Nontrivial Solutions for Asymmetric Kirchhoff Type Problems

Abstract

We consider a class of particular Kirchhoff type problems with a right-hand side nonlinearity which exhibits an asymmetric growth at and in . Namely, it is 4-linear at and 4-superlinear at . However, it need not satisfy the Ambrosetti-Rabinowitz condition on the positive semiaxis. Some existence results for nontrivial solution are established by combining Mountain Pass Theorem and a variant version of Mountain Pass Theorem with Moser-Trudinger inequality.

1. Introduction

We consider the following nonlocal Kirchhoff type problem: where is a smooth bounded domain in ) and is continuous.

It is pointed out in [1] that the problem (1) models several physical and biological systems where describes a process which depends on the average of itself (e.g., population density). Moreover, this problem is related to the stationary analogue of the Kirchhoff equation which was proposed by Kirchhoff [2] as an extension of the classical D’Alembert wave equation for free vibration of elastic strings. Kirchhoff's model takes into account the changes in length of the string produced by transverse vibrations. Some early studies of the Kirchhoff equation may be seen [3–5]. More recently, by variational methods, Alves et al. [1] and Ma and Rivera [6] studied the existence of one positive solution, and He and Zou [7] studied the existence of infinitely many positive solutions for the problem (1), respectively; Perera and Zhang [8] studied the existence of nontrivial solutions for the problem (1) via the Yang index theory; Zhang and Perera [9] and Mao and Zhang [10] studied the existence of sign-changing solutions for the problem (1) via invariant sets of descent flow. In particular, the asymptotically 4-linear case, was considered in [8]. In [9], the authors considered the 4-superlinear case: where , which implies that there exists a constant such that

Note that (AR) condition plays an important role for showing the boundedness of Palais-Smale sequences. Furthermore, by a simple calculation, it is easy to see that (AR) condition implies that Hence grows in a 4-superlinear rate as .

In the present paper, motivated by [11–14], our main purpose is to establish existence results of nontrivial solution for the problem (1) with when the nonlinearity term exhibits an asymmetric behavior as approaches and . More precisely, we assume that, for a.e. ,   grows 4-superlinear at , while at it has a 4-linear growth. To our knowledge, this asymmetric nonlocal Kirchhoff problem is rarely considered by other people.

In case of , all the above-mentioned works involve the nonlinear term of a subcritical (polynomial) growth; say,

(SCP): there exist positive constants and and such that One of the main reasons to assume this condition (SCP) is that they can use the Sobolev compact embedding , .

Over the years, many researchers studied the problem (1) by trying to drop the condition (AR); see, for instance, [8, 15].

In this paper, our first main results will be to study the problem (1) in the improved subcritical polynomial growth as follows: which is much weaker than (SCP). Note that, in this case, we do not have the Sobolev compact embedding anymore. Our work is studying the asymmetric problem (1) without the (AR) condition in the positive semiaxis. In fact, this condition was studied by Liu and Wang in [16] in the case of Laplacian (i.e., ) by the Nehari manifold approach. However, we will use the Mountain Pass Theorem and a suitable version of the Mountain Pass Theorem to get the nontrivial solution to the problem (1) in the case that . Our results are different from those in [8–10, 15].

Let us now state our results. Suppose that and satisfies() uniformly, for a.e. , where ;() uniformly, for a.e. , where ;() uniformly, for a.e. ;() is nonincreasing with respect to , for a.e. .We need the following preliminaries.

Let be the Sobolev space equipped with the inner product and the norm We denote by the usual -norm. Since is a bounded domain, continuously for , compactly for , and there exists such that

Recall that function is called a weak solution of (1) if

Seeking a weak solution of the problem (1) is equivalent to finding a critical point of functional as follows: where . Then

Definition 1. Let be a real Banach space with its dual space and . For , one says that satisfies the condition if, for any sequence with there is a subsequence such that converges strongly in . Also, one says that satisfy the condition if, for any sequence with there is subsequence such that converges strongly in .

We have the following version of the Mountain Pass Theorem (see [17, 18]).

Proposition 2. Let be a real Banach space and suppose that satisfies the condition for some , , and with . Let be characterized by where is the set of continuous paths joining and . Then, there exists a sequence such that

Lastly, we also need the following preparations.

Our assumptions lead us to the eigenvalue problem where is an eigenvalue of the problem (19) meaning that there is a nonzero such that This is called an eigenvector corresponding to eigenvalue . Set Denote by all distinct eigenvalues of the nonlinear problem (19). Then where is simple and isolated and can be achieved at some and in (see [9]).

Theorem 3. Let and assume that has the improved subcritical polynomial growth on (condition (SCPI)) and satisfies . If ( is the first eigenvalue of ) and , then the problem (1) has at least one nontrivial solution when , for all .

Theorem 4. Let and assume that has the improved subcritical polynomial growth on (condition (SCPI)) and satisfies . If ( is the first eigenvalue of ), , and uniformly, for a.e. , then the problem (1) has at least one nontrivial solution.

Here, we also give an example for . It satisfies our conditions and (SCPI).

Example A. Define where , ; , ; ; ; ; ; ; ; . Moreover, there exists such that for all .

Theorem 5. Let and assume that has the improved subcritical polynomial growth on (condition (SCPI)) and satisfies . If ( is the first eigenvalue of ) and , then the problem (1) has at least one nontrivial solution.

In case of , we have . In this case, every polynomial growth is admitted, but one knows easy examples that . Hence, one is led to look for a function with maximal growth such that

It was shown by Trudinger [19] and Moser [20] that the maximal growth is of exponential type. So we must redefine the subcritical (exponential) growth in this case as follows.

(SCE): has subcritical (exponential) growth on ; that is, uniformly on for all .

When and has the subcritical (exponential) growth (SCE), our work is again studying the asymmetric problem (1) without the (AR) condition in the positive semiaxis. Our results are as follows.

Theorem 6. Let and assume that has the subcritical exponential growth on (condition (SCE)) and satisfies . If   ( is the first eigenvalue of ) and , then the problem (1) has at least one nontrivial solution when , for all .

Theorem 7. Let and assume that has the subcritical exponential growth on (condition (SCE)) and satisfies . If   ( is the first eigenvalue of ), , and uniformly, for a.e. , then the problem (1) has at least one nontrivial solution.

Theorem 8. Let and assume that has the subcritical exponential growth on (condition (SCE)) and satisfies . If   ( is the first eigenvalue of   ) and , then the problem (1) has at least one nontrivial solution.

2. Some Lemmas

Lemma 9. Let and let be a eigenfunction with and assume that and (SCPI) hold. If   ( is the first eigenvalue of ) and , then (i)there exist , such that , for all with ;(ii) as .

Proof. By and , if , for any , there exist and such that, for all , Choose such that . By (25), the Poincaré inequality, and the Sobolev inequality, , we get So part (i) is proved if we choose small enough.
On the other hand, if , taking such that and using (26), we have Thus part (ii) is proved. By exactly slight modification to the proof above, we can prove (ii) if .

Lemma 10 (see [19, 20]). Let ; then , for all . Moreover, The inequality is optimal; for any growth with the corresponding supremum is .

Lemma 11. Let and let be a eigenfunction with and assume that and (SCE) hold. If     ( is the first eigenvalue of ) and , then(i)there exist such that , for all with ;(ii) as .

Proof. By (SCE) and , if , for any , there exist , , , and such that, for all , Choose such that . By (30), the Holder inequality, and the Moser-Trudinger embedding inequality, we get where is sufficiently close to , , and . So part (i) is proved if we choose small enough.
On the other hand, if , taking such that and using (31), we have Thus part (ii) is proved. By exactly slight modification to the proof above, we can prove (ii) if . By exactly slight modification to the proof above, we can prove (ii) if .

Lemma 12. For the functional defined by (19), if , a.e. , , and then there exists a subsequence, still denoted by , such that

Proof. Since as , for a suitable subsequence, we may assume that We claim that, for any and , Indeed, for any , at fixed and , if we set then Hence Therefore,
and our claim (37) is proved.
On the other hand, That is, Combining (37) and (43), we find that

3. Proofs of the Main Results

Proof of Theorem 3. By Lemma 9, the geometry conditions of Mountain Mass Theorem hold. So we only need to verify condition (PS). Let be a (PS) sequence such that, for every , where is a positive constant and is a sequence which converges to zero.
Step 1. In order to prove that has a convergence subsequence, we first show that it is a bounded sequence. To do this, we argue by contradiction assuming that, for a subsequence which we follow denoted by , we have Without loss of generality, we can assume that , for all , and define . Obviously, , for all , and then it is possible to extract a subsequence (denoted also by ) such that where and . Dividing both sides of (46) by , we obtain Passing to the limit we deduce from (48) that for all .
Now we claim that for a.e. . To verify this, let us observe that by choosing in (53) we have where . But, on the other hand, from and , for some positive constant . Moreover, using , for a.e. , (50), and the superlinearity of (see ), we also deduce
Therefore, if , by the Fatou lemma, we will obtain that which contradicts (54). Thus and the claim is proved.

Clearly, . By , there exists such that for a.e. . By using the Lebesgue dominated convergence theorem in (53), we have for all . This contradicts our assumption; that is, , for all .

Step 2. Now, we prove that has a convergence subsequence. In fact, we can suppose that Now, since has the subcritical growth on , for every , we can find a constant such that Then Similarly, since in , . Since is arbitrary, we can conclude that By (46), we have From (62) and (63), we obtain So we have in which means that satisfies .

Proof of Theorem 4. Since , obviously, Lemma 9(i) holds. We only need to show that Lemma 9(ii) holds. Let . Using the condition , then there exists large enough such that for all and large enough. So we have By Proposition 2, there exists a sequence such that Clearly, (68) implies that To complete our proof, we first need to verify that is bounded in . Similar to the proof of Theorem 3, we have , , , and for all . By maximum principle, is an eigenfunction of ; then for a.e. . By our assumptions, we have uniformly in , which implies that On the other hand, (69) implies that Thus which contradicts (72). Hence is bounded. According to the Step 2 proof of Theorem 3, we have in which means that satisfies .

Proof of Theorem 5. By Lemma 9 and Proposition 2, (67)–(69) hold. We still can prove that is bounded in . Assume that as . Similar to the proof of Theorem 3, we have and when , as . Let where . Since is bounded in , it is possible to extract a subsequence (denoted also by ) such that where and .
If as , then . In fact, letting and noticing , it follows from that where is a large enough constant. Therefore, by (69) and (75), we have So for a.e. . But if , then . Hence On the other hand, by as , we have as . From Lemma 12 and (67), we get Obviously, it contradicts (79). So is bounded in . According to the Step 2 proof of Theorem 3, we have in which means that satisfies .

Proof of Theorem 6. By Lemma 11, the geometry conditions of Mountain Pass Theorem hold. So we only need to verify condition . Similar to the Step 1 proof of Theorem 3, we easily know that sequence is bounded in . Next, we prove that has a convergence subsequence. Without loss of generality, suppose that Now, since has the subcritical exponential growth on , we can find a constant such that Thus, by the Moser-Trudinger inequality (see Lemma 10), Similar to the last proof of Theorem 3, we have in which means that satisfies .

Proof of Theorem 7. Combining the proof of Theorems 4 and 6, we easily prove it.

Proof of Theorem 8. Combining the proof of Theorems 5 and 6, we easily prove it.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank the referees for valuable comments and suggestions in improving this paper. This work was supported by the National Natural Science Foundation of China (Grant no. 11101319) and the Planned Projects for Postdoctoral Research Funds of Jiangsu Province (Grant no. 1301038C).