Abstract
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.
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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 .
The classical Poisson kernel in is defined by
where and .
It is well known (see [1]) that the Poisson kernel is harmonic for and has the expansion
which converges for . We define a modified Cauchy kernel of by
where m is a nonnegative integer.
To solve the Dirichlet problem in , as in [2], we use the modified Poisson kernel defined by
We remark that the modified Poisson kernel is harmonic in . About modified Poisson kernel in a cone, we refer readers to papers by I Miyamoto, H Yoshida, L Qiao and GT Deng (e.g. see [3–11]).
Put
where is a continuous function in .
For any positive real number α, We denote by the space of all measurable functions in satisfying
and by the set of all measurable functions in such that
We also denote by the set of all continuous functions in , harmonic in with and .
About the solution of the Dirichlet problem with continuous data in , we refer readers to the following result (see [12, 13]).
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 .
Theorem 2 If and , then
We say that u is of order λ if
If , then u is said to be of finite order. See Hayman-Kennedy [[14], Definition 4.1].
Our next aim is to give solutions of the Dirichlet problem for harmonic functions of infinite order in . For this purpose, we define a nondecreasing and continuously differentiable function on the interval . We assume further that
Remark For any ϵ (), there exists a sufficiently large positive number R such that , by (1.4) we have
Let be the set of continuous functions f in such that
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 ).
Corollary 1 If f is a continuous function in satisfying
then is a solution of the Dirichlet problem in with f satisfying
For harmonic functions of finite order in , we have the following integral representations.
Corollary 2 Let () and let m be an integer such that .
-
(I)
If , then is a harmonic function in and can be continuously extended to such that for . There exists a constant such that for all .
-
(II)
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.
Theorem 4 Let u be a real-valued function harmonic in and continuous in . If , then we have
for all , where is an entire function in and vanishes continuously in .
2 Main lemmas
The Carleman formula refers to holomorphic functions in (see [15, 16]).
Lemma 1 If and () is a harmonic function in with continuous boundary in , then we have
where
and
Lemma 2 For any , , and , we have
where ,
where ,
where .
Proof If and , we have , which gives
If , we obtain
If and , then we also have , which yields
Thus this lemma is proved. □
Lemma 3 (see [[17], Theorem 10])
Let be a harmonic function in such that vanishes continuously in . If
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. .
If , Lemma 1 gives
Since , we obtain
from (1.1) and hence
Then from (1.2), (3.1), and (3.2) we have
which gives
Thus from .
Case 2. .
Since , we see from (1.1) that
and we see from (1.2) that
We have from (3.3), (3.4), and Lemma 1
Set
We have
from the L’Hospital’s rule and hence we have
So
Then from . We complete the proof of Theorem 1.
4 Proof of Theorem 2
For any , there exists such that
from Theorem 1. For any fixed and , we write
where
By (2.1), (2.2), (2.3), and (4.1), we have the following estimates:
Combining (4.2)-(4.5), (1.3) holds. Thus we complete the proof of Theorem 2.
5 Proof of Theorem 3
Take a number r satisfying , where is a sufficiently large positive number. For any ϵ (), we have
from the remark, which shows that there exists a positive constant dependent only on r such that
for any .
For any and , we have and
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 .
Now we shall prove the boundary behavior of . For any fixed , we can choose a number such that . We write
where
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
We prove (II). Consider the function . Then it follows from Corollary 1 that this is harmonic in and vanishes continuously in . Since
for any and
from (1.1), for every we have
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).
If , then for . (II) shows that there exists a constant such that
Put
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 [[12], 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.
Change history
28 April 2020
The Editor-in-Chief has retracted this article [1] because it shows significant overlap with a previously published article [2]. The article also shows evidence of peer review manipulation. In addition, the identity of the corresponding author could not be verified: Stockholms Universitet have confirmed that Alexander Yamada has not been affiliated with their institution. The authors have not responded to any correspondence regarding this retraction.
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Acknowledgements
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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All authors contributed equally to the manuscript and read and approved the final manuscript.
The Editor-in-Chief has retracted this article because it shows significant overlap with a previously published article. The article also shows evidence of peer review manipulation. In addition, the identity of the corresponding author could not be verified: Stockholms Universitet have confirmed that Alexander Yamada has not been affiliated with their institution. The authors have not responded to any correspondence regarding this retraction.
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Zhao, T., Yamada, A. RETRACTED ARTICLE: A remark on the Dirichlet problem in a half-plane. J Inequal Appl 2014, 497 (2014). https://doi.org/10.1186/1029-242X-2014-497
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DOI: https://doi.org/10.1186/1029-242X-2014-497