# Dirichlet Problem for Complex Poisson Equation in a Half Hexagon Domain

Authors: Bibinur Shupeyeva
Journal: Journal of Complex Analysis
 (2016)

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## Abstract

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.


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## Body

## 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 domainD 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 asP+ with four corner points at 2, 1+i3, -1+i3, 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)-121+i3z¯+3+i3,z¯+2i3,-121-i3z¯-3+i3,-121+i3z¯-3-i3,z¯-2i3,-121-i3z¯+3-i3.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+i3), -(1/2)(1-i3), and displacement 3m+i3n, 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 2i3n in the direction of the imaginary axes. Thus the main period is μmn=6m+2i3n.Figure 1
Hexagons.Obviously, the repeated reflections of the pointz∈P+ are representable in different ways, using either of the points(2)z1=-121+i3z¯+3+i3,z2=-121+i3z+3+i3orzˇ1=-121-i3z¯-3+i3,zˇ2=-121-i3z-3+i3,which are connected by the relations zˇ2¯=z1-6-2i3 and zˇ1¯=z2-6-2i3.In general, all reflection points are either given by(3)z+ωmn,z¯1+ωmn,z2+ωmn,z¯+ωmn,z1+ωmn,z¯2+ωmnor by(4)z+ωmn,zˇ1¯+ωmn,zˇ2+ωmn,z¯+ωmn,zˇ1+ωmn,zˇ2¯+ωmn,where ωmn=3m+i3n 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 forP+ 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 byH1- being the right-hand half plane which has the boundary line passing through the points 2 and 1+i3, H2- being the upper half plane with the border line through the points ±1+i3, H3- being the left-hand half plane with the border line passing through the points -1+i3 and −2, and H4- being the half plane which is below the real axis.Let thenH1+, 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 functionsG1(z,ζ), z=x+iy, ζ=ξ+iη as described below.For the half planeH1+ with the boundary described by the relation ζ-2=-(1/2)(1+i3)(ζ¯-2) we have(5)G1z,ζ=log⁡1/21+i3ζ¯-2+z-2ζ-z2,z,ζ∈H1+,-12∂νζG1z,ζ=-3-i4z-z1ζ-z2,ζ∈∂H1+,z∈H1+,where z1=-(1/2)(1+i3)z¯+3+i3, z1∈H1+.For the half planeH2+ the relation on the boundary is given as ζ=ζ¯+2i3; then(6)G1z,ζ=log⁡ζ¯-z+2i3ζ-z2,z,ζ∈H2+,-12∂νζG1z,ζ=-1iz-z2ζ-z2,ζ∈∂H2+,z∈H2+;here z2=z¯+2i3, z∈H2+.The boundary of the half planeH3+ is described by ζ+2=-(1/2)(1-i3)(ζ¯+2) and(7)G1z,ζ=log⁡1/21-i3ζ¯+2+z+2ζ-z2,z,ζ∈H3+,-12∂νζG1z,ζ=-3+i4z-z1ˇζ-z2,ζ∈∂H3+,z∈H3+,where z1ˇ=-(1/2)(1-i3)z¯-3+i3, z∈H3+.Finally, for the half planeH4+ with the boundary described by ζ=ζ¯, we have(8)G1z,ζ=log⁡ζ¯-zζ-z2,z,ζ∈H4+,-12∂νζG1z,ζ=1iz-z¯ζ-z2,ζ∈∂H4+,z∈H4+.


## 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:(9)B1z,ζ=∏m+n∈2Zζ-z¯-ωmnζ-z1-ωmnζ-z2¯-ωmnζ-z-ωmnζ-z1¯-ωmnζ-z2-ωmn=∏m+n∈2Zζ-ωmn-23-z¯-23ζ-ωmn-23-z-23,where z1=-(1/2)(1+i3)z¯+3+i3, z2=-(1/2)(1+i3)z+3+i3, or a function(10)B2z,ζ=∏m+n∈2Zζ-ωmn+23-z¯+23ζ-ωmn+23-z+23,where z1ˇ=-(1/2)(1-i3)z¯-3+i3, z2ˇ=-(1/2)(1-i3)z-3+i3. Here z is considered as a parameter and ζ∈C is the variable.For the boundary part∂2P, the line from 1+i3 to -1+i3, 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:(11)B1-121+i3z,-121+i3ζ=∏m+n∈2Zζ-ωmn+1-i33-z¯+1+i33ζ-ωmn+1-i33-z+1-i33,which becomes 1 on the boundary ∂2P, where ζ-i3=ζ¯+i3.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(12)∏m+n∈2Zζ-ωmn-23-z¯-23ζ-ωmn-23-z-23converges, where ωmn=3m+i3n, m+n∈2Z.Lemma 2.
The equalitiesB1(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 functionsB1(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+;(30)
limζ→∂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.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 (13)G1z,ζ=log⁡B1z,ζ2=log⁡B2z,ζ2=log⁡B3z,ζ2or, by the symmetry property,(14)G1z,ζ=log⁡∏m+n∈2Zz-ωmn-23-ζ¯-23z-ωmn-23-ζ-232,(15)G1z,ζ=log⁡∏m+n∈2Zz-ωmn+23-ζ¯+23z-ωmn+23-ζ+232,(16)G1z,ζ=log⁡∏m+n∈2Zz-ωmn+1-i33-ζ¯+1+i33z-ωmn+1-i33-ζ+1-i332.Lemma 3.
The functionG1(z,ζ) has vanishing boundary values on ∂P+; that is, (17)limζ→ζ0∈∂P+⁡G1z,ζ=0.Theorem 4 (see [13]).
Anyw∈C2(P+;C)∩C1(P+¯;C) can be represented as(18)wz=-14π∫∂P+wζ∂νζG1z,ζdsζ-1π∫P+wζζ¯ζG1z,ζdξdη,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+.We consider now the different forms of the Green function and take the derivatives∂ζG1(z,ζ), ∂ζ¯G1(z,ζ).For the right-hand side, a boundary∂1P+, we choose the form (14) for ζ∈∂P+, z∈P+. Here the outward normal derivative is ∂νζ=(3/2+i/2)∂ζ+(3/2-i/2)∂ζ¯; then(19)∂νζG1z,ζ=-33+iζ-22∑m+n∈2Zz-ωmn-23-z-ωmn-2¯3z-ωmn-23-ζ-232,since ζ-2=-(1/2)(1+i3)(ζ¯-2), (ζ-2)3=(ζ¯-2)3.For the boundary part∂4P+, a line between (-2,0), (2,0) on a real axis, the outward normal derivative is ∂νζ=-i(∂ζ-∂ζ¯), ζ=ζ¯; then(20)∂νζG1z,ζ=6iζ-22∑m+n∈2Zz-ωmn-23-z-ωmn-2¯3z-ωmn-23-ζ-232.For the boundary part ∂3P+ on the left-hand side of P+, we take form (15). The outward normal derivative is ∂νζ=(3/2-i/2)∂ζ+(3/2+i/2)∂ζ¯ also here ζ=ζ1ˇ=-(1/2)(1-i3)ζ¯-3+i3 and (ζ+2)3=(ζ¯+2)3; then(21)∂νζG1z,ζ=-33-iζ+22∑m+n∈2Zz-ωmn+23-z-ωmn+2¯3z-ωmn+23-ζ+232.For the upper boundary part ∂2P+, a line joining the points ±1+i3, form (16) is valid. Here ∂νζ=i(∂ζ-∂ζ¯) and ζ-i3=ζ¯+i3; then ∂νζG1(z,ζ) is (22)-6iζ+1-i32∑m+n∈2Zz-ωmn+1-i33-z-ωmn+1-i3¯3z-ωmn+1-i33-ζ+1-i332.


## 4. Harmonic Dirichlet Problem

The representation formula in Theorem4 provides the solution to the Dirichlet problem for the Poisson equation.At first the boundary behavior of the integral is to be studied. Let forγ∈C(∂P+;R)(23)φz=-14π∫∂P+γζ∂νζGz,ζdsζ,z∈P+.Lemma 5.
Forγ∈C(∂P+;R) the function presented in (23) satisfies the relation (24)limz→ζ0⁡φz=γζ0,where ζ0 is any fixed point on ∂P+∖{±2,±1+i3}.Proof.
Letζ0 be defined on different boundary parts and consider the boundary behavior when z→ζ0.
Case 1. If ζ0 is taken on ∂1P+ so that ζ0=-(1/2)(1+i3)ζ0¯+3+i3 then(25)ζ0-22=-121-i3ζ0-22,ζ0-23=ζ0¯-23.On ∂1P+ where ζ=ζ1=-(1/2)(1+i3)ζ¯+3+i3, (ζ-2)3=(ζ¯-2)3.
Form=n=0 formula (19) gives(26)-33+iζ-22z-23-z¯-23z-23-ζ-232=-33+iζ-22z-z¯z-22+z-2z1-2+z1-22z-ζ2z-22+z-2ζ-2+ζ-222.Because z1=-(1/2)(1+i3)z¯+3+i3, then (z1-2)3=(z¯-2)3. The limit in the following ratio as z→ζ0 and ζ=ζ0 gives (27)limz→ζ0⁡-33+iζ-22z-22+z-2z1-2+z1-22z-22+z-2ζ-2+ζ-222=-3+iζ0-24ζ0-24=3-i,ζ0≠2.For the other terms of the sum,(28)z-ωmn-23=-121-i3z-ωmn-23=z1¯-ωkl-23,which follows from the rearrangement of the indices in ωmn for certain k+l∈2Z. Thus (29)∑m+n∈2Z,m2+n2>0z-ωmn-23z-ωmn-23-ζ-232=∑m+n∈2Z,m2+n2>0z1-ωmn-2¯3z1-ωmn-23-ζ-232.Hence for z→ζ0 on ∂1P+(30)∂νζG1z,ζ=3-iz-z1z-ζ21+o1.On ∂2P+ζ=ζ2=ζ¯+2i3 and ζ-i3=ζ¯+i3, for m=n=0 in (22), the formula becomes(31)-6iζ+1-i32z+1-i33-z+1-i3¯3z+1-i33-ζ+1-i332.This term is not singular for z≠ζ and the terms of the sum can be in general rewritten as (z-ωmn+1-i3)3=(z1¯-ωkl+1+i3)3 for certain k+l∈2Z. Therefore(32)∑m+n∈2Zz-ωmn+1-i33z-ωmn+1-i33-ζ+1-i332=∑m+n∈2Zz1-ωmn+1-i3¯3z-ωmn+1-i33-ζ+1-i32.Letting z→ζ0, z1→ζ0∈∂1P+ the sum (22) tends to 0.
Similarly, for the rest parts of the boundary∂3P+, ∂4P+ one can get that the sums in (21) and (20) tend to zero as we let z→ζ0∈∂1P. As a result for the case ζ0∈∂1P+(33)limz→ζ0⁡-14π∫∂P+γζ∂νζG1z,ζdsζ=limz→ζ0⁡-3-i4π∫∂1P+γζz-z1z-ζ2dsζ=γζ0on the boundary ∂1P.
Case 2. Let ζ0 be from ∂2P+, where ζ0=ζ0¯+2i3, ζ0-i3=ζ0¯+i3.
On∂2P+, ζ=ζ¯+2i3, ζ-i3=ζ¯+i3.
Form=n=0 the term in (22) is(34)-6iζ+1-i32z+1-i33-z+1-i3¯3z+1-i33-ζ+1-i332.On this boundary z=z2=z¯+2i3 and z-i3=z¯+i3 or z2-i3=z¯+i3; therefore (35)z+1-i33-z+1-i3¯3=z+1-i33-z2+1-i33=z-z2z+1-i32+z+1-i3z2+1-i3+z2+1-i32.Substituting the latter into (34) and considering z→ζ0∈∂2P+, ζ0≠1+i3, ζ=ζ0(36)limz→ζ0⁡-6iζ+1-i32z+1-i32+z+1-i3z2+1-i3+z2+1-i32z+1-i32+z+1-i3ζ+1-i3+ζ+1-i322gives(37)-2iζ0+1-i34ζ+1-i34=-2i,ζ0≠-1+i3.For m≠0, n≠0 by (38)∑m+n∈2Z,m2+n2>0z-ωmn+1-i33z-ωmn+1-i33-ζ+1-332=∑m+n∈2Z,m2+n2>0z2-ωmn+1+i3¯3z2-ωmn+1-i33-ζ+1-i332.Letting z→ζ0∈∂2P+ and since z2→ζ0, the sum tends to 0. Then (39)∂νζG1z,ζ=-2iz-z2z-ζ21+o1on ∂2P+. Similar computations on the boundary parts ∂1P+, ∂3P+, ∂4P+ give that the sums (21) and (20) tend to zero as z→ζ0. Therefore, on the boundary part ∂2P+ for ζ0∈∂2P+(40)limz→ζ0⁡-14π∫∂P+γζ∂νζG1z,ζdsζ=limz→ζ0⁡-12πi∫∂2P+γζz-z2z-ζ2dsζ=γζ0.
Case 3. Let ζ0 be defined on ∂3P+ by ζ0=-(1/2)(1-i3)ζ0¯-3+i3.
On∂3P+ with ζ=ζ1ˇ=-(1/2)(1-i3)ζ¯-3+i3, (ζ+2)3=(ζ¯+2)3.
Form=n=0 in (21) the formula becomes(41)-3-3+iζ+22z+23-z¯+23z+23-ζ+232.Since (z1ˇ+2)3=(z¯+2)3, then(42)z+23-z¯+23=z+23-z1ˇ+23=z-z1ˇz+22+z+2z1ˇ+2+z1ˇ+22.Letting z→ζ0, z1ˇ→ζ0, and ζ=ζ0 for the fraction(43)limz→ζ0⁡3-3+iζ+22z+22+z+2z1ˇ+2+z1ˇ+22z+22+z+2ζ+2+ζ+222=-3+iζ0+24ζ0+24=3+i.For the other terms of (21) (z-ωmn+2)3=(z1ˇ¯-ωkl+2)3 and(44)∑m+n∈2Z,m2+n2>0z-ωmn+23z-ωmn+23-ζ+232=∑m+n∈2Z,m2+n2>0z1ˇ-ωmn+2¯3z1ˇ-ωmn+23-ζ+232.Therefore(45)∂νζG1z,ζ=-3-iz-z3z-ζ21+o1for z→ζ0∈∂3P+. On the boundary parts ∂1P+, ∂2P+, ∂4P+ in the same manner we can prove that the sums (19), (22), and (20) tend to zero as z→ζ0∈∂3P+. Thus for this Case 3 is as follows:(46)limz→ζ0⁡-14π∫∂P+γζ∂νζG1z,ζdsζ=limz→ζ0⁡-3+i4π∫∂3P+γζz-z1ˇz-ζ2dsζ=γζ0on the boundary part ∂3P.
Case 4. Let ζ0 be from ∂4P+, 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∂4P+, where the boundary behavior is to be observed carefully.
On∂4P+ with ζ=ζ¯ for m=n=0 in formula (20) we have(47)6iζ-22z-23-z¯-23z-23-ζ-232.Here z=z4=z¯; then term (47) is(48)6iζ-22z-z4z-ζ2z-22+z-2z4-2+z4-22z-22+z-2ζ-2+ζ-222and taking the limit in the second fraction for z→ζ0∈∂4P+ and since ζ=ζ0(49)limz→ζ0⁡6iζ-22z-22+z-2z4-2+z4-22z-22+z-2ζ-2+ζ-222=2iζ0-24ζ0-24=2i.Again, the terms of the sum (20) are rewritten and it follows that(50)∑m+n∈2Z,m2+n2>0z-ωmn-23z-ωmn-23-ζ-232=∑m+n∈2Z,m2+n2>0z4-ωmn-2¯3z4-ωmn-23-ζ-232.If z→ζ0, z4→ζ0∈∂4P+, this sum (20) tends to 0 for ζ∈∂4P+. Thus on this boundary part(51)∂νζG1z,ζ=2iz-z4z-ζ21+o1,limz→ζ0⁡-14π∫∂P+γζ∂νζG1z,ζdsζ=limz→ζ0⁡12πi∫∂4P+γζz-z¯z-ζ2dsζ=γζ0on the boundary ∂4P+. Thus, equality (24) for the function φ(z) is valid.In the next lemma the boundary behavior of the functionφ(ζ) in the corner points ±2,±1+i3, 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.Lemma 6.
Ifγ∈C(∂P+;C), then (52)limz→ζ0∈∂P+⁡-14π∫∂P+γζ-γζ0∂νζG1z,ζdsζ=0,ζ0∈±2,±1+i3,z∈P+.The proof of this lemma is given in detail in [9]. We consider now the main theorem of this paper.Theorem 7.
The Dirichlet problem fo the Poisson equation(53)wzz¯=finP+,w=γon∂P+forf∈LpP+;C,2<p,γ∈C∂P+;Cis uniquely solvable in the Sobolev space W2,p(P+;C)∩C(P+¯;C) by(54)wz=-14π∫∂P+γζ∂νζG1z,ζdsζ-1π∫P+fζG1z,ζdξdη,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(fζ/ζ-z)dξdη, described in [13, 14] as ∂z¯Tf(z)=f(z), provides a weak solution of wzz¯(55)G1z,ζ=log⁡∏m+n∈2Zζ-ωmn-23-z¯-23ζ-ωmn-23-z-232and the derivative(56)∂zG1z,ζ=3z-22ζ-23-z-23-3z-22ζ¯-23-z-23+∑m+n∈2Z,m2+n2>03z-22ζ-ωmn-23-z-23-3z-22ζ-ωmn-2¯3-z-23.
In order to construct the Pompeiu-type operator we consider the following term:(57)3z-22ζ-23-z-23=1ζ-z+23-z-ζζ-22+ζ-2z-2+z-22.Define a function (58)g~ζ,z=23-z-ζζ-22+ζ-2z-2+z-22-3z-22ζ¯-23-z-23+∑m+n∈2Z,m2+n2>03z-22ζ-ωmn-23-z-23-3z-22ζ-ωmn-2¯3-z-23which is analytic with respect to z∈P+; then ∂zG1z, zeta=1/(ζ-z)+g~(ζ,z). Then, the equation wzζ is rewritten as (59)∂zz¯-1π∫P+fζG1z,ζdξdη=∂z¯-1π∫P+fζ1ζ-z+g~ζ,zdξdη=fz.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.



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