Abstract and Applied Analysis

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

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

Research Article

Existence of Positive Solutions of Semilinear Biharmonic Equations

Yajing Zhang, Yinmei Lü, and Ningning Wang

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 be such that . The uniform convergence of to on compact sets and the fact that (90) holds on compact sets imply

for large . Since the Morse index of is finite, it follows easily that the Morse index of is finite. Proposition 3 implies that is finite. We claim that there exists a such that

Then Proposition 5 yields , which contradicts .

Now, we prove (94). Applying Lemma 9 to (63) in the ball for fixed , we obtain

By (H3), we estimate

Therefore

which tends to zero as .

Using a similar argument that leads to (69) we can prove

Then its limit exists as . Using L’Hospital’s rule as (88) we get

as . Therefore, by the Lebesgue dominated convergence theorem, we obtain

Letting in (95), we get

By Lemma 7, there exists such that

Since , this implies that

Taking the limit in (101), one has

Assertion (94) follows.

Case 2. .

Two cases may occur: either or as .

If as , then for all the ball is contained in for large enough. We also obtain a contradiction as in Case 1.

If as , by the blow-up argument we get a solution of

with in , , and the Morse index being finite, where . We may deduce as Case 1 that . This is a contradiction since . Thus the proof of Proposition 10 is completed.

Conflict of Interests

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

Acknowledgments

This work is supported by the NNSFC (no. 61374089), the NSF of Shanxi Province (no. 2009011008), the Shanxi Scholarship Council of China (no. 2011-005), and the Program for the Top Young Academic Leaders of Higher Learning Institutions of Shanxi.

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