# A Liouville Type Result for Schrödinger Equation on Half-Spaces

Authors: Baiyu Liu
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 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:(1)-Δu-βz2u-u2*-1=0,x∈H,u=0,x∈∂H,
wheren≥3,2*=2n/(n-2),β>0, and
(2)H=ℝ+n={x=(x′,z)∣x′∈ℝn-1,z>0}.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,
(3)-Δu=f(v),inℝ+n,-Δv=g(u),inℝ+n,u=v=0,on∂ℝ+n,
and obtained some sufficient conditions for f and g, under which system (3) admits only trivial solution.Let𝒟01,2(H)be the space given by the completion ofC0∞(H)under the norm∥u∥=(∫H‍|∇u|2dx)1/2. We say thatuis a weak solution of (1) ifu∈𝒟01,2(H)satisfies
(4)∫H∇u·∇φdx=∫H‍βz2uφdx+∫H‍u2*-1φdx,
for allφ∈Cc∞(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∈𝒟01,2(H)be a nonnegative weak solution of (1) with0<β<1/16. Then,u≡0.Remark 2.
For a weak solutionu∈𝒟01,2(H), by using a regularity lifting method [8], we know thatu∈C2,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∈𝒟01,2(ℝ+n); then,
(5)∫ℝ+n|u|2z2dx≤4∫ℝ+n‍|∇u|2dx.This inequality plays a crucial role in estimating the singular potential term in the following proof.In the following, we assume thatu∈𝒟01,2(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
(6)Σλ={(x′,z)∣x′∈ℝn-1,z∈(0,λ)}=ℝn-1×(0,λ).
Forx∈Σλ, we writexλ=(x1,…,xn-1,2λ-z)which is the reflected point ofxwith respect to the hyperplaneTλ={x=(x′,z)∣z=λ}and define
(7)uλ(x)=u(xλ),wλ(x)=uλ(x)-u(x).Then, direct computation gives(8)-Δwλ(x)=-Δuλ(x)+Δu(x)=β(2λ-z)2uλ(x)+(uλ(x))2*-1-βz2u(x)-(u(x))2*-1=β(2λ-z)2uλ(x)-βz2uλ(x)+βz2uλ(x)-βz2u(x)+(uλ(x))2*-1-(u(x))2*-1=βz2wλ(x)+ξ(x,λ)wλ(x)+β(1(2λ-z)2-1z2)uλ(x);
hereξ(x,λ)=((uλ(x))2*-1-u(x)2*-1)/(uλ(x)-u(x)).Forx∈Σλ, we have2λ-z>z,1/(2λ-z)2<1/z2,uλ(x)≥0, andβ>0. Therefore,
(9)-Δwλ(x)≤βz2wλ(x)+ξ(x,λ)wλ(x).Definewλ+(x)=max{wλ(x),0}andwλ-(x)=-min{wλ(x),0}. Clearly,wλ+(x)≥0,wλ-(x)≥0 andwλ(x)=wλ+(x)-wλ-(x). Define
(10)Σλ-={x∈Σλ∣wλ(x)<0}.The heart of our argument is the following lemma.Lemma 4.
There exists aC0>0, such that, forλ>0, if||wλ-||L2*(Σλ)>0, then
(11)∥u(x)∥L2*(Σλ-)≥C0.Proof.
For0<ϵ<λ/4, letηϵ(z)∈C(ℝ+) be defined by
(12)ηϵ(z)={1,z>ϵ,logz-2logϵ-logϵ,ϵ2≤z≤ϵ0,z<ϵ2.,
Testing (9) inΣλwith functionηϵ2wλ-, we obtain
(13)-∫Σλ‍Δwλ(x)ηϵ2wλ-dx≤β∫Σλ‍1z2wλ(x)ηϵ2wλ-dx+∫Σλ‍ξ(x,λ)wλ(x)ηϵ2wλ-dx.
The left hand side of (13) is
(14)-∫Σλ‍Δwλ(x)ηϵ2wλ-dx=∫Σλ-‍|∇wλ-|2ηϵ2dx+2∫Σλ-‍ηϵwλ-∇wλ(x)·∇ηϵdx.
Hence, we derive
(15)∫Σλ-|∇wλ-|2ηϵ2dx≤I+II+III,
where
(16)I=β∫Σλ-‍1z2ηϵ2(wλ-)2dx,II=∫Σλ-‍ξ(x,λ)ηϵ2(wλ-)2dx,III=-2∫Σλ-‍ηϵwλ-∇wλ(x)·∇ηϵdx.
Using Lemma 3, we have
(17)I=β∫Σλ-‍1z2ηϵ2(wλ-)2dx≤4β∫Σλ-‍|∇(wλ-ηϵ)|2dx≤8β∫Σλ-‍(ηϵ2|∇wλ-|2+(wλ-)2|∇ηϵ|2)dx.
Forx∈Σλ-,0≤uλ(x)<u(x),0<ξ(x,λ)<((n+2)/(n-2)(u(x))4/(n-2), which implies
(18)II=∫Σλ-‍ξ(x,λ)ηϵ2(wλ-)2dx≤n+2n-2∫Σλ-‍(u(x))4/(n-2)ηϵ2(wλ-)2dx.
By using Hölder inequality, we verify that(19)II≤n+2n-2(∫Σλ-‍u2*ηϵn/2dx)2/n·(∫Σλ-‍(wλ-)2*ηϵn/(n-2)dx)(n-2)/n≤n+2n-2∥u∥L2*(Σλ-)4/(n-2)∥wλ-∥L2*(Σλ-)2,(20)III=-2∫Σλ-‍ηϵwλ-∇wλ·∇ηϵdx≤2∫Σλ-‍ηϵwλ-|∇wλ-||∇ηϵ|dx≤14∫Σλ-‍ηϵ2|∇wλ-|2dx+4∫Σλ-‍|∇ηϵ|2(wλ-)2dx.
Putting (17), (19), and (20) into (15) and using the assumption1<β<1/16, we then deduce that
(21)∫Σλ-|∇wλ-|2ηϵ2dx≤9∫Σλ-‍|∇ηϵ|2(wλ-)2dx+4·n+2n-2∥u∥L2*(Σλ-)4/(n-2)∥wλ-∥L2*(Σλ-)2.
Moreover, by the Sobolev inequality, we know that(22)∥wλ-ηϵ∥L2*(Σλ-)2≤C2∥∇(wλ-ηϵ)∥L2(Σλ-)≤C2∫Σλ-‍|ηϵ∇wλ-+wλ-∇ηϵ|2dx≤2C2∫Σλ-(ηϵ2|∇wλ-|2+(wλ-)2|∇ηϵ|2)dx.‍
Combine the above inequality with (21) to get
(23)∥wλ-ηϵ∥L2*(Σλ-)2≤20C2∫Σλ-|∇ηϵ|2(wλ-)2dx+8C2n+2n-2∥u∥L2*(Σλ-)(n-2)/4∥wλ-∥L2*(Σλ-)2.
Now we claim that(24)∫Σλ-|∇ηϵ|2(wλ-)2dx⟶0,asϵ⟶0.
Notice that, forx∈Σλ-,0<wλ-(x)=u(x)-uλ(x)≤u(x). Hence,
(25)0≤∫Σλ-‍|∇ηϵ|2(wλ-)2dx≤∫Σλ-‍|ηϵ′(z)|2u2dx=∫ϵ2≤z≤ϵ‍(u(x))2z2(logϵ)2dx≤1(logϵ)2∫H‍(u(x))2z2dx≤41(logϵ)2∫H‍|∇u|2dx.
Sinceu∈𝒟01,2(H),∫H‍|∇u|2dx<+∞. Thus (24) is valid.
Now, lettingϵ→0in (23), by using dominated convergence theorem, we obtain
(26)1≤8C2·n+2n-2∥u∥L2*(Σλ-)4/(n-2)
If ∥wλ-∥L2*(Σλ-)≠0.
One can chooseC0=((n-2)/8C2(n+2))(n-2)/4, whereCis the best constant in the Sobolev inequality.Using Lemma4, we now can start the moving plane process as the following Lemma.Lemma 5.
There is aλ0>0, such that, for all0<λ<λ0,
(27)wλ(x)≥0,∀x∈Σλ.Proof.
Sinceu∈𝒟01,2(H), using Sobolev inequality, we haveu(x)∈L2*(H). Chooseλ0>0small enough such that
(28)∥u∥L2*(Σλ0)<C0,
whereC0is the same as in Lemma 4.
Hence, for all0<λ<λ0,
(29)∥u∥L2*(Σλ-)≤∥u∥L2*(Σλ)≤∥u∥L2*(Σλ0)<C0,
which is a contradiction to Lemma 4, if ∥wλ-∥L2*(Σλ-)≠0. That is to say,
(30)∥wλ-∥L2*(Σλ-)=0,
which implies that wλ≥0, forx∈Σλ.Now we move the hyperplaneTλupwards by increasing the value ofλcontinuously as long as (27) holds. We will show that the hyperplane will be moved to the infinity. Precisely, define
(31)Λ=sup{λ>0∣wμ(x)≥0,∀x∈Σμ,∀0<μ≤λ}.
By the result of Lemma 5,Λ≥λ0>0.Lemma 6.
We haveΛ=+∞.Proof.
SupposeΛ<+∞.
On one hand, by continuity we know thatwΛ(x)≥0, for allx∈ΣΛ, which means
(32)ΣΛ-=∅.
On the other hand, by the definition ofΛ, there is{δi}i=1∞ that satisfy (i)δi→0, asi→∞, and (ii) ∥wΛ+δi-(x)∥L2*(ΣΛ+δi)>0, for alli. By Lemma 4, we get ∥u(x)∥L2*(ΣΛ+δi-)≥C0>0. By using the dominated convergence theorem, we obtain
(33)∥u(x)∥L2*(ΣΛ-)≥C0>0,
which is a contradiction to (32).


## 3. Proof of Theorem1

In this section, we prove Theorem1.Sinceuis a superharmonic continuous function inH(see Remark 2), we have eitheru≡0inHoru>0inH.Ifu>0inH, then there is some(x0′,z0)∈Hsatisfyingu(x0′,z0)=c>0. Moreover, by continuity, there is aδ>0, such thatu(x′,z0)>a/2, for all|x′-x0′|<δ. By using Lemma 6, we know thatu(x)is increasing with respect tozinH. Thus,u(x′,z)≥a/2for all|x′-x0′|<δandz≥z0. Hence,
(34)∫z0+∞‍∫|x′-x0′|<δ‍|u(x′,z)|2*dx′dz≥∫z0+∞‍∫|x′-x0′|<δ‍(a2)2*dx′dz=+∞,
which contradicts the fact thatu∈L2*(H).Therefore,u≡0inH.



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*Source: 957468-2013-11-28.xml*