Abstract

We are concerned with the existence of multiple solutions to the nonhomogeneous Kirchhoff type equation where are positive constants, , we can find a constant such that for all the equation has at least two radial solutions provided .

1. Introduction and Main Result

In this paper, we consider the existence of multiple solutions to the following nonhomogeneous Kirchhoff type equation: where , are positive constants and satisfies the following conditions: ( h1) and , where is the embedding coefficient of and; (), where denotes the usual inner product in .

Recently, there have been many references about the existence of nontrivial solutions to the following Kirchhoff type equation by using variational method [15]: where , are positive constants. ,,. A main tool to deal with problem (3) is the mountain pass theorem. For this purpose, one usually assumes that is subcritical, superlinear at the origin, and either 4-superlinear at infinity or satisfies the following global Ambrosetti-Rabinowitz type condition (AR in short): (AR)there exists such that for all and . Under the above assumptions, the mountain pass geometry structure and the boundedness of Palais-Smale sequence or Cerami sequence can be obtained.

For example, in [5], when satisfies above assumptions and the potential satisfies the following conditions:(), and for each , meas, where meas denotes the Lebesgue measure,

which ensure the compact imbedding of ,, the author obtained the existence of a nontrivial solution to problem (3).

The existence of infinitely many solutions was considered in [2, 3] respectively, by the fountain theorem and a variant version of the fountain theorem, where is odd on and is also subcritical, superlinear at the origin, and either 4-superlinear at infinity or satisfies AR condition or some conditions weaker than AR condition. In [2], , and in [3], , satisfies the condition (V).

The existence of ground state solutions to problem (3) was also considered in [1, 4]. In [1], the authors studied (3) under the conditions: , a positive potential satisfies satisfies (AR), , for some and increases for all . They obtained a positive ground state solution by using the Nehari manifold.

Under the same condition of in [1], the authors in [4] discussed the existence of multiple ground state solutions, where , which contains a critical growth term.

Recently, in [6], the authors studied the existence of a positive solution for the following Kirchhoff equation: where ,, is subcritical, superlinear at the origin and infinity. In order to construct the mountain pass geometry structure and obtain the bounded PS sequence, they combined a truncation argument with a monotonicity trick introduced by Jeanjean [7], and obtained that there exists such that problem (4) has at least one positive solution for .

Motivated by the aformentioned references, we consider the existence of multiple solutions to the nonhomogeneous Kirchhoff equation (1), where . By using the variational method, we obtain that the problem has at least two positive radial solutions. Under proper assumptions on , the problem has a local minimum around the origin with negative energy by Ekeland variational principle. Note that the term is neither -superlinear nor satisfies AR condition for . In order to obtain the bounded PS sequence, we also use the indirect method in [7]. Meanwhile, for , we take a transform of to construct the mountain pass geometry structure. Finally, the combination of Pohozaev identity with the method in [7] obtains the bounded PS sequence. Therefore, we obtain the second solution which has positive energy.

Let be the usual Sobolev space equipped with the inner product and norm We denote by the usual norm. Then, we have that continuously for . Hence, there exists such that Let be the subspace of containing only the radial functions. Then the imbedding is compact for [8, Corollary 1.26, page 18]. Let be the completion of with respect to the norm .

Define the energy functional by By ,, we have . And, for any ,

Furthermore, by (), , the functional is also a functional defined on . By standard argument, the weak solution of (1) is corresponding to the critical point of the functional on .

Our main result is as follows.

Theorem 1. Let and satisfy ()-(). Then, problem (1) has at least two nontrivial radial solutions and , satisfying .

The paper is organized as follows. In Section 2, we give the existence of the negative energy solution . The existence of positive energy solution and the proof of Theorem 1 are given in Section 3.

2. Existence of Negative Energy Solution

In this section, we give the existence of the negative energy solution. In order to obtain our first solution, we need the following preliminaries.

Lemma 2. Let and satisfy (). Then, there exists such that , where .

Proof. For , by (7), the Hölder inequality and the Sobolev inequality imply that Set since , by calculating directly, we see that , where ,. Then it follows that, if , that is, , there exists such that , where

Lemma 3. Let and satisfy (). Then , where is given by Lemma 2 and .

Proof. By (), ,, then for , there exists such that . Then , so . Hence, by (7), for small enough, we have Then, by the definition of ,.

Lemma 4. Let and satisfy (). The bounded PS sequence of the functional possesses a convergent subsequence.

Proof. Let be a bounded PS sequence of , that is and are bounded, in , where is the dual space of . We may assume that, up to a subsequence, It follows that By (8), we can obtain that Since is also bounded in , then Therefore, That is, in .

Theorem 5. Let and satisfy (). Then, there exists such that where is given by Lemma 2 and .

Proof. By Lemma 3, , then by Ekeland variational principle [9], there exists such that Then, by Lemma 2  , then is a bounded PS sequence of . Therefore, Lemma 4 implies that there exists such that , up to a subsequence. So and .

3. Proof of Theorem 1

In this section, we will show the existence of the second solution. Note that , when , neither satisfies (AR) condition nor is 4-superlinear. So, in order to obtain the bounded PS sequence, following the argument in [6], we also use a direct method in [7]. Firstly, we recall the following main result in [7]. The “monotonicity trick” at the core of this theorem was invented by Struwe (see [9]).

Theorem 6 (see [7]). Let be a Banach space and be an interval. Consider the family of functionals on with nonnegative and either or as . We assume that there are two points , in such that where Then, for almost every there is a sequence such that (i) is bounded;(ii);(iii) in the dual of .

In our case, ,, and define by where , Then is a family of functionals on . For any ,, and as .

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

Lemma 7. Let and satisfy ()-(). Then, the following claims hold:(i)there exist and such that for all (ii)for any , where

Proof. (i) Since for all and ,. By Lemma 2, there exist independent of such that with .
We choose a function and . Set for . Then, for all , by (7) and (), we have Noting , then there exists large enough satisfying , which is independent of , such that for all , with .
(ii) Since is nonincreasing on , then by the definition of and (i), for all , we have .

By Lemma 7 and Theorem 6, for almost every , there exists a bounded sequence such that ,. By Lemma 4, there exists such that in . Therefore, and . It follows from (ii) of Lemma 7 that .

Therefore, there exists with and a nonnegative sequence (denoted by for simplicity) satisfying

In order to obtain the boundedness of , we need the following Pohozaev identity. The proof is similar to the argument in [10].

Lemma 8. Under the conditions of () and (), if is a weak solution of (1), the following Pohozaev identity holds:

Lemma 9. Consider in (29) is bounded in .

Proof. Firstly, since , then by Lemma 8, satisfies the following Pohozaev identity: On the other hand, by and , we have that Then, combining (33) with (31), we can obtain Since by Lemma 7, and , so in order to prove the boundedness of in , we only need to prove that is bounded. By contradiction, we assume that , up to a subsequence. Let ,,,.
Note that is bounded and ,. Multiplying (31)–(33) by , we have that where denotes the quantity tends to zero as . By calculating, we obtain that Since and for , so this is a contradiction for large enough. Therefore, is bounded in .

Proof of Theorem 1. Since By Lemma 9 and , is bounded and . Thus, when , we have that is bounded and . Therefore is a bounded PS sequence of . By Lemma 4, has a convergent subsequence. We may assume that , up to a subsequence. Consequently, . According to Lemma 7, we have . Thus is a positive energy solution to problem (1). Hence, by Theorem 5 problem (1) has two solutions and satisfying .

Acknowledgments

The authors thank the anonymous referee for the careful reading and some helpful comments, which greatly improve the paper. Projects supported by the National Natural Science Foundation of China (Grant no. 11071149), Science Council of Shanxi Province (2010011001-1, 2012011004-2, 2013021001-4).