Abstract
The existence of nontrivial solutions of Kirchhoff type problems is proved by using the local linking method, and the new results do not require classical compactness conditions.
1. Introduction
This paper is concerned with the existence of nontrivial solutions of the following nonlocal Kirchhoff type problem: where is a smooth bounded domain in (), , and : is a continuous real function. Similar nonlocal problems model several biological systems where describes a process which depends on the average of itself, for example, that of the population density; see [1].
Problem (1) is related to the stationary analogue of the equation proposed by Kirchhoff [2] as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. Some early classical studies of Kirchhoff equations were those of Bernstein [3] and Pohožaev [4]. However, (2) received great attention only after Lions [5] proposed an abstract framework for the problem. Some interesting results can be found in [6] and the references theorem. In recent years, Alves et al. [7] and Ma and Rivera [8] have obtained positive solutions of such problems by variational methods.
Later Perera and Zhang [9, 10] obtained nontrivial solutions of Kirchhoff type problems with asymptotically 4-linear term via Yang index. In [9] using the invariant sets of descent flow, Perera and Zhang got a positive, a negative, and a sign-changing solution under the 4-sublinear case, asymptotically 4-linear case, and 4-superlinear case. In [10], the authors considered the 4-superlinear case: where , which implies In [11], Mao and Zhang used minimax methods and invariant sets of decent flow to prove 4-superlinear Kirchhoff type problems without the PS condition and got multiple solutions. In [12], the paper studies the problem (1) by means of the Morse theory and local linking.
Motivated by [12–15] and by the new information concerning local linking, we study Kirchhoff type problems 4-superlinear growth as . We make the following assumptions.There exist , , , and such that as uniformly in .There exist and such that
We are ready to state our main results.
Theorem 1. Assume that ()–() hold. Then problem (1) has at least one nontrivial solution.
Remark 2. Most of the results on the existence and multiplicity of solutions of (1) were obtained under the above superlinear condition () with or without the evenness assumption. Roughly speaking the role of () is to ensure the boundedness of all () (or ()) sequences for the corresponding functional. However, there are many functions which are superlinear but it is not necessary to satisfy () even if . For example, where . Then it is easy to check that () does not hold even for any . On the other hand, in order to verify (), it is usually an annoying task to compute the primitive function of and sometimes it is almost impossible. For example, where .
The aim of this paper is to deal with superlinear problems which do not satisfy (). is weaker than used in [10]. Without (), it becomes more complicated. We do not know in our situations whether the Palais-Smale sequence is bounded. We replace () (or ()) sequences with weaker Cerami sequence or -sequence (the definition given in the next section).
Remark 3. Compared with the method of invariant sets of descent flow and Morse theory, Yang index, our method is more simple and direct.
2. Preliminaries
Let be a real Banach space with a direct sum decomposition: Consider two sequences of subspace: such that , , where , , .
For every multi-index , we denote by the space We can know A sequence is admissible if, for every , there is such that For every we denote by the functional restricted to .
Definition 4 (see [14]). Let . The function satisfies condition if every sequence such that is admissible and contains a subsequence which converges to a critical point of .
Definition 5 (see [15]). Let be Banach space with a direct sum decomposition . The functional has a local linking at 0 if, for some ,
Lemma 6 (see [14]). Let be Banach space with a direct sum decomposition . Suppose that satisfies the following assumptions. has a local linking at 0 with respect to . satisfies condition. maps bounded sets into bounded sets.For every , as and .
Then has at least one nontrivial critical point.
Next we let be the Sobolev space equipped with the inner product and the norm
Denote by the distinct Dirichlet eigenvalues of on , and denote by the eigenfunction corresponding to the eigenvalues; then is achieved by . Let Recall that a function is called a weak solution of (1) if Weak solutions are the critical points of functional: Then
3. The Existence of the Solutions
Lemma 7. Let satisfies (); the function has a local linking with respect to , where , .
Proof. Let , . If , then ; and if , then . If , from , for the given , there exists a such that and since is finite ; by , we have
Noting that , we can choose small enough such that for .
For , from , , we have
then
For , choosing small enough such that for , we have for , .
Choose ; it is very clear that has a local linking at 0 with respect to .
Lemma 8. Suppose () holds. Then function satisfies condition ; that is, for every sequence has a convergent subsequence if is bounded and as , where , , and is a Hilbert basis for .
Proof. Suppose that is bounded and as ; then there exists a constant such that
for every . On the one hand, by (6), there is constant such that
then there exists , , such that
Hence
We know
So for all and a.e. , we obtain
It follows from (26) and (31)
On the other hand, by (7),
then there are constants and , , such that
Hence
By , one has
Hence we obtain that
for all and a.e. . Then we have
So is bounded. If , by (32) and Hölder inequality,
Thus we get that is bounded. If , we obtain
Then by (32) and , we know that is bounded too. Hence is bounded. Since is bounded, for a subsequence, converges to some weakly in and strongly in , , and a.e. in :
Noting that , ,
as . In addition,
Since , then . The combination of (32)–(44) implies that
Thus, in .
Lemma 9. maps bounded sets into bounded sets.
Proof. From , we know so Hence maps bounded sets into bounded sets.
Lemma 10. For every , , , .
Proof. By condition and (2), there exist enough constants , , for all ; we have So, for all , , one has Hence, for every , , we have Then the conclusion is obtained.
Proof of Theorem 1. By Lemmas 7–10 and Lemma 6, we complete the proof.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.