WANG Xinjing

School of Mathematics and Statistics,Huanghuai University,Zhumadian 463000,China.

Abstract. The aim of the paper is to study the properties of positive classical solutions to the fractional Laplace equation with the singular term.Using the extension method,we prove the nonexistence and symmetric of solutions to the singular fractional equation.

1 Introduction

In this paper,we consider the fractional problem

where 1≤α ＜2,.The fractional Laplacian (-Δ)α/2in Rnis nonlocal pseudo-differential operator defined by

wherePVis the Cauchy principal value,Cn,αis a positive constant.For the information about the fractional Laplacian can refer to[1,2].

Caffarelli and Silvestre in[3]put forward the extension method which can reduce the nonlocal problem relating to (-Δ)α/2in Rnto a local one in higher dimensions in.For a functionu:Rn →R,define the extensionU:Rn×[0,∞)→R that satisfies

they showed that

This method has been applied to deal with equations involving the fractional Laplacian and fruitful results have been obtained,see[4-7]and the references therein.

We use the above extension method to reduce the nonlocal problem (1.1) into a local one and apply the method of moving planes to the local problem

There are many results about the Laplace equation with nonlinear boundary conditions problem

In [8],Hu proved that (1.3) has no positive solution for.This nonexistence result have been extended toby Ou in [9],and the form of solution forp=also has been obtained.Zhu considered the relative problem of (1.3) in [10],given the Liouville type theorems about the nonnegative solution to some indefinite elliptic equations.In [11],Lou and Zhu shown the classification of the solutions of problem (1.3).Their ideas lead us to study the problem (1.1) by its extension problem (1.2).The main method in those above papers is the method of moving planes which comes from Gidas,Ni and Nirenberg[12],and then this method have been to handle to many other types Laplace equations to get their properties of the solutions,can refer to[13-16]and their references therein.Cai and Ma [17] studied the radially symmetry properties of positive solutions to the fractional Laplace equation with negative powers,which add the growth/decay property of the solution.In this paper,by the extension method the condition can be delete.

Our main results are the following:

Theorem 1.1.(i) For,there is no positive solution to problem(1.1).

(ii) For,non-negative u is symmetric.

The conclusions of Theorem 1.1 come from directly for the next theorem.

Theorem 1.2.(i) For,there is no positive solution to problem(1.2).

(ii) For,non-negative U is symmetric about an axis parallel to the y-axis.

The paper is organized as follows.Section 2 give the proofs of Theorem 1.1 and Theorem 1.2,the tools are extension method and the method of moving planes.As a byproduct the form of solution of problem (1.2) is shown in Section 3,when.

2 Properties of solutions

For the problem (1.2),we first perform Kelvin transform on a solution of it with respect to a point onThis gives the solution a decay at infinity at some rate as approaches infinity and satisfies a nonlinear boundary condition except at one possible singular point.

LetUbe a solution of (1.2) and letvbe its Kelvin transform with respect to the origin,

whereX=(x,y)=(x1,···,xn,y).It is easy to check thatvsatisfies (see also[4])

Forλ∈R,we define

Next,based on the Kelvin transform of (2.1),a key theorem will be given which will be used in the process of moving planes.The first part provide the begin point of the moving plane.The second part is a narrow region maximum principle.

Theorem 2.1.The following is valid for v:

The above theorem following the proofs of the next two lemmas.

Lemma 2.1.Assume that v satisfies the equation and boundary condition of problem(2.2).

by virtue of the maximum principle.It follows thatvp(P0)≥ν0,thatv(P0)＞φ1(P0),we derive a contradiction.Sov-φ1＞0 onfor some smallr.The proof of (i) is completed by taking the limit forr→0.

(ii) The (2.1) implies that there is a positive constantν1such that

Assumevλ(X)-v(X) is negative somewhere in the exterior domain of|X|≥R0,thenwλ(X) necessarily takes on its negative minimum at a point on{X|x1≤λ,y=0,|X|≥R0}.Noting that at this point|X|＞||Xλ||andvλ(X)＜v(X).

at the point,a contradiction to the Hopf maximum principle.

Similarly,for 0＜p＜1 and at the minimum point ofwλ(X),

and forp≤0 and at the minimum ofwλ(X),

we get the same contradiction.

Next we give the proof of Theorem 2.1(i) firstly.

Proof.Let

By Lemma 2.1 (i),we obtainv(X)≥ν2in the half ball which except the origin.Since limX→∞v(X)=0,there is a constantNsufficiently large such thatv(X)≤ν2=vλ(X)=v(Xλ) in the half ballfor allλ ≤-N.Consequently,by Lemma 2.1(ii),v(X)≤vλ(Xfor allXin Σλ.

Lemma 2.2.Suppose λ0＜0and

Proof.(i) For anyε＞0,consider the auxiliary function

Noting thatwεsatisfies the equation in the half ball,except the origin.And on the lower boundary,except the origin,we have

By the Hopf maximum principle,wε≥0 andvλ0(X)-v(X)≥ν3in the half ball,except the origin.

(ii) The Hopf maximum principle works directly for points on the upper half plane satisfyingy＞0.For the points on the planey=0,the principle also works.We only to check that the function

for appropriately chosen numbersε,a,ρa(bǔ)nd point (z1,0,···,0,0),is a super solution of the equation on a half annulus and satisfies

on the intersection of the annulus with.By the maximum principle,Fis nonnegative in the half annulus andat the boundary point,which implies (ii).

Now we prove Theorem 2.1(ii).

Theorem 2.1(i) shows that forλsufficiently negative,we have

This provides a starting point to move the plane.Then,we move the plane to the right as long as inequality (2.5) holds to its limiting position by the Theorem 2.1(ii).As the same process,we can move the plane from the right hand to the left and stop at the limiting position,which obtain thatvλ(X)≤v(X) in Σλ.Hence,we havevλ(X)=v(X) about the limiting plane.

In the last part of this section,we give the proof of Theorem 1.2.

Proof.(i) For,by Theorem 2.1,vis symmetric abouty-axis.Uis also symmetric about they-axis.Because we can perform the Kelvin transform with respect to any point ony=0,the same proof implies thatUis symmetric any axis parallel to they-axis.HenceUmust be dependent onyonly and be identically equal to zero,a contradiction.That is,there is no solution of (1.2).

The same proof implies that the Kelvin transform ofUwith respect to any point onis symmetric about an axis parallel to they-axis.

Because the (1.2) is the extension of (1.1),the Theorem 1.1 following from Theorem 1.2.

3 Solution of (1.2) for p=

In this section,we prove that Theorem 1.2 (ii) implies the symmetry ofUhas a specific form.

Lemma 3.1.There is a positive constantμ＞0such that

Also,U is symmetric about an axis parallel to the y-axis.

Proof.Direct calculation shows that,

We claim that (3.1) is valid forμ=v(0).Noting that (3.1) is invariant of the translation of the coordinate system,we have

for any pointZon the planey=0.Thus (3.1) is valid no matter whetherUis symmetric about they-axis or not.

By the Kelvin transform in (3.1),we can apply the method of moving planes toU,thenUis symmetric about an axis parallel to they-axis.

Now assume thatUis symmetric about they-axis.It is easy to check that

are also solutions of (1.2) forand symmetric about axes that pass through points on they-axis.

Lemma 3.2.There is an s＞0such that vs is symmetric about the axis that passes through the point(,0,···,0,0)and is parallel to the y-axis.Proof.LetXt=(t,0,···,0,0) and define

as the 1-dimensional functions generated byUandvsrespectively.Because of the monotonicity property associated with the method of moving planes,we need only to check that there is ans＞0 such that

In fact,a direct computation shows that

Sincevssatisfies the equation in the upper half space and decays at infinity at the rate|X|α-n,by the maximum principle,

That is,vsis a fundamental solution of the Laplace equation multiplied by a constant.Since the Kelvin transform of a fundamental solution of the Laplace equation is still a fundamental solution,Usis also a fundamental solution multiplied by a constant.Hence,Umust be the form in Theorem 3.1.

Acknowledgement

This work is supported by the Key Scientific Research Project of the Colleges and Universities in Henan Province (No.22A110013),the Key Specialized Research and Development Breakthrough Program in Henan Province (No.222102310265),the Natural Science Foundation of Henan Province of China (No.222300420499) and the Cultivation Foundation of National Natural Science Foundation of Huanghuai University (No.XKPY-202008).