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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 624328, 11 pages

http://dx.doi.org/10.1155/2014/624328

## Existence of Positive Solutions of Semilinear Biharmonic Equations

School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China

Received 12 January 2014; Accepted 14 March 2014; Published 9 April 2014

Academic Editor: Adrian Petrusel

Copyright © 2014 Yajing Zhang et al. 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.

#### Abstract

This paper is concerned with the existence of positive solutions of semilinear biharmonic problem whose associated functionals do not satisfy the Palais-Smale condition.

#### 1. Introduction

We consider the semilinear biharmonic problem where is a bounded domain with smooth boundary in .

Problems of this type have been studied in [1–7]. In [6] Liu and Wang studied (1) when is asymptotically linear with respect to at infinity. In order to find critical points of the functional associated with (1), one usually applies the Mountain Pass Theorem proposed by Ambrosetti and Rabinowitz [8]. For applying the theorem, one often requires the following condition, that is, for some and : where . Condition (2) is important for ensuring that each Palais-Smale sequence is bounded in . We say satisfies the Palais-Smale condition (henceforth denoted by (PS)) if any sequence for which is bounded and as possesses a convergent subsequence. Note that the nonlinear term is asymptotically linear, not superlinear, with respect to at infinity, which means that condition (2) cannot be assumed in their case. Lin et al. use some tricks and techniques to prove that the (PS) sequence is bounded. Then they use Mountain Pass Theorem to get a positive solution to (1).

In [2] Ramos and Rodrigues considered (1) with the nonlinearity , where is a real parameter, changes sign in , and is subcritical and has a superlinear behavior both at zero and at infinity. They extended for the biharmonic operator results that were obtained for the corresponding second order problem in [9]. Their assumptions on do not seem to imply suitable compactness properties (namely, the so-called Palais-Smale condition) for the corresponding functional, if one uses a variational argument. Moreover, due to the absence of sign in the nonlinear term, it is not clear whether the geometric structure of the functional associated with (1) falls into one of the usual schemes used in critical point theory.

In this paper, we suppose that satisfies the following:(H1), , and if ;(H2) for all ;(H3)there exist and such that for and ;(H4)there exist and such that for all and .

This type of hypotheses assumed here does not imply the (PS) condition and does not fit in the condition that implies a priori bounds. Recently, de Figueiredo and Yang [10], Liu et al. [11], and Ramos et al. [9] have considered semilinear second order elliptic problems without the (PS) condition. Our assumptions (H1)–(H4) exactly come from [10]. In [9, 10] the link between the Morse index and the bounds of solutions is shown. In [12] Bahri and Lions mentioned that bounds on Morse indices are useful in some problems to prove the Palais-Smale compactness condition.

For the reader's convenience, we give an example: for , where . Due to advances of our method and our interest in positive solutions, without loss of generality, we may assume that for . It is easy to see that satisfies the conditions (H1)–(H4). Moreover, it is obvious that does not satisfy the hypothetical conditions on nonlinearity in [2].

Our main result is the following.

Theorem 1. *Suppose satisfies (H1)–(H4). Problem (1) has at least a positive solution.*

The organization of the paper is as follows. In Section 2 we prove some new nonlinear Liouville type theorems which may be useful in other situations. In Section 3 we prove Theorem 1. Firstly, we apply the Mountain Pass Theorem to a suitable sequence of truncated problems. In particular, it follows that the Morse index of the solutions of the truncated problems is finite. We use this fact and the blow-up argument to show that the sequence of the truncated problems is bounded. A version of the well-known Pohozaev identity is in turn essential. Throughout this paper, the constant will denote various generic constants.

#### 2. Liouville Type Theorems

For , let be a cut-off function satisfying Define and then

Lemma 2. *Suppose that is a function satisfying , where and are constants. Let be a nonnegative solution of the following problem:
**
with finite Morse index, where . Then there exists such that , .*

The Morse index of solutions of (7) is defined as the dimension of the negative space corresponding to the spectral decomposition of the operator .

*Proof of Lemma 2. *Suppose the assertion is false. Then for , there exists such that and for , we may find such that . Then the supports of and are disjoint, so the Morse index of is larger than or equal to . Iterating the argument, we may get a contradiction since the Morse index of is supposed to be finite.

Proposition 3. *Let be a bounded nonnegative solution with finite Morse index of (7). Then both and are finite.*

*Proof. *By Lemma 2, there exists a such that
That is,
Multiplying (7) by , we find
From (9) and (10), it follows that
where is small enough. Consequently,
Using the value of , we get
Using the interpolation inequality (see [13]), we obtain
From (9) and (10), it follows that
Using the value of again, we have
Estimating the right side of (16) by the argument exactly as above and using (14), we have
where does not depend on .

Since over ,
By (17), we get
Hence
since and is bounded.

We will prove . Assume to the contrary that is not finite; by (20), we obtain
for large .

Using Hölder’s inequality, we get
Substituting (22) for the right-hand side in (21) gives.
Let , , and . Then iterating (23), we get
where , .

Since is bounded, the left side of (24) is of the order , while the right side is of the order , where
which yields a contradiction since .

As above using (14), we obtain
Combining (26) and (22), we have
Using the already proved fact that is finite, we obtain .

Using an approach similar to the method used in the proof of Proposition 3, we prove the following proposition.

Proposition 4. *Suppose that is a function satisfying , where and are constants. Let be a bounded nonnegative solution with finite Morse index of
**
where and . Then both and are finite.*

Proposition 5. *Let be as in Proposition 3. Suppose that
**
where . Then .*

*Proof. *By Proposition 3 and (7), we obtain
By (29), it yields
Then .

Similar arguments are used to prove the following proposition.

Proposition 6. *Let be as in Proposition 4. Suppose that
**
where . Then .*

For future reference, we have the following result at the end of this section.

Lemma 7. *Let be a nonnegative solution of problem (7) with finite Morse index. Then there exists such that for one has
*

*Proof. *By Proposition 3, . We proceed as in [10, 12]. Denote by the Morse index of with respect to the operator . Suppose that . Let , be smooth functions such that
Denote for . For each , there exists such that
So we deduce as (12) and (15) that
By (36), we have
where is small enough.

Thus
Integrating by parts, we obtain
Consequently,
Hence
Multiplying (7) by and integrating by parts, we obtain
Consequently,
Hence
By (38)–(44), we have
By the definition of , it follows that
This implies by Hölder’s inequality that
Hence, there exists such that
The assertion follows.

#### 3. Proof of Theorem 1

Let us first note that hypotheses (H1)–(H4) imply that there exist a sequence as and a continuous function such that Using L'Hospital's rule, we obtain Without loss of generality, we may assume that , if . Define a truncation of by Note that in the variable .

Let us consider the truncated problem

Lemma 8. *Problem (52) possesses at least one positive solution with finite Morse index.*

*Proof. *Consider the functional associated with problem (52)
where . We will use the Mountain Pass Theorem by Ambrosetti and Rabinowitz [8] to obtain existence result for problem (52). One can easily check that there exist and such that for ,
We note that condition (54) is important for ensuring that has a Mountain Pass geometry and satisfies the Palais-Smale condition. So, using the Mountain Pass Theorem, we obtain a nontrivial weak solution of (52). By Lemma B3 in [14], is a classical solution of (52). By the maximum principle for with Navier boundary conditions we get that is positive. The geometry of the Mountain Pass, described in [15, 16], implies that the Morse indices of are less than or equal to . Thus Lemma 8 is proven.

Let be a function satisfying (H1)–(H3), and consider the problem where are positive constants and .

Lemma 9. *Let be a solution of (55). Then for any ball one has
**
where and denotes the unit outward normal to .*

*Proof. *By standard procedures, one can prove the Pohozaev type identity. We give the proof for completeness and for the reader's convenience.

By Proposition 2.2 in [17], we have
It is clear that
By (55), we have
Substituting (57) and (58) into (59) and using the divergence theorem, we find
Using the Green's formula, we obtain
Substituting (61) into (60) and using on , we obtain (56).

If it happens that , for some , then is also a solution of (1), and the proof of Theorem 1 will be completed. Thus it suffices to prove the following proposition. We prove the proposition by the blow-up technique of Gidas and Spruck [18].

Proposition 10. *Suppose that is a solution of (52) with finite Morse index. Then there exists a such that .*

*Proof. *Assume by contradiction that there does not exist such a . So we should have , for all . Then as .

Let . Define
which satisfies
and in , .

Due to compactness of we may also assume that . So there are two cases to be considered, and .*Case **1*. .

Given there is an such that for all . By the -estimates due to Agmon et al. [19], we have that for all
By assumptions (H1)–(H3) and the definition of , it follows that
and then for large ,
So we obtain that
Choosing , it follows from standard embedding theorems that is uniformly bounded in . By the Schauder estimate of Agmon et al. [19] one has
Next we claim that
In order to do that we write
Then we have
According to the definition of , we divide the estimate of into three cases.(i)If , then since is we get
(ii)If , we use condition (H3) to get
(iii)If , by the definition of , we have
By (71)–(74), we obtain
which proves (69), and therefore
Using Arzela-Ascoli Theorem and (69) and (76), we obtain a subsequence of still denoted by , such that
where , as .

Assume that . By (65), we have
By (77), we have
for large . Consequently,
That is,
Then, by the definition of and the assumption (H4), we get
for large . Combining (79) and (83) and letting , we obtain
Define
Then there exist positive constants and such that , . Passing to the limit in (63) and using (77) and (78), we conclude that satisfies
By a diagonal process, it follows that
Next we claim that the Morse index of is finite. If , by the L'Hospital's rule we have
By (H3) for ,
Then, by (88), we get
If , then as . By assumption (H3), we have
as . Therefore, (90) holds for all . By the diagonal process, one knows that (90) holds also in and it converges uniformly on compact sets of as .

The uniform convergence of to on compact sets implies that the Morse index of is finite. To handle this, we set
Let