Dirichlet problem for the Schrödinger operator on a cone
© Qiao and Deng; licensee Springer 2012
Received: 16 February 2012
Accepted: 2 May 2012
Published: 18 June 2012
In this article, a solution of the Dirichlet problem for the Schrödinger operator on a cone is constructed by the generalized Poisson integral with a slowly growing continuous boundary function. 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.
KeywordsDirichlet problem stationary Schrödinger equation cone
1 Introduction and results
Let R and be the set of all real numbers and the set of all positive real numbers respectively. We denote the n-dimensional Euclidean space by (). A point in is denoted by , where . The Euclidean distance between two points P and Q in is denoted by . Also with the origin O of is simply denoted by . The boundary and the closure of a set S in are denoted by ∂ S and respectively.
We introduce a system of spherical coordinates , , in which are related to Cartesian coordinates by .
The unit sphere and the upper half unit sphere in are denoted by and , respectively. For simplicity, a point on and the set for a set Ω, , are often identified with Θ and Ω, respectively. For two sets and , the set in is simply denoted by .
For and , let denote an open ball with a center at P and radius r in . . By , we denote the set in with the domain Ω on . We call it a cone. We denote the sets and with an interval on R by and . By we denote . By we denote which is . We denote the -dimensional volume elements induced by the Euclidean metric on by .
Let denote the class of nonnegative radial potentials , i.e., , , such that with some if and with if or .
where , Δ is the Laplace operator and . These solutions called a-harmonic functions or generalized harmonic functions are associated with the operator . Note that they are (classical) harmonic functions in the case . Under these assumptions, the operator can be extended in the usual way from the space to an essentially self-adjoint operator on (see [1–3]). We will denote it as well. This last one has a Green’s function . Here is positive on and its inner normal derivative . We denote this derivative by , which is called the Poisson a-kernel with respect to . We remark that and are the Green’s function and Poisson kernel of the Laplacian in respectively.
for every . Note that h is a solution of the classical Dirichlet problem for the Laplacian in the case .
for (see Courant and Hilbert ), where .
In order to ensure the existences of (). We put a rather strong assumption on Ω: if , then Ω is a -domain () on surrounded by a finite number of mutually disjoint closed hypersurfaces (e.g., see , pp. 88-89] for the definition of -domain). Then () and on ∂ Ω (here and below, denotes differentiation along the interior normal).
where the symbol denotes a constant depending only on n.
normalized under the condition .
In the rest of the article, we assume that and we shall suppress this assumption for simplicity. Further, we use the standard notations , , is the integer part of d and , where d is a positive real number.
where and are some positive constants.
where and , is their Wronskian. The series converges uniformly if either or ().
where is a continuous function on and is a surface area element on .
With regard to classical solutions of the Dirichlet problem for the Laplacian, Yoshida and Miyamoto , Theorem 1] proved the following result.
Our first aim is to give growth properties at infinity for .
Theorem 1 Let (resp. ), (resp. ) and
Next, we are concerned with solutions of the Dirichlet problem for the Schrödinger operator on .
Theorem 2 Let γ andbe as in Theorem 1. If u is a continuous function onsatisfying (1.6), thenis a solution of the Dirichlet problem for the Schrödinger operator onwith u and (1.7) (resp. (1.8)) holds.
If we take , then we immediately have the following corollary, which is just Theorem A in the case .
By using Corollary, we can give a solution of the Dirichlet problem for any continuous function on .
Throughout this article, let M denote various constants independent of the variables in questions, which may be different from line to line.
for anyand any.
Lemma 2 (see )
for anyandsatisfying (), whereis a constant dependent of n, m and s.
Lemma 3 (see , Theorem 1])
Ifis a solution of Equation (1.1) onsatisfying
(2.6) gives that (2.5) holds, from which the conclusion immediately follows. □
3 Proof of Theorem 1
We only prove the case , the remaining case can be proved similarly.
which is similar to the estimate of .
and divide into two sets and .
which is similar to the estimate of .
where is a positive integer satisfying .
Combining (3.3)–(3.11), we obtain that if is sufficiently large and ϵ is sufficiently small, then as , where . Then we complete the proof of Theorem 1.
4 Proof of Theorem 2
Thus is finite for any . Since is a generalized harmonic function of for any fixed , is also a generalized harmonic function of . That is to say, is a solution of Equation (1.1) on .
Now we study the boundary behavior of . Let be any fixed point and l be any positive number satisfying .
Notice that is the Poisson a-integral of , we have . Since (; ) as , we have from the definition of the kernel function . , and therefore tends to zero.
for any from the arbitrariness of l. Thus we complete the proof of Theorem 2 from Theorem 1.
5 Proof of Theorem 3
From Corollary, we have the solution of the Dirichlet problem on with u satisfying (1.9). Consider the function . Then it follows that this is the solution of Equation (1.1) in and vanishes continuously on .
from (1.10) and (1.11). Then the conclusions of Theorem 3 follow immediately from Lemma 4.
The authors would like to thank anonymous reviewers for their valuable comments and suggestions about improving the quality of the manuscript. This work is supported by The National Natural Science Foundation of China under Grant 11071020 and Specialized Research Fund for the Doctoral Program of Higher Education under Grant 20100003110004.
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