#### Abstract

We consider a nonlinear Schrödinger equation with a singular potential on half spaces. Using a Hardy-type inequality and the moving plane method, we obtain a Liouville type result for its nonnegative solutions.

#### 1. Introduction

Recently, properties of nontrivial solutions for nonlinear elliptic equations on half spaces have attracted a great deal of attention from physicians and mathematicians; see, for example, [15].

In this paper, we consider nonnegative solutions of the following Schrödinger equation with a singular potential on the half-space: where,,, and

Equation (1) is related to the Grushin type equation with critical exponent and the Webster scalar curvature equation [6, 7].

We are interested in the Liouville type result for nonnegative solutions of (1). This work is motivated by some monotonicity results and Liouville type results for elliptic equations on half-spaces; see, for example, [2, 3]. In [2], Dancer found some sufficient conditions for nonlinear termsuch that the positive bounded solutionofwith Dirichlet boundary value condition is monotone increasing in. Guo [3] considered nonnegative solutions for the elliptic system, and obtained some sufficient conditions for and , under which system (3) admits only trivial solution.

Letbe the space given by the completion ofunder the norm. We say thatis a weak solution of (1) ifsatisfies for all.

Using a Hardy-type inequality and the moving plane method in integral forms [810], we obtain the following Liouville type result.

Theorem 1. Letbe a nonnegative weak solution of (1) with. Then,.

Remark 2. For a weak solution, by using a regularity lifting method [8], we know that, for all bounded smooth domain. Hence, it is a classical solution.

#### 2. Preliminary

In this section, we prepare some lemmas.

Firstly, we recall the Hardy-Sobolev inequality in the half space; see [1113].

Lemma 3. Let; then,

This inequality plays a crucial role in estimating the singular potential term in the following proof.

In the following, we assume that is a nonnegative weak solution of (1) with. We are going to use the method of moving plane in the half-space.

For each, let For, we writewhich is the reflected point ofwith respect to the hyperplaneand define

Then, direct computation gives here.

For, we have,,, and. Therefore,

Defineand. Clearly,, and. Define

The heart of our argument is the following lemma.

Lemma 4. There exists a, such that, for, if, then

Proof. For, let be defined by
Testing (9) inwith function, we obtain The left hand side of (13) is Hence, we derive where Using Lemma 3, we have
For,,, which implies
By using Hölder inequality, we verify that
Putting (17), (19), and (20) into (15) and using the assumption, we then deduce that
Moreover, by the Sobolev inequality, we know that
Combine the above inequality with (21) to get
Now we claim that Notice that, for,. Hence, Since,. Thus (24) is valid.
Now, lettingin (23), by using dominated convergence theorem, we obtain If .
One can choose, whereis the best constant in the Sobolev inequality.

Using Lemma 4, we now can start the moving plane process as the following Lemma.

Lemma 5. There is a, such that, for all,

Proof. Since, using Sobolev inequality, we have. Choosesmall enough such that whereis the same as in Lemma 4.
Hence, for all, which is a contradiction to Lemma 4, if . That is to say, which implies that , for.

Now we move the hyperplane upwards by increasing the value ofcontinuously as long as (27) holds. We will show that the hyperplane will be moved to the infinity. Precisely, define By the result of Lemma 5,.

Lemma 6. We have .

Proof. Suppose .
On one hand, by continuity we know that, for all, which means
On the other hand, by the definition of, there is that satisfy (i), as, and (ii) , for all. By Lemma 4, we get . By using the dominated convergence theorem, we obtain which is a contradiction to (32).

#### 3. Proof of Theorem 1

In this section, we prove Theorem 1.

Sinceis a superharmonic continuous function in(see Remark 2), we have eitherinorin.

Ifin, then there is somesatisfying. Moreover, by continuity, there is a, such that, for all. By using Lemma 6, we know thatis increasing with respect toin. Thus,for alland. Hence, which contradicts the fact that.

Therefore,in.

#### Acknowledgments

This project is supported by the National Natural Science Foundation of China (no. 11201025) and the Fundamental Research Funds for the Central Universities (FRF-TP-12-106A).