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# Dirichlet problems of harmonic functions

## Abstract

In this paper, a solution of the Dirichlet problem in the upper half-plane isconstructed by the generalized Dirichlet integral with a fast growing continuousboundary function.

MSC: 31B05, 31B10.

## 1 Introduction and results

Let R be the set of all real numbers, and let C denote the complexplane with points $z=x+iy$, where $x,y∈R$. The boundary and closure of an open set Ω aredenoted by Ω and $Ω ¯$ respectively. The upper half-plane is the set$C + :={z=x+iy∈C:y>0}$, whose boundary is $∂ C + =R$. Let $[d]$ denote the integer part of the positive real numberd.

Given a continuous function f in $∂ C +$, we say that h is a solution of the(classical) Dirichlet problem in $C +$ with f, if $Δh=0$ in $C +$ and $lim z ∈ C + , z → t h(z)=f(t)$ for every $t∈∂ C +$.

The classical Poisson kernel in $C +$ is defined by

$P(z,t)= y π | z − t | 2 ,$

where $z=x+iy∈ C +$ and $t∈R$.

It is well known (see [1, 2]) that the Poisson kernel $P(z,t)$ is harmonic for $z∈C−{t}$ and has the expansion

$P(z,t)= 1 π Im ∑ k = 0 ∞ z k t k + 1 ,$

which converges for $|z|<|t|$. We define a modified Cauchy kernel of$z∈ C +$ by

where m is a nonnegative integer.

To solve the Dirichlet problem in $C +$, as in , we use the following modified Poisson kernel defined by

We remark that the modified Poisson kernel $P m (z,t)$ is harmonic in $C +$.

Put

$U m (f)(z)= ∫ − ∞ ∞ P m (z,t)f(t)dt,$

where $f(t)$ is a continuous function in $∂ C +$.

We say that u is of order λ if

$λ= lim sup r → ∞ log ( sup H ∩ B ( r ) | u | ) log r .$

If $λ<∞$, then u is said to be of finite order. SeeHayman-Kennedy [, Definition 4.1].

In case $λ<∞$, about the solution of the Dirichlet problem withcontinuous data in H, we refer readers to the following result, which isdue to Nevanlinna (see ).

Theorem A Let u be a nonnegative real-valued function harmonic in$C +$and continuous in$C ¯ +$. If

$∫ − ∞ ∞ u ( t ) 1 + t 2 dt<∞,$

then there exists a nonnegative real constant d such that

$u(z)=dy+ ∫ − ∞ ∞ P(z,t)u(t)dt$

for all$z=x+iy∈ C +$.

Inspired by Theorem A, we consider the Dirichlet problem for harmonic functions ofinfinite order in $C +$. To do this, we define a nondecreasing andcontinuously differentiable function $ρ(R)≥1$ on the interval $[0,+∞)$. We assume further that

$ε 0 = lim sup R → ∞ ρ ′ ( R ) R log R ρ ( R ) <1.$
(1.1)

Remark For any ϵ ($0<ϵ<1− ϵ 0$), there exists a sufficiently large positive numberR such that $r>R$, by (1.1) we have

$ρ(r)<ρ(e) ( ln r ) ϵ 0 + ϵ .$

Let $F(ρ,α)$ be the set of continuous functions f on$∂ C +$ such that

$∫ − ∞ ∞ | f ( t ) | 1 + | t | ρ ( | t | ) + α + 1 dt<∞,$
(1.2)

where α is a positive real number.

Now we show the solution of the Dirichlet problem with continuous data in$C +$. For similar results in a cone, we refer readers tothe paper by Qiao (see [7, 8]). For similar results with respect to the Schrödinger operator in ahalf-space, we refer readers to the paper by Ren, Su and Yang (see ).

Theorem 1 If$f∈F(ρ,α)$, then the integral$U [ ρ ( | t | ) + α ] (f)(z)$is a solution of the Dirichlet problem in$C +$with f.

The following result is obtained by putting $[ρ(|t|)+α]=m$ in Theorem 1.

Corollary If f is a continuous function in $∂ C +$ satisfying

$∫ − ∞ ∞ | f ( t ) | 1 + | t | m + 2 dt<∞,$

then$U m (f)(z)$is a solution of the Dirichlet problem in$C +$with f.

Theorem 2 Let u be a real-valued function harmonic in$C +$and continuous in$C ¯ +$. If$u∈F(p,ρ,α)$, then we have$u(z)= U [ ρ ( | t | ) + α ] (u)(z)+ImΠ(z)$for all$z∈ C ¯ +$, where$Π(z)$is an entire function in$C +$and vanishes continuously in$∂ C +$.

## 2 Proof of Theorem 1

By a simple calculation, we have the following inequality:

$| C m ( z , t ) | ≤M | z | m + 1 | t | − m − 2$
(2.1)

for any $z∈ C +$ and $t∈∂ C +$ satisfying $|t|≥max{1,2|z|}$, where M is a positive constant.

Take a number r satisfying $r>R$, where R is a sufficiently large positivenumber. For any ϵ ($0<ϵ<1− ϵ 0$), from Remark we have

$ρ(r)<ρ(e) ( ln r ) ( ϵ 0 + ϵ ) ,$

which yields that there exists a positive constant $M(r)$ dependent only on r such that

$k − α / 2 ( 2 r ) ρ ( k + 1 ) + α + 1 ≤M(r)$
(2.2)

for any $k> k r =[2r]+1$.

For any $z∈ C +$ and $|z|≤r$, we have by (1.2), (2.1), (2.2),$1/p+1/q=1$ and Hölder’s inequality

$M ∑ k = k r ∞ ∫ k ≤ | t | < k + 1 | z | [ ρ ( | t | ) + α ] + 1 | t | [ ρ ( | t | ) + α ] + 2 | f ( t ) | d t ≤ M ∑ k = k r ∞ r ρ ( k + 1 ) + α + 1 k α / 2 ∫ k ≤ | t | < k + 1 | f ( t ) | | t | ρ ( | t | ) + 1 + p α / 2 d t ≤ 2 M M ( r ) ∫ | t | ≥ k r | f ( t ) | 1 + | t | ρ ( | t | ) + 1 + p α / 2 d t < ∞ .$

Thus $U [ ρ ( | t | ) + α ] (f)(z)$ is finite for any $z∈ C +$. Since $P [ ρ ( | t | ) + α ] (z,t)$ is a harmonic function of $z∈ C +$ for any fixed $t∈∂ C +$, $U [ ρ ( | t | ) + α ] (f)(z)$ is also a harmonic function of$z∈ C +$.

Now we shall prove the boundary behavior of $U [ ρ ( | t | ) + α ] (f)(z)$. For any fixed boundary point $t ′ ∈∂ C +$, we can choose a number T such that$T>| t ′ |+1$. Now we write

$U [ ρ ( | t | ) + α ] (f)(z)= I 1 (z)− I 2 (z)+ I 3 (z),$

where

$I 1 ( z ) = ∫ | t | ≤ 2 T P ( z , t ) f ( t ) d t , I 2 ( z ) = Im ∑ k = 0 [ ρ ( | t | + α ) ] ∫ 1 < | t | ≤ 2 T z k π t k + 1 f ( t ) d t , I 3 ( z ) = ∫ | t | > 2 T P [ ρ ( | t | + α ) ] ( z , t ) f ( t ) d t .$

Note that $I 1 (z)$ is the Poisson integral of $u(t) χ [ − 2 T , 2 T ] (t)$, where $χ [ − 2 T , 2 T ]$ is the characteristic function of the interval$[−2T,2T]$. So it tends to $f( t ′ )$ as $z→ t ′$. Clearly, $I 2 (z)$ vanishes on $∂ C +$. Further, $I 3 (z)=O(y)$, which tends to zero as $z→ t ′$. Thus the function $U [ ρ ( | t | ) + α ] (f)(z)$ can be continuously extended to$C ¯ +$ such that $U [ ρ ( | t | ) + α ] (f)( t ′ )=f( t ′ )$ for any $t ′ ∈∂ C +$. Theorem 1 is proved.

## 3 Proof of Theorem 2

Consider that the function $u(z)− U [ ρ ( | t | ) + α ] (u)(z)$, which is harmonic in $C +$, can be continuously extended to$C ¯ +$ and vanishes in $∂ C +$.

The Schwarz reflection principle [, p.68] applied to $u(z)− U [ ρ ( | t | ) + α ] (u)(z)$ shows that there exists an entire harmonic function$Π(z)$ in $C +$ satisfying $Π( z ¯ )= Π ( z ) ¯$ such that $ImΠ(z)=u(z)− U [ ρ ( | t | ) + α ] (u)(z)$ for $z∈ C ¯ +$.

Thus $u(z)= U [ ρ ( | t | ) + α ] (u)(z)+ImΠ(z)$ for all $z∈ C ¯ +$, where $Π(z)$ is an entire function in $C +$ and vanishes continuously in $∂ C +$. Then we complete the proof of Theorem 2.

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## Acknowledgements

The authors are thankful to the referees for their helpful suggestions andnecessary corrections in the completion of this paper.

## Author information

Correspondence to Gang Xu.

### Competing interests

The authors declare that there is no conflict of interests regarding the publicationof this article.

### Authors’ contributions

All authors contributed equally to the manuscript and read and approved the finalmanuscript.

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• #### DOI

https://doi.org/10.1186/1687-2770-2013-262

### Keywords

• Dirichlet problem
• harmonic function
• half-plane 