Hindawi Publishing Corporation
Journal of Complex Analysis
Volume 2016, Article ID 8097095, 8 pages
http://dx.doi.org/10.1155/2016/8097095
Research Article
Dirichlet Problem for Complex Poisson Equation in
a Half Hexagon Domain
Bibinur Shupeyeva
Nazarbayev University, 53 Kabanbay Batyr Avenue, Astana 010000, Kazakhstan
Correspondence should be addressed to Bibinur Shupeyeva; [email protected]
Received 21 October 2015; Revised 15 December 2015; Accepted 10 January 2016
Academic Editor: Vladislav Kravchenko
Copyright © 2016 Bibinur Shupeyeva. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The parqueting-reflection method is applied to a nonregular domain and the harmonic Green function for the half hexagon is
constructed. The related Dirichlet problem for the Poisson equation is solved explicitly.
1. Introduction
The basic boundary value problems for the second-order
complex partial differential equations are the harmonic
Dirichlet and Neumann problems for the Laplace and Poisson
equations. In order to find the solution in explicit or closed
form diverse methods have been applied. In case a given
domain 𝐷 is simply connected and has a piecewise smooth
boundary 𝜕𝐷 the tools of complex analysis such as Schwarz
reflection principle and conformal mapping serve perfectly.
When a given domain 𝐷 is piecewise smooth polygonal and
has corners the Schwarz-Christoffel formula can be used.
Difficulties arise since the elliptic integrals appearing in the
formula imply complicated computations and need to be
solved numerically. As analogue to this formula, another
method can be applied which gives the covering of the entire
complex plane C by reflection of the given domain 𝐷 at
its boundary. The method is fully described in numerous
papers of Begehr and other authors; see, for example [1–12].
Our aim is to find the solution of the Dirichlet boundary
value problem for the Poisson equation through the Poisson
integral formula. It is known that the Poisson kernel function
is an analogue of the Cauchy kernel for the analytic functions
and the Poisson integral formula solves the Dirichlet problem
for the inhomogeneous Laplace equation. One way to obtain
the Poisson kernel leads to the harmonic Green function
which is to be constructed by use of the parqueting-reflection
method.
In this paper we first consider the half hexagon domain
and implement the parqueting-reflection method. The reflection points treated in a proper way help to construct the
certain meromorphic functions needed to find the harmonic
Green function and representation formula. The later one
provides the solution to the harmonic Dirichlet problem
which is shown in the last part.
2. Half Hexagon Domain and Poisson Kernel
We consider a polygonal domain with corner points. The half
hexagon denoted as 𝑃+ with four corner points at 2, 1 + 𝑖√3,
−1 + 𝑖√3, and −2 lies in the upper half plane. A point 𝑧 ∈ 𝑃+
will later serve as a pole of the Green function. Its complex
conjugate 𝑧 does not lie in 𝑃+ . 𝑃+ is reflected at the real
axis so that the entire hexagon 𝑃 (Figure 1) is obtained. The
pole 𝑧 is reflected onto 𝑧 which will later become a zero of a
certain meromorphic function related to the Green function.
The points 𝑧 and 𝑧 from 𝑃 are reflected again through all the
sides of the hexagon, starting with the right upper side and
continuing in a positive direction. The successive reflections
of 𝑧 give the points, which will later become zeros of the
meromorphic function mentioned above. They are
1
− (1 + 𝑖√3) 𝑧 + 3 + 𝑖√3,
2
𝑧 + 2𝑖√3,
2
Journal of Complex Analysis
P2
−1 + i√3
P3
1 + i√3
P+
Z
P
−2
P1
2
Z
P4
P6
P5
Figure 1: Hexagons.
1
− (1 − 𝑖√3) 𝑧 − 3 + 𝑖√3,
2
which are connected by the relations 𝑧̌ 2 = 𝑧1 − 6 − 2𝑖√3 and
𝑧̌ 1 = 𝑧2 − 6 − 2𝑖√3.
In general, all reflection points are either given by
1
− (1 + 𝑖√3) 𝑧 − 3 − 𝑖√3,
2
𝑧 + 𝜔𝑚𝑛 ,
𝑧 − 2𝑖√3,
𝑧1 + 𝜔𝑚𝑛 ,
1
− (1 − 𝑖√3) 𝑧 + 3 − 𝑖√3.
2
𝑧2 + 𝜔𝑚𝑛 ,
(1)
𝑧 + 𝜔𝑚𝑛 ,
Reflection of the point 𝑧 ∈ 𝑃 defines the poles of the
meromorphic function in the hexagons 𝑃1 , . . . , 𝑃6 . These
points in turn are reflected through the sides of the new
hexagons, except for reflecting to the original hexagon 𝑃.
Hence each hexagon includes now 3 poles and 3 zeros.
Continuation of these operations reveals that all the points
have the same coefficients of rotation: 1, −(1/2)(1 + 𝑖√3),
−(1/2)(1 − 𝑖√3), and displacement 3𝑚 + 𝑖√3𝑛, 𝑚 + 𝑛 ∈ 2Z.
Note that reflection includes rotation and shifting and the
points from one hexagon can be expressed through the points
of another one. In general the points from the hexagons differ
by displacements 6𝑚 in the direction of the real and 2𝑖√3𝑛 in
the direction of the imaginary axes. Thus the main period is
𝜇𝑚𝑛 = 6𝑚 + 2𝑖√3𝑛.
Obviously, the repeated reflections of the point 𝑧 ∈ 𝑃+ are
representable in different ways, using either of the points
𝑧1 + 𝜔𝑚𝑛 ,
1
𝑧1 = − (1 + 𝑖√3) 𝑧 + 3 + 𝑖√3,
2
1
𝑧2 = − (1 + 𝑖√3) 𝑧 + 3 + 𝑖√3 or
2
1
𝑧̌ 1 = − (1 − 𝑖√3) 𝑧 − 3 + 𝑖√3,
2
1
𝑧̌ 2 = − (1 − 𝑖√3) 𝑧 − 3 + 𝑖√3,
2
(2)
(3)
𝑧2 + 𝜔𝑚𝑛
or by
𝑧 + 𝜔𝑚𝑛 ,
𝑧̌ 1 + 𝜔𝑚𝑛 ,
𝑧̌ 2 + 𝜔𝑚𝑛 ,
𝑧 + 𝜔𝑚𝑛 ,
(4)
𝑧̌ 1 + 𝜔𝑚𝑛 ,
𝑧̌ 2 + 𝜔𝑚𝑛 ,
where 𝜔𝑚𝑛 = 3𝑚 + 𝑖√3𝑛 such that 𝑚 + 𝑛 ∈ 2Z.
We choose zeros as direct reflection of poles and poles as
direct reflection of zeros. Then having a set of zeros and a set
of poles, one can construct the Schwarz kernel for 𝑃+ and treat
the related Schwarz problem [9] and Riemann-Hilbert-type
boundary value problem.
The half hexagon can be viewed as the complement of the
intersection of four half planes. We define them by 𝐻1− being
the right-hand half plane which has the boundary line passing
through the points 2 and 1 + 𝑖√3, 𝐻2− being the upper half
Journal of Complex Analysis
3
plane with the border line through the points ±1 + 𝑖√3, 𝐻3−
being the left-hand half plane with the border line passing
through the points −1 + 𝑖√3 and −2, and 𝐻4− being the half
plane which is below the real axis.
Let then 𝐻1+ , 𝐻2+ , 𝐻3+ , 𝐻4+ be the complementary half
planes of those listed above. The Green functions of these half
planes are, in fact, the Green functions for the complementary
half planes 𝐻1− , . . . , 𝐻4− . The outward normal derivatives of
the Green function on the boundary is the Poisson kernel.
The kernel provides the boundary condition 𝑤 = 𝛾 in the
Dirichlet problem.
The Poisson kernels can be found from the respective
Green functions 𝐺1 (𝑧, 𝜁), 𝑧 = 𝑥 + 𝑖𝑦, 𝜁 = 𝜉 + 𝑖𝜂 as described
below.
For the half plane 𝐻1+ with the boundary described by the
relation 𝜁 − 2 = −(1/2)(1 + 𝑖√3)(𝜁 − 2) we have
󵄨2
󵄨󵄨
󵄨󵄨 (1/2) (1 + 𝑖√3) (𝜁 − 2) + 𝑧 − 2 󵄨󵄨󵄨
󵄨󵄨 ,
󵄨
󵄨
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨
󵄨󵄨
𝜁−𝑧
󵄨󵄨
󵄨󵄨
󵄨
󵄨
Finally, for the half plane 𝐻4+ with the boundary described
by 𝜁 = 𝜁, we have
󵄨2
󵄨󵄨
󵄨󵄨 𝜁 − 𝑧 󵄨󵄨󵄨
󵄨󵄨 ,
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨󵄨
󵄨󵄨 𝜁 − 𝑧 󵄨󵄨󵄨
󵄨
󵄨
𝑧, 𝜁 ∈ 𝐻4+ ,
1
1 𝑧−𝑧
, 𝜁 ∈ 𝜕𝐻4+ , 𝑧 ∈ 𝐻4+ .
− 𝜕]𝜁 𝐺1 (𝑧, 𝜁) = 󵄨
2
𝑖 󵄨󵄨𝜁 − 𝑧󵄨󵄨󵄨2
󵄨
󵄨
(8)
3. Green Representation Formula
The method of reflections helps to find the harmonic Green
function; see [3–5]. The reflection points given in (3) or (4)
are used to construct a meromorphic function:
𝐵1 (𝑧, 𝜁)
(𝜁 − 𝑧 − 𝜔𝑚𝑛 ) (𝜁 − 𝑧1 − 𝜔𝑚𝑛 ) (𝜁 − 𝑧2 − 𝜔𝑚𝑛 )
𝑚+𝑛∈2Z (𝜁 − 𝑧 − 𝜔𝑚𝑛 ) (𝜁 − 𝑧1 − 𝜔𝑚𝑛 ) (𝜁 − 𝑧2 − 𝜔𝑚𝑛 ) (9)
= ∏
𝑧, 𝜁 ∈ 𝐻1+ ,
(√3 − 𝑖) 𝑧 − 𝑧1
1
− 𝜕]𝜁 𝐺1 (𝑧, 𝜁) = −
󵄨2 ,
󵄨󵄨
2
4
󵄨󵄨𝜁 − 𝑧󵄨󵄨󵄨
3
(5)
where 𝑧1 = −(1/2)(1 + 𝑖√3)𝑧 + 3 + 𝑖√3, 𝑧1 ∈
For the half plane 𝐻2+ the relation on the boundary is
given as 𝜁 = 𝜁 + 2𝑖√3; then
(6)
here 𝑧2 = 𝑧 + 2𝑖√3, 𝑧 ∈ 𝐻2+ .
The boundary of the half plane 𝐻3+ is described by 𝜁 + 2 =
−(1/2)(1 − 𝑖√3)(𝜁 + 2) and
󵄨2
󵄨󵄨
󵄨󵄨 (1/2) (1 − 𝑖√3) (𝜁 + 2) + 𝑧 + 2 󵄨󵄨󵄨
󵄨󵄨 ,
󵄨
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨󵄨
󵄨󵄨
𝜁−𝑧
󵄨󵄨
󵄨󵄨
󵄨
󵄨
√3 + 𝑖 𝑧 − 𝑧1̌
1
− 𝜕]𝜁 𝐺1 (𝑧, 𝜁) = −
,
2
4 󵄨󵄨󵄨𝜁 − 𝑧󵄨󵄨󵄨2
󵄨
󵄨
𝑚+𝑛∈2Z (𝜁
3
− 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
,
3
𝐻1+ .
𝑧, 𝜁 ∈ 𝐻3+ ,
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
where 𝑧1 = −(1/2)(1+𝑖√3)𝑧+3+𝑖√3, 𝑧2 = −(1/2)(1+𝑖√3)𝑧+
3 + 𝑖√3, or a function
𝜁 ∈ 𝜕𝐻1+ , 𝑧 ∈ 𝐻1+ ,
󵄨󵄨
󵄨2
󵄨󵄨 𝜁 − 𝑧 + 2𝑖√3 󵄨󵄨󵄨
󵄨󵄨 , 𝑧, 𝜁 ∈ 𝐻+ ,
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨󵄨
2
󵄨󵄨
󵄨󵄨
𝜁
−
𝑧
󵄨󵄨
󵄨
1 𝑧 − 𝑧2
1
, 𝜁 ∈ 𝜕𝐻2+ , 𝑧 ∈ 𝐻2+ ;
− 𝜕]𝜁 𝐺1 (𝑧, 𝜁) = − 󵄨
2
𝑖 󵄨󵄨𝜁 − 𝑧󵄨󵄨󵄨2
󵄨
󵄨
= ∏
(7)
𝐵2 (𝑧, 𝜁) = ∏
(𝜁 − 𝜔𝑚𝑛 + 2) − (𝑧 + 2)3
𝑚+𝑛∈2Z (𝜁
3
− 𝜔𝑚𝑛 + 2) − (𝑧 + 2)3
,
(10)
where 𝑧1̌ = −(1/2)(1−𝑖√3)𝑧−3+𝑖√3, 𝑧2̌ = −(1/2)(1−𝑖√3)𝑧−
3 + 𝑖√3. Here 𝑧 is considered as a parameter and 𝜁 ∈ C is the
variable.
For the boundary part 𝜕2 𝑃, the line from 1 + 𝑖√3 to
−1 + 𝑖√3, a meromorphic function 𝐵3 (𝑧, 𝜁), is deduced from
𝐵1 (𝑧, 𝜁) by rotating the variable 𝜁 and the parameter 𝑧 about
the angle 𝜋/3:
1
1
𝐵1 (− (1 + 𝑖√3) 𝑧, − (1 + 𝑖√3) 𝜁)
2
2
= ∏
3
3
3
3
(𝜁 − 𝜔𝑚𝑛 + 1 − 𝑖√3) − (𝑧 + 1 + 𝑖√3)
𝑚+𝑛∈2Z (𝜁
− 𝜔𝑚𝑛 + 1 − 𝑖√3) − (𝑧 + 1 − 𝑖√3)
(11)
,
which becomes 1 on the boundary 𝜕2 𝑃, where 𝜁 − 𝑖√3 = 𝜁 +
𝑖√3.
The following lemmas will be needed to prove the Green
representation formula below. The complete proofs of these
lemmas are given in [9].
Lemma 1. The infinite product
3
𝜁∈
𝜕𝐻3+ ,
𝑧∈
where 𝑧1̌ = −(1/2)(1 − 𝑖√3)𝑧 − 3 + 𝑖√3, 𝑧 ∈ 𝐻3+ .
𝐻3+ ,
∏
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
𝑚+𝑛∈2Z (𝜁
3
− 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
converges, where 𝜔𝑚𝑛 = 3𝑚 + 𝑖√3𝑛, 𝑚 + 𝑛 ∈ 2Z.
(12)
4
Journal of Complex Analysis
Lemma 2. The equalities 𝐵1 (𝑧, 𝜁) = 𝐵2 (𝑧, 𝜁) = 𝐵3 (𝑧, 𝜁) hold
for (𝑧, 𝜁) ∈ 𝑃+ × 𝜕𝑃+ .
The proof of this equality is based on the fact that the
functions 𝐵1 (𝑧, ⋅), 𝐵2 (𝑧, ⋅), since 𝐵3 (𝑧, ⋅) can be obtained from
𝐵1 (𝑧, ⋅), have the same poles and zeros; see [9].
The Green function must satisfy the following conditions;
see [13]:
(10 ) 𝐺1 (𝑧, 𝜁) is harmonic in 𝑃+ \ {𝑧};
(20 ) 𝐺1 (𝑧, 𝜁) + log |𝜁 − 𝑧|2 is harmonic in 𝜁 ∈ 𝑃+ for any
𝑧 ∈ 𝑃+ ;
0
(3 ) lim 𝜁→𝜕𝑃+ 𝐺(𝑧, 𝜁) = 0 for any 𝑧 ∈ 𝑃+ ;
and the additional properties:
(40 ) 𝐺1 (𝑧, 𝜁) = 𝐺1 (𝜁, 𝑧), 𝑧 and 𝜁 in 𝑃+ , 𝑧 ≠ 𝜁;
(50 ) 𝐺1 (𝑧, 𝜁) > 0, 𝑧 and 𝜁 in 𝑃+ , 𝑧 ≠ 𝜁.
We consider now the different forms of the Green function and take the derivatives 𝜕𝜁 𝐺1 (𝑧, 𝜁), 𝜕𝜁 𝐺1 (𝑧, 𝜁).
For the right-hand side, a boundary 𝜕1 𝑃+ , we choose the
form (14) for 𝜁 ∈ 𝜕𝑃+ , 𝑧 ∈ 𝑃+ . Here the outward normal
derivative is 𝜕]𝜁 = (√3/2 + 𝑖/2)𝜕𝜁 + (√3/2 − 𝑖/2)𝜕𝜁 ; then
𝜕]𝜁 𝐺1 (𝑧, 𝜁) = −3 (√3 + 𝑖) (𝜁 − 2)2
󵄨󵄨
3 󵄨2
3
󵄨󵄨
(𝑧 − 𝜔𝑚𝑛 − 2) − (𝜁 − 2) 󵄨󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨 ,
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨 ∏
󵄨󵄨𝑚+𝑛∈2Z (𝑧 − 𝜔 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑚𝑛
(14)
󵄨󵄨
3 󵄨2
3
󵄨󵄨
(𝑧 − 𝜔𝑚𝑛 + 2) − (𝜁 + 2) 󵄨󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨 ,
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨 ∏
󵄨󵄨𝑚+𝑛∈2Z (𝑧 − 𝜔 + 2)3 − (𝜁 + 2)3 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑚𝑛
(15)
since 𝜁 − 2 = −(1/2)(1 + 𝑖√3)(𝜁 − 2), (𝜁 − 2)3 = (𝜁 − 2)3 .
For the boundary part 𝜕4 𝑃+ , a line between (−2, 0), (2, 0)
on a real axis, the outward normal derivative is 𝜕]𝜁 = −𝑖(𝜕𝜁 −
𝜕]𝜁 𝐺1 (𝑧, 𝜁) = 6𝑖 (𝜁 − 2)2
3
3
(20)
(𝑧 − 𝜔𝑚𝑛 − 2) − (𝑧 − 𝜔𝑚𝑛 − 2)
⋅ ∑
.
󵄨󵄨
󵄨2
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨
𝑚+𝑛∈2Z
󵄨
󵄨
For the boundary part 𝜕3 𝑃+ on the left-hand side of 𝑃+ ,
we take form (15). The outward normal derivative is 𝜕]𝜁 =
(√3/2 − 𝑖/2)𝜕 + (√3/2 + 𝑖/2)𝜕 also here 𝜁 = 𝜁 ̌ = −(1/2)(1 −
𝜁
𝜁
𝑖√3)𝜁 − 3 + 𝑖√3 and (𝜁 + 2)3 = (𝜁 + 2)3 ; then
𝜕]𝜁 𝐺1 (𝑧, 𝜁) = −3 (√3 − 𝑖) (𝜁 + 2)2
3
3
(21)
(𝑧 − 𝜔𝑚𝑛 + 2) − (𝑧 − 𝜔𝑚𝑛 + 2)
⋅ ∑
.
󵄨󵄨
󵄨2
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 + 2)3 − (𝜁 + 2)3 󵄨󵄨󵄨
𝑚+𝑛∈2Z
󵄨
󵄨
+
For the upper boundary part 𝜕2 𝑃 , a line joining the points
±1+𝑖√3, form (16) is valid. Here 𝜕]𝜁 = 𝑖(𝜕𝜁 −𝜕𝜁 ) and 𝜁−𝑖√3 =
𝜁 + 𝑖√3; then 𝜕] 𝐺1 (𝑧, 𝜁) is
− 6𝑖 (𝜁 + 1 − 𝑖√3)
󵄨󵄨
3 󵄨2
3
(16)
󵄨󵄨
(𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3) − (𝜁 + 1 + 𝑖√3) 󵄨󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨 .
= log 󵄨󵄨 ∏
󵄨
3
3
󵄨󵄨𝑚+𝑛∈2Z (𝑧 − 𝜔 + 1 − 𝑖√3) − (𝜁 + 1 − 𝑖√3) 󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑚𝑛
Lemma 3. The function 𝐺1 (𝑧, 𝜁) has vanishing boundary
values on 𝜕𝑃+ ; that is,
(17)
Theorem 4 (see [13]). Any 𝑤 ∈ 𝐶2 (𝑃+ ; C) ∩ 𝐶1 (𝑃+ ; C) can be
represented as
𝑤 (𝑧) = −
1
∫ 𝑤 (𝜁) 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁
4𝜋 𝜕𝑃+
−
1
∫ 𝑤 (𝜁) 𝐺1 (𝑧, 𝜁) 𝑑𝜉 𝑑𝜂,
𝜋 𝑃+ 𝜁𝜁
1
𝜁
𝐺1 (𝑧, 𝜁)
lim 𝐺1 (𝑧, 𝜁) = 0.
(19)
𝜕𝜁 ), 𝜁 = 𝜁; then
By the properties (10 )–(30 ) the Green function 𝐺1 (𝑧, 𝜁) is
uniquely defined. Obviously, 𝐺1 (𝑧, 𝜁) as defined above is
harmonic in 𝜁 ∈ 𝑃+ \ {𝑧} as 𝐵1 (𝑧, 𝜁) is analytic in 𝑃+ up to a
single pole at 𝑧. Adding log |𝜁 − 𝑧|2 gives a harmonic function
of 𝜁 ∈ 𝑃+ . The symmetry property (40 ) is a consequence from
the properties (10 )–(30 ). The harmonic Green function for
the half hexagon 𝑃+ is
󵄨
󵄨2
󵄨2
󵄨
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨󵄨𝐵1 (𝑧, 𝜁)󵄨󵄨󵄨 = log 󵄨󵄨󵄨𝐵2 (𝑧, 𝜁)󵄨󵄨󵄨
(13)
󵄨2
󵄨
= log 󵄨󵄨󵄨𝐵3 (𝑧, 𝜁)󵄨󵄨󵄨
or, by the symmetry property,
𝜁→𝜁0 ∈𝜕𝑃+
3
3
(𝑧 − 𝜔𝑚𝑛 − 2) − (𝑧 − 𝜔𝑚𝑛 − 2)
⋅ ∑
,
󵄨󵄨󵄨(𝑧 − 𝜔 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨2
𝑚+𝑛∈2Z
󵄨󵄨
󵄨󵄨
𝑚𝑛
3
3
(𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3) − (𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3) (22)
⋅ ∑
.
3
3 󵄨󵄨2
󵄨󵄨󵄨
𝑚+𝑛∈2Z
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3) − (𝜁 + 1 − 𝑖√3) 󵄨󵄨󵄨
󵄨
󵄨
4. Harmonic Dirichlet Problem
The representation formula in Theorem 4 provides the solution to the Dirichlet problem for the Poisson equation.
At first the boundary behavior of the integral is to be
studied. Let for 𝛾 ∈ 𝐶(𝜕𝑃+ ; R)
𝜑 (𝑧) = −
(18)
where 𝑠𝜁 is the arc length parameter on 𝜕𝑃+ with respect to the
variable 𝜁 = 𝜉 + 𝑖𝜂 and 𝐺1 (𝑧, 𝜁) = 2𝐺(𝑧, 𝜁) is the harmonic
Green function for 𝑃+ .
2
1
∫ 𝛾 (𝜁) 𝜕]𝜁 𝐺 (𝑧, 𝜁) 𝑑𝑠𝜁 , 𝑧 ∈ 𝑃+ .
4𝜋 𝜕𝑃+
(23)
Lemma 5. For 𝛾 ∈ 𝐶(𝜕𝑃+ ; R) the function presented in (23)
satisfies the relation
lim 𝜑 (𝑧) = 𝛾 (𝜁0 ) ,
𝑧→𝜁0
where 𝜁0 is any fixed point on 𝜕𝑃+ \ {±2, ±1 + 𝑖√3}.
(24)
Journal of Complex Analysis
5
Proof. Let 𝜁0 be defined on different boundary parts and
consider the boundary behavior when 𝑧 → 𝜁0 .
+
Case 1. If 𝜁0 is taken on 𝜕1 𝑃 so that 𝜁0 = −(1/2)(1 + 𝑖√3)𝜁0 +
3 + 𝑖√3 then
1
2
(𝜁0 − 2) = − (1 − 𝑖√3) (𝜁0 − 2) ,
2
For 𝑚 = 𝑛 = 0 formula (19) gives
(𝑧 − 2)3 − (𝑧 − 2)3
− 3 (√3 + 𝑖) (𝜁 − 2)2 󵄨
󵄨󵄨(𝑧 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨2
󵄨󵄨
󵄨󵄨
= −3 (√3 + 𝑖) (𝜁 − 2)2
2
3
3
(26)
2
(25)
⋅
(𝜁0 − 2) = (𝜁0 − 2) .
On 𝜕1 𝑃 where 𝜁 = 𝜁1 = −(1/2)(1 + 𝑖√3)𝜁 + 3 + 𝑖√3, (𝜁 − 2)3 =
(𝜁 − 2)3 .
+
(𝑧 − 𝑧) [(𝑧 − 2)2 + (𝑧 − 2) (𝑧1 − 2) + (𝑧1 − 2) ]
.
󵄨󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2 󵄨󵄨󵄨󵄨(𝑧 − 2)2 + (𝑧 − 2) (𝜁 − 2) + (𝜁 − 2)2 󵄨󵄨󵄨󵄨2
󵄨 󵄨
󵄨
󵄨
Because 𝑧1 = −(1/2)(1 + 𝑖√3)𝑧 + 3 + 𝑖√3, then (𝑧1 − 2)3 =
(𝑧 − 2)3 . The limit in the following ratio as 𝑧 → 𝜁0 and 𝜁 = 𝜁0
gives
2
4
2
2
{ −3 (√3 + 𝑖) (𝜁 − 2) [(𝑧 − 2) + (𝑧 − 2) (𝑧1 − 2) + (𝑧1 − 2) ] } − (√3 + 𝑖) (𝜁0 − 2)
lim {
= (√3 − 𝑖) ,
}=
󵄨󵄨
󵄨󵄨
󵄨4
2 󵄨󵄨2
2
𝑧→𝜁0
𝜁0 − 2󵄨󵄨󵄨
󵄨
󵄨
󵄨
−
2)
+
−
2)
−
2)
+
−
2)
(𝑧
(𝑧
(𝜁
(𝜁
󵄨
󵄨
󵄨
{
}
󵄨
󵄨
(28)
3
= (𝑧1 − 𝜔𝑘𝑙 − 2) ,
3
(𝑧 − 𝜔𝑚𝑛 − 2)
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
󵄨󵄨
󵄨2
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨
󵄨
󵄨
(29)
3
=
∑
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
3
∑ 󵄨
󵄨󵄨
(𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3)
𝑚+𝑛∈2Z 󵄨󵄨󵄨(𝑧
which follows from the rearrangement of the indices in 𝜔𝑚𝑛
for certain 𝑘 + 𝑙 ∈ 2Z. Thus
∑
(𝑧1 − 𝜔𝑚𝑛 − 2)
󵄨󵄨
󵄨2 .
󵄨󵄨(𝑧1 − 𝜔𝑚𝑛 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨
󵄨
󵄨
3
3 󵄨󵄨2
− 𝜔𝑚𝑛 + 1 − 𝑖√3) − (𝜁 + 1 − 𝑖√3) 󵄨󵄨󵄨
󵄨
(32)
3
󵄨󵄨2
󵄨󵄨
󵄨󵄨
󵄨󵄨
√
(𝑧
−
𝜔
+
1
−
𝑖
3)
󵄨󵄨
󵄨󵄨
1
𝑚𝑛
󵄨󵄨 .
= ∑ 󵄨󵄨󵄨
󵄨󵄨
3
󵄨
󵄨
√
√
𝑚+𝑛∈2Z 󵄨 (𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖 3) − (𝜁 + 1 − 𝑖 3) 󵄨󵄨
󵄨󵄨
󵄨󵄨
+
Letting 𝑧 → 𝜁0 , 𝑧1 → 𝜁0 ∈ 𝜕1 𝑃 the sum (22) tends to 0.
Similarly, for the rest parts of the boundary 𝜕3 𝑃+ , 𝜕4 𝑃+
one can get that the sums in (21) and (20) tend to zero as we
let 𝑧 → 𝜁0 ∈ 𝜕1 𝑃. As a result for the case 𝜁0 ∈ 𝜕1 𝑃+
lim [−
𝑧→𝜁0
1
∫ 𝛾 (𝜁) 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁 ]
4𝜋 𝜕𝑃+
= lim [−
Hence for 𝑧 → 𝜁0 on 𝜕1 𝑃+
(√3 − 𝑖)
𝑧→𝜁0
(√3 − 𝑖) (𝑧 − 𝑧1 )
𝜕]𝜁 𝐺1 (𝑧, 𝜁) =
(1 + 𝑜 (1)) .
󵄨2
󵄨󵄨
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨
(30)
2
− 6𝑖 (𝜁 + 1 − 𝑖√3)
3
4𝜋
∫
⋅󵄨
.
3
3 󵄨󵄨2
󵄨󵄨
󵄨󵄨(𝑧 + 1 − 𝑖√3) − (𝜁 + 1 − 𝑖√3) 󵄨󵄨󵄨
󵄨
󵄨
𝑧 − 𝑧1
𝛾 (𝜁) 󵄨
𝑑𝑠 ]
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2 𝜁
󵄨
󵄨
(33)
on the boundary 𝜕1 𝑃.
Case 2. Let 𝜁0 be from 𝜕2 𝑃+ , where 𝜁0 = 𝜁0 + 2𝑖√3, 𝜁0 − 𝑖√3 =
𝜁0 + 𝑖√3.
On 𝜕2 𝑃+ , 𝜁 = 𝜁 + 2𝑖√3, 𝜁 − 𝑖√3 = 𝜁 + 𝑖√3.
For 𝑚 = 𝑛 = 0 the term in (22) is
− 6𝑖 (𝜁 + 1 − 𝑖√3)
3
𝜕1
𝑃+
= 𝛾 (𝜁0 )
On 𝜕2 𝑃+ 𝜁 = 𝜁2 = 𝜁+2𝑖√3 and 𝜁−𝑖√3 = 𝜁+𝑖√3, for 𝑚 = 𝑛 = 0
in (22), the formula becomes
(𝑧 + 1 − 𝑖√3) − (𝑧 + 1 − 𝑖√3)
(27)
This term is not singular for 𝑧 ≠ 𝜁 and the terms of the sum
can be in general rewritten as (𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3)3 = (𝑧1 −
𝜔𝑘𝑙 + 1 + 𝑖√3)3 for certain 𝑘 + 𝑙 ∈ 2Z. Therefore
For the other terms of the sum,
3
1
3
(𝑧 − 𝜔𝑚𝑛 − 2) = [− (1 − 𝑖√3) (𝑧 − 𝜔𝑚𝑛 − 2)]
2
𝜁0 ≠ 2.
(31)
2
3
(𝑧 + 1 − 𝑖√3) − (𝑧 + 1 − 𝑖√3)
3
⋅󵄨
.
3
3 󵄨󵄨2
󵄨󵄨
󵄨󵄨(𝑧 + 1 − 𝑖√3) − (𝜁 + 1 − 𝑖√3) 󵄨󵄨󵄨
󵄨
󵄨
(34)
6
Journal of Complex Analysis
+ (𝑧 + 1 − 𝑖√3) (𝑧2 + 1 − 𝑖√3)
On this boundary 𝑧 = 𝑧2 = 𝑧 + 2𝑖√3 and 𝑧 − 𝑖√3 = 𝑧 + 𝑖√3
or 𝑧2 − 𝑖√3 = 𝑧 + 𝑖√3; therefore
3
3
(𝑧 + 1 − 𝑖√3) − (𝑧 + 1 − 𝑖√3) = (𝑧 + 1 − 𝑖√3)
3
− (𝑧2 + 1 − 𝑖√3) = (𝑧 − 𝑧2 ) [(𝑧 + 1 − 𝑖√3)
2
+ (𝑧2 + 1 − 𝑖√3) ] .
3
(35)
Substituting the latter into (34) and considering 𝑧 → 𝜁0 ∈
𝜕2 𝑃+ , 𝜁0 ≠ 1 + 𝑖√3, 𝜁 = 𝜁0
2
2
2
2
{
{ −6𝑖 (𝜁 + 1 − 𝑖√3) [(𝑧 + 1 − 𝑖√3) + (𝑧 + 1 − 𝑖√3) (𝑧2 + 1 − 𝑖√3) + (𝑧2 + 1 − 𝑖√3)
lim {
󵄨󵄨
󵄨2
𝑧→𝜁0 {
󵄨󵄨(𝑧 + 1 − 𝑖√3)2 + (𝑧 + 1 − 𝑖√3) (𝜁 + 1 − 𝑖√3) + (𝜁 + 1 − 𝑖√3)2 󵄨󵄨󵄨
󵄨󵄨
{
󵄨󵄨
4
(37)
=
3
󵄨󵄨
󵄨2
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3)3 − (𝜁 + 1 − √3)3 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
(𝑧2 − 𝜔𝑚𝑛 + 1 + 𝑖√3)
∑
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
1
∫ 𝛾 (𝜁) 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁 ]
4𝜋 𝜕𝑃+
𝑧 − 𝑧2
1
= lim [−
𝑑𝑠 ] = 𝛾 (𝜁0 ) .
∫ 𝛾 (𝜁) 󵄨
+
𝑧→𝜁0
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2 𝜁
2𝜋𝑖 𝜕2 𝑃
󵄨
󵄨
(𝑧 − 𝜔𝑚𝑛 + 1 − 𝑖√3)
∑
lim [−
𝑧→𝜁0
For 𝑚 ≠ 0, 𝑛 ≠ 0 by
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
(36)
on 𝜕2 𝑃+ . Similar computations on the boundary parts 𝜕1 𝑃+ ,
𝜕3 𝑃+ , 𝜕4 𝑃+ give that the sums (21) and (20) tend to zero as
𝑧 → 𝜁0 . Therefore, on the boundary part 𝜕2 𝑃+ for 𝜁0 ∈ 𝜕2 𝑃+
gives
(𝜁0 + 1 − 𝑖√3)
−2𝑖 󵄨
= −2𝑖, 𝜁0 ≠ −1 + 𝑖√3.
󵄨󵄨𝜁 + 1 − 𝑖√3󵄨󵄨󵄨4
󵄨󵄨
󵄨󵄨
]}
}
}
}
}
3
(38)
󵄨󵄨
󵄨2 .
󵄨󵄨(𝑧2 − 𝜔𝑚𝑛 + 1 − 𝑖√3)3 − (𝜁 + 1 − 𝑖√3)3 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
Letting 𝑧 → 𝜁0 ∈ 𝜕2 𝑃+ and since 𝑧2 → 𝜁0 , the sum tends to
0. Then
𝑧 − 𝑧2
𝜕]𝜁 𝐺1 (𝑧, 𝜁) = −2𝑖 󵄨
(1 + 𝑜 (1))
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2
󵄨
󵄨
Case 3. Let 𝜁0 be defined on 𝜕3 𝑃+ by 𝜁0 = −(1/2)(1 − 𝑖√3)𝜁0 −
3 + 𝑖√3.
On 𝜕3 𝑃+ with 𝜁 = 𝜁1̌ = −(1/2)(1−𝑖√3)𝜁−3+𝑖√3, (𝜁+2)3 =
(𝜁 + 2)3 .
For 𝑚 = 𝑛 = 0 in (21) the formula becomes
(𝑧 + 2)3 − (𝑧 + 2)3
.
−3 (−√3 + 𝑖) (𝜁 + 2)2 󵄨
󵄨󵄨(𝑧 + 2)3 − (𝜁 + 2)3 󵄨󵄨󵄨2
󵄨󵄨
󵄨󵄨
Since (𝑧1̌ + 2)3 = (𝑧 + 2)3 , then
(𝑧 + 2)3 − (𝑧 + 2)3 = (𝑧 + 2)3 − (𝑧1̌ + 2)
2
(42)
Letting 𝑧 → 𝜁0 , 𝑧1̌ → 𝜁0 , and 𝜁 = 𝜁0 for the fraction
2
4
3 (−√3 + 𝑖) (𝜁 + 2)2 [(𝑧 + 2)2 + (𝑧 + 2) (𝑧1̌ + 2) + (𝑧1̌ + 2) ] (−√3 + 𝑖) (𝜁0 + 2)
lim
=
= (√3 + 𝑖) .
󵄨󵄨
󵄨2
󵄨󵄨
󵄨4
𝑧→𝜁0
󵄨󵄨(𝑧 + 2)2 + (𝑧 + 2) (𝜁 + 2) + (𝜁 + 2)2 󵄨󵄨󵄨
󵄨󵄨𝜁0 + 2󵄨󵄨󵄨
󵄨
󵄨
For the other terms of (21) (𝑧 − 𝜔𝑚𝑛 + 2)3 = (𝑧1̌ − 𝜔𝑘𝑙 + 2)3 and
(41)
3
= (𝑧 − 𝑧1̌ ) [(𝑧 + 2)2 + (𝑧 + 2) (𝑧1̌ + 2) + (𝑧1̌ + 2) ] .
(39)
(40)
(43)
Therefore
3
(𝑧 − 𝜔𝑚𝑛 + 2)
∑
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
󵄨󵄨
󵄨2
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 + 2)3 − (𝜁 + 2)3 󵄨󵄨󵄨
󵄨
󵄨
3
=
∑
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
(𝑧1̌ − 𝜔𝑚𝑛 + 2)
󵄨󵄨
󵄨2 .
󵄨󵄨(𝑧1̌ − 𝜔𝑚𝑛 + 2)3 − (𝜁 + 2)3 󵄨󵄨󵄨
󵄨
󵄨
(44)
𝑧 − 𝑧3
𝜕]𝜁 𝐺1 (𝑧, 𝜁) = (−√3 − 𝑖) 󵄨
(1 + 𝑜 (1))
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2
󵄨
󵄨
(45)
for 𝑧 → 𝜁0 ∈ 𝜕3 𝑃+ . On the boundary parts 𝜕1 𝑃+ , 𝜕2 𝑃+ , 𝜕4 𝑃+
in the same manner we can prove that the sums (19), (22), and
Journal of Complex Analysis
7
(20) tend to zero as 𝑧 → 𝜁0 ∈ 𝜕3 𝑃+ . Thus for this Case 3 is as
follows:
lim [−
𝑧→𝜁0
𝑧 − 𝑧4
𝜕]𝜁 𝐺1 (𝑧, 𝜁) = 2𝑖 󵄨
(1 + 𝑜 (1)) ,
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2
󵄨
󵄨
1
∫ 𝛾 (𝜁) 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁 ]
4𝜋 𝜕𝑃+
= lim [−
(√3 + 𝑖)
𝑧→𝜁0
4𝜋
If 𝑧 → 𝜁0 , 𝑧4 → 𝜁0 ∈ 𝜕4 𝑃+ , this sum (20) tends to 0 for
𝜁 ∈ 𝜕4 𝑃+ . Thus on this boundary part
𝑧 − 𝑧1̌
𝛾 (𝜁) 󵄨
󵄨2 𝑑𝑠𝜁 ]
+
󵄨
𝜕3 𝑃
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨
(46)
∫
lim [−
𝑧→𝜁0
= 𝛾 (𝜁0 )
= lim
Case 4. Let 𝜁0 be from 𝜕4 𝑃+ , where 𝜁0 = 𝜁4 = 𝜁.
Obviously, similar calculations on the boundary parts
imply the related sums to be convergent to zero, except for
the boundary part 𝜕4 𝑃+ , where the boundary behavior is to
be observed carefully.
On 𝜕4 𝑃+ with 𝜁 = 𝜁 for 𝑚 = 𝑛 = 0 in formula (20) we
have
(47)
𝑧−𝑧
𝛾 (𝜁) 󵄨
𝑑𝑠 = 𝛾 (𝜁0 )
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2 𝜁
󵄨
󵄨
In the next lemma the boundary behavior of the function
𝜑(𝜁) in the corner points ±2, ±1 + 𝑖√3, is observed. It is
shown that the continuity of the function is preserved at all
the corner points which are treated as an intersection of two
lines through the boundary parts.
lim
= 0,
{−
1
∫ [𝛾 (𝜁) − 𝛾 (𝜁0 )] 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁 }
4𝜋 𝜕𝑃+
(52)
𝜁0 ∈ {±2, ±1 + 𝑖√3} , 𝑧 ∈ 𝑃+ .
The proof of this lemma is given in detail in [9]. We
consider now the main theorem of this paper.
(𝑧 − 𝑧4 )
6𝑖 (𝜁 − 2)2 󵄨
󵄨󵄨𝑧 − 𝜁󵄨󵄨󵄨2
󵄨
󵄨
(48)
2
(𝑧 − 2)2 + (𝑧 − 2) (𝑧4 − 2) + (𝑧4 − 2)
⋅󵄨
󵄨󵄨(𝑧 − 2)2 + (𝑧 − 2) (𝜁 − 2) + (𝜁 − 2)2 󵄨󵄨󵄨2
󵄨󵄨
󵄨󵄨
+
4
(49)
(𝜁 − 2)
= 2𝑖 󵄨 0
󵄨4 = 2𝑖.
󵄨󵄨󵄨𝜁0 − 2󵄨󵄨󵄨
Again, the terms of the sum (20) are rewritten and it follows
that
3
(𝑧 − 𝜔𝑚𝑛 − 2)
󵄨󵄨
󵄨2
󵄨󵄨(𝑧 − 𝜔𝑚𝑛 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨
󵄨
󵄨
3
(𝑧4 − 𝜔𝑚𝑛 − 2)
.
󵄨
3
3 󵄨2
𝑚+𝑛∈2Z, 󵄨󵄨󵄨(𝑧4 − 𝜔𝑚𝑛 − 2) − (𝜁 − 2) 󵄨󵄨󵄨
󵄨
𝑚2 +𝑛2 >0 󵄨
𝑤 = 𝛾 𝑜𝑛 𝜕𝑃+
(53)
𝑓𝑜𝑟 𝑓 ∈ 𝐿 𝑝 (𝑃+ ; C) , 2 < 𝑝, 𝛾 ∈ 𝐶 (𝜕𝑃+ ; C)
is uniquely solvable in the Sobolev space 𝑊2,𝑝 (𝑃+ ; C) ∩
𝐶(𝑃+ ; C) by
2
(𝑧 − 2)2 + (𝑧 − 2) (𝑧4 − 2) + (𝑧4 − 2)
lim 6𝑖 (𝜁 − 2) 󵄨
󵄨2
𝑧→𝜁0
󵄨󵄨󵄨(𝑧 − 2)2 + (𝑧 − 2) (𝜁 − 2) + (𝜁 − 2)2 󵄨󵄨󵄨
󵄨
󵄨
2
Theorem 7. The Dirichlet problem fo the Poisson equation
𝑤𝑧𝑧 = 𝑓 𝑖𝑛 𝑃+ ,
and taking the limit in the second fraction for 𝑧 → 𝜁0 ∈ 𝜕4 𝑃
and since 𝜁 = 𝜁0
∑
𝜕4
𝑃+
on the boundary 𝜕4 𝑃+ . Thus, equality (24) for the function
𝜑(𝑧) is valid.
𝑧→𝜁0 ∈𝜕𝑃+
Here 𝑧 = 𝑧4 = 𝑧; then term (47) is
=
∫
(51)
Lemma 6. If 𝛾 ∈ 𝐶(𝜕𝑃+ ; C), then
(𝑧 − 2)3 − (𝑧 − 2)3
6𝑖 (𝜁 − 2)2 󵄨
.
󵄨󵄨(𝑧 − 2)3 − (𝜁 − 2)3 󵄨󵄨󵄨2
󵄨󵄨
󵄨󵄨
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
1
𝑧→𝜁0 2𝜋𝑖
on the boundary part 𝜕3 𝑃.
∑
1
∫ 𝛾 (𝜁) 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁 ]
4𝜋 𝜕𝑃+
(50)
𝑤 (𝑧) = −
1
∫ 𝛾 (𝜁) 𝜕]𝜁 𝐺1 (𝑧, 𝜁) 𝑑𝑠𝜁
4𝜋 𝜕𝑃+
1
− ∫ 𝑓 (𝜁) 𝐺1 (𝑧, 𝜁) 𝑑𝜉 𝑑𝜂,
𝜋 𝑃+
(54)
where 𝜁 = 𝜉 + 𝑖𝜂.
Proof. We need to prove that (54) is a solution of the Poisson
equation in problem (53). The property of the Pompeiu
operator 𝑇𝑓(𝑧) = −(1/𝜋) ∫𝐷(𝑓(𝜁)/(𝜁 − 𝑧))𝑑𝜉 𝑑𝜂, described
in [13, 14] as 𝜕𝑧 𝑇𝑓(𝑧) = 𝑓(𝑧), provides a weak solution of 𝑤𝑧𝑧
󵄨󵄨
󵄨2
3
󵄨󵄨
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3 󵄨󵄨󵄨󵄨
󵄨
𝐺1 (𝑧, 𝜁) = log 󵄨󵄨 ∏
󵄨
󵄨󵄨𝑚+𝑛∈2Z (𝜁 − 𝜔 − 2)3 − (𝑧 − 2)3 󵄨󵄨󵄨
󵄨
󵄨
𝑚𝑛
(55)
8
Journal of Complex Analysis
References
and the derivative
3 (𝑧 − 2)
𝜕𝑧 𝐺1 (𝑧, 𝜁) =
−
3
(𝜁 − 2) − (𝑧 − 2)
3
3 (𝑧 − 2)2
3
(𝜁 − 2) − (𝑧 − 2)3
∑
[
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
[
+
−
2
3 (𝑧 − 2)2
(56)
3
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
3 (𝑧 − 2)2
3
3
].
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2) ]
In order to construct the Pompeiu-type operator we
consider the following term:
3 (𝑧 − 2)2
3
(𝜁 − 2) − (𝑧 − 2)3
1
2 (3 − 𝑧) − 𝜁
=
.
+
𝜁 − 𝑧 (𝜁 − 2)2 + (𝜁 − 2) (𝑧 − 2) + (𝑧 − 2)2
(57)
Define a function
̃ (𝜁, 𝑧) =
𝑔
−
+
(𝜁 − 2) + (𝜁 − 2) (𝑧 − 2) + (𝑧 − 2)2
3 (𝑧 − 2)2
3
(𝜁 − 2) − (𝑧 − 2)3
∑
𝑚+𝑛∈2Z,
𝑚2 +𝑛2 >0
−
2 (3 − 𝑧) − 𝜁
2
(
3 (𝑧 − 2)2
(58)
3
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
3 (𝑧 − 2)2
3
(𝜁 − 𝜔𝑚𝑛 − 2) − (𝑧 − 2)3
)
which is analytic with respect to 𝑧 ∈ 𝑃+ ; then 𝜕𝑧 𝐺1 𝑧, 𝑧𝑒𝑡𝑎 =
̃ (𝜁, 𝑧). Then, the equation 𝑤𝑧𝜁 is rewritten as
1/(𝜁 − 𝑧) + 𝑔
1
𝜕𝑧𝑧 {− ∫ 𝑓 (𝜁) 𝐺1 (𝑧, 𝜁) 𝑑𝜉 𝑑𝜂}
𝜋 𝑃+
1
1
̃ (𝜁, 𝑧)] 𝑑𝜉 𝑑𝜂}
= 𝜕𝑧 {− ∫ 𝑓 (𝜁) [
+𝑔
+
𝜋 𝑃
𝜁−𝑧
(59)
= 𝑓 (𝑧) .
This provides the solution to the differential equation in
problem (53) in a weak sense. The boundary condition 𝑤 = 𝛾
on the boundary 𝜕𝑃+ holds because of Lemmas 5 and 6.
Conflict of Interests
The author declares that there is no conflict of interests
regarding the publication of this paper.
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