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Multiple Results to Some Biharmonic Problems
We study a nonlinear elliptic problem defined in a bounded domain involving biharmonic operator together with an asymptotically linear term. We establish at least three nontrivial solutions using the topological degree theory and the critical groups.
We consider the following biharmonic problem: where is a smooth bounded domain and is of class with .
In the past decades, biharmonic operators have attracted much attention of many researchers and experts. While , the solutions of (1) characterized the travelling waves in a suspension bridge; see .
In 1998, Micheletti and Pistoia  considered the following biharmonic problem: where , are constants and is a bounded smooth domain, and they established multiple results by using a minimax process.
Three years later, Zhang  considered a more general condition; that is, where and satisfies the subcritical growth; at least one nontrivial solution was obtained.
At the same time, Leray-Schauder degree as a very wonderful tool was introduced to handle biharmonic problems; see [6–9]. To our best knowledge, there are few papers considered (1) by combining the critical point theory (especially Morse theory) with Leray-Schauder degree.
Zhang  first considered the following second-order elliptic problem: where denotes Neumann operator or Dirichlet operator. Sub- and sup-solutions methods with critical point theory were used to obtain at least two distinct solutions. Also under subcritical growth condition, Chang  proved that if is an isolated critical point of , then, for all , with integral coefficients, where , denote the energy functional under and , means th critical group corresponding to (4), which inspires us to consider (1).
As far as we know, there are few papers concerned with the biharmonic problem (1) using this method; only Qian and Li  considered And they proved that if is an isolated critical point of , then, for all , with integral coefficients, where , denote the energy functional on the space and , means th critical group. In our paper, the results in  are improved, and some new results are obtained.
Let denote the eigenvalues of (counting with their multiplicity) with corresponding eigenfunctions . We may choose in . Let , , then are eigenvalues of the following biharmonic problem  corresponding eigenfunctions ;
In order to obtain nontrivial solutions, we now assume that the nonlinearity satisfies the following conditions:(f1), for all , and there exist constant numbers and with , such that (f2)there exists with , such that , for all ;(f3) uniformly for ;(f4)there is some small, such that where .
The main result of this paper is the following
Theorem 1. Suppose satisfies (f1)–(f4). Then (1) has at least three solutions.
In this section, we first recall some lemmas and preliminaries.
Let denote the set of which are -times continuous differentiable in and identically vanishing on with the norm , where , , .
Lemma 2. is a solid cone of ; that is, .
It is well known that the positive cone is a solid cone of . Our proof depends on the fact above; what is more, the technique we used here is originated from [15, page 628].
Proof. Since is a closed positive cone of , by the definition of , , for , , thus the embedding is continuous. is closed in (in fact ). Obviously ; thus has nonempty interior. The proof is finished.
Remark 3. Using the method above, it is not difficult to know that is a solid cone in , .
Remark 4. For any , if is an interior point of in , then is an interior point of in .
In what follows, we will use the Hilbert space , and the norm on is given by . It is well known that solutions of (1) are critical points of the functional where . Since , it is easy to know that , and
Corresponding to the eigenvalues we have the splitting where
Consider the problem
For all , denote , , , .
Remark 6. Actually, for all , there exists a unique weak solution of (12). Since by Riesz representation theorem, for all , there exists a unique such that ; thus is the corresponding weak solution.
Consider the Cauchy problem in ,
Lemma 7 (see ). Let be a real Hilbert space, and let satisfy the condition. Assume that where is a compact mapping, and that is an isolated critical point of . Then we have
Let be a retract of a real Banach space , let be a relatively open subset of , and let be a completely continuous operator. Suppose that has no fixed points on and that the fixed point of is bounded. The following lemma establishes the relationship of fixed point index and topological degree.
Lemma 8 (see ). If any fixed point of in is an interior point of , then there exists an open subset of with such that contains all fixed points of in and
Remark 9. Let be a bounded open subset of , and let there be no zero points of on . Since can be compactly embedded into , it follows from the bootstrap argument and the definition of Leray-Schauder degree that
In what follows, is denoted simply by .
Remark 10 (see ). Remark 9 implies that two topological degrees in both and are the same. Combining with Lemma 7, we can obtain the connection between the topological degree and the critical group:
Lemma 11. Let be the unique solution of (14) with the maximal interval . We have the following conclusions.(i)If , then , and is continuous as a function of from to .(ii)If , , , and as , then as .(iii)If for some then and is bounded in the norm.
Proof. We only need to construct the embedding chains like (5) and (6) of ; the rest can be proved similar to [19, Lemma 2].
Without loss of generality, can be assumed to satisfy . We can choose , such that Let , and define by
A direct computation shows that Hence there exists a number such that
Let and choose and such that
Let Define Then we have the following imbedding chains:What is more, we have the chains of bounded and continuous operators
Lemma 12. Suppose that (f1) and (f3) hold. Then satisfies the condition.
The proof of this lemma is similar to the proof of [5, Lemma 2.1]. We omit it here.
Since is completely continuous, then by the above bootstrap iteration, is completely continuous. For our application, sometimes we would consider the restriction of on a smaller Banach space . The functional may lose the (PS) condition. However following , the following two lemmas can be obtained.
Lemma 13 (see ). Suppose that (f1) and (f3) hold. Then possesses the following properties. (i) is a closed subset.(ii)For each pair , implies that is a strong deformation retract of , where denotes the critical set of (and also ).
Lemma 14 (see ). with integral coefficients.
Here and in what follows, we always assume that has only finitely many critical points.
Lemma 15 (see ). Let be an isolated critical point of , where . Denote , , and assume that has a local linking at with respect to a direct sum decomposition , where ; that is, there exists small such that Then
3. Calculation of Degree
Lemma 16. Suppose that (f1), (f3), and (f4) hold. Then there exists , such that, for all ,
Proof. Since is an orthogonal basis of , for , there exist , such that . Let Since is finite dimensional, we have that, for given , there exists some such that if then By (f4), for with , For with , Thus possesses a local linking at the origin. By Lemma 15, the critical groups of at the origin satisfy Then there exists small such that there is no other critical point in except , for all , and the following can be obtained by Lemma 13 and Remark 10:
Lemma 17. Suppose (f1) and (f2) hold. There exists ( is defined in Lemma 16), such that, for all ,
Proof. We only prove that . By (f1), , it follows from the condition (f2) that there exist and such that
If for some and , where is a positive number, that is then we have from (41) that Thus, and this is a contradiction. Therefore, according to the property of fixed point index, we get
Similarly, we can also show that there exists such that for all . Let . Then the conclusion holds.
Lemma 18. Suppose that (f1) and (f3) hold. Then there exists , , such that
Proof. Since uniformly for , there exist constants and such that
We will first show that any solution of (1) is bounded. Suppose is a solution; then satisfy
Multiplying by , we have
by (47), it is easy to see
Let , and using Young inequality, there exists constant such that
By Poincaré inequality, there exists only dependent on , such that
By a bootstrap argument, there exists such that .
Let , and we will show that Suppose there exists , , such that ; then . By (f3), such that , for all , Then , where . Then , where . Let ; then, for all , we have Then
From Lemma 17, we have
By Lemma 18 and the strong maximum principle of second order elliptic problem, it is easy to know that if is a solution of (1), then , and thus, by Lemma 2, . Using Lemma 8, there is a bounded open , such that Similarly, there is a bounded open subset , such that
4. Proof of Main Result
Proof. By conditions (f1) and (f3), it is easy to know that
that is, there exists large enough, such that
If has no fixed point in , then the additivity property of degree implies It follows that . This is a contradiction. Thus (1) has at least a solution in .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This research was supported by the National Natural Science Foundation of China (no. 10871096) and the Project of Graduate Education Innovation of Jiangsu Province (CXLX13_367).
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Copyright © 2014 Xingdong Tang and Jihui Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.