ON A SUPER POLYHARMONIC PROPERTY OF A HIGHER-ORDER FRACTIONAL LAPLACIAN?

(徐美清)

School of Mathematical Science, Shanghai Jiao Tong University, Shanghai 200240, China

E-mail: xmq157@sjtu.edu.cn

Abstract Let 0<α<2, p ≥1, m ∈N+.Consider the positive solution u of the PDEIn [1](Transactions of the American Mathematical Society,2021),Cao,Dai and Qin showed that,under the condition u ∈Lα,(0.1)possesses a super polyharmonic property≥0 for k=0,1,···,m-1.In this paper,we show another kind of super polyharmonic property(-?)ku >0 for k=1,···,m-1,under the conditions (-?)mu ∈Lα and (-?)mu ≥0.Both kinds of super polyharmonic properties can lead to an equivalence between (0.1) and the integral equationdy.One can classify solutions to(0.1)following the work of [2]and [3]by Chen,Li,Ou.

Key words super polyharmonic;fractional Laplacian;equivalence;classification

1 Introduction

In this paper we consider the partial differential equation

under the conditions thatn ≥2,0<α<2,p ≥1,m ∈N+.

For any real numberαbetween 0 and 2,(-?)is a nonlocal differential operator defined by

The super polyharmonic property≥0 fork=0,1,···,m-1 of (1.1) was first proven in [1](Transactions of the American mathematical society,2021) by Cao,Dai and Qin under the condition thatu ∈Lα.In fact,they worked on more general equations of the form=f(x,u,Du,···).Their idea was to use Green’s formula and the Poisson kernel to represent the PDE.

Inspired by their method,in this paper we obtain another kind of super polyharmonic property,but under a quite different definition.We writeand therefore it is natural to require that (-?)mu(x)∈Lα.Moreover,we require that(-?)mu(x)≥0.The result is a different kind of super polyharmonic property: (-?)ku(x)≥0 fork=1,···,m-1.

Theorem 1.1Letn ≥2,0<α<2,p ≥1,m ∈N+.Suppose that (-?)mu ∈Lα(Rn),u ∈C∞(Rn) and thatuis a positive solution of (1.1).Moreover,suppose that (-?)mu ≥0 in Rn.Then,fork=1,···,m-1,

The proof for (-?)mu ≥0 remains unsolved and will be worked upon in the future.

In Theorem 1.1 we require thatu ∈C∞(Rn),but this is not necessary.In fact,one only needs thatu ∈(see [4],Proposition 2.4) to make sure thatuis a continuous function and that its value is given by (1.2).Nevertheless,u ∈C∞(Rn) is enough,since the precise regularity is not the main goal in this paper.

(1.1) is closely related to the integral equation

Once super polyharmonic property holds,then the equivalence between (1.5) and (1.1) can be obtained quickly.Then one can use Chen,Li and Ou’s works,[2]and[5],about the classification of solutions to integral equations and obtain the classification of positive solutions to higherorder fractional equations.

Theorem 1.2Suppose that the conditions in Theorem 1.1 hold.And for 2m+α

Letube a positive solution of (1.1).Thenualso solves (1.5),and vice versa.

Corollary 1.3The solutionuas given in Theorem 1.2 satisfies that

1.for 0<2m+α

with some constantc=c(n,m,α) and for somet>0 andx0∈Rn;

2.for 0<2m+α

Corollary 1.3 is an immediate consequence of Theorem 1.2 and the classification results in[2,5].

The super polyharmonic property of the higher-order Laplacian has long been studied,and it leads directly to the Liouville theorem and the equivalence between integral equations like(1.5) and PDEs like (1.1).For the integer higher-order Laplacian,Wei and Xu [6]first gave the super polyharmonic property for even order equations.After a dozen years,Chen and Li[7]proved the general case and covered Wei and Xu’s work.There is also much work on other types of super polyharmonic properties;see [8–12].

For more results about the fractional equation (-?)u=up,the polyharmonic equations and the maximum principle regarding fractional Laplacians,please refer to[13–16].For general conclusions about fractional Laplacians and higher-order fractional Laplacians,please refer to[17–21].For other types of Laplacian equations,for example,p-Laplacian equations,see[22–24].

2 Proof of Theorem 1.1

Proof of Theorem 1.1Denote that (-?)ku(x) asuk(x) fork=1,···,m.Then (1.1)can be rewritten as

First,show that if (-?)muis nonnegative,then it must be positive.Indeed,if there existsx0such that (-?)mu(x0)=um(x0)=0,thenum(x0)-um(y)≤0 for anyy ∈Rn.Thus,

Now we prove thatuk(x)>0 fork=1,···,m-1 by contradiction.If we were to suppose that our assertion was not the case,then there would existt ∈{1,2,···,m-1} such thatutis less than or equal to 0 somewhere.Denotekas the largest integer among all such integers.Without loss of generality,suppose thatuk(0)≤0.Then there are two cases:

1.uk(0)=0,anduk ≥0 in Rn;

2.uk(0)<0.

Case 1 is impossible.The reason is that under this case,0 is a minimum point ofuk.However,-?uk=uk+1≥0,so by the maximum principle,ukcannot obtain a minimum.Thus we only need to consider Case 2.

We argue by two steps to see such akdoes not exist.

Step 1kcannot be even.Denote bythe spherical average ofvcentered at 0,i.e.,

Thus,

which in turn implies that

Ifkis even,repeat the process above to get that

where we know thatck>0 but do not know the sign of the otherci,i=1,···,k-1.Forrbeing sufficiently large,(r) attains a negative value,which contradicts thatu>0.

Step 2kcannot be odd.Ifkis odd,repeating a process similar to that above,we get that

where we know thatck>0,but do not know the sign of the otherci,i=1,···,k-1.Hence there existsr0>0 and a positive constantMsuch that

Letum=v1+v2withv1,v2satisfying that

for anyR>0.By [25],it follows that

andκ(n,α),c(n,α) are the constants depending only onαandn.Then

and the right-hand side is a constant.ChooseRsatisfying thatR>2r0.Then

Combining (2.18) and (2.20),it follows thatum(0)=∞,which is a contradiction.

3 Proof of Theorem 1.2

To prove the equivalence,one needs to show that any solution of one equation also solves another equation.It is easy to see that any solution of an integral equation (1.5) also solves PDE (1.1).Thus it suffices to show that any solution of a PDE solves the integral equation.

Denote the Riesz potential as

The proof relies on two properties of the Riesz potential (see [26],Chapter V,Section 1.1):

2.Iα(Iβf)=Iα+βfforα>0,β>0 andα+β

and then prove,by the Liouville theorem and induction,that

Finally,by Property 2 of the Riesz potential,it follows that

Proof of Theorem 1.2Set thatwhere(x,y) is the Green’s function for the fractional Laplacian,as stated in (2.15).Then

By the Liouville theorem (see [27],Theorem 1),w0≡C0≥0,i.e.,

In fact,C0can only be 0,and thusdy,which will be proven later.

Let

By the Liouville theorem for harmonic functions,one obtains

for someC1≥0.Moreover,(3.12) implies thatC0=0,otherwise

fork=2,3,···,m-1.Using a similar process and induction,one proves that

Applying Property 2 to{fk},one obtains that

Note thatu ≥Cm.IfCm>0,then

ThusCm=0,anduindeed solves integral equation (1.5).

Conflict of InterestThe author declares no conflict of interest.