# Dirichlet Problem for the Schrödinger Operator in a Half Space

Authors: Baiyun Su
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

For continuous boundary data, the modified Poisson integral is used to write solutions to the half space Dirichlet problem for the Schrödinger operator. Meanwhile, a solution of the Poisson integral for any continuous boundary function is also given explicitly by the Poisson integral with the generalized Poisson kernel depending on this boundary function.


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## Body

## 1. Introduction and Results

LetR and R+ be the sets of all real numbers and of all positive real numbers, respectively. Let Rn(n≥2) denote the n-dimensional Euclidean space with points x=(x',xn), where x'=(x1,x2,…,xn-1)∈Rn-1 and xn∈R. The unit sphere and the upper half unit sphere in Rn are denoted by Sn-1 and S+n-1, respectively. The boundary and closure of an open set D of Rn are denoted by ∂D and D¯, respectively. The upper half space is the set H={(x',xn)∈Rn:xn>0}, whose boundary is ∂H.For a setE, E⊂R+∪{0}, we denote {x∈H;|x|∈E} and {x∈∂H;|x|∈E} by HE and ∂HE, respectively. We identify Rn with Rn-1×R and Rn-1 with Rn-1×{0}, writing typical points x,y∈Rn as x=(x',xn),y=(y',yn), where y′=(y1,y2,…,yn-1)∈Rn-1, and putting
(1.1)x·y=∑j=1n‍xjyj,|x|=x·x,Θ=x|x|,Φ=y|y|.Forx∈Rn and r>0, let B(x,r) denote the open ball with center at x and radius r(>0) in Rn. We will say that a set E⊂H has a covering {rj,Rj} if there exists a sequence of balls {Bj} with centers in H such that E⊂⋃j=1∞Bj, where rj is the radius of Bj and Rj is the distance between the origin and the center of Bj.Let𝒜a denote the class of nonnegative radial potentials a(x), that is, 0≤a(x)=a(|x|), x∈H, such that a∈Llocb(H) with some b>n/2 if n≥4 and with b=2 if n=2 or n=3.This paper is devoted to the stationary Schrödinger equation(1.2)SSEu(x)=-Δu(x)+a(x)u(x)=0,
where x∈H, Δ is the Laplace operator and a∈𝒜a. These solutions are called a-harmonic functions or generalized harmonic functions associated with the operator SSE. Note that they are (classical) harmonic functions in the case a=0. Under these assumptions the operator SSE can be extended in the usual way from the space C0∞(H) to an essentially self-adjoint operator on L2(H) (see [1–3]). We will denote it by SSE as well. This last one has a Green function Ga(x,y). Here, Ga(x,y) is positive on H and its inner normal derivative ∂Ga(x,y)/∂n(y')≥0. We denote this derivative by Pa(x,y'), which is called the Poisson a-kernel with respect to H. We remark that G(x,y) and P(x,y') are the Green function and Poisson kernel of the Laplacian in H, respectively.LetΔ* be a Laplace-Beltrami operator (spherical part of the Laplace) on the unit sphere. It is known (see, e.g., [4, page 41]) that the eigenvalue problem
(1.3)Δ*φ(Θ)+λφ(Θ)=0,Θ∈S+n-1,φ(Θ)=0,Θ∈∂S+n-1,
has the eigenvalues λj=j(j+n-2)(j=0,1,2…). Corresponding eigenfunctions are denoted by φjv(1≤v≤vj), where vj is the multiplicity of λj. We norm the eigenfunctions in L2(S+n-1) and φ1=φ11>0.Hence, well-known estimates (see, e.g., [5, page 14]) imply the following inequality:
(1.4)∑v=1vj‍φjv(Θ)∂φjv(Φ)∂nΦ≤M(n)j2n-1,
where the symbol M(n) denotes a constant depending only on n.LetVj(r) and Wj(r) stand, respectively, for the increasing and nonincreasing, as r→+∞, solutions of the equation
(1.5)-y′′(r)-n-1ry′(r)+(λjr2+a(r))y(r)=0,0<r<∞,
normalized under the condition Vj(1)=Wj(1)=1.We will also consider the classℬa, consisting of the potentials a∈𝒜a such that there exists a finite limit limr→∞r2a(r)=k∈[0,∞). Moreover, r-1|r2a(r)-k|∈L(1,∞). If a∈ℬa, then solutions of (1.2) are continuous (see [6]).In the rest of paper, we assume thata∈ℬa, and we will suppress this assumption for simplicity. Further, we use the standard notations u+=max{u,0}, u-=-min{u,0}, [d] is the integer part of d and d=[d]+{d}, where d is a positive real number.Denote(1.6)ιj,k±=2-n±(n-2)2+4(k+λj)2(j=0,1,2,3…).Remark 1.1.
ι
j
,
0
+
=
j
(
j
=
0,1
,
2,3
,
…
) in the case a=0.It is known (see [7]) that in the case under consideration the solutions to (1.5) have the asymptotics
(1.7)Vj(r)~d1rιj,k+,Wj(r)~d2rιj,k-,asr→∞,
where d1 and d2 are some positive constants.Ifa∈𝒜a, it is known that the following expansion for the Green function Ga(x,y) (see [8, Chapter 11], [1, 9])
(1.8)Ga(x,y)=∑j=0∞‍1χ′(1)Vj(min(|x|,|y|))Wj(max(|x|,|y|))(∑v=1vj‍φjv(Θ)φjv(Φ)),
where |x|≠|y| and χ′(1)=w(W1(r),V1(r))|r=1 is its Wronskian. The series converges uniformly if either |x|≤s|y| or |y|≤s|x|(0<s<1).For a nonnegative integerm and two points x,y∈H, we put
(1.9)K(a,m)(x,y)={0if|y|<1,K~(a,m)(x,y)if1≤|y|<∞,
where
(1.10)K~(a,m)(x,y)=∑j=0m‍1χ′(1)Vj(|x|)Wj(|y|)(∑v=1vj‍φjv(Θ)φjv(Φ)).We introduce another function ofx,y∈H(1.11)G(a,m)(x,y)=Ga(x,y)-K(a,m)(x,y).The generalized Poisson kernelP(a,m)(x,y') with respect to H is defined by
(1.12)P(a,m)(x,y′)=∂G(a,m)(x,y)∂n(y′).In fact(1.13)P(a,0)(x,y′)=Pa(x,y′).We remark that the kernel functionP(0,m)(x,y') coincides with ones in Finkelstein and Scheinberg [10] and Siegel and Talvila [11] (see [8, Chapter 11]).Put(1.14)U(a,m;u)(x)=∫∂H‍P(a,m)(x,y′)u(y′)dy',
where u(y′) is a continuous function on ∂H.Ifγ is a real number and γ≥0,(resp.,γ<0), ι[γ],k++{γ}>-ι1,k++1(resp.,-ι[-γ],k+-{-γ}>-ι1,k++1) and
(1.15)ι[γ],k++{γ}-n+1≤ιm+1,k+<ι[γ],k++{γ}-n+2(resp.,-ι[-γ],k+-{-γ}-n+1≤ιm+1,k+<-ι[-γ],k+-{-γ}-n+2).If these conditions all hold, we writeγ∈𝒞(k,m,n)(resp.,γ∈𝒟(k,m,n)).Letγ∈𝒞(k,m,n)(resp.,γ∈𝒟(k,m,n)) and u be functions on ∂H satisfying
(1.16)∫∂H‍|u(y′)|1+|y′|ι[γ],k++{γ}dy′<∞(resp.∫∂H‍|u(y′)|(1+|y′|ι[-γ],k++{-γ})dy′<∞).Forγ and u, we define the positive measure μ (resp., ν) on Rn by
(1.17)dμ(y′)={|u(y′)||y′|-ι[γ],k+-{γ}dy′,y'∈∂H(1,+∞),0,y'∈Rn-∂H(1,+∞)(resp.dν(y′)={|u(y′)||y′|ι[-γ],k++{-γ}dy′,y'∈∂H(1,+∞),0,y'∈Rn-∂H(1,+∞)).We remark that the total mass ofμ and ν is finite.Letϵ>0 and ξ≥0, and let μ be any positive measure on Rn having finite mass. For each x=(x',xn)∈Rn, the maximal function is defined by
(1.18)M(x;μ,ξ)=sup0<ρ<|x|/2μ(B(x,ρ))ρξ.
The set {x=(x',xn)∈Rn;M(x;μ,ξ)|x|ξ>ϵ} is denoted by E(ϵ;μ,ξ).About classical solutions of the Dirichlet problem for the Laplacian, Siegel and Talvila (cf. [11, Corollary 2.1]) proved the following result.Theorem A.
Ifu is a continuous function on ∂H satisfying
(1.19)∫∂H‍|u(y′)|1+|y′|n+mdy'<∞,
then, the function U(0,m;u)(x) satisfies
(1.20)U(0,m;u)∈C2(H)∩C0(H¯),ΔU(0,m;u)=0inH,U(0,m;u)=uon∂H,lim|x|→∞,x∈HU(0,m;u)(x)=o(xn1-n|x|n+m).Our first aim is to give the growth properties at infinity forU(a,m;u)(x).Theorem 1.2.
If0≤ζ≤n, γ∈𝒞(k,m,n)(resp.,γ∈𝒟(k,m,n)) and u is a measurable function on ∂H satisfying (1.16), then there exists a covering {rj,Rj} of E(ϵ;μ,n-ζ)(resp.,E(ϵ;ν,n-ζ))(⊂H) satisfying
(1.21)∑j=0∞‍(rjRj)2-ζVj(Rjrj)Wj(Rjrj)<∞
such that
(1.22)lim|x|→∞,x∈H-E(ϵ;μ,n-ζ)|x|-ι[γ],k+-{γ}+n-1φ1ζ-1(Θ)U(a,m;u)(x)=0(1.23)(resp.,lim|x|→∞,x∈H-E(ϵ;ν,n-ζ)|x|ι[-γ],k++{-γ}+n-1φ1ζ-1(Θ)U(a,m;u)(x)=0).
Ifu is a measurable function on ∂H satisfying
(1.24)∫∂H‍|u(y′)|1+|y′|γdy'<∞,
where γ is a real number, for this γ and u, we define
(1.25)dμ′(y′)={|u(y′)||y′|-γdy′,y'∈∂H(1,+∞),0,y'∈Rn-∂H(1,+∞).
Obviously, the total mass ofμ' is also finite.If we takea=0 in Theorem 1.2, then we immediately have the following growth property based on (1.5) and Remark 1.1.Corollary 1.3.
Let0≤ζ≤n, γ>-(n-1)(p-1) and γ-n≤m<γ-n+1. If u is defined as previously, then the function U(0,m;u)(x) is a harmonic function on H and there exists a covering {rj,Rj} of E(ϵ;μ',n-ζ)(⊂H) satisfying
(1.26)∑j=0∞‍(rjRj)n-ζ<∞
such that
(1.27)lim|x|→∞,x∈H-E(ϵ;μ′,n-ζ)|x|n-γ-1φ1ζ-1(Θ)U(a,m;u)(x)=0.Remark 1.4.
In the caseζ=n, (1.26) is a finite sum, and the set E(ϵ;μ',0) is a bounded set and (1.27) holds in H.Next we are concerned with solutions of the Dirichlet problem for the Schrödinger operator onH. For related results, we refer the readers to the paper by Kheyfits [1].Theorem 1.5.
Ifγ∈𝒞(k,m,n)(resp.,γ∈𝒟(k,m,n)) and u is a continuous function on ∂H satisfying (1.16), then
(1.28)U(a,m;u)∈C2(H)∩C0(H¯),SSEU(a,m;u)=0inH,U(a,m;u)=uon∂H,(1.29)lim|x|→∞,x∈H|x|-ι[γ],k+-{γ}+n-1φ1n-1(Θ)U(a,m;u)(x)=0(1.30)(resp.,lim|x|→∞,x∈H|x|ι[-γ],k++{-γ}+n-1φ1n-1(Θ)U(a,m;u)(x)=0).If we takeι[γ],k++{γ}=ιm+1,k++n-1, then we immediately have the following corollary, which is just Theorem A in the case a=0.Corollary 1.6.
Ifu is a continuous function on ∂H satisfying
(1.31)∫∂H‍|u(y′)|1+|y′|ιm+1,k++n-1dy'<∞,
then (1.28) hold and
(1.32)lim|x|→∞,x∈H|x|-ιm+1,k+φ1n-1(Θ)U(a,m;u)(x)=0.As an application of Corollary1.6, we can give a solution of the Dirichlet problem for any continuous function on ∂H.Theorem 1.7.
Ifu is a continuous function on ∂H satisfying (1.31) and h(x) is a solution of the Dirichlet problem for the Schrödinger operator on H with u satisfying
(1.33)lim|x|→∞,x∈H|x|-ιm+1,k+h+(x)=0,
then
(1.34)h(x)=U(a,m;u)(x)+∑j=0m‍(∑v=1vj‍djvφjv(Θ))Vj(|x|),
where x∈H and djv are constants.


## 2.  Lemmas

Throughout this paper, letM denote various constants independent of the variables in questions, which may be different from line to line.Lemma 2.1.
If1≤|y'|<(1/2)|x|, then
(2.1)|Pa(x,y′)|≤M|x|ι1,k-|y′|ι1,k+-1φ1(Θ).
If |y'|≥1 and |y'|≥2|x|, then
(2.2)|P(a,m)(x,y′)|≤MVm+1(|x|)Wm+1(|y′|)|y′|φ1(Θ)∂φ1(Φ)∂nΦ.
If (1/2)|x|<|y'|<2|x|, then
(2.3)|P(x,y′)|≤M|x-y′|-n|x|φ1(Θ).Proof.
Equations (2.1) and (2.2) are obtained by Kheyfits (see [8, Chapter 11] or [1, Lemma 1]). Equation (2.3) follows from Hayman and Kennedy (see [12, Lemma 4.2]).Lemma 2.2 (see [2, Theorem 1]).
Ifu(x) is a solution of (1.2) on H satisfying
(2.4)lim|x|→∞,x∈H|x|-ιm+1,k+u+(x)=0,
then
(2.5)u(x)=∑j=0m‍(∑v=1vj‍djvφjv(Θ))Vj(|x|).Lemma 2.3.
Letϵ>0 and ξ≥0, and let μ be any positive measure on Rn having finite total mass. Then, E(ϵ;μ,ξ) has a covering {rj,Rj}(j=1,2,…) satisfying
(2.6)∑j=1∞‍(rjRj)2-n+ξVj(Rjrj)Wj(Rjrj)<∞.Proof.
Set(2.7)Ej(ϵ;μ,ξ)={x∈E(ϵ;μ,ξ):2j≤|x|<2j+1}(j=2,3,4,…).
Ifx∈Ej(ϵ;μ,ξ), then there exists a positive number ρ(x) such that
(2.8)(ρ(x)|x|)2-n+ξVj(|x|ρ(x))Wj(|x|ρ(x))~(ρ(x)|x|)ξ≤μ(B(x,ρ(x)))ϵ.
Here,Ej(ϵ;μ,ξ) can be covered by the union of a family of balls {B(xj,i,ρj,i):xj,i∈Ej(ϵ;μ,ξ)}(ρj,i=ρ(xj,i)). By the Vitali lemma (see [13]), there exists Λj⊂Ej(ϵ;μ,ξ), which is at most countable, such that {B(xj,i,ρj,i):xj,i∈Λj} are disjoint and Ej(ϵ;μ,ξ)⊂⋃xj,i∈ΛjB(xj,i,5ρj,i).
So(2.9)⋃j=2∞Ej(ϵ;μ,ξ)⊂⋃j=2∞⋃xj,i∈ΛjB(xj,i,5ρj,i).On the other hand, note that⋃xj,i∈ΛjB(xj,i,ρj,i)⊂{x:2j-1≤|x|<2j+2}, so that
(2.10)∑Pj,i∈Λj‍(5ρj,i|xj,i|)2-n+ξVj(|xj,i|5ρj,i)Wj(|xj,i|5ρj,i)~∑xj,i∈Λj‍(5ρj,i|xj,i|)ξ≤5ξ∑xj,i∈Λj‍μ(B(xj,i,ρj,i))ϵ≤5ξϵμ(H[2j-1,2j+2)).Hence, we obtain(2.11)∑j=1∞‍∑xj,i∈Λj‍(ρj,i|xj,i|)2-n+ξVj(|xj,i|ρj,i)Wj(|xj,i|ρj,i)~∑j=1∞‍∑xj,i∈Λj‍(ρj,i|xj,i|)ξ≤∑j=1∞‍μ(H[2j-1,2j+2))ϵ≤3μ(Rn)ϵ.SinceE(ϵ;μ,ξ)∩{x∈Rn;|x|≥4}=⋃j=2∞Ej(ϵ;μ,ξ), then E(ϵ;μ,ξ) is finally covered by a sequence of balls (B(xj,i,ρj,i),B(x1,6))(j=2,3,…;i=1,2,…) satisfying
(2.12)∑j,i‍(ρj,i|xj,i|)2-n+ξVj(|xj,i|ρj,i)Wj(|xj,i|ρj,i)~∑j,i‍(ρj,i|xj,i|)ξ≤3μ(Rn)ϵ+6ξ<+∞,
where B(x1,6)(x1=(1,0,…,0)∈Rn) is the ball that covers {x∈Rn;|x|<4}.


## 3.  Proof of Theorem1.2

We only prove the caseγ≥0, the remaining case γ<0 can be proved similarly.For anyϵ>0, there exists Rϵ>1 such that
(3.1)∫∂H(Rϵ,∞)‍|u(y′)|1+|y′|ι[γ],k++{γ}dy'<ϵ.The relationGa(x,y)≤G(x,y) implies this inequality (see [14])
(3.2)Pa(x,y′)≤P(x,y′).For any fixed pointx∈H(Rϵ,+∞)-E(ϵ;μ,n-ζ) satisfying |x|>2Rϵ, letting I1=∂H[0,1), I2=∂H[1,Rϵ], I3=∂H(Rϵ,(1/2)|x|], I4=∂H((1/2)|x|,2|x|), I5=∂H[2|x|,∞) and I6=∂H[1,2|x|), we write
(3.3)|U(a,m;u)(x)|≤∑i=16‍Ua,i(x),
where
(3.4)Ua,i(x)=∫Ii‍|Pa(x,y′)||u(y′)|dy′(i=1,2,3,4),Ua,5(x)=∫I5‍|P(a,m)(x,y′)||u(y′)|dy',Ua,6(x)=∫I6‍|∂K~(Ω,a,m)(x,y)∂n(y′)||u(y′)|dy'.Byι[γ],k++{γ}>-ι1,k++1, (1.16), (2.1), and (3.1), we have the following growth estimates
(3.5)Ua,2(x)≤M|x|ι1,k-φ1(Θ)∫I2‍|y′|ι1,k+-1|u(y′)|dy'≤M|x|ι1,k-Rϵι[γ],k++{γ}+ι1,k+-1φ1(Θ),Ua,1(x)≤M|x|ι1,k-φ1(Θ),Ua,3(x)≤Mϵ|x|ι[γ],k++{γ}-n+1φ1(Θ).Next, we will estimateUa,4(x).Take a sufficiently small positive numberd3 such that I4⊂B(x,(1/2)|x|) for any x∈Π(d3), where
(3.6)Π(d3)={x∈H;infz∈∂S+n-1|x|x|-z|z||<d3,0<|x|<∞},
and divide H into two sets Π(d3) and H-Π(d3).Ifx∈H-Π(d3), then there exists a positive d3′ such that |x-y'|≥d3′|x| for any y'∈∂H, and hence
(3.7)Ua,4(x)≤M|x|1-nφ1(Θ)∫I4‍|u(y′)|dy'≤Mϵ|x|ι[γ],k++{γ}-n+1φ1(Θ).We will consider the casex∈Π(d3). Now put
(3.8)Ξi(x)={y∈I4;2i-1δ(x)≤|x-y′|<2iδ(x)},
where δ(x)=infy'∈∂H|x-y'|.Since∂H∩{y∈Rn:|x-y|<δ(x)}=∅, we have
(3.9)Ua,4(x)=M∑i=1i(x)‍∫Ξi(x)‍|x|φ1(Θ)|u(y′)||x-y′|ndy',
where i(x) is a positive integer satisfying 2i(x)-1δ(x)≤|x|/2<2i(x)δ(x).Since|x|φ1(Θ)≤Mδ(x)(x∈H), we obtain
(3.10)∫Ξi(x)‍|x|φ1(Θ)|u(y′)||x-y′|ndy′≤2(1-i)nφ1(Θ)δ(x)ζ-n∫Ξi(x)‍|x|δ(x)-ζ|u(y′)|dy'≤Mφ11-ζ(Θ)δ(x)ζ-n∫Ξi(x)‍|x|1-ζ|u(y′)|dy'≤M|x|n-ζφ11-ζ(Θ)δ(x)ζ-n∫Ξi(x)‍|y′|1-n|u(y′)|dy'≤Mϵ|x|ι[γ],k++{γ}-ζ+1φ11-ζ(Θ)(μ(Ξi(x))(2iδ(x))n-ζ)
for i=0,1,2,…,i(x).Sincex∉E(ϵ;μ,n-ζ), we have
(3.11)μ(Ξi(x))(2iδ(x))n-ζ≤μ(B(x,2iδ(x)))(2iδ(x))n-ζ≤M(x;μ,n-ζ)≤ϵ|x|ζ-n(i=0,1,2,…,i(x)-1),μ(Λi(x)(x))(2iδ(x))n-ζ≤μ(B(x,|x|/2))(|x|/2)n-ζ≤ϵ|x|ζ-n.So(3.12)Ua,4(x)≤Mϵ|x|ι[γ],k++{γ}-n+1φ11-ζ(Θ).Byιm+1,k+≥ι[γ],k++{γ}-n+1, (1.7), (2.2), and (3.1), we have
(3.13)Ua,5(x)≤MVm+1(|x|)∫I5‍|u(y′)|Vm+1(|y′|)|y′|n-1dy'≤Mϵ|x|ι[γ],k++{γ}-n+1φ1(Θ).We only considerUa,6(x) in the case m≥1, since Ua,6(x)≡0 for m=0. By the definition of K~(a,m), (1.4), and (2.2), we see that
(3.14)Ua,6(x)≤Mχ′(1)∑j=0m‍j2n-1qj(|x|),
where
(3.15)qj(|x|)=Vj(|x|)∫I6‍Wj(|y′|)|u(y′)||y′|dy'.To estimateqj(|x|), we write
(3.16)qj(|x|)≤qj′(|x|)+qj′′(|x|),
where
(3.17)qj′(|x|)=Vj(|x|)φ1(Θ)∫I2‍Wj(|y′|)|u(y′)||y′|dy',qj′′(|x|)=Vj(|x|)φ1(Θ)∫{y′∈∂H:Rϵ<|y′|<2|x|}‍Wj(|y′|)|u(y′)||y′|dy′.Notice that(3.18)Vj(|x|)Vm+1(|y′|)Vj(|y′|)|y′|≤MVm+1(|x|)|x|≤M|x|ιm+1,k+-1(|y′|≥1,Rϵ<2|x|).
Thus, by ιm+1,k+<ι[γ],k++{γ}-n+2, (1.7), and (1.16), we conclude
(3.19)qj′(|x|)=Vj(|x|)φ1(Θ)∫I2‍|u(y′)|Vj(|y′|)|y′|n-1dy'≤MVj(|x|)φ1(Θ)∫I2‍Vm+1(|y′|)|y′|ιm+1,k+|u(y′)|Vj(|y′|)|y′|n-1dy'≤M|x|ιm+1,k+-1Rϵι[γ],k++{γ}-ιm+1,k+-n+2φ1(Θ).Analogous to the estimate ofqj′(|x|), we have
(3.20)qj′′(|x|)≤Mϵ|x|ι[γ],k++{γ}-n+1φ1(Θ).Thus, we can conclude that(3.21)qj(|x|)≤Mϵ|x|ι[γ],k++{γ}-n+1φ1(Θ),
which yields
(3.22)Ua,6(x)≤Mϵ|x|ι[γ],k++{γ}-n+1φ1(Θ).Combining (3.5)–(3.22), we obtain that if Rϵ is sufficiently large and ϵ is sufficiently small, then U(a,m;u)(x)=o(|x|ι[γ],k++{γ}-n+1φ11-ζ(Θ)) as |x|→∞, where x∈H(Rϵ,+∞)-E(ϵ;μ,n-ζ). Finally, there exists an additional finite ball B0 covering H[0,Rϵ], which together with Lemma 2.3 gives the conclusion of Theorem 1.2.


## 4.  Proof of Theorem1.5

For any fixedx∈H, take a number satisfying R>max{1,2|x|}. By ιm+1,k+≥ι[γ],k++{γ}-n+1, (1.5), (1.16), and (2.2), we have
(4.1)∫∂H(R,∞)‍|P(a,m)(x,y′)||u(y′)|dy′≤MVm+1(|x|)φ1(Θ)∫∂H(R,∞)‍|u(y′)||y′|ιm+1,k++n-1dy'≤M|x|ιm+1,k+φ1(Θ)∫∂H(2|x|,∞)‍|y′|ι[γ],k++{γ}-ιm+1,k+-n+1dy'≤M|x|ι[γ],k++{γ}-n+1φ1(Θ)<∞.
Then, U(a,m;u)(x) is absolutely convergent and finite for any x∈H. Thus U(a,m;u)(x) is a solution of (1.2) on H.Now we study the boundary behavior ofU(a,m;u)(x). Let y'∈∂H be any fixed point and l any positive number satisfying l>max{|y'|+1,(1/2)R}.SetχS(l) as the characteristic function of S(l)={y′∈∂H,|y′|≤l}, and write
(4.2)U(a,m;u)(x)=U′(x)-U′′(x)+U′′′(x),
where
(4.3)U′(x)=∫∂H[0,2l]‍Pa(x,y′)u(y′)dy',U′′(x)=∫∂H(1,2l]‍∂K(a,m)(x,y)∂n(y')u(y')dy',U′′′(x)=∫∂H(2l,∞)‍P(a,m)(x,y′)u(y′)dy'.Notice thatU'(x) is the Poisson a-integral of u(y′)χS(2l), We have limx→y',x∈HU′(x)=u(y'). Since limΘ→Φ′φjv(Θ)=0(j=1,2,3…;1≤v≤vj) as x→y′∈∂H, we have limx→y′,x∈HU′′(x)=0 from the definition of the kernel function K(a,m)(x,y). U′′′(x)=O(|x|ι[γ],k++{γ}-n+1φ1(Θ)) and therefore tends to zero.So the functionU(a,m;u)(x) can be continuously extended to H¯ such that
(4.4)limx→y′,x∈HU(a,m;u)(x)=u(y′)
for any y′∈∂H from the arbitrariness of l.Finally, (1.29) and (1.30) follow from (1.22) and (1.23), respectively, in the case ζ=n. Thus, we complete the proof of Theorem 1.5.


## 5. Proof of Theorem1.7

From Corollary1.6, we have the solution U(a,m;u)(x) of the Dirichlet problem on H with u satisfying (1.31). Consider the function h(x)-U(a,m;u)(x). Then, it follows that this is a solution of (1.2) in H and vanishes continuously on ∂H.Since(5.1)0≤(h-U(Ω,a,m;u))+(x)≤h+(x)+(U(a,m;u))-(x)
for any x∈H, we have
(5.2)lim|x|→∞,x∈H|x|-ιm+1,k+(h-U(Ω,a,m;u))+(x)=0
from (1.32) and (1.33). Then, the conclusion of Theorem 1.7 follows immediately from Lemma 2.2.



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*Source: 578197-2012-08-28.xml*