GradientEstimatesandLiouville-TypeTheoremsfor a Nonlinear Elliptic Equation

ZHU Chaona

School of Mathematical Sciences,University of Science and Technology of China,Hefei 230026,China.

Abstract.In this paper,we consider the following nonlinear elliptic equationon thecompletesmooth metricspace(Rn,g0,e?fdvg0),where g0istheEuclideanmetric on Rnand f=|x|2/4.We prove gradient estimates and Liouville-Type theorems for positive solutions of the above equation.

Key Words:Gradient estimate;Liouville-Type theorem;shrinking gradient Ricci soliton.

1 Introduction

Recall that a triple(M,g,f)is called a shrinking gradient Ricci soliton(c.f.[1])if

for some positive constant λ,where(M,g)is a Riemannian manifold and f is a smooth function on M.We say that the gradient soliton is complete if both(M,g)and the vector field?gf is complete.We call the function f a potential function.

The Bakry-Emery Ricci tensor on the Riemannian manifold(M,g)is defined by

for some smooth function f on M.Thus if Ricf=λg for some positive constant λ,then(M,g,f)is a shrinking gradient Ricci soliton.

The f-Laplacian operator is defined by

If f is constant,the f-Laplacian operator is reduced to the classical Laplacian.If a smooth function u on M satisfies△fu=0(≥0,≤0),we call it f-harmonic(f-subharmonic,fsuperharmonic).For a positive f-harmonic function,its gradient estimate and Liouville theorem have been studied by many authors.Wei and Wylie[2]proved that any positive f-harmonic function with some growth conditions must be constant if Ricf≥K＞0.Brighton[3]proved that a positive bounded f-harmonic function with Ricf≥0 is constant.For a local Cheng-Yau’s gradient estimate,Wu[4]obtained it for positive fharmonic functions with Ricf≥?(n?1)K and|?f|≤θ.Chen and Chen[5]also proved the same gradient estimate for positive f-harmonic functions with another condition Ric≥?(n?1)H.Munteanu and Sesum[6]applied the De Giorgi-Nash-Moser theory to get a global gradient estimate for a positive f-harmonic function and proved that a positive f-harmonic function with sublinear growth of f on the metric space is constant if Ricf≥0.Li[7]applied probabilistic arguments to give an alternative proof of Brighton’s gradient estimate and Liouville theorem for positive f-harmonic functions.Wu[8]proved a Liouville property for any f-harmonic function with polynomial growth on a complete noncompact smooth metric measure space on which any diameter of geodesic sphere has sublinear growth and whose Bakry-Emery Ricci curvature satisfies a quadratic decay lower bound,i.e.,Ricf≥?cr?2.

In this paper,we are interested in the following nonlinear elliptic equation

on the space(Rn,g0,e?|x|2/4dvg0)which is a complete smooth metric space and is a shrinking gradient Ricci soliton.Here g0is the Euclidean metric on Rn.For a quasiharmonic function u on Rn,u satisfies the equation(1.1)for h=0.Li and Wang[9]derived gradient estimates for positive quasi-harmonic functions and showed that there is no nonconstant positive quasi-harmonic function on Rnwith polynomial growth.Zhu and Wang[10]showed that there is neither a nonconstant positive quasi-harmonic function nor a nonconstant Lp(Rn,ds2)(p＞,n≥3)quasi-harmonic function,where ds2=e?|x|2/2(n?2)g0.But for all 1≤ p≤n/(n?2),there exists a nonconstant quasi-harmonic function in Lp(Rn,ds2)(n≥3).Ge and Zhang[11]proved that there doesn’t exist a nonconstant positive f-harmonic function on the complete gradient shrinking Ricci solitons.They also obtained Lp(p≥1 or 0＜p≤1)Liouville theorems on the complete gradient shrinking Ricci solitons.We derive the similar results for positive solutions of equation(1.1).

Here is an outline for this paper.In Section 2.1,we will prove gradient estimates of positive solutions to the equation(1.1)for 1＜α＜n/(n?2)(n≥4).When n=3,we obtain the results for 1＜α＜5/2 and when n=2,we obtain the results for 1＜α＜4.In Section 2.2,we will derive gradient estimates of positive solutions to the equation(1.1)for 0＜α＜1 whose conditions are the same as the case of 1＜α＜n/(n?2)(n≥4).In Section 2.3,we will prove gradient estimates of positive solutions to the equation(1.1)for α=1 which is more easierthanthe othertwo casesand theconditionis different from them.Besides,we should note that there is no restriction on the dimension for the latter two cases.Finally,in Section 3,we will prove a Liouville-Type theorem.

2 Gradient estimates for the equation△fu+huα=0

In this section,we derive gradient estimates for positive solutions of equation(1.1)in the cases 1＜α＜n/(n?2)(n≥4),α=1 and 0＜α＜1.

Let u be a positive solution of equation(1.1),and set

where β＞0 to be fixed later.Then

Consequently,

We set

where v is a constant to be fixed later.By straightforward computing,we get

2.1 The case of 1＜α＜.

Lemma 2.1.Assume that h(x)∈C2(Rn).For 1＜α＜,if ψ(x)is defined by(2.2)where,then

where β is a positive constant such that

Proof.By straightforward computing,we obtain

and

Substituting(2.3)and(2.6)into(2.5),we get

It follows from(2.2)that

By the fundamental inequality and(2.1),we obtain that

Substituting(2.8)and(2.9)into(2.7),we have

This completes the proof of the lemma.

where Aα,β,Bα,βand Cα,βare constants depending only on α,β,n.

Proof.Let r(x)denote the distance between x and 0.We consider the function

where ψ is defined by(2.2)with

We may assume that the function F(x)achieves its maximum at x0∈BR(0).By the maximum principle,we have that,at x0

Consequently,△fF(x0)≤0.By computing directly,we have

Since f=|x|2/4,then

Substituting(2.4)and(2.12)into(2.11),we have

By Young’s inequality,we get

By(2.2)and h≥0,we have

If△fh≥0,substituting(2.10),(2.14),(2.15)and(2.16)into(2.13)and letting β＞0 be sufficiently small such that

then we obtain

Multiplying by(R2?r2)2through(2.18),we get

By one-variable quadratic inequality theory,we immediately obtain that

Obviously,we have

where Aα,β,Bα,βand Cα,βare constants depending only on α,β,n.

Remark 2.1.For n=3,the system of inequalities(2.17)holds for 1＜α＜5/2.Thus the result is still true for n=3 when 1＜α＜5/2.For n=2,the system of inequalities(2.17)holds for 1＜α＜4.Thus the result is still true for n=2 when 1＜α＜4.

2.2 The case of 0＜α＜1.

Lemma 2.2.Assume that h(x)∈C2(Rn)and h≥0.For 0＜α＜1,if ψ(x)is defined by(2.2)where v=β2/α,then

Proof.If 0＜α＜1 and h≥0,then

So(2.7)takes the form that

By Young’s inequality,we have

Substituting(2.3),(2.8),(2.9)and(2.21)into(2.20),we obtain that

Set v=β2/α,it follows from(2.22)that

This completes the proof of the lemma.

Theorem 2.2.Let(Rn,g0,e?|x|2/4dvg0)be the n-dimensional complete smooth metric space.Assume that h(x)∈C2(Rn)satisfies h≥0 and△fh≥0.If u(x)is a positive solution of equation(1.1)for 0＜α＜1 on BR(0),then

where Aα,β,Bα,βand Cα,βare constants depending only on α,β,n.

Proof.Let r(x)denote the distance between x and 0.We consider the function

where ψ is defined by(2.2)with v=β2/α.

We may assume that the function F(x)achieves its maximum at x0∈BR(0).By the maximum principle,we have that,at x0

Consequently,△fF(x0)≤0.By computing directly,we have

Since f=|x|2/4,then

Substituting(2.19)and(2.25)into(2.24),we have

By Young’s inequality,we obtain

By(2.2)and h≥0,we get

If△fh≥0,substituting(2.23),(2.27),(2.28),(2.29)and(2.30)into(2.26)and letting β＞0 be sufficiently small such that,then we have

Multiplied by(R2?r2)2,(2.31)yields that

By one-variable quadratic inequality theory,we immediately obtain that

Obviously,we have

where Aα,β,Bα,βand Cα,βare constants depending only on α,β,n.

2.3 The case of α=1.

In this section,we will consider the following equation

i.e.the case α=1 of equation(1.1).Let u be a positive solution of equation(2.32),and set

Then

Set

Thus,

Lemma 2.3.Assume that h(x)∈C2(Rn)and|?h|≤k.If ψ(x)is defined by(2.33),then

Proof.By straightforward computing,we have

Substituting(2.34)and(2.37)into(2.36),we get

By Young’s inequality,we obtain

Substituting(2.33),(2.39)and(2.40)into(2.38),we have

If|?h|≤k,then

This completes the proof of the lemma.

Theorem 2.3.Let(Rn,g0,e?|x|2/4dvg0)be the n-dimensional complete smooth metric space.Assume that h(x)∈C2(Rn)satisfies|?h|≤k.If u(x)is a positive solution of equation(2.32),then

where A,B are constants only dependent on n.

Proof.Let r(x)denote the distance between x and 0.We consider the function

where ψ is defined by(2.33).

We may assume that the function F(x)achieves its maximum at x0∈BR(0).By the maximum principle,we have that,at x0

Consequently,△fF(x0)≤0.By computing directly,we have

Since f=|x|2/4,then

Substituting(2.35)and(2.43)into(2.42),we have

By Young’s inequality,we obtain

Substituting(2.41)and(2.45)into(2.44),we have

Multiplying(R2?r2)4ψ through(2.46),then we get

By one-variable quadratic inequality theory,we immediately obtain that

Obviously,we have

where A,B are constants depending only on n.

3 Liouville theorems

Inthis section,we derive Liouville theoremsfor Eq.(1.1).We recall some theoremswhich were proved by Ge-Zhang[11].

Theorem 3.1.([11,Theorem 0.4])Let(M,g,f)be a complete shrinking gradient Ricci soliton,u be a nonnegative smooth function on M.If

then

Theorem 3.2.([11,Corollary 0.3])Let(Rn,g0,e?|x|2/4dvg0)be the n-dimensional complete smooth metricspace.Assumethat h(x)∈C2(Rn)andsatisfies oneofthe followingthree conditions

(2)0＜α＜1,h≥0 and△fh≥0;

(3) α=1,|?h|≤k.

If u(x)is a positive solution of equation(1.1),then we have

for r(x)≥C,where C is a constant large enough.

Remark 3.1.This Theorem is derived from gradient estimates.In Section 2,we have derived similar results to in[11,Theorem 0.2].The proof is the same as in[11,Corollary 0.3],so we omit it here.

Theorem 3.3.([11,Theorem 0.5])Let(M,g,f)be a complete shrinking gradient Ricci soliton,i.e.,

then any positive f-harmonic function on M is constant.

Theorem 3.4.Let(Rn,g0,e?|x|2/4dvg0)be the n-dimensional complete smooth metric space.Assume that h(x)∈C2(Rn)and satisfies one of the following three conditions

(2)0＜α＜1,h≥0 and△fh≥0;

(3) α=1,|?h|≤k and h≥0.

If u(x)is a positive solution of equation(1.1),then u(x)is constant.Moreover,if h/≡0,then u≡0.

Proof.It is easy to see that

i.e.,(Rn,g0,e?|x|2/4dvg0)is a complete shrinking gradient Ricci soliton.

If h≥0,then a positive solution of equation(1.1)satisfies

Set G=log(1+u),then

By Harnack’s inequality(Theorem 3.2),we have

Using Theorem 3.1,we deduce that G is f-harmonic.By Theorem 3.3,we immediately get that G is constant.Thus u is constant.Moreover,if h/≡0,then u≡0.

Acknowledgement

The author thanks Professor Li Jiayu for his guidance.The research was supported by NSFC No.11526212,No.11131007.