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Growth estimates for modified Neumann integrals in a half-space
Journal of Inequalities and Applications volume 2013, Article number: 572 (2013)
Our aim in this paper is to deal with the growth properties for modified Neumann integrals in a half-space of . 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 be the sets of all real numbers and of all positive real numbers, respectively. Let () denote the n-dimensional Euclidean space with points , where and . The boundary and closure of an open set Ω of are denoted by ∂ Ω and , respectively. For and , let denote the open ball with center at x and radius r in .
The upper half-space is the set , whose boundary is ∂H. For a set F, , we denote and by HF and , respectively. We identify with and with , writing typical points as , , where . Let θ be the angle between x and , i.e., and , where is the i th unit coordinate vector and is normal to ∂H.
We shall say that a set has a covering if there exists a sequence of balls with centers in H such that , where is the radius of and is the distance between the origin and the center of .
For positive functions and , we say that if 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, is the integer part of d and , 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
for every point .
For and , consider the kernel function
where and is the surface area of the n-dimensional unit sphere. It has the expression
where is the ultraspherical (Gegenbauer) polynomials . The series converges for , and each term in it is a harmonic function of x.
The Neumann integral is defined by
where f is a continuous function on ∂H, and is the volume of the unit n-ball.
The Neumann integral is a solution of the Neumann problem on H with f if (see [, Theorem 1 and Remarks])
In this paper, we consider functions f satisfying
for and .
For this p and α, we define the positive measure μ on by
If f is a measurable function on ∂H satisfying (1.1), we remark that the total mass of μ is finite.
Let and . For each , the maximal function is defined by
The set is denoted by .
To obtain the Neumann solution for the boundary data f, as in [3–6], we use the following modified kernel function defined by
for a non-negative integer m.
For and , the generalized Neumann kernel is defined by
Since () is harmonic in H (see ), is also harmonic in H for any fixed . Also, will be of order as (see [, Theorem D]).
where f is a continuous function on ∂H. Here, note that is nothing but the Neumann integral .
The following result is due to Siegel and Talvila (see [, Corollary 2.1]). For similar results with respect to the Schrödinger operator in a half-space, we refer readers to papers by Su (see ).
Theorem A If f is a continuous function on ∂H satisfying (1.1) with and , then
The next result deals with a type of uniqueness of solutions for the Neumann problem on H (see [, 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
and h is a solution of the Neumann problem on H with f such that
for any , where is the positive part of h,
and is a polynomial of of degree less than .
Our first aim is to be concerned with the growth property of at infinity and establish the following theorem.
Theorem 1 Let , , and
If f is a measurable function on ∂ satisfying (1.1), then there exists a covering of (⊂H) satisfying
Corollary 1 Let , and
If f is a measurable function on ∂H satisfying (1.1), then
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 is a solution of the Neumann problem on H with f and (1.4) holds, where the exceptional set (⊂H) has a covering satisfying (1.3).
Remark In the case , and , then (1.3) is a finite sum and the set 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 , and
If f is a continuous function on ∂H satisfying (1.1), then the function 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 and .
Theorem 3 Let , , l be a positive integer and
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
for any and is a polynomial of of degree less than .
In our discussions, the following estimates for the kernel function are fundamental (see [, Lemma 4.2] and [, Lemmas 2.1 and 2.4]).
If , then .
If , then .
If , then .
If and , then .
The following lemma is due to Qiao (see ).
Lemma 2 If , and λ is a positive measure in satisfying , then has a covering () such that
Lemma 3 ([, 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
for any fixed .
Lemma 4 ([, Lemma 1])
If is a harmonic polynomial of of degree m and vanishes on ∂H, then there exists a polynomial of degree m such that
3 Proof of Theorem 1
For any , there exists such that
Take any point such that , and write
First note that
If and , then . By Lemma 1(1), (3.1) and the Hölder inequality, we have
If , then
Moreover, by (3.1) and (3.3), we get
By Lemma 1(3), (3.1) and the Hölder inequality, we have
If , then . We obtain, by Lemma 1(4), (3.1) and the Hölder inequality,
Finally, we shall estimate . Take a sufficiently small positive number b such that for any , where
and divide H into two sets and .
If , then there exists a positive number such that for any , and hence
which is similar to the estimate of .
We shall consider the case . Now put
Since , we have
where is a positive integer satisfying .
Similar to the estimate of , we obtain
Since , we have
Combining (3.2), (3.4)-(3.7), we obtain that if is sufficiently large and ϵ is a sufficiently small number, then as , where . Finally, there exists an additional finite ball covering , which together with Lemma 2, gives the conclusion of Theorem 1.
4 Proof of Theorem 2
For any fixed , take a number R satisfying . If , then . By (1.1), Lemma 1(4) and the Hölder inequality, we have
Hence is absolutely convergent and finite for any . Thus is harmonic on H.
for any point , we only need to apply Lemma 3 to and .
We complete the proof of Theorem 2.
5 Proof of Theorem 3
Consider the function . Then it follows from Theorems 2 and 3 that is a solution of the Neumann problem on H with f and it is an even function of (see [, p.92]).
for any , and
from Theorem 2.
Moreover, (1.6) gives that
This implies that is a polynomial of degree less than (see [, Appendix]), which gives the conclusion of Theorem 3 from Lemma 4.
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The authors are thankful to the referees for their helpful suggestions and necessary corrections in the completion of this paper.
The authors declare that there is no conflict of interests regarding the publication of this article.
All authors contributed equally to the manuscript and read and approved the final manuscript.
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Ren, Y., Yang, P. Growth estimates for modified Neumann integrals in a half-space. J Inequal Appl 2013, 572 (2013). https://doi.org/10.1186/1029-242X-2013-572
- Dirichlet problem
- harmonic function