R E S E A R C H
Open Access
Growth estimates for modified Neumann
integrals in a half-space
Yudong Ren
1*and Pai Yang
2*Correspondence:
1Department of Mathematics and
Information Science, Henan University of Economics and Law, Zhengzhou, 450002, P.R. China Full list of author information is available at the end of the article
Abstract
Our aim in this paper is to deal with the growth properties for modified Neumann integrals in a half-space ofRn. 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
LetRandR+be the sets of all real numbers and of all positive real numbers, respectively.
LetRn(n≥) denote then-dimensional Euclidean space with pointsx= (x,xn), where
x= (x,x, . . . ,xn–)∈Rn–andxn∈R. The boundary and closure of an open setofRn
are denoted by∂and, respectively. Forx∈Rnandr> , letBn(x,r) denote the open
ball with center atxand radiusrinRn.
The upper half-space is the setH={(x,xn)∈Rn:xn> }, whose boundary is∂H. For
a setF,F⊂R+∪ {}, we denote{x∈H;|x| ∈F}and{x∈∂H;|x| ∈F}byHFand∂HF,
respectively. We identifyRnwithRn–×RandRn–withRn–× {}, writing typical points
x,y∈Rnasx= (x,x
n),y= (y,yn), wherey= (y,y, . . . ,yn–)∈Rn–. Letθ be the angle
betweenxandˆen,i.e.,xn=|x|cosθ and ≤θ<π/, whereeˆnis theith unit coordinate
vector andˆenis normal to∂H.
We shall say that a setE⊂Hhas a covering{rj,Rj}if there exists a sequence of balls{Bj}
with centers inHsuch thatE⊂∞j=Bj, whererjis the radius ofBjandRjis the distance
between the origin and the center ofBj.
For positive functionsgandg, we say thatggifg≤Mgfor some positive
con-stantM. Throughout this paper, letMdenote various constants independent of the vari-ables in question. Further, we use the standard notations, [d] is the integer part ofdand d= [d] +{d}, wheredis a positive real number.
Given a continuous functionf in∂H, we say thathis a solution of the Neumann problem inHwithf, ifhis a harmonic function inHand
lim x∈H,x→y
∂ ∂xn
h(x) =fy
for every pointy∈∂H.
Forx∈Rnandy∈Rn–, consider the kernel function
Kn
x,y= – βn
|x–y|n–,
whereβn= /(n– )σnandσnis the surface area of then-dimensional unit sphere. It has
the expression
Kn
x,y=
∞
k= |x|k
|y|n+k–C
n– k
x·y
|x||y|
,
whereC n
k(t) is the ultraspherical (Gegenbauer) polynomials []. The series converges for
|y|>|x|, and each term in it is a harmonic function ofx. The Neumann integral is defined by
N[f](x) =
∂H
Kn
x,yfydy,
wheref is a continuous function on∂H,αn= /nσnandσn=π
n
/ ( +n
) is the volume of
the unitn-ball.
The Neumann integralN[f](x) is a solution of the Neumann problem onHwithf if (see [, Theorem and Remarks])
∂H
f(y) ( +|y|)n–dy
<∞.
In this paper, we consider functionsf satisfying
∂H
|f(y)|p
( +|y|)n+α–dy
<∞ (.)
for ≤p<∞andα∈R.
For thispandα, we define the positive measureμonRnby
dμy= |f(y
)|p|y|–n–α+dy, y∈∂H(, +∞),
, Q∈Rn–∂H(, +∞).
Iff is a measurable function on∂Hsatisfying (.), we remark that the total mass ofμis finite.
Let> andδ≥. For eachx∈Rn, the maximal functionM(x;μ,δ) is defined by
M(x;μ,δ) = sup <ρ<|x|
μ(Bn(x,r))
ρδ .
The set{x∈Rn;M(x;μ,δ) >}is denoted byE(;μ,δ).
To obtain the Neumann solution for the boundary dataf, as in [–], we use the follow-ing modified kernel function defined by
Ln,m
x,y=
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–βn
m–
k= | x|k
|y|n+k–C
n– k (
x·y
|x||y|), |y| ≥m≥,
, |y|< m≥,
, m=
Forx∈Rnandy∈Rn–, the generalized Neumann kernel is defined by
Kn,m
x,y=Kn
x,y–Ln,m
x,y (m≥).
Since|x|kCn– k (
x·y
|x||y|) (k≥) is harmonic inH (see []),Kn,m(·,y) is also harmonic inH
for any fixedy∈∂H. Also,Kn,m(x,y) will be of order|y|–(n+m–)asy→ ∞(see [,
The-orem D]). Put
Nm[f](x) =
∂H
Kn,m
x,yfydy,
wheref is a continuous function on∂H. Here, note thatN[f](x) is nothing but the
Neu-mann integralN[f](x).
The following result is due to Siegel and Talvila (see [, Corollary .]). 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(.)with p= andα=m,then
lim
|x|→∞,x∈HNm[f](x) =o
|x|msecn–θ. (.)
The next result deals with a type of uniqueness of solutions for the Neumann problem onH(see [, Theorem ]).
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)| ( +|y|)n+m–dy
<∞,
and h is a solution of the Neumann problem on H with f such that
lim |x|→∞,x∈Hh
+(x) =o|x|l+m,
then
h(x) =Nm[f](x) +
x+
[l+m]
j=
(–)j
(j)!x
j nj
x
for any x= (x,xn)∈H,where h+(x)is the positive part of h,
j=
∂ ∂x +
∂ ∂x+· · ·+
∂ ∂xn–
(j= , . . .)
Our first aim is to be concerned with the growth property ofNm[f] at infinity and
es-tablish the following theorem.
Theorem Let≤p<∞, ≤β≤(n– )p,n+α– > –(n– )(p– )and
– –α
p <m< – –α
p if p> ,
α≤m<α+ if p= .
If f is a measurable function on∂ satisfying(.),then there exists a covering{rj,Rj} of
E(;μ, (n– )p–β) (⊂H)satisfying
∞
j=
rj
Rj
(n–)p–β
<∞ (.)
such that
lim
|x|→∞,x∈H–E(;μ,(n–)p–β)Nm[f](x) =o
|x|+αp–secβpθ. (.)
Corollary Let <p<∞,n+α– > –(n– )(p– )and
– –α
p <m< – –α
p .
If f is a measurable function on∂H satisfying(.),then
lim
|x|→∞,x∈HNm[f](x) =o
|x|+α–p secn–θ. (.)
As an application of Theorem , we now show the solution of the Neumann problem with continuous data onH.
Theorem Let p,β,α and m be defined as in Theorem.If f is a continuous function on∂H satisfying(.),then the function Nm[f]is a solution of the Neumann problem on H
with f and(.)holds,where the exceptional set E(;μ, (n– )p–β) (⊂H)has a covering
{rj,Rj}satisfying(.).
Remark In the casep= ,α =mandβ=n– , then (.) is a finite sum and the set E(;μ, ) is a bounded set. So (.) holds inH. That is to say, (.) holds. This is just the result of Theorem A.
Corollary Let≤p<∞,n+α– > –(n– )(p– )and
– –α
p <m< – –α
p if p> ,
α≤m<α+ if p= .
If f is a continuous function on∂H satisfying(.),then the function Nm[f]is a solution of
The following result extends Theorem B, which is our result in the casep= andα=m. Theorem Let≤p<∞,α> –p,l be a positive integer and
– –α
p <m< – –α
p if p> ,
α≤m<α+ if p= .
If f is a continuous function on∂H satisfying(.)and h is a solution of the Neumann problem on H with f such that
lim |x|→∞,x∈Hh
+(x) =o|x|l+[+α–p ], (.)
then
h(x) =Nm[f](x) +
x+
[l+[+
α–
p ]
]
j=
(–)j
(j)!x
j nj
x (.)
for any x= (x,xn)∈H and(x)is a polynomial of x∈Rn–of degree less than l+ [ +α–p ].
2 Lemmas
In our discussions, the following estimates for the kernel functionKn,m(x,y) are
funda-mental (see [, Lemma .] and [, Lemmas . and .]).
Lemma
() If≤ |y| ≤|x|,then|Kn,m(x,y)||x|m–|y|–n–m+.
() If|x|<|y| ≤ |x|,then|Kn,m(x,y)||x–y|–n.
() If|x|<|y| ≤|x|,then|Kn,m(x,y)|x–n n.
() If|y| ≥|x|and|y| ≥,then|Kn,m(x,y)||x|m|y|–n–m.
The following lemma is due to Qiao (see []).
Lemma If> ,η≥andλis a positive measure inRnsatisfyingλ(Rn) <∞,then E(;λ,η)has a covering{rj,Rj}(j= , , . . .)such that
∞
j=
rj
Rj η
<∞.
Lemma ([, Lemma ]) Let p,β,αand m be defined as in Theorem.If f is a locally integral and upper semi-continuous function on∂H satisfying(.),then
lim sup x∈H,x→y
∂ ∂xn
Nm[f](x)≤f
y
Lemma ([, Lemma ]) If h(x)is a harmonic polynomial of x= (x,xn)∈H of degree m
and∂h/∂xnvanishes on∂H,then there exists a polynomial(x)of degree m such that
h(x) = (x
) +[m] j=
(–)j (j)!x
j
nj(x), m≥,
(x), m= , .
3 Proof of Theorem 1
For any> , there existsR> such that
∂H(R,∞)
|f(y)|p
( +|y|)n+α–dy
<. (.)
Take any pointx∈H(R,∞) –E(;μ, (n– )p–β) such that|x|> R, and write
Nm[f](x) = G + G + G + G + G
Kn,m
x,yfydy
=U(x) +U(x) +U(x) +U(x) +U(x),
where
G=
y∈∂H:y≤, G=
y∈∂H: <y≤|x|
,
G=
y∈∂H:|x| <y
≤
|x|
, G=
y∈∂H: |x|<y
≤|x|
G=
y∈∂H:y≥|x|.
First note that
U(x)
G
|f(y)|
|x–y|n–dy
|x|–n
G
fydy,
so that
lim |x|→∞,x∈H|x|
–+–pα
U(x) = . (.)
Ifm< ––α
p and p +
q = , then ( –n–m+ n+α–
p )q+n– > . By Lemma (), (.)
and the Hölder inequality, we have
U(x)|x|m–
G
y–n–m+fydy
|x|m–
G |f(y)|p
|y|n+α–dy
p
G
y(–n–m++ n+α–
p )qdy
q
|x|––pα
G |f(y)|p
|y|n+α–dy
p
Put
U(x) =U(x) +U(x),
where
U(x) =
G∩Bn–(R) Kn,m
x,yfydy,
U(x) =
G\Bn–(R) Kn,m
x,yfydy.
If|x| ≥R, then
U(x)R –m––pα |x|m–.
Moreover, by (.) and (.), we get
U(x)|x|– –α
p .
That is,
U(x)|x|––pα. (.)
By Lemma (), (.) and the Hölder inequality, we have
U(x)x–n n|x| n–––pα
. (.)
Ifm> ––pα, then ( –n–m+n+αp–)q+n– < . We obtain, by Lemma (), (.) and the Hölder inequality,
U(x)|x|m
G
y–n–m+fydy
|x|m
G |f(y)|p
|y|n+α– dy
p
G
y(–n–m++n+αp–)qdy
q
|x|––pα. (.)
Finally, we shall estimate U(x). Take a sufficiently small positive numberb such that
∂H[|x|,|x|]⊂B(x,|x|) for anyx∈(b), where
(b) =
x∈H; inf y∈∂H
|xx|– y
|y|
<b
Ifx∈H–(b), then there exists a positive numberbsuch that|x–y| ≥b|x|for any y∈∂H, and hence
U(x)
G
y–nfydy
|x|m
G
y–n–m
fydy
|x|––pα,
which is similar to the estimate ofU(x).
We shall consider the casex∈(b). Now put
Hi(x) =
y∈∂H
|
x| ,
|x|
; i–δ(x)≤x–y< iδ(x)
,
whereδ(x) =infy∈H|x–y|.
Since∂H∩ {y∈Rn–:|x–y|<δ(x)}=∅, we have
U(x) = i(x)
i=
Hi(x)
|g(y)|
|x–y|n–dy ,
wherei(x) is a positive integer satisfying i(x)–δ(x)≤|x|
< i(x)δ(x).
Similar to the estimate ofU(x), we obtain
Hi(x)
|g(y)|
|x–y|n–dy
Hi(x)
|g(y)|
{i–δ(x)}n–dy
δ(x)β
–(n–)p p
Hi(x)
δ(x)
(n–)p–β
p –n+gydy
cos–βpθ δ(x) β–(n–)p
p
Hi(x)
|x|–βpgydy
|x|n––βpcos– β pθ δ(x)
β–(n–)p p
Hi(x)
y–ngydy
|x|n–+α–pβ–cos–βpθ
μ(Hi(x))
iδ(x)(n–)p–β
p
fori= , , , . . . ,i(x).
Sincex∈/E(;μ, (n– )p–β), we have
μ(Hi(x))
{iδ(x)}(n–)p–β
μ(Bn–(x, iδ(x))) {iδ(x)}(n–)p–β M
x;μ, (n– )p–β|x|β–(n–)p
fori= , , , . . . ,i(x) – and
μ(Hi(x)(x)) {iδ(x)}(n–)p–β
μ(Bn–(x,|x|))
So
U(x)|x|+
α–
p secβpθ. (.)
Combining (.), (.)-(.), we obtain that ifRis sufficiently large andis a sufficiently small number, then Nm[f](x) =o(|x|+
α–
p sec β
pθ) as |x| → ∞, wherex∈H(R, +∞) – E(;μ, (n– )p–β). Finally, there exists an additional finite ballBcoveringH(,R], which
together with Lemma , gives the conclusion of Theorem .
4 Proof of Theorem 2
For any fixedx∈H, take a numberRsatisfyingR>max{, |x|}. Ifm>–pα, then ( –n– m+n+αp–)q+n– < . By (.), Lemma () and the Hölder inequality, we have
∂H(R,∞) Kn,m
x,yfydy
|x|m
∂H(R,∞)
y–n–mfydy
|x|m
∂H(R,∞)
|f(y)|p
|y|n+α–dy
p
∂H(R,∞)
y(–n–m++ n+α–
p )qdy
q
<∞.
HenceNm[f](x) is absolutely convergent and finite for anyx∈H. ThusNm[f](x) is
har-monic onH. To prove
lim x→y,x∈H
∂ ∂xn
Nm[f](x) =f
y
for any pointy∈∂H, we only need to apply Lemma tof(y) and –f(y). We complete the proof of Theorem .
5 Proof of Theorem 3
Consider the functionh(x) =h(x) –Nm[f](x). Then it follows from Theorems and that
h(x) is a solution of the Neumann problem onHwithf and it is an even function ofxn
(see [, p.]). Since
≤h–Nm[f] +
(x)≤h+(x) +Nm[f] –
(x)
for anyx∈H, and
lim
|x|→∞,x∈HNm[f](x) =o
|x|+α–p
from Theorem .
Moreover, (.) gives that
lim |x|→∞,x∈H
h–Nm[f]
This implies thath(x) is a polynomial of degree less thanl+ [ +αp–] (see [, Appendix]), which gives the conclusion of Theorem from Lemma .
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Authors’ contributions
All authors contributed equally to the manuscript and read and approved the final manuscript.
Author details
1Department of Mathematics and Information Science, Henan University of Economics and Law, Zhengzhou, 450002, P.R.
China.2College of Applied Mathematics, Chengdu University of Information Technology, Chengdu, 610225, P.R. China.
Acknowledgements
The authors are thankful to the referees for their helpful suggestions and necessary corrections in the completion of this paper.
Received: 13 August 2013 Accepted: 23 October 2013 Published:04 Dec 2013
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