#### Abstract

We study the following Kirchhoff-type equations , in , , in , where is a bounded smooth domain of , , , , and . Under some suitable conditions, we prove that the equation has three solutions of mountain pass type: one positive, one negative, and sign-changing. Furthermore, if is odd with respect to its second variable, this problem has infinitely many sign-changing solutions.

#### 1. Introduction and Preliminaries

In this paper, we study the following Kirchhoff-type equations: where is a bounded smooth domain of , , , and with , , and .

When , problem (1) is related to the stationary analogue of the Kirchhoff equation proposed by Kirchhoff in [1] as an existence of the classical D’Alembert’s 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 interesting studies by variational methods can be found in [2–11]. In these papers, many achievements had been obtained on the existence of ground states, infinitely many radial solutions, soliton solutions, and high energy solution for (1) by using the Fountain Theorem, the mountain pass theorem, using the variational methods and the local minimum methods, and the invariant sets of descent flow. Particularly, in [12], the authors consider the following problem: where is a bounded smooth domain of ( or ), , , and is a continuous function which is 3-superlinear. The unbounded sequence of sign-changing solutions of (3) is obtained by using some variants of the mountain pass theorem. In [13], authors considered the following -Laplacian equation coupled with the Dirichlet boundary condition:where , the parameter , is a nonzero potential, and with . By using variational method, they proved that for every problem (4) has at least two nonzero, nonnegative weak solutions, while there exists such that problem (4) has at least three nonzero, nonnegative weak solutions. In [14], Ricceri proved that there were at least three distinct weak solutions in for the following equation: by using and improving the three critical points’ theorem, where ; let be an open interval with , .

In this paper, we study the sign-changing solutions of problem (1). We need the following assumptions: , . , and for some , where is a positive constant, for , and for . uniformly in , as . There exist and such that , for all and .

We need the following several notations. Let with the inner produce and norm

Recall that a function is called a weak solution of problem (1) if Seeking a weak solution of problem (1) is equivalent to finding a critical point of the -functional

Since is a bounded domain, it is well known that the embedding is continuous for all and the embedding is compact for all . Furthermore, there is another norm and we know that and are equivalent on ; that is, there exist constants , such thatBy Lemma 1 in [9], we know that, under the conditions , , and , and for each , for all .

Our main result of this paper is the following.

Theorem 1. *Suppose that and are satisfied. Then (1) has three solutions of mountain pass type: one positive, one negative, and one sign-changing. If moreover is odd with respect to its second variable (i.e., holds), then problem (1) has infinitely many sign-changing solutions.*

Throughout the paper, and denote the strong and weak convergence, respectively. , , , and express distinct constants. For , the usual Lebesgue space is endowed with the norm

The paper is organized as follows. In Section 2, we introduce some notions and results of some critical theorem. In Section 3, we complete the proof of Theorem 1.

#### 2. Some Critical Point Theorems

Let us begin by recalling some notions and results of some critical point theorems (see [15]).

In the following, will denote a Hilbert space endowed with the norm , which is a closed convex cone.

For , we denote by the -neighborhood of ; that is, Define

Let . We denote by the set of critical points of and .

For , we consider the following situation.

: there exists a locally Lipschitz continuous vector field ( odd if is even) such that(i), ;(ii)there exists a constant such that(iii)for and , there exists such that if is such that and .

*Definition 2. *Let , . One says that satisfies the condition if each sequence with and in has a convergent subsequence.

Theorem 3 (see [15]). *Let with . Assume there exists such that is satisfied. Assume also that there exist and such that Then there exist sequences such that where If in addition satisfies the condition for any , then has critical point .*

Theorem 4 (see [15]). *Let . Assume there exists such that is satisfied. Assume also that there exists a continuous map such that, for any , the following conditions are satisfied: *(1)* and .*(2)*.*(3)*,**whereThen there exists a sequence such that where If in addition satisfies the condition for any , then has a sign-changing critical point.*

In the following, we assume that is of the form and that there is another norm on such that embeds continuously into .

We introduce the following notations:

Notice that Assume there exist constants and and numbers such that and define

In the following, we introduce a sign-changing critical points theorem.

Theorem 5 (see [15]). *Let be an even functional. Assume that there exist and such that (26) holds. Assume also that there exists such that and the following conditions are satisfied: * * and , as .* *, , where .**Then, for large enough there exists a sequence such thatwhereIf in addition satisfies the condition for any , then it possesses a sequence of sign-changing critical points such that , as .*

#### 3. Proof of Theorem 1

We divide the proof of Theorem 1 into the following lemmas.

For fixed, we consider the functional It is easy to prove that is of class , coercive, bounded below, weakly lower semicontinuous, and strictly convex in . Therefore, by Theorem 1.1 in [16], admits a unique global minimizer in which is the unique solution to the problem

Here we introduce an auxiliary operator , which will be used to construct the descending flow for the functional . We define an operator : for , is the unique solution of (31). Then the set of fixed points of coincide with the set of critical point of .

Furthermore, the operator has the following important properties.

Lemma 6. *(1) is continuous and maps bounded sets to bounded sets.**(2) For any , one has where is defined in (11).**(3) There exists for enough small such that .*

*Proof. *(1) Let such that in . For any , by the definition of , we have Let and . Taking in (33) and (34), we obtain Using the Hölder inequality and the Sobolev embedding theorem, we have where is a constant. By and Theorem A.1 in [17], one has in . Because in as , then as . By (36), we obtain in , which implies that is continuous on .

On the other hand, for any , taking in (34), we obtain Using the Hölder inequality, the Sobolev embedding theorem, , and the fact , we obtain where is constant. This shows that is bounded in whenever is bounded in .

(2) Taking in (34), we have thus for all . Moreover, using again (34), we have By the Hölder inequality, we conclude that where is defined in (11).

(3) From and , for any , there exists such thatSet and . We denote and , for any . Taking in (34) and using the Hölder inequality, we obtain which implies Since , for all , , and , it follows from the Sobolev embedding theorem that there is a constant such that . Moreover, one can easily verify that . Consequently, by (46) and the Sobolev embedding theorem, we have where . Therefore, Similarly, we can prove that for some constant .

Hence where . We can choose small enough so that, for all , It then follows that , .

Notice that the vector field is not locally Lipschitz. However, it can be used as [18] to construct a locally Lipschitz vector field which will satisfy condition . More precisely, we have the following result.

Lemma 7 (see [19, Lemma 3.4]). *There exists a locally Lipschitz continuous operator ( odd when is even) such that *(1)*, for any .*(2)*, for any .*(3)* for any , where is obtained in Lemma 6(3).*

*Remark 8. *Lemmas 6 and 7 imply that

Lemma 9. *Let and . Then there exists such that if is such that and .*

*Proof. *By the definition of the operator , we have It follows that If , using Lemma 7(2) and the Hölder inequality, one has for (see (11)). Suppose that there exists a sequence such that , and . By (55), we see that is bounded. It follows from Remark 8 above that , which is a contradiction.

If , the conclusion is concluded by Remark 8.

Lemma 10. *The functional satisfies the condition at any level .*

*Proof. *In view of (9) and (12), , for any , one has Let be a sequence such that and in as . Inequality (56) implies that is bounded in . Then, up to a subsequence, we have in and for . Using a standard argument, one has . Notice that Consequently, by (43), the Hölder inequality, and the boundness of in , we know that, for any , there is a constant such that By the arbitrariness of and which is compact for , we have in .

Lemma 11. *For , there exists such that for any *

*Proof. *For any , this follows from the fact that where is the Sobolev constant in the continuous embedding for all .

Lemma 12. *For small enough, one has *

*Proof. *Let . It is clear that Using Lemma 11, we obtain where .

Let us denote by the distinct eigenvalues of the problemIt is well known that each has finite multiplicity, the principle eigenvalue is simple with positive eigenfunction , and the eigenfunctions corresponding to are sign-changing. Let be the eigenspace associated with . We set Note that any element of is sign-changing.

We define where

Lemma 13. *One has for any fixed *

*Proof. *By the definition , we have For each fixed , for small enough, by (44) and (67), we have Notice that where is the Sobolev constant in the embedding .

For any , one has This implies that where .

Furthermore, , one has where is the Sobolev constant in the embedding . By (74), one has , for . Hence, For any , we deduce where is a constant. From (71), one hasBy (68), one has . From Lemma 3.8 in [17], we know that , as . Set in (78); we have

Lemma 14. *For any , one has*

*Proof. *The proof is similar to the proof of Lemma 5.4 in [20].

*Proof of Theorem 1. * *Step **1 (the existence of a positive and a negative solution)*. By (44) and the Sobolev embedding theorem, , there exists constant such that Consequently, there exists (small enough) such that From and (44), we have for some positive constants and . Thus, by (83) and the Sobolev embedding theorem, one has where , . Fixing , it is easy to prove thatTherefore, we can find such that This shows that condition of Theorem 3 is satisfied. By Lemmas 6, 7, and 9, condition is satisfied for small enough. By Lemma 10, satisfies the condition at any positive level . Hence, the existence of a positive and a negative solution follows from Theorem 3.

In the following proof, we adopt the notations of Theorem 4.*Step **2 (the existence of a sign-changing solution)*. Using the the main idea of [21], we will verify the assumptions of Theorem 4.

Let be such that