Liouville Theorem for Some Elliptic Equations with Weights and Finite Morse Indices

Abstract

We establish the nonexistence of solution for the following nonlinear elliptic problem with weights: in , where is a positive parameter. Suppose that , for or , for ; we will show that this equation does not possess nontrivial bounded solution with finite Morse index.

1. Introduction

Liouville theorems are very important in proving a priori bound of solutions for elliptic equations. As far as we know, the most powerful tool for proving the a priori bound is the blow-up method. During the blow-up process, we first suppose that the solutions are unbounded; then we can scale the sequence of solutions. Finally, we get a nontrivial solution for the limit equation. On the other hand, if we can prove that the limit equation does not possess a nontrivial solution, then we get a contradiction. So the solutions must be bounded. From the above statements, it is easy to see that the most important ingredient in proving the a priori bound is the nonexistence result for the limit equation. These kinds of nonexistence results are usually called Liouville type theorems.

For elliptic equations, the first Liouville theorem was proved by Gidas and Spruck in [1], in which the authors proved that the following equationdoes not possess positive solutions provided . Moreover, it was also proved that the exponent is optimal in the sense that problem (1) indeed possesses a positive solution for . So the exponent is usually called the critical exponent for problem (1). However, this Liouville theorem is not sufficient for proving the a priori bound since the solutions may blow up on the boundary of the domain. In order to overcome this difficulty, the authors studied the limit equation in the half spacein [2]. They proved the above equation also does not possess positive solutions provided . The above two Liouville theorems are what we need to prove a priori bound for positive solutions of nonlinear elliptic equations in bounded domain. Later, Chen and Li obtained similar nonexistence results for the above two equations in [3] by using the moving plane method.

At the same time, elliptic equations with weights were also widely studied and there are many existence and nonexistence results for problem (3). If , we say this problem is a Hardy type equation, while for , we say this problem is a Henon type equation. For the Hardy type problem, it can be proved that this problem does not possess positive solution provided by using the moving plane method as in [3]. However, for the Henon type equation, this proof of the nonexistence result is completely open up to now. The main difference between the two cases lies in that, for , the weight is decreasing in , so the moving plane method works. However, for , the weight is increasing in , so the moving plane method does not work.

On the other hand, we note that the above-mentioned results only claim that the above equations do not possess positive solution. A natural and more difficult question is whether the above equations possess sign-changing solution. However, this question is also completely open up to now. A partial answer was given in [4] in which the authors assume the solution has finite Morse index; then they proved the nonexistence result for this kind of solution. To prove this result, the author first deduced some integrable conditions on the solution based on finite Morse index; then they use the Pohozaev identity to prove the nonexistence result. After this work, there are many extensions on similar problems. For example, Harrabi et al. extended these results to more general nonlinear problems in [5, 6]. The corresponding Neumann boundary value problems were studied in [7]; Yu studied the mixed boundary problems, the nonlinear boundary value problem, and the fractional Laplacian equation in [8], [9], and [10], respectively.

In this paper, inspired by the above works, we study another problem, that is, the following elliptic equation with weight: where is a positive parameter. We are mainly concerned with the nonexistence of solution with finite Morse index. Because of the interaction of and , in order to prove the nonexistence result, we need to add a new bound for the exponent of the weight. More precisely, we have the following result.

Theorem 1. Suppose that , for or , for ; let be a bounded solution for problem (4) with ; then , where is the Morse index of .

The rest of this paper is devoted to the proof of the above theorem. We first deduce some inequality based on finite Morse index; then we derive some integral conditions on this solution. Finally, we use the Pohozaev inequality to prove the above theorem. In the following, we denote by a positive constant, which may vary from line to line.

2. Proof of Theorem 1

In this section, we always assume the conditions in Theorem 1 hold. We establish the nonexistence of finite Morse index solution for problem (4). For this purpose, we first recall the definition of Morse index. Let be a solution of problem (4); we defineas the Morse index for , whereand henceAlso, for further use, we need to define some cut-off function. Let ; we define as follows:Moreover, we assume that for and for . In the same spirit of [8–10], we have the following result.

Lemma 2. Let be a solution of (4) with finite Morse index; then there exists such thatfor any .

Proof. The proof is the same as [8–10]; we omit it.

The next lemma is the key ingredient in the proof of Theorem 1.

Lemma 3. Let be a bounded solution of problem (4) with finite Morse index and let , satisfy the assumptions in Theorem 1; then one has

Proof. We will use the information of finite Morse index of to prove our result. We first prove that . By Lemma 2, there exists a positive constant , such thatfor any . Then by the definition of , we conclude thatA direct calculation shows that the right hand side of (12) equalsOn the other hand, if we multiply (4) by and integrate by parts, then we haveInserting (14) into (12), then we getIn particular, we haveIf , we already have and since the right hand sides of (16) and (17) are bounded by a positive constant independent of . This proves the result for . So in the following, we always assume that . We deduce from (16) and the Holder inequality thatSuppose on the contrary that is infinite; then we havefor some and large enough. Denote , , and ; if we iterate the above inequality times, then we getwith . Since is bounded, a direct calculation shows that the right hand side of (20) is of order withas . In particular, we can choose large enough, such that . Then it follows from (20) thatas , which is impossible. So we get .
Next, we show that . By the same spirit as the above, we deduce from (17) thatSuppose on the contrary that is infinite; then we havefor some and large enough. Denote and ; similarly, if we iterate the above inequality times, then we getwith . By the boundedness of , a direct calculation shows that the right hand side of (25) is of order withas . So we can still choose large enough, such that . Then it follows from (25) thatas , which is a contradiction. So we get that .
Finally, we show that . For this purpose, we first choose a cut-off function such that for and for . For any , we multiply (4) by and integrate by parts; then we getSincewe infer from the above two equations thatBy the assumption for and the fact that , we getby letting in the above equation. This finishes the proof of this lemma.

In order to complete the proof of Theorem 1, we need the following Pohozaev identity for problem (4).

Lemma 4. Suppose that is a solution of (4); then the following identity holds:

Proof. The proof of this lemma is standard; we give the details to keep this paper self-contained.
Multiplying (4) by and integrating in , then we getThe left hand side of (33) equals while the right hand side of (33) equals Combining the above two equations together, then we get the above local Pohozaev identity for problem (4).

With the above preparations, we can prove Theorem 1 now.

Proof of Theorem 1. First, since , , and by Lemma 3, then there exists a sequence such thatas . Let in the local Pohozaev identity and let ; then we getNext, multiplying (4) by and integrating by parts, then we getWe infer from the above two identities thatBy the assumptions on the exponents and in Theorem 1, we haveFor (39) to hold, the only possibility iswhich finally implies .

Competing Interests

The authors declare that there are no competing interests regarding the publication of this paper.