Harnack inequality for polyharmonic equations

Jiamin Zeng, Runjie Zheng, and Yi Fang

Department of Mathematics, Anhui University of Technology, Ma’anshan 243002, China

Abstract: Some new types of mean value formulas for the polyharmonic functions were established.Based on the formulas, the Harnack inequality for the nonnegative solutions to the polyharmonic equations was proved.

Keywords: Harnack inequality; mean value formulas; polyharmonic equations; polyharmonic functions; Hopf’s lemma

1 Introduction

The Harnack estimates for the harmonic equation have been investigated profoundly[1,2].Let Ω?Rn(n≥2) be a connected domain.In 2006, Caristi and Mitidieri[3]considered the Harnack inequality for nonnegative solutions to the biharmonic equation

They used the mean value formulas for the biharmonic functions, which are the solutions to the biharmonic equations and the maximum principle, to prove the Harnack inequality.Motivated by the approaches and results in their work, we will consider the Harnack inequality for the nonnegative weak solutions to thek-harmonic (k≥3) equation

The functionuthat satisfies Eq.(2) is calledk-harmonic or polyharmonic.We shall focus on the casek=3 and prove the mean value formulas for the 3-harmonic function and for the generalk-harmonic function cases by induction argument.Then, we will give the proof of the Harnack inequality for Eq.(2).

Theorem 1.1.Assume thatuis a nonnegative weak solution of Eq.(2) such that -Δu≥0 in Ω.Then there existsC=C(n)＞0, such that for anyx∈Ω and eachRsatisfying 0＜2R＜dist(x,?Ω)andB2R(x)??Ω, it holds that

Remark 1.1.The assumption -Δu≥0 in Ω is necessary.Since if we letu(x)=x21forx=(x1,···,xn)∈Rn, then(-Δ)ku(x)=0fork≥3, -Δu(x)=-2＜0.However, for anyR＞0,udoes not satisfy the Harnack inequality inBR(0).

The remaining part of this paper is organized as follows: In Section 2, we figure out the mean value formulas for polyharmonic functions, and in Section 3, we give the proof of Theorem 1.1.

2 The mean value formulas

In this section, we first prove the mean value formulas for 3-harmonic functions, and then extend the mean value formulas to the general polyharmonic function cases by the induction argument.The mean value formulas we consider here are different from those in Refs.[4, 5] and references therein.

Definition 2.1.For anyx∈Rn,r＞0, the spherical average ofuis defined as

Remark 2.1.When there is no ambiguity, we will simply writeinstead ofObviously, we have

where ωnis the measure of the unit sphere ?B1(0).

The following lemma was very useful in the study of higher-order conformally invariant elliptic equations[6].For the convenience of the readers, we will give a proof.

Lemma 2.1.For any integerk≥1, it holds that

Notice that

where ξ is the outward unit normal vector to the boundary?B1(0).Then the divergence theorem[2]implies that

Since

then

Therefore, for any integerk≥1, it can be easily concluded thatby induction.

Now, we give the proof of the mean value formula for 3-harmonic functions by the approach in Ref.[7].

Lemma 2.2.Assume thatuis a weak solution to(-Δ)3u=0in Ω.For anyx∈Ω, denoted bydx=dist(x,?Ω),then for any 0＜R＜dx, the following mean value formula holds

Proof.By Weyl’s lemma[7], we can prove thatu∈C∞(Ω).For any fixed pointx∈Ω, we denote by

and

Note that (-Δ)3u=0, so Δ2uis harmonic.By the mean value formula, we havefor any 0＜r＜dx.Then, it is easy to check thatis a special solution to Eq.(4).Therefore, the general solutions to Eq.(4) can be given by

wherecij(i,j=1,···,4) are all constant.Next, we calculate these constants.

Consider the casen≥5 first.Sinceuˉ is continuous in[0,dx]anduˉ(0)=u(x)=c11, which implies thatc12=c13=0,

Then, taking the Laplacian operatoron both sides of Eq.(5), we obtain

which yields

For the casesn=2,3,4, we can use the same strategy as the casesn≥5.Therefore, forn≥2, we have the following uniform formula

For anyR∈[0,dx], multiplying ωnrn-1on both sides of Eq.(6) and integrating with respect tor∈[0,R], we get

FixingR=rin Eq.(7) and combining it with Eq.(6), we obtain

Again, for anyR∈[0,dx], multiplyingrn+k(k≥0) on both sides of Eq.(8) and integrating with respect tor∈[0,R], we obtain

It follows that

Therefore,

Pluggingk=0 andk=1 into Eq.(9), we have two special equalities as follows:

and

which implies that

By the similar arguments, generally, we can obtain the mean value formulas fork-harmonic functions.

Lemma 2.3.Assume thatuis a weak solution to(-Δ)ku=0in Ω,k≥3.For anyx∈Ω, denoted bydx=dist(x,?Ω), then for any 0＜R＜dx, the following mean value equality holds

Proof.If we denote by

and

which finishes the proof.

3 Proof of Theorem 1.1

Now, we can give the proof of Theorem 1.1.

Proof.On the one hand, for anyx∈Ω,0＜2R＜dist(x,?Ω), by Lemma 2.3 we have

Ifz∈BR/2(x), we haveBR/2(z)?BR(x) and

Therefore,

The value of the above positive constantsC(n,k) may vary in different places.

On the other hand, sinceu≥0 and -Δu≥0 in Ω, by the mean value inequality, for anyx∈Ω,t＞0, ifBt(x)??Ω,then

By Hopf’s lemma[1,2], we haveWithout loss of generality, we assume the minimum pointx0∈?BR/2(x),thenu(x)≥u(x0).Obviously,BR/2(x)?BR(x)?B3R/2(x0)??Ω,so we have

By (10) and (11), the proof is completed.

Acknowledgements

The authors are grateful to Dr.Bo Xia at the University of Science and Technology of China for useful discussions and comments on this work.This work was supported by the National Natural Science Foundation of China (11801006,12071489).

Conflict of interest

The authors declare that they have no conflict of interest.

Biographies

Jiamin Zengis currently a postgraduate student at Anhui University of Technology.His research mainly focuses on elliptic partial differential equations.

Yi Fangis an Associate Professor at Anhui University of Technology.He received his Ph.D.degree from the University of Science and Technology of China in 2015.His research mainly focuses on elliptic partial differential equations.