Open Access

Dirichlet problems of harmonic functions

Boundary Value Problems20132013:262

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

Received: 6 September 2013

Accepted: 5 November 2013

Published: 2 December 2013

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.

Keywords

Dirichlet problem harmonic function half-plane

1 Introduction and results

Let R be the set of all real numbers, and let C denote the complexplane with points z = x + i y , 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 + i y 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 Δ h = 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 + i y 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
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 ) d t ,

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 d t < ,
then there exists a nonnegative real constant d such that
u ( z ) = d y + P ( z , t ) u ( t ) d t

for all z = x + i y 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 d t < ,
(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 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 d t < ,

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 ) + Im Π ( 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 = [ 2 r ] + 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 [ 2 T , 2 T ] . 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 function Π ( z ) in C + satisfying Π ( z ¯ ) = Π ( z ) ¯ such that Im Π ( z ) = u ( z ) U [ ρ ( | t | ) + α ] ( u ) ( z ) for z C ¯ + .

Thus u ( z ) = U [ ρ ( | t | ) + α ] ( u ) ( z ) + Im Π ( 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.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
College of Computer and Information Engineering, Henan University of Economics and Law
(2)
College of Applied Mathematics, Chengdu University of Information Technology
(3)
Department of Mathematics, Henan University of Economics and Law

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Copyright

© Xu et al.; licensee Springer. 2013

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