Open Access

Growth estimates for modified Neumann integrals in a half-space

Journal of Inequalities and Applications20132013:572

https://doi.org/10.1186/1029-242X-2013-572

Received: 13 August 2013

Accepted: 23 October 2013

Published: 4 December 2013

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.

Keywords

Dirichlet problem harmonic function half-space

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 ( n 3 ) 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 { x H ; | x | F } and { x H ; | x | F } by HF and H F , 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 | cos θ and 0 θ < 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 E H 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 / ( n 2 ) 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 1 p < and α R .

For this p and α, we define the positive measure μ on R n by
d μ ( 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 δ 0 . 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 ) ( m 0 ) .

Since | x | k C k n 2 2 ( x y | x | | y | ) ( k 0 ) 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 1 p < , 0 β ( n 2 ) p , n + α 2 > ( n 1 ) ( p 1 ) 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 , ( n 2 ) p β ) (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 + α 2 > ( n 1 ) ( p 1 ) 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 , ( n 2 ) p β ) (H) has a covering { r j , R j } satisfying (1.3).

Remark In the case p = 1 , α = m and β = n 2 , 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 1 p < , n + α 2 > ( n 1 ) ( p 1 ) 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 1 p < , α > 1 p , 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 , η 0 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 x H ( R ϵ , ) E ( ϵ ; μ , ( n 2 ) p β ) 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 ( 3 n m + n + α 2 p ) q + n 1 > 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 ( 2 n m + n + α 2 p ) q + n 1 < 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 x Π ( b ) , where
Π ( b ) = { x H ; inf y H | x | x | y | y | | < b }

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

If x H Π ( 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 x Π ( 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 x E ( ϵ ; μ , ( n 2 ) p β ) , 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 x H ( R ϵ , + ) E ( ϵ ; μ , ( n 2 ) p β ) . 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 x H , take a number R satisfying R > max { 1 , 2 | x | } . If m > 1 α p , then ( 2 n m + n + α 2 p ) q + n 1 < 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 x H . 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 x H , 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.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics and Information Science, Henan University of Economics and Law
(2)
College of Applied Mathematics, Chengdu University of Information Technology

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© Ren and Yang; licensee Springer. 2013

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