A Liouville Type Result for Schrödinger Equation on Half-Spaces

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, [1–5].

In this paper, we consider nonnegative solutions of the following Schrödinger equation with a singular potential on the half-space: 𝛽−Δ𝑢−𝑧²𝑢−𝑢^2∗−1=0,𝑥∈𝐻,𝑢=0,𝑥∈𝜕𝐻, (1) where𝑛≥3,2^∗=2𝑛/(𝑛−2),𝛽>0, and 𝐻=ℝ^𝑛₊=𝑥=𝑥^′,𝑧∣𝑥^′∈ℝ^𝑛−1,𝑧>0. (2)

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 term𝑓(𝑢)such that the positive bounded solution𝑢of−Δ𝑢=𝑓(𝑢)with Dirichlet boundary value condition is monotone increasing in𝑧. Guo [3] considered nonnegative solutions for the elliptic system, −Δ𝑢=𝑓(𝑣),inℝ^𝑛₊,−Δ𝑣=𝑔(𝑢),inℝ^𝑛₊,𝑢=𝑣=0,on𝜕ℝ^𝑛₊, (3) and obtained some sufficient conditions for 𝑓 and 𝑔, under which system (3) admits only trivial solution.

Let𝒟₀^1,2(𝐻)be the space given by the completion of𝐶^∞₀(𝐻)under the norm‖𝑢‖=(∫_𝐻|∇𝑢|²𝑑𝑥)^1/2. We say that𝑢is a weak solution of (1) if𝑢∈𝒟₀^1,2(𝐻)satisfies _𝐻∇𝑢⋅∇𝜑𝑑𝑥=_𝐻𝛽𝑧²𝑢𝜑𝑑𝑥+_𝐻𝑢^2∗−1𝜑𝑑𝑥, (4) for all𝜑∈𝐶^∞_𝑐(𝐻).

Using a Hardy-type inequality and the moving plane method in integral forms [8–10], we obtain the following Liouville type result.

Theorem 1. Let𝑢∈𝒟₀^1,2(𝐻)be a nonnegative weak solution of (1) with0<𝛽<1/16. Then,𝑢≡0.

Remark 2. For a weak solution𝑢∈𝒟₀^1,2(𝐻), by using a regularity lifting method [8], we know that𝑢∈𝐶^2,𝛼(Ω), for all bounded smooth domainΩ⊂𝐻. 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 [11–13].

Lemma 3. Let𝑢∈𝒟₀^1,2(ℝ^𝑛₊); then, _ℝ𝑛+|𝑢|²𝑧²𝑑𝑥≤4_ℝ𝑛+|∇𝑢|²𝑑𝑥. (5)

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

In the following, we assume that 𝑢∈𝒟₀^1,2(𝐻)is a nonnegative weak solution of (1) with0<𝛽<1/16. We are going to use the method of moving plane in the half-space.

For each𝜆>0, let Σ_𝜆=𝑥^′,𝑧∣𝑥^′∈ℝ^𝑛−1,𝑧∈(0,𝜆)=ℝ^𝑛−1×(0,𝜆). (6) For𝑥∈Σ_𝜆, we write𝑥^𝜆=(𝑥₁,…,𝑥_𝑛−1,2𝜆−𝑧)which is the reflected point of𝑥with respect to the hyperplane𝑇_𝜆={𝑥=(𝑥^′,𝑧)∣𝑧=𝜆}and define 𝑢_𝜆(𝑥)=𝑢𝑥^𝜆,𝑤_𝜆(𝑥)=𝑢_𝜆(𝑥)−𝑢(𝑥). (7)

Then, direct computation gives −Δ𝑤_𝜆(𝑥)=−Δ𝑢_𝜆(𝑥)(𝑥)=𝛽+Δ𝑢(2𝜆−𝑧)²𝑢_𝜆(𝑥)+𝑢_𝜆(𝑥)^2∗−1−𝛽𝑧²𝑢(𝑥)−(𝑢(𝑥))^2∗−1=𝛽(2𝜆−𝑧)²𝑢_𝜆(𝑥)−𝛽𝑧²𝑢_𝜆(𝑥)+𝛽𝑧²𝑢_𝜆(𝑥)−𝛽𝑧²𝑢(𝑥)+𝑢_𝜆(𝑥)^2∗−1−(𝑢(𝑥))^2∗−1=𝛽𝑧²𝑤_𝜆(𝑥)+𝜉(𝑥,𝜆)𝑤_𝜆(𝑥)1+𝛽(2𝜆−𝑧)²−1𝑧²𝑢_𝜆(𝑥); (8) here𝜉(𝑥,𝜆)=((𝑢_𝜆(𝑥))^2∗−1−𝑢(𝑥)^2∗−1)/(𝑢_𝜆(𝑥)−𝑢(𝑥)).

For𝑥∈Σ_𝜆, we have2𝜆−𝑧>𝑧,1/(2𝜆−𝑧)²<1/𝑧²,𝑢_𝜆(𝑥)≥0, and𝛽>0. Therefore, −Δ𝑤_𝜆(𝑥)≤𝛽𝑧²𝑤_𝜆(𝑥)+𝜉(𝑥,𝜆)𝑤_𝜆(𝑥). (9)

Define𝑤⁺_𝜆(𝑥)=max{𝑤_𝜆(𝑥),0}and𝑤⁻_𝜆(𝑥)=−min{𝑤_𝜆(𝑥),0}. Clearly,𝑤⁺_𝜆(𝑥)≥0,𝑤⁻_𝜆(𝑥)≥0 and𝑤_𝜆(𝑥)=𝑤⁺_𝜆(𝑥)−𝑤⁻_𝜆(𝑥). Define Σ⁻_𝜆=𝑥∈Σ_𝜆∣𝑤_𝜆(𝑥)<0. (10)

The heart of our argument is the following lemma.

Lemma 4. There exists a𝐶₀>0, such that, for𝜆>0, if||𝑤⁻_𝜆||_𝐿2^∗_(Σ𝜆)>0, then ‖𝑢(𝑥)‖_𝐿2^∗_(Σ−𝜆)≥𝐶₀. (11)

Proof. For0<𝜖<𝜆/4, let𝜂_𝜖(𝑧)∈𝐶(ℝ⁺) be defined by 𝜂_𝜖(𝑧)=1,𝑧>𝜖,log𝑧−2log𝜖−log𝜖,𝜖²≤𝑧≤𝜖0,𝑧<𝜖²., (12)
Testing (9) inΣ_𝜆with function𝜂²_𝜖𝑤⁻_𝜆, we obtain −_Σ𝜆Δ𝑤_𝜆(𝑥)𝜂²_𝜖𝑤⁻_𝜆𝑑𝑥≤𝛽_Σ𝜆1𝑧²𝑤_𝜆(𝑥)𝜂²_𝜖𝑤⁻_𝜆+𝑑𝑥_Σ𝜆𝜉(𝑥,𝜆)𝑤_𝜆(𝑥)𝜂²_𝜖𝑤⁻_𝜆𝑑𝑥. (13) The left hand side of (13) is −_Σ𝜆Δ𝑤_𝜆(𝑥)𝜂²_𝜖𝑤⁻_𝜆=𝑑𝑥_Σ−𝜆∇𝑤⁻_𝜆²𝜂²_𝜖𝑑𝑥+2_Σ−𝜆𝜂_𝜖𝑤⁻_𝜆∇𝑤_𝜆(𝑥)⋅∇𝜂_𝜖𝑑𝑥. (14) Hence, we derive _Σ−𝜆∇𝑤⁻_𝜆²𝜂²_𝜖𝑑𝑥≤𝐼+𝐼𝐼+𝐼𝐼𝐼, (15) where 𝐼=𝛽_Σ−𝜆1𝑧²𝜂²_𝜖𝑤⁻_𝜆²𝑑𝑥,𝐼𝐼=_Σ−𝜆𝜉(𝑥,𝜆)𝜂²_𝜖𝑤⁻_𝜆²𝑑𝑥,𝐼𝐼𝐼=−2_Σ−𝜆𝜂_𝜖𝑤⁻_𝜆∇𝑤_𝜆(𝑥)⋅∇𝜂_𝜖𝑑𝑥. (16) Using Lemma 3, we have 𝐼=𝛽_Σ−𝜆1𝑧²𝜂²_𝜖𝑤⁻_𝜆²𝑑𝑥≤4𝛽_Σ−𝜆∇𝑤⁻_𝜆𝜂_𝜖²𝑑𝑥≤8𝛽_Σ−𝜆𝜂²_𝜖∇𝑤⁻_𝜆²+𝑤⁻_𝜆²∇𝜂_𝜖²𝑑𝑥. (17)
For𝑥∈Σ⁻_𝜆,0≤𝑢_𝜆(𝑥)<𝑢(𝑥),0<𝜉(𝑥,𝜆)<((𝑛+2)/(𝑛−2)(𝑢(𝑥))^4/(𝑛−2), which implies 𝐼𝐼=_Σ−𝜆𝜉(𝑥,𝜆)𝜂²_𝜖𝑤⁻_𝜆²≤𝑑𝑥𝑛+2𝑛−2_Σ⁻_𝜆(𝑢(𝑥))^4/(𝑛−2)𝜂²_𝜖𝑤⁻_𝜆²𝑑𝑥. (18)
By using Hölder inequality, we verify that 𝐼𝐼≤𝑛+2𝑛−2_Σ⁻_𝜆𝑢^2∗𝜂_𝜖^𝑛/2𝑑𝑥^2/𝑛⋅_Σ−𝜆𝑤⁻_𝜆^2∗𝜂_𝜖^{𝑛/(𝑛−2)}𝑑𝑥^{(𝑛−2)/𝑛}≤𝑛+2‖𝑢‖𝑛−2^{𝐿4/(𝑛−2)2∗Σ−}_𝜆𝑤⁻_𝜆²_𝐿2^∗_Σ−𝜆,𝐼𝐼𝐼=−2_Σ−𝜆𝜂_𝜖𝑤⁻_𝜆∇𝑤_𝜆⋅∇𝜂_𝜖𝑑𝑥≤2_Σ−𝜆𝜂_𝜖𝑤⁻_𝜆∇𝑤⁻_𝜆∇𝜂_𝜖≤1𝑑𝑥4_Σ⁻_𝜆𝜂²_𝜖∇𝑤⁻_𝜆²𝑑𝑥+4_Σ−𝜆∇𝜂_𝜖²𝑤⁻_𝜆²𝑑𝑥. (19)(20)
Putting (17), (19), and (20) into (15) and using the assumption1<𝛽<1/16, we then deduce that _Σ−𝜆∇𝑤⁻_𝜆²𝜂²_𝜖𝑑𝑥≤9_Σ−𝜆∇𝜂_𝜖²𝑤⁻_𝜆²𝑑𝑥+4⋅𝑛+2‖𝑢‖𝑛−2^{𝐿4/(𝑛−2)2∗Σ−}_𝜆𝑤⁻_𝜆²_𝐿2^∗_Σ−𝜆. (21)
Moreover, by the Sobolev inequality, we know that 𝑤⁻_𝜆𝜂_𝜖²_𝐿2^∗_Σ−𝜆≤𝐶²∇𝑤⁻_𝜆𝜂_{𝜖𝐿2(Σ−𝜆)}≤𝐶²_Σ−𝜆𝜂_𝜖∇𝑤⁻_𝜆+𝑤⁻_𝜆∇𝜂_𝜖²𝑑𝑥≤2𝐶²_Σ−𝜆𝜂²_𝜖∇𝑤⁻_𝜆²+𝑤⁻_𝜆²∇𝜂_𝜖²𝑑𝑥. (22)
Combine the above inequality with (21) to get 𝑤⁻_𝜆𝜂_𝜖²_𝐿2^∗_Σ−𝜆≤20𝐶²_Σ−𝜆∇𝜂_𝜖²𝑤⁻_𝜆²𝑑𝑥+8𝐶²𝑛+2‖𝑢‖𝑛−2^{𝐿(𝑛−2)/42∗Σ−}_𝜆𝑤⁻_𝜆²_𝐿2^∗_Σ−𝜆. (23)
Now we claim that _Σ−𝜆∇𝜂_𝜖²𝑤⁻_𝜆²𝑑𝑥⟶0,as𝜖⟶0. (24) Notice that, for𝑥∈Σ⁻_𝜆,0<𝑤⁻_𝜆(𝑥)=𝑢(𝑥)−𝑢_𝜆(𝑥)≤𝑢(𝑥). Hence, 0≤_Σ−𝜆∇𝜂_𝜖²𝑤⁻_𝜆²≤𝑑𝑥_Σ−𝜆𝜂^′_𝜖(𝑧)²𝑢²=𝑑𝑥_{𝜖2≤𝑧≤𝜖}(𝑢(𝑥))²𝑧²log𝜖²≤1𝑑𝑥log𝜖^2𝐻(𝑢(𝑥))²𝑧²1𝑑𝑥≤4log𝜖^2𝐻|∇𝑢|²𝑑𝑥. (25) Since𝑢∈𝒟₀^1,2(𝐻),∫_𝐻|∇𝑢|²𝑑𝑥<+∞. Thus (24) is valid.
Now, letting𝜖→0in (23), by using dominated convergence theorem, we obtain 1≤8𝐶²⋅𝑛+2‖𝑢‖𝑛−2^{𝐿4/(𝑛−2)2∗Σ−}_𝜆 (26) If ‖𝑤⁻_𝜆‖_𝐿2^∗_(Σ−𝜆)≠0.
One can choose𝐶₀=((𝑛−2)/8𝐶²(𝑛+2))^(𝑛−2)/4, where𝐶is 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 a𝜆₀>0, such that, for all0<𝜆<𝜆₀, 𝑤_𝜆(𝑥)≥0,∀𝑥∈Σ_𝜆. (27)

Proof. Since𝑢∈𝒟₀^1,2(𝐻), using Sobolev inequality, we have𝑢(𝑥)∈𝐿^2∗(𝐻). Choose𝜆₀>0small enough such that ‖𝑢‖_𝐿2^∗_(Σ𝜆0)<𝐶₀, (28) where𝐶₀is the same as in Lemma 4.
Hence, for all0<𝜆<𝜆₀, ‖𝑢‖_𝐿2^∗_(Σ−𝜆)≤‖𝑢‖_𝐿2^∗_(Σ𝜆)≤‖𝑢‖_𝐿2^∗_(Σ𝜆0)<𝐶₀, (29) which is a contradiction to Lemma 4, if ‖𝑤⁻_𝜆‖_𝐿2^∗_(Σ−𝜆)≠0. That is to say, 𝑤⁻_𝜆𝐿2^∗_(Σ−𝜆)=0, (30) which implies that 𝑤_𝜆≥0, for𝑥∈Σ_𝜆.

Now we move the hyperplane 𝑇_𝜆upwards by increasing the value of𝜆continuously as long as (27) holds. We will show that the hyperplane will be moved to the infinity. Precisely, define Λ=sup𝜆>0∣𝑤_𝜇(𝑥)≥0,∀𝑥∈Σ_𝜇,∀0<𝜇≤𝜆. (31) By the result of Lemma 5,Λ≥𝜆₀>0.

Lemma 6. We have Λ=+∞.

Proof. Suppose Λ<+∞.
On one hand, by continuity we know that𝑤_Λ(𝑥)≥0, for all𝑥∈Σ_Λ, which means Σ⁻_Λ=∅. (32)
On the other hand, by the definition ofΛ, there is{𝛿_𝑖}^∞_𝑖=1 that satisfy (i)𝛿_𝑖→0, as𝑖→∞, and (ii) ‖𝑤⁻_Λ+𝛿𝑖(𝑥)‖_{𝐿2∗(Σ𝑖Λ+𝛿)}>0, for all𝑖. By Lemma 4, we get ‖𝑢(𝑥)‖_𝐿2^∗_{(Σ−𝑖Λ+𝛿)}≥𝐶₀>0. By using the dominated convergence theorem, we obtain ‖𝑢(𝑥)‖_𝐿2^∗_(Σ−Λ)≥𝐶₀>0, (33) which is a contradiction to (32).

3. Proof of Theorem 1

In this section, we prove Theorem 1.

Since𝑢is a superharmonic continuous function in𝐻(see Remark 2), we have either𝑢≡0in𝐻or𝑢>0in𝐻.

If𝑢>0in𝐻, then there is some(𝑥^′₀,𝑧₀)∈𝐻satisfying𝑢(𝑥^′₀,𝑧₀)=𝑐>0. Moreover, by continuity, there is a𝛿>0, such that𝑢(𝑥^′,𝑧₀)>𝑎/2, for all|𝑥^′−𝑥^′₀|<𝛿. By using Lemma 6, we know that𝑢(𝑥)is increasing with respect to𝑧in𝐻. Thus,𝑢(𝑥^′,𝑧)≥𝑎/2for all|𝑥^′−𝑥^′₀|<𝛿and𝑧≥𝑧₀. Hence, _𝑧^+∞_{0|𝑥′−𝑥′0|<𝛿}𝑢𝑥^′,𝑧^2∗𝑑𝑥^′𝑑𝑧≥_𝑧^+∞_{0|𝑥′−𝑥′0|<𝛿}𝑎2^2∗𝑑𝑥^′𝑑𝑧=+∞, (34) which contradicts the fact that𝑢∈𝐿^2∗(𝐻).

Therefore,𝑢≡0in𝐻.

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).