Multiplicity of Positive Solutions for a Second-Order Elliptic System of Kirchhoff Type

Abstract

We study elliptic problems of Kirchhoff type in . Using variational tools, we establish the existence of at least two nontrivial and nonnegative solutions.

1. Introduction and Preliminaries

In this paper, we are concerned with the following problem: where    is a bounded domain with the smooth boundary such that , is the Laplacian operator, , ,   , and , the function , is positively homogeneous of degree which is the Sobolev critical exponent; that is, , holds for all , .

In recent years, there have been many papers concerned with the existence of the positive solutions for Kirchhoff equation which is related to the stationary analogue of the Kirchhoff equation where , . It was proposed by Kirchhoff [1] as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings.

Some interesting studies on these problems by variational methods can be found in [2–6]. As for perturbed fourth-order Kirchhoff-type elliptic problems, in [7] the following equation, where , , is an -Carathéodory function, and is a continuous function, has been investigated. The authors proved (4) has multiple nontrivial weak solutions.

In [8] the authors established the existence of a weak solution for the following system equation: where .

Motivated by the results of the above cited papers, we will attempt to treat problem (1) and extend the results for our problem.

In this paper we make the following assumptions.

Let be the best Sobolev embedding constant defined by and let be the Lebesgue measure of ; denotes the norm, , and where .

Also the following hold:(F1) is a function and ,(F2) where ,(F3) are strictly increasing function about , for all , .

In addition, using assumption (F1), we have the so-called Euler identity and, for a positive constant ,

Let be the completion of with respect to the norm .

It is easy to show that, for every , the above norm is equivalent with . Problem (1) is posed in the framework of the Sobolev space with the standard norm We will look for solutions of (1) by finding critical points of the energy functional given by where is the functional defined by It is well known that the functional . For any , there holds Consider the Nehari manifold Note that if and only if So contains all nontrivial weak solutions of (1).

Define . Then, for , Now, we split into three parts:

2. Statement of the Main Results

Let us first define and the main results read as follows.

Theorem 1. If satisfy and (F1)–(F3) hold, then problem (1) has at least one positive solution.

Theorem 2. If satisfy and (F1)–(F3) hold, then problem (1) has at least two positive solutions.

Note that, using assumption (F3), we have that are positively homogeneous of degree . This implies that for some positive constant . Similar to Willem [9, Theorem ], we consider the continuity of the superposition operator

Lemma 3. Assume that , , , and Then, for every , and the operator is continuous.

Now, we consider the functional ; then we have the following result.

Lemma 4. Assume that , satisfying (F3); then the functional is of class and where .

Proof. The proof is almost the same as that in [10].

Lemma 5. The energy functional is coercive and bounded below on .

Proof. If , then by the Hölder inequality and the Sobolev embedding theorem Thus, is coercive and bounded below on .

Lemma 6. Suppose that is a local minimizer for on and that . Then in (the dual space of the Sobolev space ).

Proof. If is a local minimizer for on , then is a solution of the optimization problem minimizer subject to . Hence, by the theory of Lagrange multipliers, there exists , such that and thus, Since , we have . Moreover, , so . This completes the proof.

Lemma 7. If , then .

Proof. Suppose otherwise that such that .
Then, for , By the Holder inequality and the Sobolev embedding theorem So, Thus, and, by the Minkowski inequality, the Sobolev embedding theorem, and (9), Thus, This implies that which is a contradiction. Thus, we can conclude that if we have .

By Lemma 7, we write and define Then we have the following result.

Lemma 8. Consider the following.(i)If , then one has .(ii)If , then for some constant

Proof. (i) Let . Then and so Thus, from the definition of and , we can deduce that .
(ii) Let . Then By (8) we have This implies that By (23) in the proof of Lemma 5, Thus, if , then for some   . This completes the proof.

For each with , set Then we have the following.

Lemma 9 (see [11, Lemma 2.6]). For each with , there are unique such that and ; .

3. Proof of the Main Theorems

We will need the following lemma.

Lemma 10 (see [12]). Consider the following. (i)If , then there exists a -sequence in for .(ii)If , then there exists a -sequence in for .

Theorem 11. If and (F1)–(F3) hold, then has a minimizer in and it satisfies the following: (i);(ii) is a positive solution of (1).

Proof. By Lemma 10(i), there exists a minimizing sequence for on such that Then by Lemma 5 and the compact imbedding theorem, there exist a subsequence and such that This implies that as . By (44) and (45), it is easy to prove that is weak solution of (1). Since and by Lemma 8(i), Letting , we see that . Thus, is a nontrivial solution of problem (1). Now it follows that strongly in and strongly in and . By and applying Fatou’s lemma, we get This implies that Let , ; then by the Brezis-Lieb lemma [13], this implies Therefore, strongly in and strongly in . Moreover, we have . In fact, if , by Lemma 9, there are unique and such that and . In particular, we have . Since there exists such that . By Lemma 9, which is a contradiction. It follows from the maximum principle that is a positive solution of problem (1). This completes the proof.

The following two lemmas are similar to those in [14].

Lemma 12. If is a -sequence for with in , then , and there exists a positive constant , such that .

Lemma 13. If is a -sequence for , then is bounded in .

Define In addition, we need the following version of the Brezis-Lieb lemma [13].

Lemma 14. Consider with and for some . Let be a bounded sequence in such that weakly in . Then as ,

Lemma 15. satisfies the condition with satisfying

Proof. Let be a -sequence for with . It follows from Lemma 13 that is bounded in , and then up to a subsequence, where is a critical point of . Furthermore, we may assume Hence we have that and Let , . Then by the Brezis-Lieb lemma [13], we obtain and by Lemma 14, Since , and (58)–(60), we can deduce that So Hence, we may assume that If , the proof is complete. Assume ; then from (63), we obtain which implies that . In addition, from Lemma 12, (61), and (63), we get which contradicts .

Lemma 16. There exists a nonnegative function and such that, for , one has In particular, for all .

Proof. Since , there is such that . Now, we consider the functional defined by and define a cut-off function such that for , for , , and . For , let
From [14], we have the following estimates: where is a minimizer of that is, Set , and , where and . Then, by (F1), (9), the definition of , and (69), we obtain that where the following fact has been used: We can choose such that, for all , we have Using the definitions of and , we get which implies that there exists satisfying On the other hand, where , , and on .
Let ; we get Combining with (77) and the above inequality, for all , we have the following.
According to properties of and , we can conclude that there exists the positive constant such that , so we have There exists a constant , such that .
Hence, we can choose such that, for all , we obtain If we see and , then for we have Finally, we prove that for all . Recall that . It is easy to see that Combining this with Lemma 9, from the definition of and (81), we get that there exists such that and for all .

Theorem 17. If and (F1)–(F3) hold, then has a minimizer in and it satisfies the following: (i),(ii) is a positive solution of problem (1),
where .

Proof. By Lemma 10(ii), there is a -sequence in for for all . From Lemmas 15, 16, and 8(ii), for satisfies condition and . Since is coercive on , we get that is bounded in . Therefor, there exists a subsequence still denoted by and such that strongly in and for all . Finally, by the same arguments as in the proof of Theorem 11, for all , we have that is a positive solution of problem (1).

Now, we complete the proof of Theorems 1 and 2. By Theorem 11, we obtain that, for all , problem (1) has a positive solution . On the other hand, from Theorem 17, we get the second positive solution for all . Since , this implies that and are distinct. This completes the proof of Theorems 1 and 2.

Conflict of Interests

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