for all . Together with Lemma 15, we only need to verify that the energy functional satisfies all conditions (B1)–(B3) of Theorem 10.
Step 1. Verification of condition (B1). According to (f1) and (f2), there exist , such that For any with , combining with (98) Theorem 6 and Lemma 8, there exist positive constants such that we obtain According to Lemma 19 and the conditions (see (12)), it follows that since and (see (12)) imply that .
Hence, we deduce Choosing , we have Obviously, there exists a large enough such that We choose by Lemma 19, and we know that as . So, there exists , such that and with , and we have . Therefore, the condition (B1) of Theorem 10 holds.
Step 2. Verification of condition (B2). By (f2) and (f3), there exist such that For , when with , we have Since all norms on the finite-dimensional space are equivalent, we obtain According to the condition (see (12)), for some that is small enough, we deduce that Therefore, the condition (B2) of Theorem 10 holds.
Step 3. Verification of condition (B3). By verification of condition (B1), one has that for and with , since as . Therefore, the condition (B3) of Theorem 10 also holds.
By above three steps and Lemma 15, satisfies all conditions of Theorem 10. The proof of Theorem 3 is completed.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no competing interests.

Authors’ Contributions

Each of the authors contributed to each part of this study equally, and all authors read and approved the final manuscript.

Acknowledgments

This work is supported by the Fundamental Research Funds for the Central Universities (2019B44914), Natural Science Foundation of Jiangsu Province (BK20180500), the National Key Research and Development Program of China (2018YFC1508100), and Natural Science Foundation of China (11701595).

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Copyright © 2021 Weichun Bu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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