Abstract and Applied Analysis

Abstract and Applied Analysis / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 578197 | https://doi.org/10.1155/2012/578197

Baiyun Su, "Dirichlet Problem for the Schrödinger Operator in a Half Space", Abstract and Applied Analysis, vol. 2012, Article ID 578197, 14 pages, 2012. https://doi.org/10.1155/2012/578197

Dirichlet Problem for the Schrödinger Operator in a Half Space

Academic Editor: Jean Pierre Gossez
Received22 Apr 2012
Accepted12 Jul 2012
Published28 Aug 2012


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.

1. Introduction and Results

Let 𝐑 and 𝐑+ be the sets of all real numbers and of all positive real numbers, respectively. Let 𝐑𝑛(𝑛2) denote the 𝑛-dimensional Euclidean space with points 𝑥=(𝑥,𝑥𝑛), where 𝑥=(𝑥1,𝑥2,,𝑥𝑛1)𝐑𝑛1 and 𝑥𝑛𝐑. The unit sphere and the upper half unit sphere in 𝐑𝑛 are denoted by 𝐒𝑛1 and 𝐒+𝑛1, 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)𝐑𝑛𝑥𝑛>0}, whose boundary is 𝜕𝐻.

For a set 𝐸, 𝐸𝐑+{0}, we denote {𝑥𝐻;|𝑥|𝐸} and {𝑥𝜕𝐻;|𝑥|𝐸} by 𝐻𝐸 and 𝜕𝐻𝐸, respectively. We identify 𝐑𝑛 with 𝐑𝑛1×𝐑 and 𝐑𝑛1 with 𝐑𝑛1×{0}, writing typical points 𝑥,𝑦𝐑𝑛 as 𝑥=(𝑥,𝑥𝑛),𝑦=(𝑦,𝑦𝑛), where 𝑦=(𝑦1,𝑦2,,𝑦𝑛1)𝐑𝑛1, and putting 𝑥𝑦=𝑛𝑗=1𝑥𝑗𝑦𝑗,|𝑥|=𝑥𝑥𝑥,Θ=𝑦|𝑥|,Φ=||𝑦||.(1.1)

For 𝑥𝐑𝑛 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 𝐸𝐻 has a covering {𝑟𝑗,𝑅𝑗} if there exists a sequence of balls {𝐵𝑗} with centers in 𝐻 such that 𝐸𝑗=1𝐵𝑗, where 𝑟𝑗 is the radius of 𝐵𝑗 and 𝑅𝑗 is the distance between the origin and the center of 𝐵𝑗.

Let 𝒜𝑎 denote the class of nonnegative radial potentials 𝑎(𝑥), that is, 0𝑎(𝑥)=𝑎(|𝑥|), 𝑥𝐻, such that 𝑎𝐿𝑏loc(𝐻) with some 𝑏>𝑛/2 if 𝑛4 and with 𝑏=2 if 𝑛=2 or 𝑛=3.

This paper is devoted to the stationary Schrödinger equation SSE𝑢(𝑥)=Δ𝑢(𝑥)+𝑎(𝑥)𝑢(𝑥)=0,(1.2) where 𝑥𝐻, Δ is the Laplace operator and 𝑎𝒜𝑎. 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 [13]). 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.

Let Δ 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 Δ𝜑(Θ)+𝜆𝜑(Θ)=0,Θ𝐒+𝑛1,𝜑(Θ)=0,Θ𝜕𝐒+𝑛1,(1.3) has the eigenvalues 𝜆𝑗=𝑗(𝑗+𝑛2)(𝑗=0,1,2). Corresponding eigenfunctions are denoted by 𝜑𝑗𝑣(1𝑣𝑣𝑗), where 𝑣𝑗 is the multiplicity of 𝜆𝑗. We norm the eigenfunctions in 𝐿2(𝐒+𝑛1) and 𝜑1=𝜑11>0.

Hence, well-known estimates (see, e.g., [5, page 14]) imply the following inequality: 𝑣𝑗𝑣=1𝜑𝑗𝑣(Θ)𝜕𝜑𝑗𝑣(Φ)𝜕𝑛Φ𝑀(𝑛)𝑗2𝑛1,(1.4) where the symbol 𝑀(𝑛) denotes a constant depending only on 𝑛.

Let 𝑉𝑗(𝑟) and 𝑊𝑗(𝑟) stand, respectively, for the increasing and nonincreasing, as 𝑟+, solutions of the equation 𝑦(𝑟)𝑛1𝑟𝑦𝜆(𝑟)+𝑗𝑟2+𝑎(𝑟)𝑦(𝑟)=0,0<𝑟<,(1.5) normalized under the condition 𝑉𝑗(1)=𝑊𝑗(1)=1.

We will also consider the class 𝑎, consisting of the potentials 𝑎𝒜𝑎 such that there exists a finite limit lim𝑟𝑟2𝑎(𝑟)=𝑘[0,). Moreover, 𝑟1|𝑟2𝑎(𝑟)𝑘|𝐿(1,). If 𝑎𝑎, then solutions of (1.2) are continuous (see [6]).

In the rest of paper, we assume that 𝑎𝑎, 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 𝜄±𝑗,𝑘=2𝑛±(𝑛2)2+4𝑘+𝜆𝑗2(𝑗=0,1,2,3).(1.6)

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 𝑉𝑗(𝑟)𝑑1𝑟𝜄+𝑗,𝑘,𝑊𝑗(𝑟)𝑑2𝑟𝜄𝑗,𝑘,as𝑟,(1.7) where 𝑑1 and 𝑑2 are some positive constants.

If 𝑎𝒜𝑎, it is known that the following expansion for the Green function 𝐺𝑎(𝑥,𝑦) (see [8, Chapter 11], [1, 9]) 𝐺𝑎(𝑥,𝑦)=𝑗=01𝜒𝑉(1)𝑗||𝑦||𝑊min|𝑥|,𝑗||𝑦||max|𝑥|,𝑣𝑗𝑣=1𝜑𝑗𝑣(Θ)𝜑𝑗𝑣,(Φ)(1.8) 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 integer 𝑚 and two points 𝑥,𝑦𝐻, we put 𝐾||𝑦||||𝑦||(𝑎,𝑚)(𝑥,𝑦)=0if<1,𝐾(𝑎,𝑚)(𝑥,𝑦)if1<,(1.9) where 𝐾(𝑎,𝑚)(𝑥,𝑦)=𝑚𝑗=01𝜒𝑉(1)𝑗(|𝑥|)𝑊𝑗||𝑦||𝑣𝑗𝑣=1𝜑𝑗𝑣(Θ)𝜑𝑗𝑣.(Φ)(1.10)

We introduce another function of 𝑥,𝑦𝐻𝐺(𝑎,𝑚)(𝑥,𝑦)=𝐺𝑎(𝑥,𝑦)𝐾(𝑎,𝑚)(𝑥,𝑦).(1.11)

The generalized Poisson kernel 𝑃(𝑎,𝑚)(𝑥,𝑦) with respect to 𝐻 is defined by 𝑃(𝑎,𝑚)𝑥,𝑦=𝜕𝐺(𝑎,𝑚)(𝑥,𝑦)𝜕𝑛(𝑦).(1.12)

In fact 𝑃(𝑎,0)𝑥,𝑦=𝑃𝑎𝑥,𝑦.(1.13)

We remark that the kernel function 𝑃(0,𝑚)(𝑥,𝑦) coincides with ones in Finkelstein and Scheinberg [10] and Siegel and Talvila [11] (see [8, Chapter 11]).

Put 𝑈(𝑎,𝑚;𝑢)(𝑥)=𝜕𝐻𝑃(𝑎,𝑚)𝑥,𝑦𝑢𝑦𝑑𝑦,(1.14) where 𝑢(𝑦) is a continuous function on 𝜕𝐻.

If 𝛾 is a real number and 𝛾0,(resp.,𝛾<0), 𝜄+[𝛾],𝑘+{𝛾}>𝜄+1,𝑘+1(resp.,𝜄+[𝛾],𝑘{𝛾}>𝜄+1,𝑘+1) and 𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜄+𝑚+1,𝑘<𝜄+[𝛾],𝑘+{𝛾}𝑛+2resp.,𝜄+[]𝛾,𝑘{𝛾}𝑛+1𝜄+𝑚+1,𝑘<𝜄+[]𝛾,𝑘.{𝛾}𝑛+2(1.15)

If these conditions all hold, we write 𝛾𝒞(𝑘,𝑚,𝑛)(resp.,𝛾𝒟(𝑘,𝑚,𝑛)).

Let 𝛾𝒞(𝑘,𝑚,𝑛)(resp.,𝛾𝒟(𝑘,𝑚,𝑛)) and 𝑢 be functions on 𝜕𝐻 satisfying 𝜕𝐻||𝑢𝑦||||𝑦1+||𝜄+[𝛾],𝑘+{𝛾}𝑑𝑦<resp.𝜕𝐻||𝑢𝑦||||𝑦1+||𝜄+[𝛾],𝑘+{𝛾}𝑑𝑦.<(1.16)

For 𝛾 and 𝑢, we define the positive measure 𝜇 (resp., 𝜈) on 𝐑𝑛 by 𝑦𝑑𝜇=||𝑢𝑦||||𝑦||𝜄+[𝛾],𝑘{𝛾}𝑑𝑦,𝑦𝜕𝐻(1,+),0,𝑦𝐑𝑛𝑦𝜕𝐻(1,+)resp.𝑑𝜈=||𝑢𝑦||||𝑦||𝜄+[𝛾],𝑘+{𝛾}𝑑𝑦,𝑦𝜕𝐻(1,+),0,𝑦𝐑𝑛.𝜕𝐻(1,+)(1.17)

We remark that the total mass of 𝜇 and 𝜈 is finite.

Let 𝜖>0 and 𝜉0, and let 𝜇 be any positive measure on 𝐑𝑛 having finite mass. For each 𝑥=(𝑥,𝑥𝑛)𝐑𝑛, the maximal function is defined by 𝑀(𝑥;𝜇,𝜉)=sup0<𝜌<|𝑥|/2𝜇(𝐵(𝑥,𝜌))𝜌𝜉.(1.18) 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. If 𝑢 is a continuous function on 𝜕𝐻 satisfying 𝜕𝐻||𝑢𝑦||||𝑦1+||𝑛+𝑚𝑑𝑦<,(1.19) then, the function 𝑈(0,𝑚;𝑢)(𝑥) satisfies 𝑈(0,𝑚;𝑢)𝐶2(𝐻)𝐶0𝐻,Δ𝑈(0,𝑚;𝑢)=0in𝐻,𝑈(0,𝑚;𝑢)=𝑢on𝜕𝐻,lim|𝑥|,𝑥𝐻𝑥𝑈(0,𝑚;𝑢)(𝑥)=𝑜𝑛1𝑛|𝑥|𝑛+𝑚.(1.20)

Our first aim is to give the growth properties at infinity for 𝑈(𝑎,𝑚;𝑢)(𝑥).

Theorem 1.2. If 0𝜁𝑛, 𝛾𝒞(𝑘,𝑚,𝑛)(resp.,𝛾𝒟(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,𝑛𝜁)(resp.,𝐸(𝜖;𝜈,𝑛𝜁))(𝐻) satisfying 𝑗=0𝑟𝑗𝑅𝑗2𝜁𝑉𝑗𝑅𝑗𝑟𝑗𝑊𝑗𝑅𝑗𝑟𝑗<(1.21) such that lim|𝑥|,𝑥𝐻𝐸(𝜖;𝜇,𝑛𝜁)|𝑥|𝜄+[𝛾],𝑘{𝛾}+𝑛1𝜑1𝜁1(Θ)𝑈(𝑎,𝑚;𝑢)(𝑥)=0(1.22)resp.,lim|𝑥|,𝑥𝐻𝐸(𝜖;𝜈,𝑛𝜁)|𝑥|𝜄+[𝛾],𝑘+{𝛾}+𝑛1𝜑1𝜁1(Θ)𝑈(𝑎,𝑚;𝑢)(𝑥)=0.(1.23)
If 𝑢 is a measurable function on 𝜕𝐻 satisfying 𝜕𝐻||𝑢𝑦||||𝑦1+||𝛾𝑑𝑦<,(1.24) where 𝛾 is a real number, for this 𝛾 and 𝑢, we define 𝑑𝜇𝑦=||𝑢𝑦||||𝑦||𝛾𝑑𝑦,𝑦𝜕𝐻(1,+),0,𝑦𝐑𝑛𝜕𝐻(1,+).(1.25)
Obviously, the total mass of 𝜇 is also finite.

If we take 𝑎=0 in Theorem 1.2, then we immediately have the following growth property based on (1.5) and Remark 1.1.

Corollary 1.3. Let 0𝜁𝑛, 𝛾>(𝑛1)(𝑝1) and 𝛾𝑛𝑚<𝛾𝑛+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,𝑛𝜁)(𝐻) satisfying 𝑗=0𝑟𝑗𝑅𝑗𝑛𝜁<(1.26) such that lim|𝑥|,𝑥𝐻𝐸𝜖;𝜇,𝑛𝜁|𝑥|𝑛𝛾1𝜑1𝜁1(Θ)𝑈(𝑎,𝑚;𝑢)(𝑥)=0.(1.27)

Remark 1.4. In the case 𝜁=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 on 𝐻. For related results, we refer the readers to the paper by Kheyfits [1].

Theorem 1.5. If 𝛾𝒞(𝑘,𝑚,𝑛)(resp.,𝛾𝒟(𝑘,𝑚,𝑛)) and 𝑢 is a continuous function on 𝜕𝐻 satisfying (1.16), then 𝑈(𝑎,𝑚;𝑢)𝐶2(𝐻)𝐶0𝐻,SSE𝑈(𝑎,𝑚;𝑢)=0in𝐻,𝑈(𝑎,𝑚;𝑢)=𝑢on𝜕𝐻,(1.28)lim|𝑥|,𝑥𝐻|𝑥|𝜄+[𝛾],𝑘{𝛾}+𝑛1𝜑1𝑛1(Θ)𝑈(𝑎,𝑚;𝑢)(𝑥)=0(1.29)resp.,lim|𝑥|,𝑥𝐻|𝑥|𝜄+[𝛾],𝑘+{𝛾}+𝑛1𝜑1𝑛1(Θ)𝑈(𝑎,𝑚;𝑢)(𝑥)=0.(1.30)

If we take 𝜄+[𝛾],𝑘+{𝛾}=𝜄+𝑚+1,𝑘+𝑛1, then we immediately have the following corollary, which is just Theorem A in the case 𝑎=0.

Corollary 1.6. If 𝑢 is a continuous function on 𝜕𝐻 satisfying 𝜕𝐻||𝑢𝑦||||𝑦1+||𝜄+𝑚+1,𝑘+𝑛1𝑑𝑦<,(1.31) then (1.28) hold and lim|𝑥|,𝑥𝐻|𝑥|𝜄+𝑚+1,𝑘𝜑1𝑛1(Θ)𝑈(𝑎,𝑚;𝑢)(𝑥)=0.(1.32)

As an application of Corollary 1.6, we can give a solution of the Dirichlet problem for any continuous function on 𝜕𝐻.

Theorem 1.7. If 𝑢 is a continuous function on 𝜕𝐻 satisfying (1.31) and (𝑥) is a solution of the Dirichlet problem for the Schrödinger operator on 𝐻 with 𝑢 satisfying lim|𝑥|,𝑥𝐻|𝑥|𝜄+𝑚+1,𝑘+(𝑥)=0,(1.33) then (𝑥)=𝑈(𝑎,𝑚;𝑢)(𝑥)+𝑚𝑗=0𝑣𝑗𝑣=1𝑑𝑗𝑣𝜑𝑗𝑣𝑉(Θ)𝑗(|𝑥|),(1.34) where 𝑥𝐻 and 𝑑𝑗𝑣 are constants.

2. Lemmas

Throughout this paper, let 𝑀 denote various constants independent of the variables in questions, which may be different from line to line.

Lemma 2.1. If 1|𝑦|<(1/2)|𝑥|, then ||𝑃𝑎𝑥,𝑦||𝑀|𝑥|𝜄1,𝑘||𝑦||𝜄+1,𝑘1𝜑1(Θ).(2.1) If |𝑦|1 and |𝑦|2|𝑥|, then ||𝑃(𝑎,𝑚)𝑥,𝑦||𝑀𝑉𝑚+1𝑊(|𝑥|)𝑚+1||𝑦||||𝑦||𝜑1(Θ)𝜕𝜑1(Φ)𝜕𝑛Φ.(2.2) If (1/2)|𝑥|<|𝑦|<2|𝑥|, then ||𝑃𝑥,𝑦||||𝑀𝑥𝑦||𝑛|𝑥|𝜑1(Θ).(2.3)

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]). If 𝑢(𝑥) is a solution of (1.2) on 𝐻 satisfying lim|𝑥|,𝑥𝐻|𝑥|𝜄+𝑚+1,𝑘𝑢+(𝑥)=0,(2.4) then 𝑢(𝑥)=𝑚𝑗=0𝑣𝑗𝑣=1𝑑𝑗𝑣𝜑𝑗𝑣𝑉(Θ)𝑗(|𝑥|).(2.5)

Lemma 2.3. Let 𝜖>0 and 𝜉0, 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 𝑗=1𝑟𝑗𝑅𝑗2𝑛+𝜉𝑉𝑗𝑅𝑗𝑟𝑗𝑊𝑗𝑅𝑗𝑟𝑗<.(2.6)

Proof. Set 𝐸𝑗(𝜖;𝜇,𝜉)=𝑥𝐸(𝜖;𝜇,𝜉)2𝑗|𝑥|<2𝑗+1(𝑗=2,3,4,).(2.7)
If 𝑥𝐸𝑗(𝜖;𝜇,𝜉), then there exists a positive number 𝜌(𝑥) such that 𝜌(𝑥)|𝑥|2𝑛+𝜉𝑉𝑗|𝑥|𝑊𝜌(𝑥)𝑗|𝑥|𝜌(𝑥)𝜌(𝑥)|𝑥|𝜉𝜇(𝐵(𝑥,𝜌(𝑥)))𝜖.(2.8)
Here, 𝐸𝑗(𝜖;𝜇,𝜉) can be covered by the union of a family of balls {𝐵(𝑥𝑗,𝑖,𝜌𝑗,𝑖)𝑥𝑗,𝑖𝐸𝑗(𝜖;𝜇,𝜉)}(𝜌𝑗,𝑖=𝜌(𝑥𝑗,𝑖)). By the Vitali lemma (see [13]), there exists Λ𝑗𝐸𝑗(𝜖;𝜇,𝜉), which is at most countable, such that {𝐵(𝑥𝑗,𝑖,𝜌𝑗,𝑖)𝑥𝑗,𝑖Λ𝑗} are disjoint and 𝐸𝑗(𝜖;𝜇,𝜉)𝑥𝑗,𝑖Λ𝑗𝐵(𝑥𝑗,𝑖,5𝜌𝑗,𝑖).
So 𝑗=2𝐸𝑗(𝜖;𝜇,𝜉)𝑗=2𝑥𝑗,𝑖Λ𝑗𝐵𝑥𝑗,𝑖,5𝜌𝑗,𝑖.(2.9)

On the other hand, note that 𝑥𝑗,𝑖Λ𝑗𝐵(𝑥𝑗,𝑖,𝜌𝑗,𝑖){𝑥2𝑗1|𝑥|<2𝑗+2}, so that 𝑃𝑗,𝑖Λ𝑗5𝜌𝑗,𝑖||𝑥𝑗,𝑖||2𝑛+𝜉𝑉𝑗||𝑥𝑗,𝑖||5𝜌𝑗,𝑖𝑊𝑗||𝑥𝑗,𝑖||5𝜌𝑗,𝑖𝑥𝑗,𝑖Λ𝑗5𝜌𝑗,𝑖||𝑥𝑗,𝑖||𝜉5𝜉𝑥𝑗,𝑖Λ𝑗𝜇𝐵𝑥𝑗,𝑖,𝜌𝑗,𝑖𝜖5𝜉𝜖𝜇𝐻2𝑗1,2𝑗+2.(2.10)

Hence, we obtain 𝑗=1𝑥𝑗,𝑖Λ𝑗𝜌𝑗,𝑖||𝑥𝑗,𝑖||2𝑛+𝜉𝑉𝑗||𝑥𝑗,𝑖||𝜌𝑗,𝑖𝑊𝑗||𝑥𝑗,𝑖||𝜌𝑗,𝑖𝑗=1𝑥𝑗,𝑖Λ𝑗𝜌𝑗,𝑖||𝑥𝑗,𝑖||𝜉𝑗=1𝜇𝐻2𝑗1,2𝑗+2𝜖3𝜇(𝐑𝑛)𝜖.(2.11)

Since 𝐸(𝜖;𝜇,𝜉){𝑥𝐑𝑛;|𝑥|4}=𝑗=2𝐸𝑗(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 𝑗,𝑖𝜌𝑗,𝑖||𝑥𝑗,𝑖||2𝑛+𝜉𝑉𝑗||𝑥𝑗,𝑖||𝜌𝑗,𝑖𝑊𝑗||𝑥𝑗,𝑖||𝜌𝑗,𝑖𝑗,𝑖𝜌𝑗,𝑖||𝑥𝑗,𝑖||𝜉3𝜇(𝐑𝑛)𝜖+6𝜉<+,(2.12) where 𝐵(𝑥1,6)(𝑥1=(1,0,,0)𝐑𝑛) is the ball that covers {𝑥𝐑𝑛;|𝑥|<4}.

3. Proof of Theorem 1.2

We only prove the case 𝛾0, the remaining case 𝛾<0 can be proved similarly.

For any 𝜖>0, there exists 𝑅𝜖>1 such that 𝜕𝐻(𝑅𝜖,)||𝑢𝑦||||𝑦1+||𝜄+[𝛾],𝑘+{𝛾}𝑑𝑦<𝜖.(3.1)

The relation 𝐺𝑎(𝑥,𝑦)𝐺(𝑥,𝑦) implies this inequality (see [14]) 𝑃𝑎𝑥,𝑦𝑃𝑥,𝑦.(3.2)

For any fixed point 𝑥𝐻(𝑅𝜖,+)𝐸(𝜖;𝜇,𝑛𝜁) 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 ||||𝑈(𝑎,𝑚;𝑢)(𝑥)6𝑖=1𝑈𝑎,𝑖(𝑥),(3.3) where 𝑈𝑎,𝑖(𝑥)=𝐼𝑖||𝑃𝑎𝑥,𝑦||||𝑢𝑦||𝑑𝑦(𝑈𝑖=1,2,3,4),𝑎,5(𝑥)=𝐼5||𝑃(𝑎,𝑚)𝑥,𝑦||||𝑢𝑦||𝑈𝑑𝑦,𝑎,6(𝑥)=𝐼6||||𝜕𝐾(Ω,𝑎,𝑚)(𝑥,𝑦)𝜕𝑛(𝑦)||||||𝑢𝑦||𝑑𝑦.(3.4)

By 𝜄+[𝛾],𝑘+{𝛾}>𝜄+1,𝑘+1, (1.16), (2.1), and (3.1), we have the following growth estimates 𝑈𝑎,2(𝑥)𝑀|𝑥|𝜄1,𝑘𝜑1(Θ)𝐼2||𝑦||𝜄+1,𝑘1||𝑢𝑦||𝑑𝑦𝑀|𝑥|𝜄1,𝑘𝑅𝜄+[𝛾],𝑘+{𝛾}+𝜄+1,𝑘𝜖1𝜑1𝑈(Θ),𝑎,1(𝑥)𝑀|𝑥|𝜄1,𝑘𝜑1𝑈(Θ),𝑎,3(𝑥)𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ).(3.5)

Next, we will estimate 𝑈𝑎,4(𝑥).

Take a sufficiently small positive number 𝑑3 such that 𝐼4𝐵(𝑥,(1/2)|𝑥|) for any 𝑥Π(𝑑3), where Π𝑑3=𝑥𝐻;inf𝑧𝜕𝐒+𝑛1||||𝑥𝑧|𝑥||||||𝑧|<𝑑3,0<|𝑥|<,(3.6) and divide 𝐻 into two sets Π(𝑑3) and 𝐻Π(𝑑3).

If 𝑥𝐻Π(𝑑3), then there exists a positive 𝑑3 such that |𝑥𝑦|𝑑3|𝑥| for any 𝑦𝜕𝐻, and hence 𝑈𝑎,4(𝑥)𝑀|𝑥|1𝑛𝜑1(Θ)𝐼4||𝑢𝑦||𝑑𝑦𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ).(3.7)

We will consider the case 𝑥Π(𝑑3). Now put Ξ𝑖(𝑥)=𝑦𝐼4;2𝑖1𝛿||(𝑥)𝑥𝑦||<2𝑖𝛿,(𝑥)(3.8) where 𝛿(𝑥)=inf𝑦𝜕𝐻|𝑥𝑦|.

Since 𝜕𝐻{𝑦𝐑𝑛|𝑥𝑦|<𝛿(𝑥)}=, we have 𝑈𝑎,4(𝑥)=𝑀𝑖(𝑥)𝑖=1Ξ𝑖(𝑥)|𝑥|𝜑1||𝑢𝑦(Θ)||||𝑥𝑦||𝑛𝑑𝑦,(3.9) where 𝑖(𝑥) is a positive integer satisfying 2𝑖(𝑥)1𝛿(𝑥)|𝑥|/2<2𝑖(𝑥)𝛿(𝑥).

Since |𝑥|𝜑1(Θ)𝑀𝛿(𝑥)(𝑥𝐻), we obtain Ξ𝑖(𝑥)|𝑥|𝜑1||𝑢𝑦(Θ)||||𝑥𝑦||𝑛𝑑𝑦2(1𝑖)𝑛𝜑1(Θ)𝛿(𝑥)𝜁𝑛Ξ𝑖(𝑥)|𝑥|𝛿(𝑥)𝜁||𝑢𝑦||𝑑𝑦𝑀𝜑11𝜁(Θ)𝛿(𝑥)𝜁𝑛Ξ𝑖(𝑥)|𝑥|1𝜁||𝑢𝑦||𝑑𝑦𝑀|𝑥|𝑛𝜁𝜑11𝜁(Θ)𝛿(𝑥)𝜁𝑛Ξ𝑖(𝑥)||𝑦||1𝑛||𝑢𝑦||𝑑𝑦𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝜁+1𝜑11𝜁𝜇Ξ(Θ)𝑖(𝑥)2𝑖𝛿(𝑥)𝑛𝜁(3.10) for 𝑖=0,1,2,,𝑖(𝑥).

Since 𝑥𝐸(𝜖;𝜇,𝑛𝜁), we have 𝜇Ξ𝑖(𝑥)2𝑖𝛿(𝑥)𝑛𝜁𝜇𝐵𝑥,2𝑖𝛿(𝑥)2𝑖𝛿(𝑥)𝑛𝜁𝑀(𝑥;𝜇,𝑛𝜁)𝜖|𝑥|𝜁𝑛𝜇Λ(𝑖=0,1,2,,𝑖(𝑥)1),𝑖(𝑥)(𝑥)2𝑖𝛿(𝑥)𝑛𝜁𝜇(𝐵(𝑥,|𝑥|/2))(|𝑥|/2)𝑛𝜁𝜖|𝑥|𝜁𝑛.(3.11)

So 𝑈𝑎,4(𝑥)𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑11𝜁(Θ).(3.12)

By 𝜄+𝑚+1,𝑘𝜄+[𝛾],𝑘+{𝛾}𝑛+1, (1.7), (2.2), and (3.1), we have 𝑈𝑎,5(𝑥)𝑀𝑉𝑚+1(|𝑥|)𝐼5||𝑢𝑦||𝑉𝑚+1||𝑦||||𝑦||𝑛1𝑑𝑦𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ).(3.13)

We only consider 𝑈𝑎,6(𝑥) in the case 𝑚1, since 𝑈𝑎,6(𝑥)0 for 𝑚=0. By the definition of 𝐾(𝑎,𝑚), (1.4), and (2.2), we see that 𝑈𝑎,6𝑀(𝑥)𝜒(1)𝑚𝑗=0𝑗2𝑛1𝑞𝑗(|𝑥|),(3.14) where 𝑞𝑗(|𝑥|)=𝑉𝑗(|𝑥|)𝐼6𝑊𝑗||𝑦||||𝑢𝑦||||𝑦||𝑑𝑦.(3.15)

To estimate 𝑞𝑗(|𝑥|), we write 𝑞𝑗(|𝑥|)𝑞𝑗(|𝑥|)+𝑞𝑗(|𝑥|),(3.16) where 𝑞𝑗(|𝑥|)=𝑉𝑗(|𝑥|)𝜑1(Θ)𝐼2𝑊𝑗||𝑦||||𝑢𝑦||||𝑦||𝑞𝑑𝑦,𝑗(|𝑥|)=𝑉𝑗(|𝑥|)𝜑1(Θ){𝑦𝜕𝐻𝑅𝜖<|𝑦|<2|𝑥|}𝑊𝑗||𝑦||||𝑢𝑦||||𝑦||𝑑𝑦.(3.17)

Notice that 𝑉𝑗𝑉(|𝑥|)𝑚+1||𝑦||𝑉𝑗||𝑦||||𝑦||𝑉𝑀𝑚+1(|𝑥|)|𝑥|𝑀|𝑥|𝜄+𝑚+1,𝑘1||𝑦||1,𝑅𝜖.<2|𝑥|(3.18) Thus, by 𝜄+𝑚+1,𝑘<𝜄+[𝛾],𝑘+{𝛾}𝑛+2, (1.7), and (1.16), we conclude 𝑞𝑗(|𝑥|)=𝑉𝑗(|𝑥|)𝜑1(Θ)𝐼2||𝑢𝑦||𝑉𝑗||𝑦||||𝑦||𝑛1𝑑𝑦𝑀𝑉𝑗(|𝑥|)𝜑1(Θ)𝐼2𝑉𝑚+1||𝑦||||𝑦||𝜄+𝑚+1,𝑘||𝑢𝑦||𝑉𝑗||𝑦||||𝑦||𝑛1𝑑𝑦𝑀|𝑥|𝜄+𝑚+1,𝑘1𝑅𝜄+[𝛾],𝑘+{𝛾}𝜄+𝑚+1,𝑘𝜖𝑛+2𝜑1(Θ).(3.19)

Analogous to the estimate of 𝑞𝑗(|𝑥|), we have 𝑞𝑗(|𝑥|)𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ).(3.20)

Thus, we can conclude that 𝑞𝑗(|𝑥|)𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ),(3.21) which yields 𝑈𝑎,6(𝑥)𝑀𝜖|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ).(3.22)

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+1𝜑11𝜁(Θ)) as |𝑥|, where 𝑥𝐻(𝑅𝜖,+)𝐸(𝜖;𝜇,𝑛𝜁). 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 Theorem 1.5

For any fixed 𝑥𝐻, take a number satisfying 𝑅>max{1,2|𝑥|}. By 𝜄+𝑚+1,𝑘𝜄+[𝛾],𝑘+{𝛾}𝑛+1, (1.5), (1.16), and (2.2), we have 𝜕𝐻(𝑅,)||𝑃(𝑎,𝑚)𝑥,𝑦||||𝑢𝑦||𝑑𝑦𝑀𝑉𝑚+1(|𝑥|)𝜑1(Θ)𝜕𝐻(𝑅,)||𝑢𝑦||||𝑦||𝜄+𝑚+1,𝑘+𝑛1𝑑𝑦𝑀|𝑥|𝜄+𝑚+1,𝑘𝜑1(Θ)𝜕𝐻(2|𝑥|,)||𝑦||𝜄+[𝛾],𝑘+{𝛾}𝜄+𝑚+1,𝑘𝑛+1𝑑𝑦𝑀|𝑥|𝜄+[𝛾],𝑘+{𝛾}𝑛+1𝜑1(Θ)<.(4.1) Then, 𝑈(𝑎,𝑚;𝑢)(𝑥) is absolutely convergent and finite for any 𝑥𝐻. Thus 𝑈(𝑎,𝑚;𝑢)(𝑥) is a solution of (1.2) on 𝐻.

Now we study the boundary behavior of 𝑈(𝑎,𝑚;𝑢)(𝑥). Let 𝑦𝜕𝐻 be any fixed point and 𝑙 any positive number satisfying 𝑙>max{|𝑦|+1,(1/2)𝑅}.

Set 𝜒𝑆(𝑙) as the characteristic function of 𝑆(𝑙)={𝑦𝜕𝐻,|𝑦|𝑙}, and write 𝑈(𝑎,𝑚;𝑢)(𝑥)=𝑈(𝑥)𝑈(𝑥)+𝑈(𝑥),(4.2) where 𝑈(𝑥)=𝜕𝐻[0,2𝑙]𝑃𝑎𝑥,𝑦𝑢𝑦𝑈𝑑𝑦,(𝑥)=𝜕𝐻(1,2𝑙]𝜕𝐾(𝑎,𝑚)(𝑥,𝑦)𝑈𝜕𝑛(𝑦)𝑢(𝑦)𝑑𝑦,(𝑥)=𝜕𝐻(2𝑙,)𝑃(𝑎,𝑚)𝑥,𝑦𝑢𝑦𝑑𝑦.(4.3)

Notice that 𝑈(𝑥) is the Poisson 𝑎-integral of 𝑢(𝑦)𝜒𝑆(2𝑙), We have lim𝑥𝑦,𝑥𝐻𝑈(𝑥)=𝑢(𝑦). Since limΘΦ𝜑𝑗𝑣(Θ)=0(𝑗=1,2,3;1𝑣𝑣𝑗) as 𝑥𝑦𝜕𝐻, we have lim𝑥𝑦,𝑥𝐻𝑈(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 function 𝑈(𝑎,𝑚;𝑢)(𝑥) can be continuously extended to 𝐻 such that lim𝑥𝑦,𝑥𝐻𝑈𝑦(𝑎,𝑚;𝑢)(𝑥)=𝑢(4.4) 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 Theorem 1.7

From Corollary 1.6, we have the solution 𝑈(𝑎,𝑚;𝑢)(𝑥) of the Dirichlet problem on 𝐻 with 𝑢 satisfying (1.31). Consider the function (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 0(𝑈(Ω,𝑎,𝑚;𝑢))+(𝑥)+(𝑥)+(𝑈(𝑎,𝑚;𝑢))(𝑥)(5.1) for any 𝑥𝐻, we have lim|𝑥|,𝑥𝐻|𝑥|𝜄+𝑚+1,𝑘(𝑈(Ω,𝑎,𝑚;𝑢))+(𝑥)=0(5.2) from (1.32) and (1.33). Then, the conclusion of Theorem 1.7 follows immediately from Lemma 2.2.


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Copyright © 2012 Baiyun Su. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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