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Growth estimates for modified Neumann integrals in a half-space

Abstract

Our aim in this paper is to deal with the growth properties for modified Neumann integrals in a half-space of R n . As an application, the solutions of Neumann problems in it for a slowly growing continuous function are also given.

1 Introduction and main results

Let R and R + be the sets of all real numbers and of all positive real numbers, respectively. Let R n (n3) denote the n-dimensional Euclidean space with points x=( x , x n ), where x =( x 1 , x 2 ,, x n 1 ) R n 1 and x n R. The boundary and closure of an open set Ω of R n are denoted by Ω and Ω ¯ , respectively. For x R n and r>0, let B n (x,r) denote the open ball with center at x and radius r in R n .

The upper half-space is the set H={( x , x n ) R n : x n >0}, whose boundary is ∂H. For a set F, F R + {0}, we denote {xH;|x|F} and {xH;|x|F} by HF and HF, respectively. We identify R n with R n 1 ×R and R n 1 with R n 1 ×{0}, writing typical points x,y R n as x=( x , x n ), y=( y , y n ), where y =( y 1 , y 2 ,, y n 1 ) R n 1 . Let θ be the angle between x and e ˆ n , i.e., x n =|x|cosθ and 0θ<π/2, where e ˆ n is the i th unit coordinate vector and e ˆ n is normal to ∂H.

We shall say that a set EH has a covering { r j , R j } if there exists a sequence of balls { B j } with centers in H such that E j = 0 B j , where r j is the radius of B j and R j is the distance between the origin and the center of B j .

For positive functions g 1 and g 2 , we say that g 1 g 2 if g 1 M g 2 for some positive constant M. Throughout this paper, let M denote various constants independent of the variables in question. Further, we use the standard notations, [d] is the integer part of d and d=[d]+{d}, where d is a positive real number.

Given a continuous function f in ∂H, we say that h is a solution of the Neumann problem in H with f, if h is a harmonic function in H and

lim x H , x y x n h(x)=f ( y )

for every point y H.

For x R n and y R n 1 , consider the kernel function

K n ( x , y ) = β n | x y | n 2 ,

where β n =2/(n2) σ n and σ n is the surface area of the n-dimensional unit sphere. It has the expression

K n ( x , y ) = k = 0 | x | k | y | n + k 2 C k n 2 2 ( x y | x | | y | ) ,

where C k n 2 (t) is the ultraspherical (Gegenbauer) polynomials [1]. The series converges for | y |>|x|, and each term in it is a harmonic function of x.

The Neumann integral is defined by

N[f](x)= H K n ( x , y ) f ( y ) d y ,

where f is a continuous function on ∂H, α n =2/n σ n and σ n = π n 2 /Γ(1+ n 2 ) is the volume of the unit n-ball.

The Neumann integral N[f](x) is a solution of the Neumann problem on H with f if (see [[2], Theorem 1 and Remarks])

H f ( y ) ( 1 + | y | ) n 2 d y <.

In this paper, we consider functions f satisfying

H | f ( y ) | p ( 1 + | y | ) n + α 2 d y <
(1.1)

for 1p< and αR.

For this p and α, we define the positive measure μ on R n by

dμ ( y ) = { | f ( y ) | p | y | n α + 2 d y , y H ( 1 , + ) , 0 , Q R n H ( 1 , + ) .

If f is a measurable function on ∂H satisfying (1.1), we remark that the total mass of μ is finite.

Let ϵ>0 and δ0. For each x R n , the maximal function M(x;μ,δ) is defined by

M(x;μ,δ)= sup 0 < ρ < | x | 2 μ ( B n ( x , r ) ) ρ δ .

The set {x R n ;M(x;μ,δ)>ϵ} is denoted by E(ϵ;μ,δ).

To obtain the Neumann solution for the boundary data f, as in [36], we use the following modified kernel function defined by

L n , m ( x , y ) = { β n k = 0 m 1 | x | k | y | n + k 2 C k n 2 2 ( x y | x | | y | ) , | y | 1 m 1 , 0 , | y | < 1 m 1 , 0 , m = 0

for a non-negative integer m.

For x R n and y R n 1 , the generalized Neumann kernel is defined by

K n , m ( x , y ) = K n ( x , y ) L n , m ( x , y ) (m0).

Since |x | k C k n 2 2 ( x y | x | | y | ) (k0) is harmonic in H (see [4]), K n , m (, y ) is also harmonic in H for any fixed y H. Also, K n , m (x, y ) will be of order | y | ( n + m 2 ) as y (see [[7], Theorem D]).

Put

N m [f](x)= H K n , m ( x , y ) f ( y ) d y ,

where f is a continuous function on ∂H. Here, note that N 0 [f](x) is nothing but the Neumann integral N[f](x).

The following result is due to Siegel and Talvila (see [[5], Corollary 2.1]). For similar results with respect to the Schrödinger operator in a half-space, we refer readers to papers by Su (see [8]).

Theorem A If f is a continuous function on ∂H satisfying (1.1) with p=1 and α=m, then

lim | x | , x H N m [f](x)=o ( | x | m sec n 2 θ ) .
(1.2)

The next result deals with a type of uniqueness of solutions for the Neumann problem on H (see [[9], Theorem 3]).

Theorem B Let l be a positive integer and m be a non-negative integer. If f is a continuous function on ∂H satisfying

H | f ( y ) | ( 1 + | y | ) n + m 2 d y <,

and h is a solution of the Neumann problem on H with f such that

lim | x | , x H h + (x)=o ( | x | l + m ) ,

then

h(x)= N m [f](x)+Π ( x ) + j = 1 [ l + m 2 ] ( 1 ) j ( 2 j ) ! x n 2 j Δ j Π ( x )

for any x=( x , x n )H, where h + (x) is the positive part of h,

Δ j = ( 2 x 1 2 + 2 x 2 2 + + 2 x n 1 2 ) (j=1,2)

and Π( x ) is a polynomial of x R n 1 of degree less than l+m.

Our first aim is to be concerned with the growth property of N m [f] at infinity and establish the following theorem.

Theorem 1 Let 1p<, 0β(n2)p, n+α2>(n1)(p1) and

1 1 α p < m < 2 1 α p if  p > 1 , α m < α + 1 if  p = 1 .

If f is a measurable function on satisfying (1.1), then there exists a covering { r j , R j } of E(ϵ;μ,(n2)pβ) (H) satisfying

j = 0 ( r j R j ) ( n 2 ) p β <
(1.3)

such that

lim | x | , x H E ( ϵ ; μ , ( n 2 ) p β ) N m [f](x)=o ( | x | 1 + α 1 p sec β p θ ) .
(1.4)

Corollary 1 Let 1<p<, n+α2>(n1)(p1) and

1 1 α p <m<2 1 α p .

If f is a measurable function on ∂H satisfying (1.1), then

lim | x | , x H N m [f](x)=o ( | x | 1 + α 1 p sec n 2 θ ) .
(1.5)

As an application of Theorem 1, we now show the solution of the Neumann problem with continuous data on H.

Theorem 2 Let p, β, α and m be defined as in Theorem  1. If f is a continuous function on ∂H satisfying (1.1), then the function N m [f] is a solution of the Neumann problem on H with f and (1.4) holds, where the exceptional set E(ϵ;μ,(n2)pβ) (H) has a covering { r j , R j } satisfying (1.3).

Remark In the case p=1, α=m and β=n2, then (1.3) is a finite sum and the set E(ϵ;μ,0) is a bounded set. So (1.4) holds in H. That is to say, (1.2) holds. This is just the result of Theorem A.

Corollary 2 Let 1p<, n+α2>(n1)(p1) and

1 1 α p < m < 2 1 α p if  p > 1 , α m < α + 1 if  p = 1 .

If f is a continuous function on ∂H satisfying (1.1), then the function N m [f] is a solution of the Neumann problem on H with f and (1.5) holds.

The following result extends Theorem B, which is our result in the case p=1 and α=m.

Theorem 3 Let 1p<, α>1p, l be a positive integer and

1 1 α p < m < 2 1 α p if  p > 1 , α m < α + 1 if  p = 1 .

If f is a continuous function on ∂H satisfying (1.1) and h is a solution of the Neumann problem on H with f such that

lim | x | , x H h + (x)=o ( | x | l + [ 1 + α 1 p ] ) ,
(1.6)

then

h(x)= N m [f](x)+Π ( x ) + j = 1 [ l + [ 1 + α 1 p ] 2 ] ( 1 ) j ( 2 j ) ! x n 2 j Δ j Π ( x )
(1.7)

for any x=( x , x n )H and Π( x ) is a polynomial of x R n 1 of degree less than l+[1+ α 1 p ].

2 Lemmas

In our discussions, the following estimates for the kernel function K n , m (x, y ) are fundamental (see [[10], Lemma 4.2] and [[4], Lemmas 2.1 and 2.4]).

Lemma 1

  1. (1)

    If 1| y | | x | 2 , then | K n , m (x, y )||x | m 1 | y | n m + 3 .

  2. (2)

    If | x | 2 <| y | 3 2 |x|, then | K n , m (x, y )||x y | 2 n .

  3. (3)

    If 3 2 |x|<| y |2|x|, then | K n , m (x, y )| x n 2 n .

  4. (4)

    If | y |2|x| and | y |1, then | K n , m (x, y )||x | m | y | 2 n m .

The following lemma is due to Qiao (see [4]).

Lemma 2 If ϵ>0, η0 and λ is a positive measure in R n satisfying λ( R n )<, then E(ϵ;λ,η) has a covering { r j , R j } (j=1,2,) such that

j = 1 ( r j R j ) η <.

Lemma 3 ([[9], Lemma 4])

Let p, β, α and m be defined as in Theorem  1. If f is a locally integral and upper semi-continuous function on ∂H satisfying (1.1), then

lim sup x H , x y x n N m [f](x)f ( y )

for any fixed y H.

Lemma 4 ([[2], Lemma 1])

If h(x) is a harmonic polynomial of x=( x , x n )H of degree m and h/ x n vanishes on ∂H, then there exists a polynomial Π( x ) of degree m such that

h(x)= { Π ( x ) + j = 1 [ m 2 ] ( 1 ) j ( 2 j ) ! x n 2 j Δ j Π ( x ) , m 2 , Π ( x ) , m = 0 , 1 .

3 Proof of Theorem 1

For any ϵ>0, there exists R ϵ >1 such that

H ( R ϵ , ) | f ( y ) | p ( 1 + | y | ) n + α 2 d y <ϵ.
(3.1)

Take any point xH( R ϵ ,)E(ϵ;μ,(n2)pβ) such that |x|>2 R ϵ , and write

N m [ f ] ( x ) = ( G 1 + G 2 + G 3 + G 4 + G 5 ) K n , m ( x , y ) f ( y ) d y = U 1 ( x ) + U 2 ( x ) + U 3 ( x ) + U 4 ( x ) + U 5 ( x ) ,

where

G 1 = { y H : | y | 1 } , G 2 = { y H : 1 < | y | | x | 2 } , G 3 = { y H : | x | 2 < | y | 3 2 | x | } , G 4 = { y H : 3 2 | x | < | y | 2 | x | } G 5 = { y H : | y | 2 | x | } .

First note that

| U 1 ( x ) | G 1 | f ( y ) | | x y | n 2 d y | x | 2 n G 1 | f ( y ) | d y ,

so that

lim | x | , x H |x | 1 + 1 α p U 1 (x)=0.
(3.2)

If m<2 1 α p and 1 p + 1 q =1, then (3nm+ n + α 2 p )q+n1>0. By Lemma 1(1), (3.1) and the Hölder inequality, we have

| U 2 ( x ) | | x | m 1 G 2 | y | n m + 3 | f ( y ) | d y | x | m 1 ( G 2 | f ( y ) | p | y | n + α 2 d y ) 1 p ( G 2 | y | ( n m + 3 + n + α 2 p ) q d y ) 1 q | x | 1 1 α p ( G 2 | f ( y ) | p | y | n + α 2 d y ) 1 p .
(3.3)

Put

U 2 (x)= U 21 (x)+ U 22 (x),

where

U 21 ( x ) = G 2 B n 1 ( R ϵ ) K n , m ( x , y ) f ( y ) d y , U 22 ( x ) = G 2 B n 1 ( R ϵ ) K n , m ( x , y ) f ( y ) d y .

If |x|2 R ϵ , then

| U 21 (x)| R ϵ 2 m 1 α p |x | m 1 .

Moreover, by (3.1) and (3.3), we get

| U 22 (x)|ϵ|x | 1 1 α p .

That is,

| U 2 (x)|ϵ|x | 1 1 α p .
(3.4)

By Lemma 1(3), (3.1) and the Hölder inequality, we have

| U 4 (x)|ϵ x n 2 n |x | n 1 1 α p .
(3.5)

If m>1 1 α p , then (2nm+ n + α 2 p )q+n1<0. We obtain, by Lemma 1(4), (3.1) and the Hölder inequality,

| U 5 ( x ) | | x | m G 5 | y | n m + 2 | f ( y ) | d y | x | m ( G 5 | f ( y ) | p | y | n + α 2 d y ) 1 p ( G 5 | y | ( n m + 2 + n + α 2 p ) q d y ) 1 q ϵ | x | 1 1 α p .
(3.6)

Finally, we shall estimate U 3 (x). Take a sufficiently small positive number b such that H[ | x | 2 , 3 2 |x|]B(x, | x | 2 ) for any xΠ(b), where

Π(b)= { x H ; inf y H | x | x | y | y | | < b }

and divide H into two sets Π(b) and HΠ(b).

If xHΠ(b), then there exists a positive number b such that |x y | b |x| for any y H, and hence

| U 3 ( x ) | G 3 | y | 2 n | f ( y ) | d y | x | m G 3 | y | 2 n m | f ( y ) | d y ϵ | x | 1 1 α p ,

which is similar to the estimate of U 5 (x).

We shall consider the case xΠ(b). Now put

H i (x)= { y H [ | x | 2 , 3 2 | x | ] ; 2 i 1 δ ( x ) | x y | < 2 i δ ( x ) } ,

where δ(x)= inf y H |x y |.

Since H{ y R n 1 :|x y |<δ(x)}=, we have

U 3 (x)= i = 1 i ( x ) H i ( x ) | g ( y ) | | x y | n 2 d y ,

where i(x) is a positive integer satisfying 2 i ( x ) 1 δ(x) | x | 2 < 2 i ( x ) δ(x).

Similar to the estimate of U 5 (x), we obtain

H i ( x ) | g ( y ) | | x y | n 2 d y H i ( x ) | g ( y ) | { 2 i 1 δ ( x ) } n 2 d y δ ( x ) β ( n 2 ) p p H i ( x ) δ ( x ) ( n 2 ) p β p n + 2 | g ( y ) | d y cos β p θ δ ( x ) β ( n 2 ) p p H i ( x ) | x | β p | g ( y ) | d y | x | n 2 β p cos β p θ δ ( x ) β ( n 2 ) p p H i ( x ) | y | 2 n | g ( y ) | d y | x | n 1 + α β 1 p cos β p θ ( μ ( H i ( x ) ) 2 i δ ( x ) ( n 2 ) p β ) 1 p

for i=0,1,2,,i(x).

Since xE(ϵ;μ,(n2)pβ), we have

μ ( H i ( x ) ) { 2 i δ ( x ) } ( n 2 ) p β μ ( B n 1 ( x , 2 i δ ( x ) ) ) { 2 i δ ( x ) } ( n 2 ) p β M ( x ; μ , ( n 2 ) p β ) ϵ|x | β ( n 2 ) p

for i=0,1,2,,i(x)1 and

μ ( H i ( x ) ( x ) ) { 2 i δ ( x ) } ( n 2 ) p β μ ( B n 1 ( x , | x | 2 ) ) ( | x | 2 ) ( n 2 ) p β ϵ|x | β ( n 2 ) p .

So

| U 3 (x)|ϵ|x | 1 + α 1 p sec β p θ.
(3.7)

Combining (3.2), (3.4)-(3.7), we obtain that if R ϵ is sufficiently large and ϵ is a sufficiently small number, then N m [f](x)=o(|x | 1 + α 1 p sec β p θ) as |x|, where xH( R ϵ ,+)E(ϵ;μ,(n2)pβ). Finally, there exists an additional finite ball B 0 covering H(0, R ϵ ], which together with Lemma 2, gives the conclusion of Theorem 1.

4 Proof of Theorem 2

For any fixed xH, take a number R satisfying R>max{1,2|x|}. If m> 1 α p , then (2nm+ n + α 2 p )q+n1<0. By (1.1), Lemma 1(4) and the Hölder inequality, we have

H ( R , ) | K n , m ( x , y ) | | f ( y ) | d y | x | m H ( R , ) | y | 2 n m | f ( y ) | d y | x | m ( H ( R , ) | f ( y ) | p | y | n + α 2 d y ) 1 p ( H ( R , ) | y | ( n m + 2 + n + α 2 p ) q d y ) 1 q < .

Hence N m [f](x) is absolutely convergent and finite for any xH. Thus N m [f](x) is harmonic on H.

To prove

lim x y , x H x n N m [f](x)=f ( y )

for any point y H, we only need to apply Lemma 3 to f(y) and f(y).

We complete the proof of Theorem 2.

5 Proof of Theorem 3

Consider the function h (x)=h(x) N m [f](x). Then it follows from Theorems 2 and 3 that h (x) is a solution of the Neumann problem on H with f and it is an even function of x n (see [[2], p.92]).

Since

0 { h N m [ f ] } + (x) h + (x)+ { N m [ f ] } (x)

for any xH, and

lim | x | , x H N m [f](x)=o ( | x | 1 + α 1 p )

from Theorem 2.

Moreover, (1.6) gives that

lim | x | , x H ( h N m [ f ] ) (x)=o ( | x | l + [ 1 + α 1 p ] ) .

This implies that h (x) is a polynomial of degree less than l+[1+ α 1 p ] (see [[11], Appendix]), which gives the conclusion of Theorem 3 from Lemma 4.

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Acknowledgements

The authors are thankful to the referees for their helpful suggestions and necessary corrections in the completion of this paper.

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Correspondence to Yudong Ren.

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All authors contributed equally to the manuscript and read and approved the final manuscript.

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Keywords

  • Dirichlet problem
  • harmonic function
  • half-space