A remark on the Dirichlet problem in a half-plane
© Zhao and Yamada Jr.; licensee Springer. 2014
Received: 21 September 2014
Accepted: 27 November 2014
Published: 12 December 2014
In this paper, we prove that if the positive part of a harmonic function in a half-plane satisfies a slowly growing condition, then its negative part can also be dominated by a similarly growing condition. Further, a solution of the Dirichlet problem in a half-plane for a fast growing continuous boundary function is constructed by the generalized Dirichlet integral with this boundary function.
Keywordsharmonic function Dirichlet problem half-plane
1 Introduction and main theorem
Let R be the set of all real numbers and let C denote the complex plane with points , where . The boundary and closure of an open set Ω are denoted by ∂ Ω and , respectively. The upper half-plane is the set , whose boundary is .
We use the standard notations , , and is the integer part of the positive real number d. For positive functions and , we say that if for some positive constant M.
Given a continuous function f in , we say that h is a solution of the (classical) Dirichlet problem in with f, if in and for every .
where and .
where m is a nonnegative integer.
where is a continuous function in .
We also denote by the set of all continuous functions in , harmonic in with and .
Theorem A Let u be a real-valued function harmonic in and continuous in . If , then there exists a constant such that for all .
Inspired by Theorem A, we first prove the following.
Theorem 1 If and , then .
Then we are concerned with the growth property of at infinity in .
If , then u is said to be of finite order. See Hayman-Kennedy [, Definition 4.1].
where β is a positive real number.
Theorem 3 If , then the integral is a solution of the Dirichlet problem in with f.
The following result immediately follows from Theorem 2 (the case ) and Theorem 3 (the case ).
For harmonic functions of finite order in , we have the following integral representations.
If , then is a harmonic function in and can be continuously extended to such that for . There exists a constant such that for all .
If , then is a harmonic function in and can be continuously extended to such that for . There exists a harmonic polynomial of degree at most which vanishes in such that for all .
Finally, we prove the following.
for all , where is an entire function in and vanishes continuously in .
2 Main lemmas
Thus this lemma is proved. □
Lemma 3 (see [, Theorem 10])
then in , where is a polynomial of of degree less than m and even with respect to the variable y.
3 Proof of Theorem 1
We distinguish the following two cases.
Case 1. .
Thus from .
Case 2. .
Then from . We complete the proof of Theorem 1.
4 Proof of Theorem 2
Combining (4.2)-(4.5), (1.3) holds. Thus we complete the proof of Theorem 2.
5 Proof of Theorem 3
for any .
from (1.5), (2.2), and (5.1). Thus is finite for any . is a harmonic function of for any fixed . is also a harmonic function of .
Since is the Poisson integral of , it tends to as . Clearly, vanishes in . Further, , which tends to zero as . Thus the function can be continuously extended to such that for any . Then Theorem 3 is proved.
6 Proof of Corollary 2
from (6.1), (6.2), Corollary 1, and Lemma 3, where is a polynomial in of degree at most and even with respect to the variable y. From this we evidently obtain (II).
It immediately follows that for every , which is the conclusion of (I). Thus we complete the proof of Corollary 2.
7 Proof of Theorem 4
Consider the function , which is harmonic in , can be continuously extended to and vanishes in .
The Schwarz reflection principle [, p.68] applied to shows that there exists a harmonic function in satisfying such that for . Thus for all , where is an entire function in and vanishes continuously in . Thus we complete the proof of Theorem 4.
The authors are very thankful to the anonymous referees for their valuable comments and constructive suggestions, which helped to improve the quality of the paper. This work is supported by the Academy of Finland Grant No. 176512.
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