Harmonic Green and Neumann functions for domains bounded by two intersecting circular arcs

TL;DR

This paper develops explicit harmonic Green and Neumann functions for domains bounded by two intersecting circular arcs with a specific angle, enabling solutions to boundary value problems in these geometries.

Contribution

It extends the parqueting-reflection principle to construct harmonic functions in domains with intersecting circular arcs, providing explicit formulas for boundary value problems.

Findings

Explicit Green functions derived for domains with intersecting circular arcs.

Solutions to Dirichlet and Neumann problems obtained using Green and Neumann functions.

Application of reflection principle to complex domain geometries.

Abstract

The parqueting-reflection principle is shown to also work for constructing harmonic Green functions and harmonic Neumann functions for a class of domains, which are bounded by two arcs in $\mathbb{C}$ with a special intersecting angle $\pi/n, n\in\mathbb{N}^*$. Applying the Green representation formula and the Neumann representation formula we solve the Dirichlet and Neumann boundary problem to the Poisson equation in these domains.

Equations

$$az\bar{z}+\bar{b}z+b\bar{z}+c=0,$$

$$\left(\begin{array}{cc}a& \bar{b}\\ b& c\end{array}\right),\text{where}a,c\in \mathbb{R},b\in \mathbb{C},\mathrm{and}ac-b\bar{b}\ne 0,$$

$$\begin{array}[]{cl}\phi\,:&\mathbb{C}_{\infty}\longrightarrow\mathbb{C}_{\infty},\\ &z\mapsto-\frac{b\bar{z}+c}{a\bar{z}+\overline{b}}.\end{array}$$

$$[z:w]\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\stackrel{\mathrm{def}}{=}[az+cw:bz+dw].$$

$$R_A([z:w])=\overline{[z:w]\left(\begin{array}{cc}a& \bar{b}\\ b& c\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)}$$

$$[z:w]B\left[\begin{array}{c}\bar{z}\\ \overline{w}\end{array}\right]=0,$$

$$[u:v]AP\bar{B}{\bar{P}}^T{\bar{A}}^T\left[\begin{array}{c}\bar{u}\\ \bar{v}\end{array}\right]=0.$$

$$D_0=\{z\in \mathbb{C}:z\bar{z}-1\le 0,z\bar{z}\sin (\alpha -\theta )+z\sin \theta +\bar{z}\sin \theta -\sin (\alpha +\theta )\ge 0\},$$

$$M_{C_k}=\left(\begin{array}{cc}-\sin (\alpha +(k-1)\theta )& \sin (k-1)\theta \\ \sin (k-1)\theta & \sin (\alpha -(k-1)\theta )\text{par}\end{array}\right).$$

$$M_{C_0}=\left(\begin{array}{cc}\sin (\alpha -\theta )& \sin \theta \\ \sin \theta & -\sin (\alpha +\theta )\end{array}\right)\sim \left(\begin{array}{cc}-\sin (\alpha -\theta )& \sin (-\theta )\\ \sin (-\theta )& \sin (\alpha +\theta )\end{array}\right),$$

$$M_{C_1}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\sim \left(\begin{array}{cc}-\sin \alpha & 0\\ 0& \sin \alpha \end{array}\right),$$

$$M_{C_{k-2}}=\left(\begin{array}{cc}-\sin (\alpha +(k-3)\theta )& \sin (k-3)\theta \\ \sin (k-3)\theta & \sin (\alpha -(k-3)\theta )\end{array}\right),$$

$$M_{C_{k-1}}=\left(\begin{array}{cc}-\sin (\alpha +(k-2)\theta )& \sin (k-2)\theta \\ \sin (k-2)\theta & \sin (\alpha -(k-2)\theta )\end{array}\right),$$

$$\begin{array}{rl}{M}_{C_k}& =M_{C_{k-1}}{M}_{C_{k-2}}^{-1}{M}_{C_{k-1}}\\ & \sim \left(\begin{array}{cc}-\sin (\alpha +(k-1)\theta )& \sin (k-1)\theta \\ \sin (k-1)\theta & \sin (\alpha -(k-1)\theta )\end{array}\right)\end{array}$$

$$\begin{array}{rl}{z}_{2k}& =\overline{[1:\bar{z}]M_{C_{k+1}}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)}\\ & =\frac{-z\sin (\alpha -k\theta )-\sin k\theta }{z\sin k\theta -\sin (\alpha +k\theta )},\end{array}$$

$$\begin{array}{rl}{z}_{2k+1}& =\overline{[z:1]M_{C_{k+1}}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)}\\ & =\frac{\bar{z}\sin k\theta +\sin (\alpha -k\theta )}{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }.\end{array}$$

$$F(z,\zeta )=\prod _{k=0}^{n-1}\frac{\zeta -z_{2k+1}}{\zeta -z_{2k}}=P(z,\zeta )\times \prod _{k=0}^{n-1}\frac{z\sin k\theta -\sin (\alpha +k\theta )}{\bar{z}\sin (\alpha +k\theta )-\sin k\theta },$$

$$\begin{array}{rl}P(z,\zeta )& =\prod _{k=0}^{n-1}\frac{\bar{z}\zeta \sin (\alpha +k\theta )-(\bar{z}+\zeta )\sin k\theta -\sin (\alpha -k\theta )}{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }\\ & =\frac{\bar{z}\zeta -1}{\zeta -z}\prod _{k=1}^{n-1}\frac{\bar{z}\zeta \sin (\alpha +k\theta )-(\bar{z}+\zeta )\sin k\theta -\sin (\alpha -k\theta )}{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }.\end{array}$$

$$\begin{array}{rl}& \prod _{k=0}^{n-1}\left|\frac{z\sin k\theta -\sin (\alpha +k\theta )}{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }\right|\\ =& \prod _{k=0}^{n-1}\left|\frac{z\sin k\theta -\sin (\alpha +k\theta )}{\frac{-z\sin \theta +\sin (\alpha +\theta )}{z\sin (\alpha -\theta )+\sin \theta }\sin (\alpha +k\theta )-\sin k\theta }\right|\\ =& \prod _{k=0}^{n-1}\left|\frac{z\sin k\theta -\sin (\alpha +k\theta )}{z\sin (k+1)\theta -\sin (\alpha +(k+1)\theta )}\frac{z\sin (\alpha -\theta )+\sin \theta }{\sin \alpha }\right|\\ =& \left|\frac{-\sin \alpha }{z\sin n\theta -\sin (\alpha +n\theta )}\right|\\ =& \left|\frac{-\sin \alpha }{z\sin \pi -\sin (\alpha +\pi )}\right|\\ =& 1.\end{array}$$

$$\prod _{k=0}^{n-1}\left|\frac{z\sin k\theta -\sin (\alpha +k\theta )}{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }\right|=\prod _{k=0}^{n-1}\left|\frac{z\sin k\theta -\sin (\alpha +k\theta )}{\sin (\alpha +k\theta )-z\sin k\theta }\right|=1.$$

$$\begin{array}{rl}{\partial }_{\zeta }{G}_1(z,\zeta )=& -\sum _{k=0}^{n-1}\frac{z\sin k\theta -\sin (\alpha +k\theta )}{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }\\ & +\sum _{k=0}^{n-1}\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\zeta \sin (\alpha +k\theta )-\bar{z}\sin k\theta -\zeta \sin k\theta -\sin (\alpha -k\theta )}.\end{array}$$

$$-\frac{\zeta +\frac{\sin \theta }{\sin (\alpha -\theta )}}{\frac{\sin \alpha }{\sin (\alpha -\theta )}},$$

$${\partial }_{{\nu }_{\zeta }}=-\left(\frac{\zeta \sin (\alpha -\theta )+\sin \theta }{\sin \alpha }{\partial }_{\zeta }+\frac{\bar{\zeta }\sin (\alpha -\theta )+\sin \theta }{\sin \alpha }{\partial }_{\bar{\zeta }}\right).$$

$$\begin{array}{rl}& \frac{\zeta \sin (\alpha -\theta )+\sin \theta }{\sin \alpha }\frac{z\sin k\theta -\sin (\alpha +k\theta )}{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }\\ =& \frac{\sin (\alpha -\theta )}{\sin \alpha }-\frac{z\sin (\alpha -(k+1)\theta )+\sin (k+1)\theta }{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }.\end{array}$$

$$\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\zeta \sin (\alpha +k\theta )-(\bar{z}+\zeta )\sin k\theta -\sin (\alpha -k\theta )}$$

$$\begin{array}{rl}& -\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\bar{\zeta }\sin (k+1)\theta -\bar{z}\sin (\alpha +(k+1)\theta )+\bar{\zeta }\sin (\alpha -(k+1)\theta )+\sin (k+1)\theta }\\ & \times \frac{\bar{\zeta }\sin (\alpha -\theta )+\sin \theta }{\sin \alpha },\end{array}$$

$$\begin{array}{rl}& \frac{\zeta \sin (\alpha -\theta )+\sin \theta }{\sin \alpha }\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\zeta \sin (\alpha +k\theta )-(\bar{z}+\zeta )\sin k\theta -\sin (\alpha -k\theta )}\\ =& -\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\bar{\zeta }\sin (k+1)\theta -\bar{z}\sin (\alpha +(k+1)\theta )+\bar{\zeta }\sin (\alpha -(k+1)\theta )+\sin (k+1)\theta }\\ & \times \frac{|\zeta \sin (\alpha -\theta )+\sin \theta {|}^2}{{\sin }^2\alpha }\\ =& -\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\bar{\zeta }\sin (k+1)\theta -\bar{z}\sin (\alpha +(k+1)\theta )+\bar{\zeta }\sin (\alpha -(k+1)\theta )+\sin (k+1)\theta }.\end{array}$$

$$\begin{array}{rl}& \frac{\zeta \sin (\alpha -\theta )+\sin \theta }{\sin \alpha }\frac{\bar{z}\sin (\alpha +k\theta )-\sin k\theta }{\bar{z}\zeta \sin (\alpha +k\theta )-(\bar{z}+\zeta )\sin k\theta -\sin (\alpha -k\theta )}\\ =& \frac{\bar{z}\sin (\alpha -(l+1)\theta )+\sin (l+1)\theta }{\bar{z}\bar{\zeta }\sin l\theta +\bar{z}\sin (\alpha -l\theta )-\bar{\zeta }\sin (\alpha +l\theta )+\sin l\theta }.\end{array}$$

$$\begin{array}{rl}& \frac{\zeta \sin (\alpha -\theta )+\sin \theta }{\sin \alpha }{\partial }_{\zeta }{G}_1(z,\zeta )\\ =& -\frac{n\sin (\alpha -\theta )}{\sin \alpha }+2\sum _{k=0}^{n-1}\mathrm{Re}\left(\frac{z\sin (\alpha -(k+1)\theta )+\sin (k+1)\theta }{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }\right).\end{array}$$

$$\begin{array}{rl}& p(z,\zeta )\\ =& -\frac{1}{2}{\partial }_{{\nu }_{\zeta }}{G}_1(z,\zeta )\\ =& \mathrm{Re}\left(\frac{\zeta \sin (\alpha -\theta )+\sin \theta }{\sin \alpha }{\partial }_{\zeta }{G}_1(z,\zeta )\right)\\ =& -\frac{n\sin (\alpha -\theta )}{\sin \alpha }+2\sum _{k=0}^{n-1}\mathrm{Re}\left(\frac{z\sin (\alpha -(k+1)\theta )+\sin (k+1)\theta }{z\zeta \sin k\theta +z\sin (\alpha -k\theta )-\zeta \sin (\alpha +k\theta )+\sin k\theta }\right).\end{array}$$

Full text

Harmonic Green and Neumann functions for domains bounded by two intersecting circular arcs

Hanxing Lin

Math. Institute, FU Berlin, Arnimallee 3, D-14195 Berlin, Germany

hxlin@zedat.fu-berlin.de

*Dedicated to Professor Dr. Heinrich Begehr        

on the occasion of his $80^{\textrm{th}}$ birthday*

Abstract. The parqueting-reflection principle is shown to also work for constructing harmonic Green functions and harmonic Neumann functions for a class of domains, which are bounded by two arcs in $\mathbb{C}_{\infty}$ with a special intersecting angle $\pi/n,\,n\in\mathbb{N}^{*}$. Applying the Green representation formula and the Neumann representation formula we solve the Dirichlet and Neumann boundary problem to the Poisson equation in these domains.

Keywords. Poisson equation, harmonic Green function, harmonic Neumann function, parqueting-reflection principle

Mathematics Subject Classifications: 31A05 35A08 35J08 35J25

1 Circle Reflection in $\mathbb{C}_{\infty}$

Any circle or straight line in $\mathbb{C}$ can be expressed uniformly by an equation of the form

[TABLE]

where $a,\,c\in\mathbb{R},\,b\in\mathbb{C}$, and $ac-b\bar{b}<0$. It is a straight line if $a=0$, and a circle whose center is $-{b}/{a}$ with a radius ${\sqrt{b\bar{b}-ac}}/{|a|}$ if $a\neq 0$. In both cases we just call them circles in the extended complex plane $\mathbb{C}_{\infty}$, which is $\mathbb{C}\cup\{\infty\}$.

We see that a circle in $\mathbb{C}_{\infty}$ is determined by a matrix

[TABLE]

a $2\times 2$ Hermitian matrix with negative determinant. We call it a matrix of a circle in $\mathbb{C}_{\infty}$. A matrix of a circle is unique up to a nonzero real scalar multiple. Let $\mathrm{H}^{-}=\{A\in\mathrm{GL}_{2}(\mathbb{C}):A=A^{*},\,\mathrm{det}(A)<0\}$ be all the $2\times 2$ Hermitian matrices with negative determinant. We define an equivalence relation, denoted by $\sim$, in $\mathrm{H}^{-}$: $A\sim B$ in $\mathrm{H}^{-}$ if and only if $A=\lambda B$ for some $0\neq\lambda\in\mathbb{R}$. There exists one to one correspondence between the collection of all circles in $\mathbb{C}_{\infty}$ and all the equivalent classes in $\mathrm{H}^{-}$.

Next discussion is about reflections at circles in $\mathbb{C}_{\infty}$. A reflection at the circle $\{az\bar{z}+\bar{b}z+b\bar{z}+c=0\}$ is defined by the mapping

[TABLE]

When the mirror is a straight line, it is just the reflection at the line in typical sense. When the mirror is the circle $\{|z-z_{0}|=r\}$, it is the circle reflection, which sends $z$ to $z_{0}+{r^{2}}/(\bar{z}-\bar{z_{0}})$.

Since $\mathbb{C}_{\infty}$ is the same as the complex projective line $\mathbb{C}\mathrm{P}^{1}$, a point in $\mathbb{C}_{\infty}$ can be represented by a homogeneous coordinate $[z:w]$, where $z,\,w\in\mathbb{C}$ but cannot both be [math]. It corresponds to ${z}/{w}$ if $w\neq 0$, and to $\infty$ if $w=0$. The multiplication of a homogeneous coordinate and an invertible matrix is similarly defined as the normal matrices multiplication, namely as

[TABLE]

The conjugate of a homogeneous coordinate $\overline{[z:w]}$ can be naturally defined by $[\bar{z}:\bar{w}]$. Then a reflection through a circle $A=\left(\begin{smallmatrix}a&\bar{b}\\ b&c\end{smallmatrix}\right)$, denoted by $R_{A}$, can be given by

[TABLE]

We close this section with a small but useful result.

Lemma 1.1**.**

Reflecting circle $B$ at circle $A$ gives the circle $AB^{-1}A$.

Proof.

Suppose $[z:w]$ is a point on the circle $B$ and its reflection at the circle $A$ is $[u:v]$. Because $[z:w]$ satisfies the equality

[TABLE]

substituting $[z:w]$ by $\overline{[u:v]AP}$, where $P=\left(\begin{smallmatrix}0&1\\ -1&0\end{smallmatrix}\right)$, gives

[TABLE]

Direct calculation shows that $P\bar{B}\bar{P}^{T}\sim B^{-1}$. Therefore $[u:v]$ is on the circle $AB^{-1}A$. ∎

2 Parqueting reflections for domains bounded by two circular arcs in $\mathbb{C}_{\infty}$

Based on the conformal invariant property of harmonic Green function, one way to obtain harmonic Green functions for domains is to construct the conformal mappings from these domains onto a domain whose Green function is clearly known. But unfortunately it is normally not easy to get conformal mappings, and sometime conformal mappings may be too complicated. In these cases the parqueting-reflection principle is employed, which provides a method to construct harmonic Green functions, as well as harmonic Neumann functions. It has been verified working for several special domains whose boundary is composed of circular arcs or straight lines, for instance discs, disc sectors, half planes, strips, half strips, cones, concentric rings, hyperbolic strips and many more, see [1-8].

We aim to apply the Parqueting reflection principle to a special class of domains in the complex plane, which are bounded by two circular arcs that intersect at a special angle $\pi/n$, $n\in\mathbb{N}$. In this article an arc means specially a segment of a circle or straight line. There exists four cases for this class of domains, as shown in Figure 1.

Since every disc can be mapped conformally onto the unit disc, without loss of generality, we assume that one arc is a part of the unit circle. Suppose that

[TABLE]

where $\partial{D_{0}}=C_{0}\cup C_{1}$, $C_{1}$ is a segment of the unit circle. $C_{0}$ and $C_{1}$ intersect each other at two points, $e^{i\alpha}$ and $e^{-i\alpha}$ where $0<\alpha<\pi$, the angle between the two arcs is $\theta$, $0<\theta\leq\pi$. Figure 2 shows this domain in detail.

Reflecting $D_{0}$ at $C_{1}$ generates a new domain, denoted by $D_{2}$, $\partial D_{2}=C_{1}\cup C_{2}$. Carrying out successive steps of reflection, each time reflecting the new domain at its new generated boundary, produces a sequence of arcs $C_{k}$ and domains $D_{k}$. See that $C_{k}$ is the reflection of $C_{k-2}$ at $C_{k-1}$, $D_{k}$ is the reflection of $D_{k-1}$ at $C_{k}$, and $D_{k}$ is bounded by $C_{k-1}$ and $C_{k}$, i.e. $\partial D_{k}=C_{k-1}\cup C_{k}$. Since angles are preserved under reflections, $C_{k-1}$ and $C_{k}$ always intersect at the angle $\theta$. If $\theta=\pi/n$, after operating $2n-1$ times of reflections we see that $C_{2n}$ coincides with $C_{0}$, then these circular arcs $C_{0},C_{1},\cdots,C_{2n-1}$ divide the plane into $2n$ domains, which are just $D_{0},D_{1},\cdots,D_{2n-1}$. Thereby the sequence of domains $\{D_{0},D_{1},\cdots,D_{2n-1}\}$ provides a parqueting of the complex plane, i.e. $\mathbb{C}=\bigcup_{k=0}^{2n-1}\overline{D_{k}}$.

The following lemma provides the matrices of the boundary arcs.

Lemma 2.1**.**

The matrix of $C_{k}$ is

[TABLE]

Proof.

Verify this conclusion by induction on $k$.

[TABLE]

and

[TABLE]

they both coincide with the formula. Assume that

[TABLE]

and

[TABLE]

since $C_{k}$ is obtained by reflecting $C_{k-2}$ through $C_{k-1}$, via Lemma 1.1 we validate that

[TABLE]

∎

Let $z_{0}=z\in D_{0}$. The successive reflections which generate the domains $\{D_{k}\}$ also produce a series of reflection points generated by $z_{0}$, denoted by $\{z_{k}\}$. It means that $z_{k}\in D_{k}$ and reflecting $z_{k}$ at the circle $C_{k+1}$ gives $z_{k+1}\in D_{k+1}$. Especially, reflecting $z_{0}$ at the unit circle $C_{1}$ gives $z_{1}=1/\bar{z}\in D_{1}$. Notice that $z_{k}=z_{2n-k-1}$ holds when $z$ belongs to circle $C_{0}$, while $z_{2k}=z_{2k+1}$ holds when $z$ lies on the circle $C_{1}$. Actually, there are varieties of viewpoints to investigate the generation of $\{D_{k}\}$ and $\{z_{k}\}$. Reflecting $D_{0}$ at $C_{k}$ also produces $D_{2k-1}$, while reflecting $D_{1}$ at $C_{k+1}$ turns out to be $D_{2k}$. It means that $z_{2k}$ is the image of $z_{1}$ under reflection through $C_{k+1}$, while $z_{2k+1}$ be the image of $z_{0}$ under the reflection through $C_{k+1}$. Via the discussion of reflections in Section 1, we deduce that

[TABLE]

and

[TABLE]

3 Harmonic Green function of $D_{0}$

In this section we apply the parqueting-reflection principle to construct the harmonic Green function for $D_{0}$ and then solve the Dirichlet problem of Poisson equation in $D_{0}$.

Consider

[TABLE]

where

[TABLE]

It is obvious that $\log|P(z,\zeta)|^{2}$ as a function in variable $z$ is harmonic in $D_{0}\setminus\{\zeta\}$, and $\log|P(z,\zeta)|^{2}+\log|\zeta-z|^{2}$ is harmonic in $D_{0}$. Before showing that $\log|P(z,\zeta)|^{2}$ vanishes on the boundary $\partial D_{0}$, firstly we study some properties of $F(z,\zeta)$.

Lemma 3.1**.**

If $\theta=\frac{\pi}{n}$, then $\prod_{k=0}^{n-1}\left|\frac{z\sin k\theta-\sin(\alpha+k\theta)}{\bar{z}\sin(\alpha+k\theta)-\sin k\theta}\right|=1$ on the boundary of $D_{0}$.

Proof.

If $z\in C_{0}$, $z\bar{z}\sin(\alpha-\theta)+z\sin\theta+\bar{z}\sin\theta-\sin(\alpha+\theta)=0$, then $|z\sin(\alpha-\theta)+\sin\theta|=\sin\alpha$. Substituting $\bar{z}$ by $\frac{-z\sin\theta+\sin(\alpha+\theta)}{z\sin(\alpha-\theta)+\sin\theta}$, we get

[TABLE]

In the other case, $z\in C_{1}$, $z\bar{z}=1$, then

[TABLE]

The conclusion holds for both cases. ∎

Noticing that $\lim\limits_{z\to\partial D_{0}}F(z,\zeta)=1$, it implies immediately $\lim\limits_{z\to\partial D_{0}}|P(z,\zeta)|=1$ via Lemma 3.1. Then $\log|P(z,\zeta)|^{2}$ vanishes on the boundary. Therefore we have verified the harmonic Green function of $D_{0}$, claimed as follows.

Theorem 3.2**.**

$G_{1}(z,\zeta)=\log|P(z,\zeta)|^{2}$* is the harmonic Green function of the domain $D_{0}$.*

Next we discuss the Poisson kernel for $D_{0}$. Let $p(z,\zeta)=-\frac{1}{2}\partial_{\nu_{\zeta}}G_{1}(z,\zeta)$ be the Poisson kernel for $D_{0}$, where $z\in D_{0}$, $\zeta\in\partial D_{0}$. We calculate $p(z,\zeta)$ directly.

[TABLE]

On the boundary $C_{0}\subset\{\zeta\bar{\zeta}\sin(\alpha-\theta)+\zeta\sin\theta+\bar{\zeta}\sin\theta-\sin(\alpha+\theta)=0\}$, the outward normal vector $\nu_{\zeta}$ is

[TABLE]

while the outward normal derivative is

[TABLE]

(When $\alpha=\theta$, $C_{0}$ is just a line segment, in this case $\nu_{\zeta}=-1$ and $\partial_{\nu_{\zeta}}=-\partial_{\zeta}-\partial_{\bar{\zeta}}$.) Obviously

[TABLE]

Replacing $\zeta$ by $\frac{-\bar{\zeta}\sin\theta+\sin(\alpha+\theta)}{\overline{\zeta}\sin(\alpha-\theta)+\sin\theta}$ in

[TABLE]

gives

[TABLE]

which implies that

[TABLE]

Let $l:=n-k-1$, since $n\theta=\pi$, we have

[TABLE]

Then

[TABLE]

Thus in the case of $\zeta\in C_{0}$, $z\in D_{0}$, the Poisson kernel

[TABLE]

On the other boundary $C_{1}$, the outward normal derivative is $\partial_{\nu_{\zeta}}=\zeta\partial_{\zeta}+\bar{\zeta}\partial_{\bar{\zeta}}$.

[TABLE]

Thus in the case of $\zeta\in C_{1}$, $z\in D_{0}$, the Poisson kernel

[TABLE]

**Remark 1. **The definition of $p(z,\zeta)$ can be extended for $z\in\partial D_{0}$. When $z\in\partial D_{0},$ $\partial_{\zeta}G_{1}(z,\zeta)=0$ because of $G_{1}(z,\zeta)=0$. Therefore if $z,\zeta\in\partial D_{0}$ and $z\neq\zeta$, then $p(z,\zeta)=0$.

Further investigation about the boundary behavior of the Poisson kernel is necessary. Let $g_{0}(z,\zeta)$ be the Green function for the domain

[TABLE]

and $p_{0}(z,\zeta)$ be the corresponding Poisson kernel. It is easy to check that

[TABLE]

[TABLE]

Denote $g_{1}(z,\zeta)$ and $p_{1}(z,\zeta)$ respectively as the Green function and the Poisson kernel for the unit disc $\{|\zeta|\leq 1\}$. It is well known that

[TABLE]

[TABLE]

With these notations, an boundary property of $p_{(}z,\zeta)$ is shown in next lemma.

Lemma 3.3**.**

$\lim\limits_{z\to C_{0}}p(z,\zeta)=\lim\limits_{z\to C_{0}}p_{0}(z,\zeta)$* if $\zeta\in C_{0}$, while $\lim\limits_{z\to C_{1}}p(z,\zeta)=\lim\limits_{z\to C_{1}}p_{1}(z,\zeta)$ if $\zeta\in C_{1}$.*

Proof.

If $z$ goes to $C_{0}$, replacing $z$ by $\frac{-\bar{z}\sin\theta+\sin(\alpha+\theta)}{\bar{z}\sin(\alpha-\theta)+\sin\theta}$, we have

[TABLE]

Hence $\partial_{\nu_{\zeta}}G_{1}(z,\zeta)-\partial_{\nu_{\zeta}}g_{0}(z,\zeta)=0$ when $z$ goes to $C_{0}$. Likewise, if $z$ goes to $C_{1}$, substituting $\bar{z}$ by $1/z$, then we have

[TABLE]

Hence $\partial_{\nu_{\zeta}}G_{1}(z,\zeta)-\partial_{\nu_{\zeta}}g_{1}(z,\zeta)=0$ when $z$ goes to $C_{1}$.

By the above discussion, we deduce that, for $\zeta\in C_{0}$

[TABLE]

while for $\zeta\in C_{1}$

[TABLE]

∎

Then the following result is obtained on the basis of Remark 1 and Lemma 3.3.

Theorem 3.4**.**

For $\gamma\in\mathrm{C}(\partial D_{0};\mathbb{C})$ the boundary behavior

[TABLE]

for any $\zeta\in\partial D_{0}$ holds.

Green representation formula shows that, see [9], for any $w(z)\in\mathrm{C}^{2}(D,\mathbb{C})\cap\mathrm{C}(\bar{D},\mathbb{C})$,

[TABLE]

With the properties of Green function and the result of Theorem 3.4, this Green representation formula succeeds in giving the solution of Dirichlet boundary problem for the Poisson equation on the domain $D_{0}$, as shown in the next theorem.

Theorem 3.5**.**

The Dirichlet problem $w_{z\bar{z}}=f$ in $D_{0}$, $w=\gamma$ on $D_{0}$ for $f\in\mathrm{L}_{p}(D_{0};\mathbb{C})$, $p>2$, $\gamma\in\mathrm{C}(\partial D_{0},\mathbb{C})$ has a unique solution, which is provided by

[TABLE]

4 Harmonic Neumann function of $D_{0}$

To construct the harmonic Neumann function for $D_{0}$, we start with a rational function in variables $z$ and $\zeta$, denoted by $H(z,\zeta)=\prod_{k=0}^{n-1}(\zeta-z_{2k})(\zeta-z_{2k+1})$. Multiplying $H(z,\zeta)$ by the product of all the denominators appearing in the $z_{k}$ terms, it turns out to be a polynomial function in variables $z$ and $\zeta$, denoted by

[TABLE]

$|P(z,\zeta)|$ is symmetric in variables $z$ and $\zeta$ based on the fact that Green functions are symmetric. We also see that $|Q(z,\zeta)|$ is symmetric in $z$ and $\zeta$ by comparing $Q(z,\zeta)$ with $P(z,\zeta)$. So $Q(z,\zeta)$ will be a candidate for creating the harmonic Neumann function of $D_{0}$.

Lemma 4.1**.**

Let $N_{1}(z,\zeta)=-\log|Q(z,\zeta)|^{2}$. Then

[TABLE]

Proof.

The outward normal derivative is calculated straightforwardly.

[TABLE]

In the case of $z\in C_{1}$, $z\bar{z}=1$,

[TABLE]

Thereby for $z\in C_{1}$ and $\zeta\in\overline{D_{0}}\setminus\{z\}$,

[TABLE]

In the other case of $z\in C_{0}$, the outward normal derivative is

[TABLE]

Directly check that

[TABLE]

replacing $z$ by $\frac{-\bar{z}\sin\theta+\sin(\alpha+\theta)}{\bar{z}\sin(\alpha-\theta)+\sin\theta}$ in

[TABLE]

gives

[TABLE]

Then

[TABLE]

Hence

[TABLE]

from which we know that, for $z\in C_{0}$ and $\zeta\in\overline{D_{0}}\setminus\{z\}$,

[TABLE]

∎

Remark 2. Denote $\sigma(s)=\partial_{\nu_{z}}N_{1}(z,\zeta)$ for $z=z(s)\in\partial D_{0}$ and $\zeta\in\overline{D_{0}}\setminus\{z\}$, where $s$ is the arc length parameter, then restating the result of Lemma 4.1 gives

[TABLE]

Moreover

[TABLE]

From the construction of $N_{1}(z,\zeta)$ and Remark 2, it is seen that as a function of variable $z$ for any $\zeta\in\overline{D_{0}}$

• •

$N_{1}(z,\zeta)$ is harmonic in $D_{0}\setminus\{\zeta\}$ and continuously differentiable in $\overline{D_{0}}\setminus\{\zeta\}$,

• •

$N_{1}(z,\zeta)+\log|\zeta-z|^{2}$ is harmonic in $D_{0}$,

• •

$\partial_{\nu_{z}}N_{1}(z,\zeta)=\sigma(s)$ for $z=z(s)\in\partial D_{0}$, where $s$ is the arc length parameter. The density function $\sigma$ is a real-valued, piecewise constant function of $s$ with finite mass $\int_{\partial D_{0}}\sigma(s)\mathrm{d}s.$

We call $N_{1}(z,\zeta)$ a Neumann function for the domain $D_{0}$.

Remark 3. We conjecture that the term $\int_{\partial D_{0}}\sigma(z)N_{1}(z,\zeta)\mathrm{d}s_{z}$ is constant for $\zeta\in D_{0}$ . If this conjecture can be verified, suppose the constant is $K$, then modifying $N_{1}(z,\zeta)$ by $N_{1}(z,\zeta)-K$ will give an unique Neumann function for domain $D_{0}$ which satisfies an extra condition

• •

$\int_{\partial D_{0}}\sigma(z)N_{1}(z,\zeta)\mathrm{d}s_{z}=0$ (normalization condition).

The missing of normalization condition for the Neumann function doesn’t affect the following discussion.

Lemma 4.2**.**

For $\zeta\in C_{0}$,

[TABLE]

while for $\zeta\in C_{1}$,

[TABLE]

Proof.

For $\zeta\in C_{0}$,

[TABLE]

where

[TABLE]

From the proof of Lemma 4.1, we see that $T_{0}(z,\zeta)$ goes to

[TABLE]

when $z$ goes to $C_{0}$. Hence

[TABLE]

For $\zeta\in C_{1}$,

[TABLE]

where

[TABLE]

When $z$ goes to $C_{1}$, $T_{1}(z,\zeta)$ turns out to be

[TABLE]

which is also shown in the proof of Lemma 4.1. Therefore

[TABLE]

∎

Neumann Boundary Problem: Find a solution to the Poisson equation $w_{z\bar{z}}=f$ in $D_{0}$, satisfying $\partial_{\nu_{z}}w=\gamma$ on $\partial D_{0}$ except for the two corner points.

We recall the Neumann representation formula here, see [9]. For a regular domain $D$ in $\mathbb{C}$, any function $w(z)\in\mathrm{C}^{2}(D,\mathbb{C})\cap\mathrm{C}(\bar{D},\mathbb{C})$ can be represented by

[TABLE]

where $N_{1}(z,\zeta)$ is the Neumann function of $D$, $\sigma(\zeta)=\partial_{\nu_{z}}N_{1}(z,\zeta)$ for $z\in\partial D$. This formula is applied to provide the solutions to the Neumann boundary problem, shown as follows.

Theorem 4.3**.**

The Neumann boundary problem is solvable if and only if

[TABLE]

all the solutions are of the following form

[TABLE]

where $c$ is an arbitrary constant in $\mathbb{C}$.

Proof.

The Neumann function is a fundamental solution to the Poisson equation, and the boundary integral $\int_{\partial D_{0}}\gamma(\zeta)N_{1}(z,\zeta)\mathrm{d}s_{\zeta}$ is harmonic in $D_{0}$. It immediately implies that $w_{z\bar{z}}=f$.

[TABLE]

On the basis of Lemma 4.1 and Lemma 4.2, if $\zeta\in C_{0}$,

[TABLE]

if $\zeta\in C_{1}$,

[TABLE]

Then $\partial_{\nu_{z}}w(z)=\gamma(z)$ on $\partial D_{0}$ if and only if

[TABLE]

Hence function $w(z)$ solves the Neumann problem if the solubility condition is satisfied.

Suppose $\phi(z)$ also solves the Neumann problem, then $\phi(z)-w(z)$ is harmonic in $D_{0}$ and its normal derivative vanishes on $\partial D_{0}$. It implies that $\phi(z)-w(z)$ must be a constant.

To sum up, if the Neumann problem is solvable, all the solutions are of the form as

[TABLE]

∎