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Research | Open | Published:

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,yR. 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+iyC: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 tR.

It is well known (see [1, 2]) that the Poisson kernel P(z,t) is harmonic for zC{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 ofz C + by

C m (z,t)= { 1 π 1 t z when  | t | 1 , 1 π 1 t z 1 π k = 0 m z k t k + 1 when  | t | > 1 ,

where m is a nonnegative integer.

To solve the Dirichlet problem in C + , as in [3], we use the following modified Poisson kernel defined by

P m (z,t)=Im C m (z,t)= { P ( z , t ) when  | t | 1 , P ( z , t ) 1 π Im k = 0 m z k t k + 1 when  | t | > 1 .

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 [[1], 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 [46]).

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 allz=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 [912]).

Theorem 1 IffF(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 ¯ + . IfuF(p,ρ,α), then we haveu(z)= U [ ρ ( | t | ) + α ] (u)(z)+ImΠ(z)for allz C ¯ + , whereΠ(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 ofz 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 thatT>| 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 [[4], p.68] applied to u(z) U [ ρ ( | t | ) + α ] (u)(z) shows that there exists an entire harmonic functionΠ(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.

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Correspondence to Gang Xu.

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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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Keywords

  • Dirichlet problem
  • harmonic function
  • half-plane