# Discrete harmonic functions for non-symmetric Laplace operators in the
quarter plane

Viet Hung Hoang  CNRS and Institut Denis Poisson, UMR CNRS 7013, Université de
Tours et Université d’Orléans, Parc de Grandmont, 37200 Tours, France,
Institut für Mathematische Stochastik, Universität Münster, Orléans-Ring 10,
48149 Münster, Germany<EMAIL_ADDRESS>
###### Abstract.

We construct harmonic functions in the quarter plane for discrete Laplace
operators. In particular, the functions are conditioned to vanish on the
boundary and the Laplacians admit coefficients associated with transition
probabilities of non-symmetric random walks. By solving a boundary value
problem for generating functions of harmonic functions, we deduce explicit
expressions for the generating functions in terms of conformal mappings. These
mappings are yielded from a conformal welding problem with quasisymmetric
shift and contain information about the growth of harmonic functions. Further,
we describe the set of harmonic functions as a vector space isomorphic to the
space of formal power series.

###### Key words and phrases:

Discrete harmonic functions, conformal mappings, conformal welding

###### 2010 Mathematics Subject Classification:

Primary 31C35, 60G50; Secondary 60J45, 60J50, 31C20

This project has received funding from the European Research Council (ERC)
under the European Union’s Horizon 2020 research and innovation programme
under the Grant Agreement No. 759702.

###### Contents

  1. 1 Introduction
  2. 2 Study of the kernel
  3. 3 Conformal welding
  4. 4 Proof of the main theorems
  5. 5 Small jump random walks

##  1\. Introduction

### Discrete harmonic functions

Let us first recall some definitions. A real-valued function $h$ defined on a
domain $D\subset\mathbb{Z}^{2}$ is called discrete harmonic (or preharmonic)
with respect to the discrete Laplace operator

(1.1) $\Delta(h)(i,j)=\sum_{k,\ell}p_{k,\ell}h(i+k,j+\ell)-h(i,j),$

with $\\{p_{k,\ell}\\}$ fixed, if $\Delta(h)(i,j)=0$ for all $(i,j)\in D$. For
the sake of brevity, the words “harmonic function” and “Laplace operator” will
always be understood in the discrete context. The word “discrete” will only be
used in case where we want to emphasize it in relation to the continuous
counterpart.

During the second half of the 20th century, queuing theory was of interest to
many probabilists. Discrete harmonicity appeared vastly under the form of
balance equations to model the stationary distribution of many quantities (see
the book [25] of Kleinrock for an introduction of various models or the book
[9] of Cohen for single server queues). Another important application is the
connection to the concept of Doob’s transform. Biane in [4], Eichelsbacher and
König in [16] constructed processes conditioned to stay in cones from positive
harmonic functions. Later, Denisov and Wachtel in [13] used the functions to
find the tail asymptotics for the exit time of such conditioned walks.
Harmonic functions are also found in some problems of population dynamics to
model the extinction probabilities of the population (see [29, 2]).

The references listed above mostly concern positive harmonic functions on
account of their evident links to probabilistic quantities. However, some
authors continued to extend the problem further to signed harmonic functions
(namely, the sign may vary in some domains). Almansi in [1] proved that in the
continuous context, polyharmonic functions can be decomposed into signed
harmonic functions. The discrete analog of this result was achieved in [12,
37, 8]. Further in [8], Chapon, Fusy and Raschel unveiled some connections
between discrete polyharmonic functions and complete asymptotic expansions in
walk enumeration problems. Therefore, studies of polyharmonic and/or signed
harmonic functions become more relevant. Last but not least, one of our goals
is to study the space of harmonic functions (see [24]), which gives us a
general viewpoint on this discrete potential theory problem.

### Overview of our main model and some studies of large jump random walks

Hoang, Raschel and Tarrago in [24] studied discrete harmonic functions for
symmetric Laplace operators in the quarter plane. More precisely, they
considered the Laplacian (1.1) with $p_{k,\ell}=p_{\ell,k}$ for all
$k,\ell\in\mathbb{Z}$. The present work aims at extending their results to a
non-symmetric context. We therefore reintroduce our main problem and point out
key differences between the models of this work and of [24].

In this article, we consider the problem of finding discrete harmonic
functions in the quarter plane (i.e., $D=\mathbb{N}^{2}=\\{1,2,3,...\\}^{2}$)
which vanish on the boundary axes. The weights $\\{p_{k,\ell}\\}$ of the
Laplacian (1.1) will be modeled by transition probabilities of random walks.
This is motivated by the links between harmonic functions and various problems
of queuing systems, population dynamics,… which can be reformulated as
problems of random walks. In particular, we consider walks with arbitrary
large negative jumps, small positive jumps, and moreover not being constrained
by any symmetry conditions (see Table 1). This is the main difference of this
paper’s model in comparison with [24]. Due to the high order of the
characteristic polynomial associated with large jumps, such walks are
undoubtedly challenging to study. We refer the readers to some achievements in
the subject: an analytic technique was developed in [11, 10] by Cohen and
Boxma to study walks with two-dimensional state space; asymptotics of
multidimensional walks are investigated in [13] by Denisov and Wachtel, [15]
by Duraj and Wachtel; and recently in combinatorics, walks with steps in
$\\{-2,-1,0,1\\}^{2}$ are systematically classified in [5] by Bostan,
Bousquet-Mélou and Melczer, alongside with some related conjectures. In brief,
the study of large step walks is still largely open for further development.

Symmetric small jump walks | Non-symmetric small jump walks
---|---
|
Symmetric large negative jump walks | Non-symmetric large negative jump walks
|
Symmetric large positive jump walks | Non-symmetric large positive jump walks
|
Symmetric transition probabilities | Non-symmetric transition probabilities
$1/4$$1/4$$1/4$$1/4$ $1/6$$1/6$$1/6$$1/6$$1/6$$1/6$ | $1/6$$1/6$$1/3$$1/3$ $1/8$$1/8$$1/6$$1/6$$1/8$$1/8$
Table 1. Various examples of symmetric and non-symmetric walks

### Analytic approach to find harmonic functions

We present here a complex analytic approach which will be the backbone of our
work. The technique was pioneered by Malyshev in [32], Fayolle and
Iasnogorodski in [17], to study the stationary distribution of small jump
random walks in a two-dimensional state space. The global idea is as follows.
The balance equations for the stationary probabilities lead to a functional
equation for their generating functions. An important bivariate polynomial
characterized by the transition probabilities of the walk appears in the
equation and is usually called the kernel function. By studying the Riemann
surface associated with this kernel, then constructing a uniformizing
variable, one can solve the equation in certain regions, then continue
meromorphically the solutions into a bigger region. In another direction
proposed in [17] by Fayolle and Iasnogorodski, the algebraic curve associated
with the kernel offers a large choice of branches, and with an appropriate
subset, the functional equation can be reduced to boundary value problems
(BVP). Such a technique is also discussed in deeper insights by the same
authors in the book [18].

Since then, the technique has been successfully applied to various areas, such
as queuing systems [21, 20, 28], potential theory [35, 36, 30] and enumerative
combinatorics (counting walks in the quarter plane) [7, 27, 14], leading to
highly precise results (both exact and asymptotic results).

We also mention here the work [3] by Banderier and Flajolet. The authors
considered combinatorial aspects of one-dimensional large jump walks as
directed (large jump) two-dimensional walks. By the analytic approach, the
counting generating functions are shown to have algebraic forms constructed by
the branches of the associated algebraic curve.

And lastly, an alternative approach introduced by Cohen and Boxma in [11]
contributes largely to the development of this technique (see also [10] by
Cohen). The overall idea is the same as Fayolle and Iasnogorodski’s. However,
the subset of the kernel’s zero set is chosen differently, in a way that this
choice does not depend on the branches of the algebraic curve. Their
construction overcomes the obstacle of studying precisely the Riemann surface,
therefore it can cover a wide class of large jump random walks. This approach
will also be the central point of our present analysis.

### Non-symmetric walks with large negative jumps and harmonic functions

Consider a lattice walk on $\mathbb{Z}^{2}$ with transition probabilities (or
non-negative weights) $\\{p_{k,\ell}\\}_{(k,\ell)\in\mathbb{Z}^{2}}$ summing
to $1$ such that:

  1. (H1)

The walk has small positive jumps (at most $1$ unit) and arbitrary large
negative jumps, i.e., $p_{k,\ell}=0$ if $k\geqslant 2$ or $\ell\geqslant 2$
(see Table 1);

  2. (H2)

If $\big{|}\sum p_{k,\ell}x^{k}y^{\ell}\big{|}=1$ and $|x|=|y|=1$, then
$x=y=1$ (which is a necessary condition for the irreducibility of the walk);

  3. (H3)

$\\{p_{k,\ell}\\}_{(k,\ell)\in\mathbb{Z}^{2}}$ have finite moments of order
$2$, i.e., $\sum(k^{2}+\ell^{2})p_{k,\ell}<\infty$;

  4. (H4)

The drift is zero, i.e., $\sum kp_{k,\ell}=\sum\ell p_{k,\ell}=0$.

Now let $\Delta$ be the discrete Laplacian (1.1) associated with
$\\{p_{k,\ell}\\}_{(k,\ell)\in\mathbb{Z}^{2}}$ and acting on functions
$h:\mathbb{Z}^{2}\to\mathbb{R}$. Our goal is to describe real-valued functions
$\\{h(i,j)\\}_{i,j\in\mathbb{Z}^{2}}$ satisfying the following conditions:

  1. (H5)

$h(i,j)$ is harmonic with respect to the operator $\Delta$ on the positive
quadrant, i.e., for all $i,j\geqslant 1$, $\Delta(h)(i,j)=0$;

  2. (H6)

$h(i,j)$ vanishes elsewhere, i.e., for all $(i,j)\in\mathbb{Z}^{2}$ with
$i\leqslant 0$ and/or $j\leqslant 0$, $h(i,j)=0$.

Random walks with such non-symmetric step sets have appeared throughout the
literature. In [7, 6], the authors study the problem of counting walks for
numerous small jump random walks. In [4, 16, 26, 35], walks staying in Weyl
chambers are investigated. For example, walks in the duals of SU(3) and Sp(4)
can be viewed as non-symmetric walks in the quadrant.

### Continuous counterpart of the main problem

For future use, we recall some facts about continuous harmonic functions in
cones. A real-valued function $f$ defined on a domain $D\subset\mathbb{R}^{2}$
is called harmonic with respect to the (continuous) Laplace operator

(1.2) $\nabla^{2}:=\partial_{xx}+\partial_{yy},$

if $\nabla^{2}f=0$ on $D$.

Consider the following Dirichlet problem on the cone
$\mathscr{E}:=\\{(r\cos\phi,r\sin\phi):r>0,0<\phi<\theta\\}$ of angle
$\theta\in(0,\pi]$: find functions $f:\mathscr{E}\to\mathbb{R}$ such that

  1. (i)

$f$ is harmonic with respect to $\nabla^{2}$ on $\mathscr{E}$;

  2. (ii)

$f$ is continuous on the closure $\overline{\mathscr{E}}$ of $\mathscr{E}$ and
vanishes on the boundary $\partial\mathscr{E}$.

If the cone is the upper half-plane $\mathbb{R}\times\mathbb{R}^{+}$, then the
problem can easily be solved by complex analysis: by Schwarz reflection
principle, the function $f$ can be extended to a harmonic function on
$\mathbb{R}^{2}$ by the formula $f(x,y)=-f(x,-y)$ for all
$(x,y)\in\mathbb{R}\times\mathbb{R}_{-}$. Moreover, a function
$f:\mathbb{R}^{2}\to\mathbb{R}$ is harmonic on $\mathbb{R}^{2}$ if and only if
there exists uniquely a complex analytic function $g$ (up to an additive
constant) on $\mathbb{C}$ such that $f(x,y)=\Im g(x+yi)$ (with $\Im$ denoting
the imaginary part) for any $(x,y)\in\mathbb{R}^{2}$. Therefore,
$g(\overline{z})=\overline{g(z)}$ for $z\in\mathbb{C}$ by Schwarz reflection
principle. In other words, $g(z)$ admits the form $g(z)=\sum_{n\geqslant
0}a_{n}z^{n}$ with $(a_{n})_{n}\subset\mathbb{R}$ and $|a_{n}|^{1/n}\to 0$
(which is a necessary and sufficient condition for $g(z)$ to be analytic on
$\mathbb{C}$). One can describe the set of its solutions as the vector space

$\displaystyle H(\mathbb{R}\times\mathbb{R}^{+})=\\{\sum_{n\geqslant
1}a_{n}\Im\big{(}(x+iy)^{n}\big{)}:(a_{n})_{n}\subset\mathbb{R}\text{ and
}|a_{n}|^{1/n}\to 0\\}.$

By the mapping $z\mapsto z^{\theta/\pi}$ which maps conformally the upper
half-plane to a cone of angle $\theta$ (in the complex plane), we then obtain
a similar description for the solution set of the problem on the cone
$\mathscr{E}$ of angle $\theta$ as

(1.3) $\displaystyle H(\mathscr{E})=\\{\sum_{n\geqslant
1}a_{n}\Im\big{(}(x+iy)^{n\pi/\theta}\big{)}:(a_{n})_{n}\subset\mathbb{R}\text{
and }|a_{n}|^{1/n}\to 0\\}.$

Through the linear transform $(x,y)\mapsto(x\sin\theta-y\cos\theta,y)$ which
maps the cone of angle $\theta$ to the first quadrant
$\mathscr{Q}:=\mathbb{R}_{+}^{2}$, the problem on the cone can be reformulated
into the one on the first quadrant: find functions
$f:\mathscr{Q}\to\mathbb{R}$ such that

  1. (i)

$f$ is analytic and satisfies the equation

(1.4) $\partial_{xx}f-2\cos(\theta)\partial_{xy}f+\partial_{yy}f=0$

on the first quadrant $\mathscr{Q}$;

  2. (ii)

$f$ is continuous on $\overline{\mathscr{Q}}$ and vanishes on the boundary
$\partial\mathscr{Q}$.

As a result, the corresponding solution set of this problem is also isomorphic
to the solution set $H(\mathscr{E})$ and is described as:

(1.5) $H(\mathscr{Q})=\\{\sum_{n\geqslant
1}a_{n}h_{\theta,n}:(a_{n})_{n}\subset\mathbb{R}\text{ and }|a_{n}|^{1/n}\to
0\\},$

where

(1.6)
$h_{\theta,n}(x,y):=\Im\bigl{(}(x/\sin\theta+y\cot\theta+{i}y)^{n\pi/\theta}\bigr{)}.$

The reason for mentioning these facts about the continuous problem is twofold:
on one hand, we will see later the analogous description to (1.5) of the
discrete counterpart; on the other hand, we will also emphasize a convergence
of the discrete harmonic functions to the continuous ones.

### Generating functions and the kernel

Let $h(i,j)$ be any harmonic function satisfying the conditions (H5) and (H6).
We now introduce its generating function

(1.7) $H(x,y)=\sum_{i,j\geqslant 1}h(i,j)x^{i-1}y^{j-1},$

and its sections

(1.8) $H(x,0)=\sum_{i\geqslant 1}h(i,1)x^{i-1}\quad\text{and}\quad
H(0,y)=\sum_{j\geqslant 1}h(1,j)y^{j-1}.$

From the Assumptions (H1), (H5), and (H6), one has

$\displaystyle H(x,y)$
$\displaystyle=\sum_{k,\ell}p_{k,\ell}x^{-k}y^{-\ell}\sum_{i,j\geqslant
1}h(i+k,j+\ell)x^{i+k-1}y^{j+\ell-1}$ $\displaystyle=\sum_{k,\ell\leqslant
1}p_{k,\ell}x^{-k}y^{-\ell}H(x,y)-\sum_{k\leqslant
1}p_{k,1}H(x,0)-\sum_{\ell\leqslant
1}p_{1,\ell}H(0,y)+p_{1,1}x^{-1}y^{-1}H(0,0).$

In other words, $H(x,y)$ satisfies the functional equation

(1.9) $K(x,y)H(x,y)=K(x,0)H(x,0)+K(0,y)H(0,y)-K(0,0)H(0,0),$

where $K(x,y)$ is called the kernel and is characterized by the weights
$\\{p_{k,\ell}\\}_{(k,\ell)\in\mathbb{Z}^{2}}$:

(1.10) $K(x,y)=xy\left(1-\sum p_{k,\ell}x^{-k}y^{-\ell}\right)=xy-\sum
p_{k,\ell}x^{-k+1}y^{-\ell+1}.$

It is easily verified that solutions of Equation (1.9) take the form

(1.11) $H(x,y)=\frac{F(x)+G(y)}{K(x,y)},$

for any power series $F,G\in\mathbb{C}[[t]]$ (formal ring of power series with
coefficients in $\mathbb{C}$). However, not every quotient
$\big{(}F(x)+G(y)\big{)}/K(x,y)$ defines a bivariate power series around the
center $(0,0)$. This raises the task of finding appropriate
$F,G\in\mathbb{C}[[t]]$ such that the quotient is a power series, which also
leads to the study of the kernel’s zero set, particularly near the point
$(0,0)$.

### A subset of the kernel’s zero set

We now introduce the subset of the kernel’s zero set under the approach of
Cohen and Boxma in [11]:

(1.12) $\mathscr{K}:=\\{(x,y)\in\mathbb{C}^{2}:K(x,y)=0\text{ and
}|x|=|y|\leqslant 1\\}.$

As will be shown in Section 2, $\mathscr{K}$ behaves differently in three
subcases and consists of a unique connected set, except in the case
$p_{1,1}=0$, $p_{1,0}\neq p_{0,1}$, where it also consists of an isolated
point $(0,0)$. The projections of its connected subsets along the first and
second variables are respectively denoted by $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$. Let $X:\mathscr{S}_{2}\to\mathscr{S}_{1}$ be the mapping
such that for all $y\in\mathscr{S}_{2}$, $(X(y),y)\in\mathscr{K}$. Under
Assumption (K1), $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are Jordan curves,
$X(y)$ is a well defined and one-to-one mapping from $\mathscr{S}_{2}$ onto
$\mathscr{S}_{1}$. Further, as $y$ moves on $\mathscr{S}_{2}$, $X(y)$ moves on
$\mathscr{S}_{1}$ in different orientation. Let
$Y:\mathscr{S}_{1}\to\mathscr{S}_{2}$ be the inverse of $X$. Under
Assumption(K1), we can first deduce further insights of the kernel’s zero set
in the bidisk (Proposition 1), and later define conformal mappings from the
interior of $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ onto disjoint,
complementary domains, which help constructing appropriate $F(x)$ and $G(y)$
in (1.11).

### Main results

In order to state our theorems properly, we introduce here some new notations.
The interior (resp. exterior) of the Jordan curve $\mathscr{S}_{1}$ is denoted
by $\mathscr{S}_{1}^{+}$ (resp. $\mathscr{S}_{1}^{-}$). $\mathscr{S}_{2}^{+}$
and $\mathscr{S}_{2}^{-}$ are denoted similarly. By analytic continuation, we
first have the following result concerning solutions of the kernel in the
bidisk.

######  Proposition 1.

Under Assumption (K1),

  1. (i)

The function $X:\mathscr{S}_{2}\to\mathscr{S}_{1}$, defined such that
$(X(y),y)\in\mathscr{K}$ for all $y\in\mathscr{S}_{2}$, can be continued
analytically and is uni-valued on $\mathscr{S}_{2}^{-}\cap\mathscr{C}^{+}$. In
other words, the extension of $X(y)$ has no branch point on
$\mathscr{S}_{2}^{-}\cap\mathscr{C}^{+}$. We have an analogous statement for
$Y(x)$, which is the inverse of $X(y)$;

  2. (ii)

$K(x,y)$ has no root on $\mathscr{S}_{1}^{+}\times\mathscr{S}_{2}^{+}$.

Proposition 1 may seem technical but it is crucial to indicate domains where
$H(x,y)$ in (1.11) is well defined. Particularly in the case $p_{1,1}=0$, we
will show that the generating function $H(x,y)$, initially defined on
$\mathscr{S}_{1}^{+}\times\mathscr{S}_{2}^{+}$, also admits an extension
around $(0,0)$. It is also worth mentioning that its item (i) is a new
statement, whereas its item (ii) has been stated for the symmetric cases of
[24], and will be proven in Section 2 for non-symmetric cases.

Our main theorems also are the main results in [24], but now applied to non-
symmetric cases under two technical assumptions (K4) and (K5). Roughly
speaking, (K4) ensures that $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ do not
admit a cusp at $1$, and (K5) ensures that a “shift” function, appearing in a
conformal welding problem (Section 3), has power growth derivatives. Under the
assumptions (K1)–(K5), $\mathscr{S}_{1}^{+}$ and $\mathscr{S}_{2}^{+}$ are
mapped conformally onto two disjoint complementary domains, respectively by
two mappings $\psi_{1}$ and $\psi_{2}$ such that:

  1. (i)

$\psi_{1}^{+}(X(y))=\psi_{2}^{+}(y)$ for all $y\in\mathscr{S}_{2}$, where
$\psi_{1}^{+}$ and $\psi_{2}^{+}$ are respectively the continuous extension of
$\psi_{1}$ and $\psi_{2}$ to $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$;

  2. (ii)

$\psi_{1}^{+}(1)=\psi_{2}^{+}(1)=\infty$;

  3. (iii)

$\psi_{1}(\overline{z})=\overline{\psi_{1}(z)}$ for all
$z\in\mathscr{S}_{1}^{+}$, and $\psi_{2}(\overline{z})=\overline{\psi_{2}(z)}$
for all $z\in\mathscr{S}_{2}^{+}$.

$\psi_{1}$ and $\psi_{2}$ will be showed to be well defined around $0$ whether
or not $0$ is in $\mathscr{S}_{1}^{+},\,\mathscr{S}_{2}^{+}$.

We also introduce a related angle:

(1.13) $\theta:=\arccos\frac{-\sum k\ell p_{k,\ell}}{\sqrt{\sum
k^{2}p_{k,\ell}}\sqrt{\ell^{2}p_{k,\ell}}}.$

It turns out that $\theta$ is the arithmetic mean of the angles at $1$ of
$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$, and contains information about the
growth of harmonic functions.

Finally, we introduce the following polynomials forming a basis of the space
$\mathbb{R}[X]$:

(1.14)
$\begin{split}&P_{2n+1}=(X-\psi_{1}(0))^{n+1}(X-\psi_{2}(0))^{n},n\geqslant
0,\\\ &P_{2n}=(X-\psi_{1}(0))^{n}(X-\psi_{2}(0))^{n},n\geqslant 1.\end{split}$

We then have our first theorem.

######  Theorem 2.

Under the assumptions (K1)–(K5), for any $n\geqslant 1$, the function

(1.15) $H_{n}(x,y):=\frac{P_{n}(\psi_{1}(x))-P_{n}(\psi_{2}(y))}{K(x,y)}$

defines the generating function of a harmonic function $h_{n}(i,j)$ satisfying
(H5) and (H6). Moreover, $H_{n}$ is analytic in
$\mathscr{S}_{1}^{+}\times\mathscr{S}_{2}^{+}$.

In particular, the discrete Laplace transform of normalized discrete harmonic
functions converges pointwise to the continuous Laplace transform of
continuous harmonic functions, that is,

(1.16) $\lim_{m\to\infty}\frac{c}{m^{n\pi/\theta+1}}\mathscr{L}h_{n}(\lfloor
mx\rfloor,\lfloor my\rfloor)=\mathscr{L}h_{\theta,n}(x,y),$

for any $x,y>0$, where $c$ is a non-vanishing constant and $h_{n,\theta}$ is
defined in (1.6), with $\theta$ in (1.13), and the Laplace transform
$\mathscr{L}$ is defined as follows

(1.17)
$\mathscr{L}f(x,y):=\begin{cases}\sum_{u,v=0}^{\infty}f(u,v)e^{-(ux+vy)},&f:\mathbb{N}^{2}\to\mathbb{C},\\\
\int_{0}^{\infty}\int_{0}^{\infty}f(u,v)e^{-(ux+vy)}dudv,&f:\mathscr{Q}\to\mathbb{C}.\end{cases}$

From the harmonic functions constructed in Theorem 2, we can describe the set
of harmonic functions as a vector space.

######  Theorem 3.

Under the assumptions (K1)–(K5), the space of discrete harmonic functions is
isomorphic to the vector space of formal power series $\mathbb{R}_{0}[[t]]$
with vanishing constant term, through the isomorphism

$\Phi:\left\\{\begin{matrix}\mathbb{R}[[t]]&\longrightarrow&H(\mathbb{N}^{2})\\\
\sum_{n\geqslant 1}a_{n}t^{n}&\longmapsto&\sum_{n\geqslant
1}a_{n}h_{n}.\end{matrix}\right.$

Among the additional assumptions (K1)–(K5), the condition (K5) is particularly
technical and difficult to verify, since it concerns the growth of a function
appearing in the analysis. Therefore, we also introduce a sufficient condition
asserting the moments of order 4 of the weights $\\{p_{k,\ell}\\}_{k,\ell}$
(Assumption (H7)), under which the assumption (K5) holds true. Then Theorems 2
and 3 are also valid under Assumptions (H1)–(H7), (K1) and (K4).

Although the results are similar to that of [24], the arguments for non-
symmetric cases require more assumptions and the analysis of quasiconformal
mapping. We therefore dedicate a table of comparison between the two models to
emphasize their differences (see Table 2).

Symmetric random walk in [24] | Non-symmetric random walk in this paper
---|---
$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are non-self-intersecting. |  $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ can be self-intersecting.
| (requires Assumption (K1) to exclude
| the cases of self-intersecting curves)
$X(y)\in\mathscr{S}_{1}$ and $y\in\mathscr{S}_{2}$ |  $X(y)\in\mathscr{S}_{1}$ and $y\in\mathscr{S}_{2}$
have different orientations. | can have the same orientations.
| (requires Assumption (K1) to exclude
| the cases of the same orientations)
Angles at $1$ of $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are equal | Angles at $1$ of $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ can be different
and different from $0$ | or equal to $0$.
| (requires assumption (K4) to exclude
| the cases of the cusps at $1$)
Case $p_{1,1}=0$: $\mathscr{S}_{1}^{+}$ and $\mathscr{S}_{2}^{+}$ contain $0$.
Case $p_{1,1}=0$, $p_{1,0}=p_{0,1}$: $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$
contain $0$.
(same proof of Proposition 1)
| Case $p_{1,1}=0$, $p_{1,0}\not=p_{0,1}$:
(Case $p_{1,1}=0$, $p_{1,0}\not=p_{0,1}$ |  $0$ lies in the interior of one curve
does not exist) | and the exterior of the other curve.
| (requires analytic continuations
| to properly define functions at $0$)
$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ coincide, |  $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ do not coincide.
$X(y)=\overline{y}$ for all $y\in\mathscr{S}_{2}$. |
(Conformal welding is naturally | (Conformal welding is obtained
obtained on the unit circle) | by solving a BVP with quasisymmetric shift
| under Assumption (K5))
Table 2. Comparison between symmetric and non-symmetric cases

### Acknowledgements

We would like to thank Kilian Raschel and Pierre Tarrago for useful
discussions. We also thank an anonymous referee for very constructive and
detailed remarks: he/she made us realize that one of our technical assumptions
can be replaced by a moment assumption.

##  2\. Study of the kernel

Subsection 2.1 aims at describing the set $\mathscr{K}$ in (1.12) through a
parameterisation, and presenting its characteristics under Assumption (K1)
about the path of its connected subset. We emphasize the behavior of
$\mathscr{K}$ at $(1,1)$, as it admits a corner point, whose associated angles
are strongly related to the growth of harmonic functions. These properties are
roughly similar to those in [24] and admit vastly the same proof schemes. In
Subsection 2.2, we show some behaviors of the kernel’s solutions near
$\mathscr{K}$ by analytic continuation. Finally, Subsection 2.3 will present
the proof of Proposition 1.

We introduce some notations. Let $\mathscr{C}$, $\mathscr{C}^{+}$, and
$\overline{\mathscr{C}^{+}}$ respectively denote the unit circle, the open and
closed unit disk. We also denote $\overline{z}$ as the complex conjugate of
any point $z\in\mathbb{C}$. Finally, the abbreviation “const” will indicate a
non-vanishing constant in $\mathbb{C}$.

###  2.1. Characteristics of $\mathscr{K}$

Adapting the analysis of Boxma and Cohen in [11, Sec. II.3.2], we first
parameterise $(x,y)$ with $|x|=|y|\leqslant 1$ as

(2.1) $(x,y)=(\eta s,\eta s^{-1}),$

with $\eta,s\in\mathbb{C}$, $|\eta|\leqslant 1$ and $|s|=1$. This
interpretation comes in handy since the trajectory of the variable $s$ is
already known. The kernel (1.10) then becomes

(2.2) $K(\eta s,\eta s^{-1})=\eta^{2}-\sum
p_{k,\ell}\eta^{-k-\ell+2}s^{-k+\ell}.$

Before presenting our analysis, we make two remarks as they will be needed for
future use:

  1. (i)

Since $\sum p_{k,\ell}=1$, then $K(x,y)$ converges for all $|x|,|y|\leqslant
1$, and thus it is analytic in the open bidisk
$\mathscr{C}^{+}\times\mathscr{C}^{+}$. In particular, $K(x,y)$ is a
polynomial if the jumps are bounded;

  2. (ii)

Assumption (H2) (irreducible random walks) implies that $p_{1,1}$, $p_{0,1}$
and $p_{1,0}$ cannot simultaneously vanish.

Figure 1. Solutions of $K(\eta s,\eta s^{-1})=0$, $|s|=1$ for some walks, from
left to right: $p_{0,1}=p_{1,0}=p_{-1,1}=p_{0,-1}/2=1/5$;
$p_{0,1}=p_{-1,0}=p_{1,-1}=1/3$; $p_{1,1}=p_{-1,1}=p_{0,-1}/2=1/4$.

The following lemma concerns the number of solutions of $K(\eta s,\eta
s^{-1})$ as a function of $\eta$ with parameter $s\in\mathscr{C}$.

######  Lemma 4.

Under Assumptions (H1)–(H4), for any $s\in\mathscr{C}$, the map $\eta\mapsto
K(\eta s,\eta s^{-1})$ admits exactly two roots in
$\overline{\mathscr{C}^{+}}$ (see Figure 1). In particular,

  1. (i)

If $p_{1,1}=0$, one root is identically zero, whereas the other one, denoted
by $\eta(s)$, is a continuous function of $s$ on $\mathscr{C}$ and varies as
follows: $\eta(s)\in\mathscr{C}^{+}$ for all $s\in\mathscr{C}\setminus\\{\pm
1\\}$ and $\eta(1)=-\eta(-1)=1$. Besides,
$\eta(-s)=\overline{\eta(-\overline{s})}=-\eta(s)$ for all $s\in\mathscr{C}$;

  2. (ii)

If $p_{1,1}\neq 0$, the two roots, denoted by $\eta(s)$ and $\eta_{2}(s)$, are
continuous functions of $s$ on $\mathscr{C}$ and vary in $\mathscr{C}^{+}$ for
all $s\in\mathscr{C}\setminus\\{1\\}$. Further, $K(\eta,1)$ has two distinct
roots in $[-1,1]$, which are $1$ and another root in $(-1,0)$. Let $\eta(s)$
be the root such that $\eta(1)=1$. Besides,
$\eta(s)=\overline{\eta(\overline{s})}=-\eta_{2}(-s)$ for all $|s|=1$.

We refer the readers to the proof of [24, Lem. 1 and Lem. 2], in which the
arguments are based on Rouché’s theorem and can be applied for Lemma 4. The
only difference is that in [24], $\eta(s)$ and $\eta_{2}(s)$ are real and vary
in $[-1,1]$, whereas in the present work, we can only state that they vary in
$\overline{\mathscr{C}^{+}}$.

Notice that if $p_{1,1}=0$, then one can factorize (2.2) as $K(\eta s,\eta
s^{-1})=\eta\widetilde{K}(\eta,s)$, where

(2.3) $\widetilde{K}(\eta,s):=\eta-\sum
p_{k,\ell}\eta^{-k-\ell+1}s^{-k+\ell},$

and $\widetilde{K}(\cdot,s)$ admits $\eta(s)$ as the unique root for any
$|s|=1$. Since $\widetilde{K}(0,s)=-p_{1,0}s-p_{0,1}s^{-1}$, then $\eta(s)$
never meets $0$ if $p_{1,1}=0$, $p_{1,0}\not=p_{0,1}$, whereas $\eta(s)$ meets
$0$ only at $s=\pm i$ if $p_{1,1}=0$, $p_{1,0}=p_{0,1}$.

In the case $p_{1,1}\not=0$, since
$(\eta_{2}(s)s,\eta_{2}(s)s^{-1})=(\eta(-s)(-s),\eta(-s)(-s)^{-1})$ for all
$s\in\mathscr{C}$, then $\mathscr{K}$ can be described by either $\eta(s)$ or
$\eta_{2}(s)$. Moreover, at this step, it is not clear that $\eta(s)$ and
$\eta_{2}(s)$ are distinct roots for all $|s|=1$. In fact, the assertion holds
true if and only if $\partial_{\eta}K(\eta s,\eta s^{-1})\not=0$ for all
$|s|=1$ and $\eta=\eta(s)$, i.e.,

$\sum p_{k,\ell}(-k-\ell)\eta(s)^{-k-\ell+1}s^{-k+\ell}\not=0,$

for all $|s|=1$.

We can now properly define $\mathscr{K}$ as follows:

(2.4) $\mathscr{K}=\begin{cases}\\{(\eta(s)s,\eta(s)s^{-1}):|s|=1\\},&\text{
if }p_{1,1}=0,\,p_{1,0}=p_{0,1},\\\
\\{(\eta(s)s,\eta(s)s^{-1}):|s|=1\\}\cup\\{(0,0)\\},&\text{ if
}p_{1,1}=0,\,p_{1,0}\neq p_{0,1},\\\
\\{(\eta(s)s,\eta(s)s^{-1}):|s|=1\\},&\text{ if }p_{1,1}\neq 0.\end{cases}$

We can also properly describe the projections of the connected subset of
$\mathscr{K}$ along the first and second variables:

$\mathscr{S}_{1}=\\{\eta(s)s:|s|=1\\}\quad\text{and}\quad\mathscr{S}_{2}=\\{\eta(s)s^{-1}:|s|=1\\}.$

We now introduce the following assumption:

  1. (K1)

$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are non-self-intersecting; $\eta(s)s$
and $\eta(s)s^{-1}$ move in opposite orientations and admit non-vanishing
derivatives for any $s\in\mathscr{C}\setminus\\{\pm 1\\}$ (resp.
$s\in\mathscr{C}\setminus\\{1\\}$) in the case $p_{1,1}=0$ (resp.
$p_{1,1}\not=0$), given that the derivatives exist.

For some examples not satisfying Assumption (K1), see Figure 2. Assumption
(K1) ensures that some characteristics of the curves $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$ match those of the symmetric case in [24], which suggests
that certain of their results and arguments may be reapplied in the present
work.

Figure 2. $\mathscr{S}_{1}$ (red) and $\mathscr{S}_{2}$ (blue) for some models
not satisfying Assumption (K1), from left to right:
$p_{0,1}/8=p_{1,0}=p_{-1,0}/3=p_{1,-4}/2=1/14$ ($\mathscr{S}_{2}$ is self-
intersecting); $p_{0,1}=p_{0,-1}=p_{-1,1}=p_{1,-1}=1/4$ ($\mathscr{S}_{2}$ is
a segment); $p_{-1,0}=p_{1,-3}=p_{0,1}/3=1/5$ ($\eta(s)s$ and $\eta(s)s^{-1}$
move in the same orientations).

Lemma 5 below presents some fundamental properties of $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$, and emphasizes differences between the subcases of the
description (2.4).

Figure 3. $\mathscr{S}_{1}$ (red) and $\mathscr{S}_{2}$ (blue) for some walks,
from left to right: $p_{0,1}=p_{1,0}=p_{-1,1}=p_{0,-1}/2=1/5$;
$p_{0,1}=p_{-1,0}=p_{1,-1}=1/3$; $p_{1,1}=p_{-1,1}=p_{0,-1}/2=1/4$. The right
figure also shows that one of the curve does not necessarily lie inside the
other.

######  Lemma 5.

Under Assumptions (H1)–(H4) and (K1), the following assertions hold (see
Figure 3):

  1. (i)

If $p_{1,1}=0$ (resp. $p_{1,1}\neq 0$), then the mappings $s\mapsto\eta(s)s$
and $s\mapsto\eta(s)s^{-1}$ are two-to-one (resp. one-to-one) from
$\mathscr{C}$ onto $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$;

  2. (ii)

$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are symmetric with respect to the real
axis;

  3. (iii)

If $s$ traverses $\mathscr{C}$ counterclockwise, then $\eta(s)s$ traverses
$\mathscr{S}_{1}$ counterclockwise and $\eta(s)s^{-1}$ traverses
$\mathscr{S}_{2}$ clockwise;

  4. (iv)

The relative position of $0$ with respect to $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$ varies in different cases:

    1. (a)

If $p_{1,1}=0$, $p_{1,0}=p_{0,1}$, then
$0\in\mathscr{S}_{1},\,\mathscr{S}_{2}$;

    2. (b)

If $p_{1,1}=0$, $p_{0,1}>p_{1,0}$ (resp. $p_{0,1}<p_{1,0}$), then
$0\in\mathscr{S}_{1}^{+},\,\mathscr{S}_{2}^{-}$ (resp.
$0\in\mathscr{S}_{1}^{-},\,\mathscr{S}_{2}^{+}$);

    3. (c)

If $p_{1,1}\neq 0$, then $0\in\mathscr{S}_{1}^{+},\,\mathscr{S}_{2}^{+}$.

###### Proof.

By Lemma 4, $\eta(e^{i\phi})e^{i\phi}$ and $\eta(e^{i\phi})e^{-i\phi}$ are
periodic functions of $\phi$ with period $\pi$ in the case $p_{1,1}=0$ or
period $2\pi$ in the case $p_{1,1}\not=0$. Since $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$ are non-self-intersecting (by Assumption (K1)), then these
functions are one-to-one mappings from $[0,\pi)$ (resp. $[0,2\pi)$) onto
$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ in the case $p_{1,1}=0$ (resp.
$p_{1,1}\not=0$), which also implies Item (i). Further, recall that for all
$|s|=1$, $\eta(s)=-\overline{\eta(-\overline{s})}$ if $p_{1,1}=0$, whereas
$\eta(s)=\overline{\eta(\overline{s})}$ if $p_{1,1}\not=0$. This infers Item
(ii). Item (iii) follows easily from Assumption (K1).

We now prove Item (iv) by studying the intersection of
$\mathscr{S}_{1}\setminus\\{1\\}$ (or $\mathscr{S}_{2}\setminus\\{1\\}$) and
$\mathbb{R}$. In the case $p_{1,1}=0$, this leads us to study the roots of
$K(x,-x)$ in $[-1,1]$. One can write $K(x,-x)=-x\widehat{K}(x)$, where
$\widehat{K}$ is the series

(2.5) $\widehat{K}(x)=x+\sum p_{k,\ell}(-1)^{-\ell+1}x^{-k-\ell+1}.$

Notice that $\widehat{K}^{\prime}(x)>0$ on $(-1,1)$, and by (H2) one has
$\widehat{K}(-1)=-1+\sum p_{k,\ell}(-1)^{-k}<0$,
$\widehat{K}(0)=p_{0,1}-p_{1,0}$, $\widehat{K}(1)=1+\sum
p_{k,\ell}(-1)^{-\ell+1}>0$. Hence, if $p_{0,1}=p_{1,0}$, then $0$ is the
unique root of $\widehat{K}(x)$ on $[-1,1]$. Thus $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$ pass through $0$, which implies Item (iv)a; If
$p_{0,1}>p_{1,0}$, then $\widehat{K}(x)$ has a root in $(-1,0)$, i.e.,
$\mathscr{S}_{1}$ (resp. $\mathscr{S}_{2}$) passes through a negative (resp.
positive) point; If $p_{0,1}<p_{1,0}$, then $\widehat{K}(x)$ has a root in
$(0,1)$, i.e., $\mathscr{S}_{1}$ (resp. $\mathscr{S}_{2}$) passes through a
positive (resp. negative) point. This implies Item (iv)b.

We move to the case $p_{1,1}\not=0$ and study the root of $K(x,x)$ in
$[-1,0]$. Since $K(-1,-1)=1-\sum p_{k,\ell}(-1)^{-k-\ell}>0$ and
$K(0,0)=-p_{1,1}<0$, then $K(x,x)$ has a root in $(-1,0)$, and
$\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ pass through a same negative point,
which implies Item (iv)c. ∎

The next lemma concerns the smoothness of $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$, particularly their shapes at $1$. Let us denote by $V[K]$
the zero set of the kernel on the closed bidisk:

$V[K]:=\\{(x,y)\in\mathbb{C}^{2}:K(x,y)=0,|x|\leqslant 1,|y|\leqslant 1\\}.$

Let us recall some key definitions from complex analysis. Firstly, a curve
parametrized by a function $z:[a,b]\to\mathbb{C}$ is smooth if $z^{\prime}(t)$
exists, is continuous and does not vanish on $[a,b]$. Secondly, a singular
point of $V[K]$ is defined as a solution of the system
$\partial_{x}K(x,y)=\partial_{y}K(x,y)=K(x,y)=0$. Otherwise, any point of
$V[K]$ is called non-singular or regular. Finally, the angle at a corner point
formed by the two tangents will be simply referred to as the angle (of the
curve) at that point.

######  Lemma 6.

Under Assumptions (H1)–(H4) and (K1), the curves $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$ are smooth everywhere except at $1$, where they admit corner
points (see Figure 3) with the angles

(2.6)
$\theta_{1}=\arccos\frac{AC-A^{2}-2B^{2}+2AB}{A(A-2B+C)}\quad\text{and}\quad\theta_{2}=\arccos\frac{AC-A^{2}-2B^{2}-2AB}{A(A+2B+C)},$

where $A=\sum p_{k,\ell}(k+\ell)^{2}$, $B=\sum p_{k,\ell}(k^{2}-\ell^{2})$ and
$C=\sum p_{k,\ell}(k-\ell)^{2}$. Moreover, $\mathscr{K}\setminus\\{(1,1)\\}$
consists of non-singular points of $V[K]$.

###### Proof.

We first study the differentiability of $\eta(s)$. By Assumption (H4),
$\partial_{x}K$ and $\partial_{y}K$ are well defined for all $|x|,|y|<1$, then
$\partial_{\eta}K(\eta s,\eta s^{-1})$ and $\partial_{s}K(\eta s,\eta s^{-1})$
are also well defined for all $|\eta|<1$, $|s|=1$.

In the case $p_{1,1}=0$, recall that $K(\eta s,\eta
s^{-1})=\eta\widetilde{K}(\eta,s)$, where $\widetilde{K}$ is defined in (2.3).
By Lemma 4, for any $s_{0}\in\mathscr{C}\setminus\\{\pm 1\\}$, $\eta(s_{0})$
is the unique root of $\widetilde{K}(\eta,s_{0})$ in $\mathscr{C}^{+}$, i.e.,
there exists a continuous function $u(\eta,s)$ defined in a neighborhood of
$(\eta(s_{0}),s_{0})$ such that
$\widetilde{K}(\eta,s)=(\eta-\eta(s_{0}))u(\eta,s)$ and $u(\eta(s),s)\not=0$.
This is equivalent to
$\partial_{\eta}\widetilde{K}(\eta(s_{0}),s_{0})=u(\eta(s_{0}),s_{0})\not=0$.
Thus, $\eta^{\prime}(s)$ is well defined for all
$s\in\mathscr{C}\setminus\\{\pm 1\\}$ under the form

(2.7)
$\eta^{\prime}(s)=-\frac{\partial_{s}\widetilde{K}(\eta(s),s)}{\partial_{\eta}\widetilde{K}(\eta(s),s)},$

which is obtained by differentiating the identity
$\widetilde{K}(\eta(s),s)=0$. Since $(\eta(s)s)^{\prime}$ and
$(\eta(s)s^{-1})^{\prime}$ are non-vanishing on $\mathscr{C}\setminus\\{\pm
1\\}$ (by Assumption (K1)), then $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are
smooth everywhere except at $1$.

In the case $p_{1,1}\not=0$, notice that for all $s\in\mathscr{C}$, $\eta(s)$
and $\eta_{2}(s)$ are distinct roots of $K(\eta s,\eta s^{-1})$. Indeed, if
there exists $s_{0}\in\mathscr{C}$ such that $\eta(s_{0})=\eta_{2}(s_{0})$,
then Lemma 4 implies $\eta(s_{0})s_{0}=\eta(-s_{0})(-s_{0})$, i.e.,
$\mathscr{S}_{1}$ is self-intersecting, which contradicts Assumption (K1).
Hence, for any $s\in\mathscr{C}\setminus\\{1\\}$, $\eta(s)$ is the unique root
of $K(\eta s,\eta s^{-1})$ in a neighborhood of $\eta(s)$, which is equivalent
to $\partial\eta K(\eta(s)s,\eta(s)s^{-1})\not=0$ by similar arguments as in
the case $p_{1,1}=0$. Thus, $\eta^{\prime}(s)$ is well defined for all
$s\in\mathscr{C}\setminus\\{1\\}$. By Assumption (K1), then $\mathscr{S}_{1}$
and $\mathscr{S}_{2}$ are smooth everywhere except at $1$.

We now prove that $\mathscr{K}\setminus\\{(1,1)\\}$ consists of non-singular
points of $V[K]$. Consider first the point $(0,0)\in\mathscr{K}$, which can
only happen in the case $p_{1,1}=0$. Since $\partial_{x}K(0,0)$ and
$\partial_{y}K(0,0)$, which are respectively equal to $p_{0,1}$ and $p_{1,0}$,
cannot simultaneously vanish, then $(0,0)$ is a non-singular point of $V[K]$.
Now move to the other points of $\mathscr{K}\setminus\\{(1,1)\\}$ and let
$(\eta,s)$ be a root of $K(\eta s,\eta s^{-1})$ such that
$\eta\in\overline{\mathscr{C}}\setminus\\{0\\}$ and $s\in\mathscr{C}$, then as
shown in the beginning of the proof, $\partial_{\eta}K(\eta,s)\neq 0$. Since

$\partial_{\eta}K=\partial_{\eta}x\partial_{x}K+\partial_{\eta}y\partial_{y}K=s\partial_{x}K+s^{-1}\partial_{y}K,$

we must have either $\partial_{x}K(\eta,s)$ or $\partial_{y}K(\eta,s)$ non-
zero. The claim then follows.

We now investigate the shapes of $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ at
$1$. In the case $p_{1,1}=0$, notice that one cannot evaluate the limit of
$\eta^{\prime}(s)$ as $s\to 1$ directly from (2.7) since the quotient is
undefined at $1$. We then differentiate the identity
$\widetilde{K}(\eta(s),s)=0$ twice and evaluate its limit as $s\to 1$. The
limit $\lim_{s\to 1}\eta^{\prime}(s)$ satisfies the equation

(2.8) $AX^{2}+2BX+C=0,$

where $A$, $B$ and $C$ are defined in Lemma 6 and are finite by Assumption
(H3). Since the above equation has two distinct roots, which indicates that
$\eta(s)$ is semi-differentiable at $s=1$ and admits left and right
derivatives

$\partial_{\pm}\eta(1):=\lim_{\begin{subarray}{c}s=\exp(i\phi)\\\ \phi\to
0^{\pm}\end{subarray}}\frac{\eta(s)-1}{s-1}\in\left\\{\frac{-B\pm{i}\sqrt{AC-B^{2}}}{A}\right\\}.$

In order to know which root of (2.8) corresponds to $\partial_{+}\eta(1)$, we
take a look at the following form:

$\partial_{+}(\eta(s)s)\big{|}_{s=1}=\partial_{+}\eta(1)+\eta(1)\in\left\\{\frac{A-B\pm{i}\sqrt{AC-B^{2}}}{A}\right\\}.$

Since $\arg\partial_{+}(\eta(s)s)\big{|}_{s=1}$ indicates the oriented angle
between the tangent of $\mathscr{S}_{1}$ and $\mathscr{C}$ at $1$, which is
non-negative and smaller than or equal to $\pi/2$ by Assumption (K1), then

$\partial_{+}(\eta(s)s)\big{|}_{s=1}=\frac{A-B+{i}\sqrt{AC-B^{2}}}{A}\quad\text{and}\quad\partial_{-}(\eta(s)s)\big{|}_{s=1}=\frac{A-B-{i}\sqrt{AC-B^{2}}}{A}.$

The angle $\theta_{1}$ of $\mathscr{S}_{1}$ at $1$ is implied from

$\cos\theta_{1}=2\cos^{2}\left(\frac{\theta_{1}}{2}\right)-1=2\frac{AC-B^{2}}{(A-B)^{2}+(AC-B^{2})}-1.$

Similarly, we have:

$\displaystyle\partial_{\pm}(\eta(s)s^{-1})\big{|}_{s=1}=\frac{-A-B\pm{i}\sqrt{AC-B^{2}}}{A},$
$\displaystyle\cos\theta_{2}=2\cos^{2}\left(\frac{\theta_{2}}{2}\right)-1=2\frac{AC-B^{2}}{(A+B)^{2}+(AC-B^{2})}-1.$

In the case $p_{1,1}\not=0$, by differentiating the identity
$K(\eta(s)s,\eta(s)s^{-1})=0$ twice and evaluating its limit as $s\to 1$, we
also obtain Eq. (2.8). The rest of the proof then follows the case
$p_{1,1}=0$. ∎

The following remark shows the relation between the angles $\theta_{1}$,
$\theta_{2}$ in Lemma 6 and $\theta$ introduced in (1.13). We also show how
the weights $\\{p_{k,\ell}\\}_{k,\ell}$ impact upon $\theta_{1}$ and
$\theta_{2}$.

######  Remark 7.

The angle $\theta$ defined in (1.13) is the arithmetic mean of $\theta_{1}$
and $\theta_{2}$ in (2.6). Further, under Assumption (K1), we have the
following trichotomy:

  1. (i)

$\theta_{1},\,\theta_{2}\not=0$ (see Figure 3) if

$\begin{cases}-\sum p_{k,\ell}k\ell<\sum p_{k,\ell}k^{2},\\\ -\sum
p_{k,\ell}k\ell<\sum p_{k,\ell}\ell^{2};\end{cases}$

  2. (ii)

$\theta_{1}\not=0$, $\theta_{2}=0$ (i.e., $\mathscr{S}_{2}$ has a cusp at $1$,
see Figure 4) if

$\sum p_{k,\ell}k^{2}=-\sum p_{k,\ell}k\ell<\sum p_{k,\ell}\ell^{2};$

  3. (iii)

$\theta_{1}=0$ (i.e., $\mathscr{S}_{1}$ has a cusp at $1$), $\theta_{2}\not=0$
if

$\sum p_{k,\ell}\ell^{2}=-\sum p_{k,\ell}k\ell<\sum p_{k,\ell}k^{2}.$

The above statements may not hold true if the model does not satisfy
Assumption (K1). For example, the model $p_{-1,0}=p_{1,-3}=p_{0,1}/3=1/5$ (see
Figure 2) does not satisfy any of the statements.

###### Proof.

We have:

$\displaystyle\cos\frac{\theta_{1}+\theta_{2}}{2}$
$\displaystyle=\cos\frac{\theta_{1}}{2}\cos\frac{\theta_{2}}{2}-\sin\frac{\theta_{1}}{2}\sin\frac{\theta_{2}}{2}$
$\displaystyle=\frac{(AC-B^{2})-(A-B)(A+B)}{\sqrt{(A^{2}-2AB+AC)(A^{2}+2AB+AC)}}$
$\displaystyle=\frac{C-A}{\sqrt{(A-2B+C)(A+2B+C)}}=\cos\theta,$

which implies that $\theta=(\theta_{1}+\theta_{2})/2$.

Moving to the proof of the trichotomy, we first recall from Lemma 5(iii) that
$\eta(s)s$ and $\eta(s)s^{-1}$ move respectively clockwise and
counterclockwise as $s$ moves clockwise on $\mathscr{C}$ under Assumption
(K1). This implies that

$\Re\left(\partial_{+}(\eta(s)s)\big{|}_{s=1}\right)\geqslant
0\quad\text{and}=\frac{A-B}{A}\quad\Re\left(\partial_{+}(\eta(s)s^{-1})\big{|}_{s=1}\right)\frac{-A-B}{A}\leqslant
0,$

where $\Re$ denotes the real part of a complex number. Hence, Assumption (K1)
implies $-A\leqslant B\leqslant A$. The case $A=B=0$ cannot hold. Otherwise,
we have: $\theta_{1},\theta_{2}\not=0$ if $-A<B<A$; $\theta_{1}\not=0$,
$\theta_{2}=0$ if $-A<B=A$; $\theta_{1}=0$, $\theta_{2}\not=0$ if $-A=B<A$. ∎

Figure 4. $\mathscr{S}_{1}$ (red) and $\mathscr{S}_{2}$ (blue) for the walk
$p_{0,1}=1/2$, $p_{-1,0}=1/4$, $p_{1,0}=p_{1,-4}=1/8$. Such a model has $\sum
p_{k,\ell}k^{2}=-\sum p_{k,\ell}k\ell(=1/2)<\sum p_{k,\ell}\ell^{2}(=1)$, and
thus the angle $\theta_{2}=0$.

###  2.2. Solutions of the kernel in a neighborhood of $\mathscr{K}$

We first properly define the functions $X$ and $Y$ mentioned in Proposition 1.
Let $X:\mathscr{S}_{2}\to\mathscr{S}_{1}$ be the mapping $s\mapsto\eta(s)s$
composed with the compositional inverse of $s\mapsto\eta(s)s^{-1}$. Similarly,
$Y:\mathscr{S}_{1}\to\mathscr{S}_{2}$ denotes the mapping
$s\mapsto\eta(s)s^{-1}$ composed with the inverse of $s\mapsto\eta(s)s$. In
other words, $Y:\mathscr{S}_{1}\to\mathscr{S}_{2}$ is the inverse of
$X:\mathscr{S}_{2}\to\mathscr{S}_{1}$. The following lemma presents some
crucial properties of these functions.

######  Lemma 8.

Under Assumptions (H1)–(H4) and (K1), we have:

  1. (i)

$X(y)$ is a one-to-one mapping from $\mathscr{S}_{2}$ onto $\mathscr{S}_{1}$;

  2. (ii)

For all $y\in\mathscr{S}_{2}$, $(X(y),y)\in\mathscr{K}$;

  3. (iii)

$X(y)$ is admits non-vanishing derivatives on
$\mathscr{S}_{2}\setminus\\{1\\}$.

  4. (iv)

$X(y)$ admits non-vanishing left and right derivatives at $1$.

We have analogous statements for $Y(x)$.

###### Proof.

Since $s\mapsto\eta(s)s$ and $s\mapsto\eta(s)s^{-1}$ are one-to-one mappings
from the half-circle $\\{{e}^{{i}\phi}:\phi\in[0,\pi)\\}$ (resp.
$\mathscr{C}$) onto $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ in the case
$p_{1,1}=0$ (resp. $p_{1,1}\neq 0$) (see Lemma 5), then Item (i) follows. By
the construction of $X(y)$, for all $y\in\mathscr{S}_{2}$, there exists
$s\in\mathscr{C}$ such that $(X(y),y)=(\eta(s)s,\eta(s)s^{-1})\in\mathscr{K}$,
which implies Item (ii). Moreover, $\eta(s)s$ and $\eta(s)s^{-1}$ admit non-
vanishing derivatives on $\mathscr{C}\setminus\\{\pm 1\\}$ (resp.
$\mathscr{C}\setminus\\{1\\}$) and admit non-vanishing left/right derivatives
at $\pm 1$ (resp. $1$) in the case $p_{1,1}=0$ (resp. $p_{1,1}\neq 0$) (see
Lemma 6), then Items (iii) and (iv) follow. ∎

We formulate the next lemma to extend the domains of definition of $X(y)$ and
$Y(x)$.

######  Lemma 9.

Under Assumptions (H1)–(H4) and (K1), $X(y)$ can be continued analytically on
a neighborhood $V\subset\mathbb{C}\setminus\\{1\\}$ of
$\mathscr{S}_{2}\setminus\\{1\\}$ such that:

  1. (i)

$K(X(y),y)=0$ for all $y\in V$;

  2. (ii)

$X(y)\in\mathscr{S}_{1}^{-}$ and $|X(y)|>|y|$ for all $y\in
V\cap\mathscr{S}_{2}^{+}$;

  3. (iii)

$X(y)\in\mathscr{S}_{1}^{+}$ and $|X(y)|<|y|$ for all $y\in
V\cap\mathscr{S}_{2}^{-}$.

We have an analogous statement for $Y(x)$.

###### Proof.

We only prove the lemma for $X(y)$. The result for $Y(x)$ is deduced
similarly. For any $y\in\mathscr{S}_{2}\setminus\\{1\\}$, since $(X(y),y)$ is
a non-singular point of $V[K]$ (Lemma 6), $X^{\prime}(y)\not=0$ (Lemma 8), and
$\partial_{x}K(X(y),y)X^{\prime}(y)+\partial_{y}K(X(y),y)=0$ (by
differentiating $K(X(y),y)=0$), then $\partial_{x}K(X(y),y)$ and
$\partial_{y}K(X(y),y)$ are simultaneously non-vanishing. Thus, by the
implicit function theorem (see [23, Sec. B4]), $X(y)$ can be extended as an
analytic bijection in a neighborhood $V\subset\mathbb{C}\setminus\\{1\\}$
small enough of $\mathscr{S}_{2}\setminus\\{1\\}$ such that $K(X(y),y)=0$. We
remark that with $V$ small enough, $\mathscr{S}_{2}$ divides $V$ into two
halves, on which the sign of $(|X(y)|-|y|)$ does not change and $X(y)$ does
not pass through $\mathscr{S}_{1}$. Hence, it suffices to prove the assertions
in the lemma for some $y$ close to the intersection of
$\mathscr{S}_{2}\setminus\\{1\\}$ and $\mathbb{R}$.

Consider first the case $p_{1,1}=0$, $p_{0,1}=p_{1,0}$, where $0$ is the
intersection of $\mathscr{S}_{2}$ and $\mathbb{R}$. Differentiating once and
twice the identity $K(X(y),y)=0$ at $y=0$, one has:

$\displaystyle
X^{\prime}(0)=-\frac{\partial_{y}K(0,0)}{\partial_{x}K(0,0)}=-1,$
$\displaystyle
X^{\prime\prime}(0)=-\frac{\partial_{xx}K(0,0)X^{\prime}(0)^{2}+2\partial_{xy}K(0,0)X^{\prime}(0)+\partial_{yy}K(0,0)}{\partial_{x}K(0,0)}=-\frac{2(1+p_{-1,1}+p_{1,-1})}{p_{0,1}}.$

We then obtain the asymptotics of $X(y)$ around $0$:

$X(y)=-y-\frac{1+p_{-1,1}+p_{1,-1}}{p_{0,1}}y^{2}+o(y^{2}).$

We further point out that any high order derivative of $X(y)$ at $0$ also
admits a real value, since it is formed by high order partial derivatives of
$K(x,y)$ at $(0,0)$. This indicates that $X(y)$ is real on a neighborhood
$V_{0}\subset\mathbb{R}$ of $0$. Hence, for all $y>0$ close to $0$ (that is,
$y\in\mathscr{S}_{2}^{+}$), one has $X(y)<X(0)$ (that is,
$X(y)\in\mathscr{S}_{1}^{-}$) and $|X(y)|>|y|$. On the other hand, for all
$y<0$ close to $0$, $X(y)>X(0)$ and $|X(y)|<|y|$. The lemma is then proven in
this case.

We move to the other cases and let $y_{0}$ be the intersection of
$\mathscr{S}_{2}\setminus\\{1\\}$ and $\mathbb{R}$. In the case $p_{1,1}=0$,
$p_{0,1}>p_{1,0}$, one has $X(y_{0})=-y_{0}<0$ and the asymptotics of $X(y)$
around $y_{0}$:

$X(y)=-y_{0}+X^{\prime}(y_{0})(y-y_{0})+o(y-y_{0}),$

where

$X^{\prime}(y_{0})=-\frac{\partial_{y}K(-y_{0},y_{0})}{\partial_{x}K(-y_{0},y_{0})},$

obtained by differentiating $K(X(y),y)=0$ at $y=y_{0}$. We now prove that
$X^{\prime}(y_{0})<-1$. One has:

$\partial_{y}K(-y_{0},y_{0})=-p_{1,0}-y_{0}\left(1+\sum_{(k,\ell)\neq(1,0)}p_{k,\ell}(-\ell+1)(-1)^{-k+1}y_{0}^{-k-\ell}\right).$

Since

$\left|\sum_{(k,\ell)\neq(1,0)}p_{k,\ell}(-\ell+1)(-1)^{-k+1}y_{0}^{-k-\ell}\right|<\sum_{(k,\ell)\neq(1,0)}p_{k,\ell}(-\ell+1)=1-p_{1,0}\leqslant
1,$

then $\partial_{y}K(-y_{0},y_{0})<0$. We now have:

$\displaystyle\partial_{x}K(-y_{0},y_{0})-\partial_{y}K(-y_{0},y_{0})$
$\displaystyle=\left(y_{0}-\sum
p_{k,\ell}(-k-\ell+1)(-1)^{-k}y_{0}^{-k-\ell+1}\right)-\frac{K(-y_{0},y_{0})}{y_{0}}$
$\displaystyle=y_{0}\left(1-\sum
p_{k,\ell}(-k-\ell+1)(-1)^{-k}y_{0}^{-k-\ell}\right).$

Since

$\left|\sum p_{k,\ell}(-k-\ell+1)(-1)^{-k}y_{0}^{-k-\ell}\right|<\sum
p_{k,\ell}(-k-\ell+1)=1,$

then $\partial_{x}K(-y_{0},y_{0})-\partial_{y}K(-y_{0},y_{0})>0$. Now notice
that by Lemma 5(iii), one has $X^{\prime}(y_{0})<0$. Thus,
$\partial_{y}K(-y_{0},y_{0})<\partial_{x}K(-y_{0},y_{0})<0$ and
$X^{\prime}(y_{0})<-1$. We also notice that $X(y)$ is real for any
$y\in\mathbb{R}$ close enough to $y_{0}$, since the Taylor expansion of $X(y)$
at $y_{0}$ admits all real coefficients. This implies that for all $y>y_{0}$
close to $y_{0}$, one has $X(y)<X(y_{0})$ and $|X(y)|>|y|$. Similarly, for all
$y<y_{0}$ close to $y_{0}$, then $X(y)>X(y_{0})$ and $|X(y)|<|y|$. The lemma
is proven for the case $p_{1,1}=0$, $p_{0,1}>p_{1,0}$.

Now in the case $p_{1,1}=0$, $p_{0,1}<p_{1,0}$, one has $X(y_{0})=-y_{0}>0$
and the asymptotics of $X(y)$ around $y_{0}$:

$X(y)=-y_{0}+X^{\prime}(y_{0})(y-y_{0})+o(y-y_{0}),$

where $X^{\prime}(y_{0})=1/Y^{\prime}(-y_{0})\in(-1,0)$. The rest of the proof
in this case follows the case $p_{1,1}=0$, $p_{0,1}>p_{1,0}$.

In the case $p_{1,1}\not=0$, it is seen that $X(y_{0})=y_{0}<0$ and as $y$ is
close to $y_{0}$,

$X(y)=y_{0}+X^{\prime}(y_{0})(y-y_{0})+o(y-y_{0}),$

where $X^{\prime}(y_{0})<0$ by Lemma 5(iii). Moreover, $X(y)$ is real for any
$y\in\mathbb{R}$ close enough to $y_{0}$, since the Taylor expansion of $X(y)$
at $y_{0}$ admits all real coefficients. Hence, for all $y>y_{0}$ close to
$y_{0}$, one has $X(y)<X(y_{0})$ and $|X(y)|>|y|$. Similarly, for all
$y<y_{0}$ close to $y_{0}$, then $X(y)>X(y_{0})$ and $|X(y)|<|y|$. The proof
is then complete. ∎

######  Lemma 10.

Under Assumptions (H1)–(H4) and (K1), in the case $p_{1,1}=0$,
$p_{0,1}>p_{1,0}$, where $0$ lies in $\mathscr{S}_{1}^{+}$ and
$\mathscr{S}_{2}^{-}$, the roots of $K(x,y)$ around $(0,0)$ can be represented
by an analytic continuation of $X(y)$ along a path in
$\overline{\mathscr{S}_{2}^{-}}\cap\mathscr{C}^{+}$ under the form $(X(y),y)$.
In particular for $y$ close to $0$,

(2.9)
$X(y)=-\frac{p_{1,0}}{p_{0,1}}y-\frac{p_{1,0}/p_{0,1}+p_{-1,1}+p_{1,-1}}{p_{0,1}}y^{2}+o(y^{2}).$

The statement in Lemma 10 may not hold for $Y(x)$ in the case $p_{1,1}=0$,
$p_{0,1}>p_{1,0}$, since $p_{1,0}$ may be equal to $0$, and $x=0$ then becomes
a branch point of $Y(x)$.

###### Proof.

Construct a path
$\gamma:[0,1]\to\overline{\mathscr{S}_{2}^{-}}\cap\mathscr{C}^{+}$ starting at
$\gamma(0)\in\mathscr{S}_{2}\setminus\\{1\\}$ and ending at $\gamma(1)=0$.
Further, $\gamma((0,1))$ is included in $\mathscr{S}_{2}^{+}$ and avoids any
point in $V[K]$ such that $\partial_{x}K=0$. Then, $X(y)$ can be analytically
extended along $\gamma$. By Lemma 9, $|X(y)|<|y|$ for any $y\in\gamma$ close
enough to $\gamma(0)$. Since $\gamma((0,1))$ does not pass through
$\mathscr{S}_{2}$, then we also have $|X(y)|<|y|$ for all $y\in\gamma((0,1))$.
Thus, $X(y)$ converges to $0$ as $y$ tends to $0$ along $\gamma$. Further,
since $\partial_{x}K(0,0)=-p_{0,1}\not=0$, then $y=0$ is not a branch point of
$X(y)$.

Now differentiating the identity $K(X(y),y)=0$ once and twice at $y=0$, we
have:

$X^{\prime}(0)=-\frac{p_{1,0}}{p_{0,1}}\quad\text{and}\quad
X^{\prime\prime}(0)=-\frac{2(p_{1,0}/p_{0,1}+p_{-1,1}+p_{1,-1})}{p_{0,1}},$

and (2.9) then follows. ∎

###  2.3. Proof of Proposition 1

The proof scheme relies heavily on analytic continuations along paths and
comparison between the modulus of two coordinates of the kernel’s solutions.

###### Proof of Proposition 1(i).

Consider first the case $p_{1,1}=0$, $p_{0,1}=p_{1,0}$, in which
$0\in\mathscr{S}_{1},\,\mathscr{S}_{2}$. Construct a closed path
$\gamma:[0,1]\to\overline{\mathscr{S}_{2}^{-}}\cap\mathscr{C}^{+}$ such that
$\gamma(0)=\gamma(1)=0$, $\gamma\big{(}(0,1)\big{)}\in\mathscr{S}_{2}^{-}$. We
further assume that $\gamma((0,1))$ stays in $\mathscr{S}_{2}^{-}$ and avoids
any branch point of $X(y)$. Extend $X(y)$ along $\gamma$, then by Lemma 9,
$|X(y)|<|y|$ for any $y\in\gamma\big{(}(0,1)\big{)}$ close enough to
$\gamma(0)$. Since $\gamma((0,1))$ never passes through $\mathscr{S}_{2}$,
then we also have $X(y)|<|y|$ for all $y\in\gamma\big{(}(0,1)\big{)}$. Hence,
as $y$ tends to $\gamma(1)$ along $\gamma$, $X(y)$ also returns to $0$, which
implies that $X(y)$ is analytic in a neighborhood of $\gamma$. The extension
of $X(y)$ on the interior of $\gamma$ can be then expressed by Cauchy’s
integral formula. Further, Morera’s theorem implies that $X(y)$ is analytic in
the interior of $\gamma$, since $\oint_{\gamma}X(y)dy=0$. Therefore, $X(y)$
does not have any branch point in $\mathscr{S}_{2}^{-}\cap\mathscr{C}$ and
similarly for $Y(x)$ on $\mathscr{S}_{1}^{-}\cap\mathscr{C}$.

We move to the case $p_{1,1}=0$, $p_{0,1}>p_{1,0}$, as we shall study
separately $X(y)$ and $Y(x)$. Construct a closed path
$\gamma:[0,1]\to\mathscr{S}_{2}^{-}\cap\mathscr{C}^{+}$ such that
$\gamma(0)=\gamma(1)=0$ and $\gamma$ avoids any branch point of $X(y)$. One
can extend $X(y)$ along $\gamma$ since $X(y)$ is already well defined around
$0$ by Lemma 10. Further, (2.9) implies that $|X(y)|<|y|$ for any
$y\in\gamma\big{(}(0,1)\big{)}$ close enough to $\gamma(0)$. Thus,
$|X(y)|<|y|$ for all $y\in\gamma\big{(}(0,1)\big{)}$, since $\gamma$ does not
pass through $\mathscr{S}_{2}$. Then, as $y$ tends to $0$ along $\gamma$,
$X(y)$ must return to $0$. By the same argument as in the preceding case,
$X(y)$ can be analytically extended on the interior of $\gamma$. It follows
that $X(y)$ does not have any branch point in
$\mathscr{S}_{2}^{-}\cap\mathscr{C}$.

We proceed with the proof for $Y(x)$ still in the case $p_{1,1}=0$,
$p_{0,1}>p_{1,0}$. Let $x_{0}$ (resp. $y_{0}=-x_{0}$) be the intersection of
$\mathscr{S}_{1}\setminus\\{1\\}$ (resp. $\mathscr{S}_{2}\setminus\\{1\\}$)
and $\mathbb{R}$. Construct a closed path
$\gamma:[0,1]\to\overline{\mathscr{S}_{1}^{-}}\cap\mathscr{C}^{+}$ such that
$\gamma(0)=\gamma(1)=x_{0}$,
$\gamma\big{(}(0,1)\big{)}\in\mathscr{S}_{1}^{-}$. Extend $Y(x)$ along
$\gamma$, then by Lemma 9 and the construction of $\gamma$ ($\gamma$ does not
pass through $\mathscr{S}_{1}$), $|Y(x)|<|x|$ for all
$x\in\gamma\big{(}(0,1)\big{)}$. We notice that as $x$ tends to $x_{0}$ along
$\gamma$, then $Y(x)$ must return to $y_{0}$. Indeed, assuming that
$|Y(\gamma(1))|<|y_{0}|$, we continue to extend $Y(x)$ along a path in
$\mathscr{S}_{1}^{+}$ from $x_{0}$ to $0$. Then as $x$ tends to $0$,
$|Y(x)|<|x|$ (since the path stays in $\mathscr{S}_{1}^{+}$), and thus $Y(x)$
must converge to $0$. However, the behavior of the kernel’s solutions around
$(0,0)$ (see Lemma 10) suggests that $|Y(x)|>|x|$ as $x$ close to $0$, which
yields a contradiction. Hence, $|Y(\gamma(1))|$ must be equal to $|y_{0}|$,
and thus $\left(\gamma(1),Y(\gamma(1))\right)$ is in $\mathscr{K}$ and
coincides with $(x_{0},y_{0})$, thanks to the injectivity of the mappings
$s\mapsto(\eta(s)s,\eta(s)s^{-1})$ and $Y:\mathscr{S}_{1}\to\mathscr{S}_{2}$
(see Lemma 5(i) and Lemma 8). We then have $Y(\gamma(1))=y_{0}$. By the
reasoning in the first case, $Y(x)$ does not have any branch point in
$\mathscr{S}_{1}^{-}\cap\mathscr{C}^{+}$.

We move to the case $p_{1,1}\not=0$. Let $x_{0}$ (resp. $y_{0}=x_{0}$) be the
intersection of $\mathscr{S}_{1}\setminus\\{1\\}$ (resp.
$\mathscr{S}_{2}\setminus\\{1\\}$) and $\mathbb{R}$. Construct a closed path
$\gamma:[0,1]\to\overline{\mathscr{S}_{2}^{-}}\cap\mathscr{C}^{+}$ such that
$\gamma(0)=\gamma(1)=y_{0}$, $\gamma\big{(}(0,1)\big{)}\in\mathscr{S}_{2}^{-}$
and $\gamma$ avoids any branch point of $X(y)$. Extend $X(y)$ along $\gamma$,
then by Lemma 9 and the construction of $\gamma$ ($\gamma$ does not pass
through $\mathscr{S}_{2}$), $|X(y)|<|y|$ for all
$y\in\gamma\big{(}(0,1)\big{)}$. We notice that as $y$ tends to $y_{0}$ along
$\gamma$, then $X(y)$ must return to $x_{0}$. Indeed, assuming that
$|X(\gamma(1))|<|x_{0}|$, we continue to extend $X(y)$ along a path in
$\mathscr{S}_{2}^{+}$ from $y_{0}$ to $0$. Then as $y$ tends to $0$,
$|X(y)|<|y|$ (since the path stays in $\mathscr{S}_{1}^{+}$), and thus $Y(x)$
must converge to $0$, which causes a contradiction since $(0,0)$ is not a root
of $K(x,y)$. Hence, $|X(\gamma(1))|$ must be equal to $|x_{0}|$, and thus
$X(\gamma(1))=x_{0}$. By the reasoning in the first case, $X(y)$ does not have
any branch point in $\mathscr{S}_{2}^{-}\cap\mathscr{C}$ and similarly for
$Y(x)$ on $\mathscr{S}_{1}^{-}\cap\mathscr{C}$. ∎

Hereafter, the function $X(y)$ (resp. $Y(x)$), originally defined on
$\mathscr{S}_{2}$ (resp. $\mathscr{S}_{1}$), will also indicate its extension
on $\mathscr{S}_{2}^{-}\cap\mathscr{C}^{+}$ (resp.
$\mathscr{S}_{1}^{-}\cap\mathscr{C}^{+}$).

###### Proof of Proposition 1(ii).

We recall that $0\in\mathscr{S}_{1}^{+},\mathscr{S}_{2}^{+}$ in the case
$p_{1,1}\not=0$, and $0\in\mathscr{S}_{1},\mathscr{S}_{2}$ in the case
$p_{1,1}=0$, $p_{1,0}=p_{0,1}$. Since these properties are similar to that of
the symmetric models of [24], we then refer the readers to [24, Lem. 4] for
the proof of these cases. We now prove the lemma for the case $p_{1,1}=0$,
$p_{1,0}<p_{0,1}$.

We first introduce some notations. Let $\mathscr{V}_{x}$, $\mathscr{V}_{y}$
respectively denote $\\{(x,y)\in V[K]:|y|<|x|<1\\}$, $\\{(x,y)\in
V[K]:|x|<|y|<1\\}$, and $P_{x},P_{y}:\mathbb{C}^{2}\to\mathbb{C}$ respectively
denote the projections along the first and second variable.

Reasoning by contradiction, we assume that there exists
$(x_{1},y_{1})\in\mathscr{V}_{x}$ such that $x_{1}\in\mathscr{S}_{1}^{+}$.
Construct a path $\gamma:[0,1]\to\mathscr{S}_{1}^{+}$ having two endpoints
$\gamma(0)=x_{1}$ and $\gamma(1)=0$, and avoiding any points of $V[K]$ such
that $\partial_{x}K=0$. There thus exists a path $\gamma^{\prime}:[0,1]\to
V[K]$ such that $\gamma^{\prime}(0)=(x_{1},y_{1})$ and
$P_{x}\circ\gamma^{\prime}=\gamma$. Since $\gamma$ stays in
$\mathscr{S}_{1}^{+}$, then $\gamma^{\prime}$ does not meet
$\mathscr{K}\setminus\\{(0,0)\\}$. Consequently, $\gamma^{\prime}((0,1))$
stays in $\mathscr{V}_{x}$ and $P_{y}\circ\gamma^{\prime}(1)=0$. On the other
hand, by (2.9), one must have
$|P_{x}\circ\gamma^{\prime}(t)|<|P_{y}\circ\gamma^{\prime}(t)|$ for $t$ close
to $1$, which generates a contradiction. Hence, there is no
$(x_{1},y_{1})\in\mathscr{V}_{x}$ such that $x_{1}\in\mathscr{S}_{1}^{+}$.

Now let $y_{0}$ be the intersection of $\mathscr{S}_{2}\setminus\\{1\\}$ and
$\mathbb{R}$. Assuming that there exists $(x_{1},y_{1})\in\mathscr{V}_{y}$
such that $y_{1}\in\mathscr{S}_{2}^{+}$, we construct a path
$\gamma:[0,1]\to\mathscr{C}^{+}$ such that:

  * •

It has two endpoints $\gamma(0)=y_{1}$ and $\gamma(1)=0$, and passes through
$\gamma(\tau)=y_{0}$ with $\tau\in(0,1)$;

  * •

$\gamma((0,\tau))\in\mathscr{S}_{2}^{+}$ and
$\gamma((\tau,1))\in\mathscr{S}_{2}^{-}\cap\mathscr{C}^{+}$;

  * •

$\gamma$ avoids any point of $V[K]$ such that $\partial_{y}K=0$.

There thus exists a path $\gamma^{\prime}:[0,1]\to V[K]$ such that
$\gamma^{\prime}(0)=(x_{1},y_{1})$ and $P_{y}\circ\gamma^{\prime}=\gamma$.
Since $\gamma((0,\tau))$ stays in $\mathscr{S}_{2}^{+}$, then
$\gamma^{\prime}((0,\tau))$ does not meet $\mathscr{K}$ and thus stays in
$\mathscr{V}_{y}$.

We prove that with the above constructions, the path $\gamma^{\prime}$ has to
pass through $\mathscr{K}$ at $(X(y_{0}),y_{0})$. Indeed, we assume the
opposite, that $\gamma^{\prime}$ does not meet
$\mathscr{K}\setminus\\{(0,0)\\}$, then
$\gamma^{\prime}((0,1))\subset\mathscr{V}_{y}$ and thus
$\gamma^{\prime}(1)=(0,0)=(X(0),0)$. The analytic continuation of $X(y)$
implies that $\gamma^{\prime}([\tau,1])=\\{(X(y),y):y\in\gamma([\tau,1])\\}$
and thus, $\gamma^{\prime}(\tau)=(X(y_{0}),y_{0})$, which contradicts the
assumption.

Since $\gamma^{\prime}$ passes through $(X(y_{0}),y_{0})$, then by Lemma 9,
for any $t\in(0,\tau)$ close enough to $\tau$,
$|P_{x}\circ\gamma^{\prime}(t)|>|P_{y}\circ\gamma^{\prime}(t)|$, which
contradicts the assertion that $\gamma^{\prime}((0,\tau))$ stays in
$\mathscr{V}_{y}$. We thus conclude that there is no
$(x_{1},y_{1})\in\mathscr{V}_{y}$ such that $y_{1}\in\mathscr{S}_{2}^{+}$. The
proof is then complete. ∎

##  3\. Conformal welding

This section aims at constructing conformal mappings from
$\mathscr{S}_{1}^{+}$ and $\mathscr{S}_{2}^{+}$ onto two disjoint,
complementary domains such that the two coordinates of any point in
$\mathscr{K}$ are joined together by the boundary value of these mappings. The
problem can be reduced to that of finding two conformal mappings from the
upper and lower half-planes onto disjoint, complementary domains such that the
extensions of the mappings on $\mathbb{R}$ differ by a quasisymmetric
homeomorphism (also called a shift function). Such a problem is usually
referred to as conformal welding.

We introduce some notations. Let $\mathscr{A}$ denote a Jordan curve (that is,
a non-self-intersecting closed curve) or an infinite non-self-intersecting
curve dividing the complex plane into two disjoint domains $\mathscr{A}^{+}$
and $\mathscr{A}^{-}$ (if $\mathscr{A}$ is a Jordan curve, then
$\mathscr{A}^{-}$ conventionally indicates the domain containing infinity).
Let also $f$ be a function defined on $\mathscr{A}^{+}$ (resp.
$\mathscr{A}^{-}$), then for any $t\in\mathscr{A}$, $f^{+}(t)$ (reps.
$f^{-}(t)$) will denote the limit value of $f(x)$ as $x$ tends to $t$,
provided it exists.

###  3.1. Preliminary construction of conformal mappings

In this subsection, we construct conformal mappings from $\mathscr{S}_{1}^{+}$
and $\mathscr{S}_{2}^{+}$ respectively onto the upper and lower half-planes,
and introduce the corresponding shift functions on the real line.

$\pi_{1}(z)$$\pi_{2}(z)$$\phi(z)$$-\phi(z)$$\phi^{-1}(-z)$$(\pi_{2}^{-1})^{+}(z)$$X(y)$$\pi_{1}^{+}(z)$$\phi(z)$$\alpha$(z)
Figure 5. Some curves and conformal mappings in the model
$p_{0,1}=p_{0,-1}=3/8$, $p_{1,0}=p_{-1,0}=1/8$: $\pi_{1}(z)$ maps conformally
$\mathscr{S}_{1}^{+}$ (red domain in the top left figure) onto the unit disk;
$\pi_{2}(z)$ maps conformally $\mathscr{S}_{2}^{+}$ (blue domain in the bottom
left figure) onto the unit disk; $\phi(z)$ (resp. $-\phi(z)$) maps conformally
the unit disk onto the upper (resp. lower) half-plane; The right figure shows
the shift function $\alpha(z)$ constructed by composing $\phi^{-1}(-z)$,
$(\pi_{2}^{-1})^{+}(z)$, $X(y)$, $\pi_{1}^{+}(z)$ and $\phi(z)$ (for example,
along the gray dashed arrows in the figure).

The following lemma is implied by the classical Riemann mapping theorem.

######  Lemma 11.

Under Assumption (K1), for any $r\in\mathscr{S}_{1}^{+}\cap\mathbb{R}$, there
exists uniquely a conformal mapping $\pi_{1}$ from $\mathscr{S}_{1}^{+}$ onto
$\mathscr{C}^{+}$ such that $\pi_{1}(r)=0$ and $\pi_{1}^{\prime}(r)>0$.
Further, for all $z\in\mathscr{S}_{1}^{+}$,
$\pi_{1}(z)=\overline{\pi_{1}(\overline{z})}$ and $\pi_{1}^{+}(1)=1$.

A detailed proof of Lemma 11 can be found in [24, Lem. 5], since under
Assumption (K1), $\mathscr{S}_{1}$ is a Jordan curve and symmetric with
respect to the real axis, which is similar to the symmetric case
$p_{k,\ell}=p_{\ell,k}$ for all $k,\ell$.

We now construct a conformal mapping $\pi_{2}$ from $\mathscr{S}_{2}^{+}$ onto
$\mathscr{C}^{+}$ with analogous properties to $\pi_{1}$ and introduce a
Möbius transformation:

$\phi(z):=-i\frac{z+1}{z-1},$

which maps conformally $\mathscr{C}^{+}$ onto the upper half-plane
$\mathscr{H}^{+}$. Thus, the composition $\phi\circ\pi_{1}$ (resp.
$-\phi\circ\pi_{2}$) maps conformally $\mathscr{S}_{1}^{+}$ (resp.
$\mathscr{S}_{2}^{+}$) onto $\mathscr{H}^{+}$ (resp. $\mathscr{H}^{-}$).

Carathéodory’s theorem (see [22, Thm. 3.1]) ensures that the limit values of
such conformal mappings exist and form continuous homeomorphisms on the
boundaries of the domains. We then introduce the shift function

(3.1) $\alpha(z):=\phi\circ\pi_{1}^{+}\circ
X\circ(\pi_{2}^{-1})^{+}\circ\phi^{-1}(-z),\quad z\in\mathbb{R},$

(see Figure 5).

Hereafter in this section, we will consider the following assumption:

  1. (K4)

The set of weights $\\{p_{k,\ell}\\}_{k,\ell}$ satisfies

$\begin{cases}-\sum p_{k,\ell}k\ell<\sum p_{k,\ell}k^{2},\\\ -\sum
p_{k,\ell}k\ell<\sum p_{k,\ell}\ell^{2}.\end{cases}$

We recall from Remark 7 that Assumption (K4) is equivalent to
$\theta_{1},\,\theta_{2}\not=0$. The following lemma presents some crucial
properties of $\alpha(z)$.

######  Lemma 12.

Under Assumptions (K1) and (K4), $\alpha(z)$ is a one-to-one, strictly
increasing and odd function from $\mathbb{R}\cup\\{\infty\\}$ onto itself.
Further, it is analytic and possesses non-vanishing derivatives on
$\mathbb{R}$ with the asymptotic behavior as $z\to+\infty$

(3.2) $\alpha(z)\sim\text{const}\cdot z^{\theta_{2}/\theta_{1}},$

where $\theta_{1}$, $\theta_{2}$ are defined in (2.6).

###### Proof.

Since all the functions $\phi^{-1}|_{\mathbb{R}}$, $(\pi_{2}^{-1})^{+}$,
$X|_{\mathscr{S}_{2}}$, $\pi_{1}^{+}$, and $\phi|_{\mathscr{C}}$ are one-to-
one on their domains of definition, then $\alpha(z)$ is also one-to-one on
$\mathbb{R}\cup\\{\infty\\}$. Further, $\pi_{1}^{+}(z)$ and
$(\pi_{2}^{-1})^{+}(z)$ preserve (resp. $X(y)$ inverses) the orientation as
$z$ moves counterclockwise, then $\alpha(z)$ is strictly increasing on
$\mathbb{R}$.

Now notice that $\phi(z)=-\phi(1/z)=-\phi(\overline{z})$ for all
$z\in\mathscr{C}$, or equivalently,
$\phi^{-1}(z)=1/\phi^{-1}(-z)=\overline{\phi^{-1}(-z)}$ for all
$z\in\mathbb{R}$. Besides, one has:

$\pi_{1}^{+}\circ
X\circ(\pi_{2}^{-1})^{+}(\overline{z})=\overline{\pi_{1}^{+}\circ
X\circ(\pi_{2}^{-1})^{+}(z)},$

for all $z\in\mathscr{C}$. Thus, it is easily checked that $\alpha(z)$ is an
odd function.

Since $\pi_{1}^{+}:\mathscr{S}_{1}\to\mathscr{C}$ is the extension of the
conformal mapping $\pi_{1}$ to the boundary, and
$\mathscr{S}_{1}\setminus\\{1\\}$ and $\mathscr{C}\setminus\\{1\\}$ are
analytic and smooth, then $\pi_{1}^{+}$ is analytic (see [33, p. 186]) and
possesses non-vanishing derivatives on $\mathscr{S}_{1}\setminus\\{1\\}$ (see
[34, Thm. 3.9]). In particular, $\mathscr{S}_{1}$ admits a corner point of
angle $\theta_{1}$ at $1$, whereas $\mathscr{C}$ is smooth at $1$, then by
[34, Thm. 3.11], one has the asymptotic behavior as $z\to 1$:

$\pi_{1}^{+}(z)=1+\text{const}\cdot(z-1)^{\pi/\theta_{1}}+o((z-1)^{\pi/\theta_{1}}).$

Similarly, $(\pi_{2}^{-1})^{+}$ is analytic, possesses non-vanishing
derivatives on $\mathscr{C}\setminus\\{1\\}$, and as $z\to 1$,

$(\pi_{2}^{-1})^{+}(z)=1+\text{const}\cdot(z-1)^{\theta_{2}/\pi}+o((z-1)^{\theta_{2}/\pi}).$

We recall that by Lemma 8, $X(y)$ is analytic, possesses non-vanishing
derivatives on $\mathscr{S}_{2}\setminus\\{1\\}$, and left and right
derivatives at $1$. Since $\alpha(z)$ is composed of the above functions and
$\phi(z)$, then $\alpha(z)$ is also analytic, possesses non-vanishing
derivatives on $\mathbb{R}$, and admits the asymptotic behavior (3.2) as
$z\to\infty$. ∎

We would like to specify some facts about the shift function $\alpha(z)$ in
the symmetric case in [24], where $\mathscr{S}_{1}$ and $\mathscr{S}_{2}$ are
Jordan curves and coincide, and $X(y)=\overline{y}$ for all
$y\in\mathscr{S}_{2}$. As a result, the shift function $\alpha(z)$ degenerates
into the identity on $\mathbb{R}$ in the symmetric case. We notice that the
assumptions (K1) and (K4) always hold true in this case. Hence, by also
considering the assumptions (K1) and (K4) in the non-symmetric case, we
restrict our analysis to models that share some critical properties with the
symmetric one.

###  3.2. Conformal welding problem with quasisymmetric shift

This subsection aims at solving the following BVP: Find mappings $\chi_{1}$
and $\chi_{2}$ such that $\chi_{1}$ and $\chi_{2}$ map respectively
$\mathscr{H}^{+}$ and $\mathscr{H}^{-}$ conformally onto two disjoint,
complementary domains, satisfying the boundary condition

$\chi_{1}^{+}(\alpha(z))=\chi_{2}^{-}(z),$

for all $z\in\mathbb{R}$. We will show that the shift function $\alpha(z)$ is
quasisymmetric under some assumptions.

$\chi_{1}(z)$$\chi_{2}(z)$$\phi(z)$$\alpha$(z) Figure 6. Some curves and
conformal mappings concerning the BVP with shift in the model
$p_{0,1}=p_{0,-1}=3/8$, $p_{1,0}=p_{-1,0}=1/8$: the first figure shows the
shift function $\alpha(z)$ on $\mathbb{R}$; the second figure shows the Jordan
curve $\mathscr{G}$ constructed in Lemma 16; $\chi_{1}(z)$ maps conformally
$\mathscr{H}^{+}$ onto $\mathscr{G}^{+}$, $\chi_{2}(z)$ maps conformally
$\mathscr{H}^{-}$ onto $\mathscr{G}^{-}$; for any $z\in\mathbb{R}$ (blue
point), its image (black point in the second figure) under $\chi_{2}^{-}$
coincides with the image of $\alpha(z)$ (red point) under $\chi_{1}^{+}$; the
third figure shows the images of $\mathscr{G}$, $\mathscr{G}^{+}$ and
$\mathscr{G}^{-}$ under the Möbius transformation $\phi$ defined in (3.1).

We first recall the definition of a quasisymmetric function on the real line.
A strictly increasing function $f$ defined on the real line is quasisymmetric
if and only if there exists $k>0$ such that

(3.3) $\frac{1}{k}\leqslant\frac{f(x+t)-f(x)}{f(x)-f(x-t)}\leqslant k,$

for all $x\in\mathbb{R}$ and $t>0$.

The following lemma presents a result of quasisymmetric functions.

######  Lemma 13.

For all $p>0$, the function

(3.4) $P(x):=\text{sign}(x)|x|^{p}$

is quasisymmetric on $\mathbb{R}$.

###### Proof.

We first notice that

$\lim_{t\to 0}\frac{P(x+t)-P(x)}{P(x)-P(x-t)}=1,$

for all $x\in\mathbb{R}$. Now for all
$(x,t)\in\mathbb{R}\times\mathbb{R}^{+}$,

$\frac{P(x+t)-P(x)}{P(x)-P(x-t)}=\frac{\text{sign}(x/t+1)|x/t+1|^{p}-\text{sign}(x/t)|x/t|^{p}}{\text{sign}(x/t)|x/t|^{p}-\text{sign}(x/t-1)|x/t-1|^{p}}.$

Since the function

$\frac{\text{sign}(z+1)|z+1|^{p}-\text{sign}(z)|z|^{p}}{\text{sign}(z)|z|^{p}-\text{sign}(z-1)|z-1|^{p}}$

is continuous, strictly positive on $\mathbb{R}$, and converges to $1$ as $z$
tends to $\pm\infty$, then it is bounded in $\mathbb{R}^{+}$ and thus, there
exists a suitable number $k>0$ satisfying

$\frac{1}{k}\leqslant\frac{P(x+t)-P(x)}{P(x)-P(x-t)}\leqslant k,$

for all $(x,t)\in\mathbb{R}\times\mathbb{R}^{+}$. The proof is then complete.
∎

To show that $\alpha(z)$ is quasisymmetric, we introduce the following
assumption:

  1. (K5)

$\alpha^{\prime}(z)=\text{const}\cdot|z|^{q}+o(z^{q})$ as $z\to\pm\infty$,
with $q\in\mathbb{R}$.

Assumption (K5) is particularly technical as it is stated for the shift
function $\alpha(z)$, which is difficult to verify, instead of stated for the
weights $\\{p_{k,\ell}\\}_{k,\ell}$ or the curves $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$. Therefore, we also introduce in Lemma 14 a sufficient
condition on the moments of the weights $\\{p_{k,\ell}\\}_{k,\ell}$, under
which Assumption (K5) always holds true.

######  Lemma 14.

Assume that

  1. (H7)

The set of weights $\\{p_{k,\ell}\\}_{k,\ell}$ has finite moments of order 4,
i.e., $\sum p_{k,\ell}k^{4}+\sum p_{k,\ell}\ell^{4}<\infty$, and the one-sided
second order derivatives of $X(y)$ at $1$, which are fully characterized by
the moments of $\\{p_{k,\ell}\\}_{k,\ell}$ up to order 3, do not vanish.

Then under Assumptions (K1), (K4), and (H7), we have

$\alpha^{\prime}(z)\sim\text{const}\cdot|z|^{\theta_{2}/\theta_{1}-1},$

as $z\to\pm\infty$, where $\theta_{1}$, $\theta_{2}$ are defined in (2.6).

###### Proof.

By differentiating three times the identity $K(X(y),y)=0$ and evaluating it as
$y\to 1$, it is easily seen that the one-sided second order derivatives of
$X(y)$ at $1$ are characterized by the moments of $\\{p_{k,\ell}\\}_{k,\ell}$
up to order 3. By the assumption, we then have

$X^{\prime}(y)\sim\text{const}\cdot(y-1)+o(y-1),$

as $y\to 1$.

We now study the asymptotic behavior of the derivatives of $\pi_{2}^{-1}(z)$
as $z\to 1$. By the assumption on the moments of order 4, the parametrization
of $\mathscr{S}_{2}$, which is $s\mapsto\eta(s)s^{-1}$ on $\mathscr{C}$,
admits finite one-sided derivatives up to order 3 at $s=1$. Hence, the mapping
$s\mapsto\eta(s)s^{-1}$ and its first and second derivatives are Lipschitz
continuous on $\mathscr{C}$. Recall that as $z\to 1$,

$\pi_{2}^{-1}(z)=1+\text{const}\cdot(z-1)^{\theta_{2}/\pi}+o\left((z-1)^{\theta_{2}/\pi}\right).$

Applying a result from [38, Theorem 1], one may obtain the expansion of
$(\pi_{2}^{-1})^{\prime}$ by differentiating formally the expansion of
$\pi_{2}^{-1}$:

$\pi_{2}^{-1}(z)^{\prime}\sim\text{const}\cdot(z-1)^{\theta_{2}/\pi-1}+o\left((z-1)^{\theta_{2}/\pi-1}\right),$

as $z\to 1$. We remark that the result in [38, Theorem 1] is stated for
conformal mappings from the upper half-plane onto a domain with a corner point
at $0$, such that $0$ is mapped to $0$. However, it can be also applied to the
current context, as one can use translation and Möbius transformation to
return to the original setting of the theorem. We also obtain a similar result
for the mapping $\pi_{1}^{-1}$ and deduce the expansion of the derivatives of
its inverse by [38, Theorem 3]:

$\pi_{1}(z)^{\prime}\sim\text{const}\cdot(z-1)^{\pi/\theta_{1}-1}+o\left((z-1)^{\pi/\theta_{1}-1}\right),$

as $z\to 1$. Since $\alpha(z)$ is the composition of $\phi(z)$,
$\pi_{1}^{+}(z)$, $X(y)$, $(\pi_{2}^{-1})^{+}(z)$, and $\phi^{-1}(-z)$, all of
which have the asymptotic expansions for the derivatives, then we also obtain
the asymptotics for $\alpha^{\prime}(z)$:

$\alpha^{\prime}(z)\sim\text{const}\cdot|z|^{\theta_{2}/\theta_{1}-1},$

as $z\to\pm\infty$. ∎

$x$$t$$0$$x_{0}$$-x_{0}$$t_{0}$$t=\varepsilon_{0}x$$t=-\varepsilon_{0}x$
Figure 7. The region $\mathbb{R}\times\mathbb{R}^{+}$ is divided into smaller
domains: $R_{1}\cap\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:t\geqslant
t_{0}\\}$ (in green),
$R_{2}\cap\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:|x|\geqslant x_{0}\\}$ (in
yellow), and $R_{3}$ (in red).

######  Lemma 15.

Under Assumptions (K1)–(K5), $\alpha(z)$ is a quasisymmetric function from
$\mathbb{R}$ onto itself.

###### Proof.

Without loss of generality, we omit the constant coefficients of $\alpha(z)$
in (3.2) and $\alpha^{\prime}(z)$ in Assumption (K5). We rewrite $\alpha(z)$
as

$\alpha(z)=P(z)+r(z),$

where $P(z)=\text{sign}(z)|z|^{p}$, $p:=\theta_{2}/\theta_{1}$ and
$r(x)=o(|x|^{p})$ as $x\to\pm\infty$. We now prove that there exist
$k_{1}\in(0,1)$ and $k_{2}>1$ such that

(3.5) $F(\alpha)(x,t,k_{2})\leqslant 0\leqslant F(\alpha)(x,t,k_{1}),$

for all $(x,t)\in\mathbb{R}\times\mathbb{R}^{+}$, where $F(f)$ is an operator
acting on any function $f:\mathbb{R}\to\mathbb{R}$ and defined as follows:

(3.6) $F(f)(x,t,k):=f(x+t)-f(x)-k[f(x)-f(x-t)].$

One can rewrite $F(\alpha)$ as

(3.7)
$F(\alpha)(x,t,k)=F(P)(x,t,k)+F(r)(x,t,k)=t^{p}\left[F(P)(\frac{x}{t},1,k)+\frac{F(r)(x,t,k)}{t^{p}}\right].$

Fix $\varepsilon_{0}\in(0,1)$ and set

$R_{1}:=\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:-\frac{1}{\varepsilon_{0}}\leqslant\frac{x}{t}\leqslant\frac{1}{\varepsilon_{0}}\\},$

(see Figure 7). Since $x/t$ is bounded on $R_{1}$, then for any $k$ fixed,
$F(P)(x/t,1,k)$ possesses a maximum and a minimum on $R_{1}$. Further, since
$P(z)$ is quasisymmetric on $\mathbb{R}$ (Lemma 13), then there exist
$k_{1,1}\in(0,1)$ and $k_{2,1}>0$ such that

$F(P)(x/t,1,k_{2,1})<0<F(P)(x/t,1,k_{1,1}),$

for all $(x,t)\in R_{1}$. And thus,

(3.8) $\max_{(x,t)\in R_{1}}F(P)(x/t,1,k_{2,1})<0<\min_{(x,t)\in
R_{1}}F(P)(x/t,1,k_{1,1}).$

We now prove the following assertion: For any $k$ fixed and $\varepsilon>0$,
there exists $t_{0}>0$ such that

(3.9) $\left|\frac{F(r)(x,t,k)}{t^{p}}\right|<\varepsilon,$

for all $(x,t)\in
R_{1}\cap\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:t\geqslant t_{0}\\}$. We
have:

$\frac{F(r)(x,t,k)}{t^{p}}=\frac{r(x+t)}{t^{p}}-(1+k)\frac{r(x)}{t^{p}}+k\frac{r(x-t)}{t^{p}}.$

Looking at the first term $r(x+t)/t^{p}$, we notice that for $|x+t|$ large,
one can write:

$\left|\frac{r(x+t)}{t^{p}}\right|=\frac{|r(x+t)|}{|x+t|^{p}}\left|\frac{x}{t}+1\right|^{p}.$

Since $x/t$ is bounded on $R_{1}$ and $r(x)=o(|x|^{p})$ as $x\to\infty$, then
there exists $n_{1}$ such that for any $(x,t)\in R_{1}$, $|x+t|>n_{1}$, one
has:

(3.10) $\left|\frac{r(x+t)}{t^{p}}\right|<\frac{\varepsilon}{3}.$

Moreover, there exists $t_{1}>0$ such that (3.10) holds true for all $(x,t)\in
R_{1}$, $|x+t|\leqslant n_{1}$, $t\geqslant t_{1}$. This implies that (3.10)
also holds true for all $(x,t)\in R_{1}$, $t\geqslant t_{1}$. By similar
arguments applied to the terms $r(x)/t^{p}$ and $r(x-t)/t^{p}$, we conclude
that for any $k$ and $\varepsilon>0$, there exists $t_{0}>0$ such that (3.9)
holds true for all $(x,t)\in R_{1}$, $t\geqslant t_{0}$. Eq. (3.8) and (3.9)
imply that there exists $t_{0}>0$ such that (3.5) holds for
$(k_{1},k_{2})=(k_{1,1},k_{2,1})$ and for all $(x,t)\in R_{1}$, $t\geqslant
t_{0}$.

Now set

$R_{2}:=\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:0\leqslant
t\leqslant\varepsilon_{0}|x|\\},$

(see Figure 7). We first have:

$\partial_{t}F(\alpha)(x,t,k)=\alpha^{\prime}(x+t)-k\alpha^{\prime}(x-t).$

By Assumption (K5), $\alpha^{\prime}(z)=|z|^{q}+u(z)$, where $u(z)=o(|z|^{q})$
as $z\to\pm\infty$, and thus,

$\frac{\alpha^{\prime}(x+t)}{\alpha^{\prime}(x-t)}=\frac{|x+t|^{q}+u(x+t)}{|x-t|^{q}+u(x-t)}=\left|\frac{x+t}{x-t}\right|^{q}\frac{1+\frac{u(x+t)}{|x+t|^{q}}}{1+\frac{u(x-t)}{|x-t|^{q}}}.$

Notice that on $R_{2}$, $\big{|}(x+t)/(x-t)\big{|}$ is bounded by
$(1-\varepsilon_{0})/(1+\varepsilon_{0})$ and
$(1+\varepsilon_{0})/(1-\varepsilon_{0})$, and
$(1+\frac{u(x+t)}{|x+t|^{q}})/(1+\frac{u(x-t)}{|x-t|^{q}})$ is close to $1$ as
$|x|$ large enough. Hence, there exist $k_{1,2}\in(0,1)$, $k_{2,2}>1$, and
$x_{0}$ large enough such that

$k_{1,2}<\frac{\alpha^{\prime}(x+t)}{\alpha^{\prime}(x-t)}<k_{2,2},\quad\text{or
equivalently,}\quad\partial_{t}F(\alpha)(x,t,k_{2,2})<0<\partial_{t}F(\alpha)(x,t,k_{1,2}),$

for all $(x,t)\in R_{2}$, $|x|\geqslant x_{0}$. This implies that for any
$|x|\geqslant x_{0}$, $F(\alpha)(x,t,k_{1,2})$ and $F(\alpha)(x,t,k_{2,2})$,
as functions of $t$, are strictly monotonic on $[0,\varepsilon_{0}|x|]$. Since
$F(\alpha)(x,0,k_{1,2})=F(\alpha)(x,0,k_{2,2})=0$, then (3.5) holds true for
$(k_{1},k_{2})=(k_{1,2},k_{2,2})$ and $(x,t)\in R_{2},|x|\geqslant x_{0}$.

We now consider the set

$(\mathbb{R}\times\mathbb{R}^{+})\setminus[(R_{1}\cap\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:t\geqslant
t_{0}\\})\cup(R_{2}\cap\\{(x,t)\in\mathbb{R}\times\mathbb{R}^{+}:|x|\geqslant
x_{0}\\})],$

and let $R_{3}$ denote its closure (see Figure 7). Let

$k_{1,3}:=\min_{(x,t)\in
R_{3}}\frac{\alpha(x+t)-\alpha(x)}{\alpha(x)-\alpha(x-t)}\quad\text{and}\quad
k_{2,3}:=\max_{(x,t)\in
R_{3}}\frac{\alpha(x+t)-\alpha(x)}{\alpha(x)-\alpha(x-t)},$

which exist and are positive since $R_{3}$ is bounded and $\alpha(z)$ is
differentiable on $\mathbb{R}$. Thus (3.5) holds true for
$(k_{1},k_{2})=(k_{1,3},k_{2,3})$ and $(x,t)\in R_{3}$.

In conclusion, by choosing $k_{1}=\min_{1\leqslant n\leqslant 3}k_{1,n}$ and
$k_{2}=\max_{1\leqslant n\leqslant 3}k_{1,n}$, we deduce that (3.5) holds for
$(x,t)\in\mathbb{R}\times\mathbb{R}^{+}$, i.e. $\alpha(z)$ is quasisymmetric
on $\mathbb{R}$. ∎

The following lemma presents the existence of the BVP’s solutions introduced
at the beginning of Subsection 3.2.

######  Lemma 16.

Under Assumptions (K1)–(K5), there exist a pair of conformal mappings
$\chi_{1}$, $\chi_{2}$ and a Jordan curve $\mathscr{G}$ such that

  1. (i)

$\chi_{1}$ maps conformally $\mathscr{H}^{+}$ onto $\mathscr{G}^{+}$;

  2. (ii)

$\chi_{2}$ maps conformally $\mathscr{H}^{-}$ onto $\mathscr{G}^{-}$;

  3. (iii)

For all $z\in\mathbb{R}$,

(3.11) $\chi_{1}^{+}(\alpha(z))=\chi_{2}^{-}(z).$

We can further assume that

  1. (iv)

$\lim_{z\to\infty}\chi_{1}(z)=1$;

  2. (v)

$\mathscr{G}$ is symmetric with respect to the real axis;

  3. (vi)

$\chi_{1}(-\overline{z})=\overline{\chi_{1}(z)}$ for all
$z\in\mathscr{H}^{+}$, $\chi_{2}(-\overline{z})=\overline{\chi_{2}(z)}$ for
all $z\in\mathscr{H}^{-}$.

In particular, the curve $\mathscr{G}$ is smooth everywhere except at $1$,
where it admits a corner point with the angle

$\frac{\theta_{1}}{\theta_{1}+\theta_{2}}2\pi.$

###### Proof.

Since the shift function $\alpha(z)$ is quasisymmetric (Lemma 15), then there
exist $\widetilde{\chi}_{1}$, $\widetilde{\chi}_{2}$ and
$\widetilde{\mathscr{G}}$ satisfying Items (i)–(iii) by [31, Chap. II, Sec.
7.5].

We now construct conformal mappings $\chi_{1}$ and $\chi_{2}$ satisfying Items
(i)–(vi) from $\widetilde{\chi}_{1}$ and $\widetilde{\chi}_{2}$. By the
construction of $\widetilde{\chi}_{1}$ and $\widetilde{\chi}_{2}$, the union
of $\widetilde{\chi}_{1}(\\{a+bi:a,b>0\\})$,
$\widetilde{\chi}_{2}(\\{a-bi:a,b>0\\})$ and
$\widetilde{\chi}_{1}^{+}(\mathbb{R}_{+})$ is indeed a domain and denoted by
$\mathscr{D}^{+}$. Let $a=\widetilde{\chi}_{1}^{+}(0)$,
$b=\widetilde{\chi}_{1}^{+}(\infty)$, and $\zeta$ be a conformal mapping from
$\mathscr{D}^{+}$ onto $\mathscr{H}^{+}$ such that $\zeta^{+}(a)\not=\infty$
and $\zeta^{+}(b)=1$. Such a mapping $\zeta$ exists and can be constructed as
follows:

  * •

Let $\delta$ be a conformal mapping from $\mathscr{D}^{+}$ onto
$\mathscr{C}^{+}$ such that $\delta^{+}(a),\delta^{+}(b)\not=1$ and
$1\in\delta^{+}\circ\widetilde{\chi}_{2}^{+}(i\mathbb{R}_{-})$;

  * •

Recall that $\phi(z)$ defined in (3.1) maps conformally $\mathscr{C}^{+}$ onto
$\mathscr{H}^{+}$;

  * •

Define $\zeta(z):=\phi\circ\delta(z)-\phi\circ\delta^{+}(b)+1$.

Now define

$\displaystyle\chi_{1}(z):=\begin{cases}\zeta\circ\widetilde{\chi}_{1}(z),&\quad\text{if
}z\in\\{a+bi:a,b>0\\},\\\ \zeta^{+}\circ\widetilde{\chi}_{1}(z),&\quad\text{if
}z\in i\mathbb{R}_{+},\\\\[1.5pt]
\overline{\zeta\circ\widetilde{\chi}_{1}(-\overline{z})},&\quad\text{if
}z\in\\{-a+bi:a,b>0\\},\end{cases}$
$\displaystyle\chi_{2}(z):=\begin{cases}\zeta\circ\widetilde{\chi}_{2}(z),&\quad\text{if
}z\in\\{a-bi:a,b>0\\},\\\ \zeta^{+}\circ\widetilde{\chi}_{2}(z),&\quad\text{if
}z\in i\mathbb{R}_{-},\\\\[1.5pt]
\overline{\zeta\circ\widetilde{\chi}_{2}(-\overline{z})},&\quad\text{if
}z\in\\{-a-bi:a,b>0\\}.\end{cases}$

Put $\mathscr{G}=\chi_{1}^{+}(\mathbb{R})$. By Schwarz reflection principle,
$\chi_{1}$ and $\chi_{2}$ are analytic respectively on $\mathscr{H}^{+}$ and
$\mathscr{H}^{-}$, and $\mathscr{G}$ is a Jordan curve. It is easy to check
that $\chi_{1}$, $\chi_{2}$ and $\mathscr{G}$ satisfy Items (i)–(vi).

We now show the smoothness of $\mathscr{G}$. Let $z_{0}$ be any point on
$\mathbb{R}\cup\\{\infty\\}$ and $\lambda$ denote the angle of $\mathscr{G}$
at $\chi_{2}^{-}(z)$. Consider first the case where $z_{0}\in\mathbb{R}$.
Recall that $\alpha(z)$ is analytic and possesses non-vanishing derivatives on
$\mathbb{R}$, then by [34, Thm. 3.11], one has:

$\displaystyle\chi_{1}^{+}(\alpha(z))=\chi_{1}^{+}(\alpha(z_{0}))+\text{const}\cdot(z-z_{0})^{\lambda/\pi}+o\big{(}(z-z_{0})^{\lambda/\pi}\big{)},$
$\displaystyle\chi_{2}^{-}(z)=\chi_{2}^{-}(z_{0})+\text{const}\cdot(z-z_{0})^{(2\pi-\lambda)/\pi}+o\big{(}(z-z_{0})^{(2\pi-\lambda)/\pi}\big{)},$

as $z\in\mathbb{R},z\to z_{0}$. Item (iii) implies that
$\lambda/\pi=(2\pi-\lambda)/\pi$, and thus $\lambda=\pi$, i.e., $\mathscr{G}$
is smooth at $\chi_{2}^{-}(z_{0})$.

We now consider the case where $z_{0}=\infty$. Notice that
$\alpha(z)=z^{\theta_{2}/\theta_{1}}+o(z^{\theta_{2}/\theta_{1}})$ as
$z\to\infty$, then also by [34, Thm. 3.11], one has:

$\displaystyle\chi_{1}^{+}(\alpha(z))=1+\frac{\text{const}}{z^{(\lambda/\pi)(\theta_{2}/\theta_{1})}}+o\left(\frac{1}{z^{(\lambda/\pi)(\theta_{2}/\theta_{1})}}\right),$
$\displaystyle\chi_{2}^{-}(z)=1+\frac{\text{const}}{z^{(2\pi-\lambda)/\pi}}+o\left(\frac{1}{z^{(2\pi-\lambda)/\pi}}\right),$

as $z\in\mathbb{R},z\to\infty$. Item (iii) implies that

$\frac{\lambda}{\pi}\frac{\theta_{2}}{\theta_{1}}=\frac{2\pi-\lambda}{\pi},\text{
i.e. }\lambda=\frac{\theta_{1}}{\theta_{1}+\theta_{2}}2\pi.$

The proof is then complete. ∎

At this point, we already have enough material to build a BVP concerning the
generating function $H(x,y)$ in (1.7), with a boundary condition on the Jordan
curve $\mathscr{G}$. However, the corner point $1$ of $\mathscr{G}$ may
provoke some difficulties on the regularity of solutions around this point. To
overcome such problems, we want to transform the problem to a new one with a
boundary condition on an infinite curve, where the non-regularity problem may
happen at infinity. Therefore, we introduce the following mappings:

(3.12)
$\psi_{1}:=-i\phi\circ\chi_{1}\circ\phi\circ\pi_{1}\quad\text{and}\quad\psi_{2}:=-i\phi\circ\chi_{2}\circ(-\phi)\circ\pi_{2},$

where $\phi$ is defined in (3.1). Let $\mathscr{L}$, $\mathscr{L}^{+}$ and
$\mathscr{L}^{-}$ respectively denote $\phi(\mathscr{G})$,
$\phi(\mathscr{G}^{+})$ and $\phi(\mathscr{G}^{-})$. By this construction,
$\mathscr{L}$ is an open infinite curve, and $\psi_{1}$ (resp. $\psi_{2}$)
maps conformally $\mathscr{S}_{1}^{+}$ (resp. $\mathscr{S}_{2}^{+}$) onto
$\mathscr{L}^{+}$ (resp. $\mathscr{L}^{-}$).

The following lemma presents some crucial properties of $\psi_{1}$ and
$\psi_{2}$.

######  Lemma 17.

We have:

  1. (i)

$\psi_{1}(\overline{z})=\overline{\psi_{1}(z)}$ for all
$z\in\mathscr{S}_{1}^{+}$, and $\psi_{2}(\overline{z})=\overline{\psi_{2}(z)}$
for all $z\in\mathscr{S}_{2}^{+}$;

  2. (ii)

$(\psi_{1}^{-1})^{+}\times(\psi_{2}^{-1})^{-}(\mathscr{L})=\begin{cases}\mathscr{K}\setminus\\{(0,0)\\},&\quad\text{if
}p_{1,1}=0,p_{0,1}\not=p_{1,0},\\\
\mathscr{K},&\quad\text{otherwise};\end{cases}$

  3. (iii)

The asymptotic behaviors of $\psi_{1}$ and $\psi_{2}$ around $1$ are

$\psi_{1}(x)\sim\frac{\text{const}}{(1-x)^{2\pi/(\theta_{1}+\theta_{2})}}\quad\text{and}\quad\psi_{2}(y)\sim\frac{\text{const}}{(1-y)^{2\pi/(\theta_{1}+\theta_{2})}}.$

###### Proof.

We first prove Item (i). For any $z\in\mathscr{S}_{1}^{+}$, we have:

$\displaystyle\psi_{1}(\overline{z})$
$\displaystyle=-i\phi\circ\chi_{1}\circ\phi\circ\pi_{1}(\overline{z})=-i\phi\circ\chi_{1}\circ\phi\circ\overline{\pi_{1}(z)}=-i\phi\circ\chi_{1}\circ\overline{(-\phi)\circ\pi_{1}(z)}$
$\displaystyle=-i\phi\circ\overline{\chi_{1}\circ\phi\circ\pi_{1}(z)}=\overline{-i\phi\circ\chi_{1}\circ\phi\circ\pi_{1}(z)}=\overline{\psi_{1}(z)}.$

Similarly, $\psi_{2}(\overline{z})=\overline{\psi_{2}(z)}$ for any
$z\in\mathscr{S}_{2}^{+}$.

Item (ii) is a direct consequence of Eq. (3.11). We move to the proof of Item
(iii). Recall that

$\displaystyle\pi_{1}(z)=1+\text{const}\cdot(z-1)^{\pi/\theta_{1}}+o\big{(}(z-1)^{\pi/\theta_{1}}\big{)},\quad\text{as}\quad
z\to 1,$
$\displaystyle\chi_{1}(z)=1+\frac{\text{const}}{z^{2\theta_{1}/(\theta_{1}+\theta_{2})}}+o\left(\frac{1}{z^{2\theta_{1}/(\theta_{1}+\theta_{2})}}\right),\quad\text{as}\quad
z\to\infty,$ $\displaystyle\phi(z)=-i\frac{z+1}{z-1}.$

We then have:

$\displaystyle\phi\circ\pi_{1}(z)=-i+\frac{\text{const}}{(z-1)^{\pi/\theta_{1}}}+o\left(\frac{1}{(z-1)^{\pi/\theta_{1}}}\right),$
$\displaystyle\chi_{1}\circ\phi\circ\pi_{1}(z)=1+\text{const}\cdot(z-1)^{2\pi/(\theta_{1}+\theta_{2})}+o\left((z-1)^{2\pi/(\theta_{1}+\theta_{2})}\right),$
$\displaystyle\psi_{1}(z)=-i\phi\circ\chi_{1}\circ\phi\circ\pi_{1}(z)=\frac{\text{const}}{(z-1)^{2\pi/(\theta_{1}+\theta_{2})}}+o\left(\frac{1}{(z-1)^{2\pi/(\theta_{1}+\theta_{2})}}\right),$

as $z\to 1$. The asymptotic behavior of $\psi_{2}$ around $1$ is deduced
similarly. The proof is then complete. ∎

### On the regularity and analyticity of $\psi_{1}$ and $\psi_{2}$ around zero

Since the extensions of $\psi_{1}$ and $\psi_{2}$ around zero will play a
major role in defining the bivariate series $H(x,y)$, we spend a small part in
the article to discuss it.

We first recap some information about the point zero. In the case $p_{1,1}\neq
0$, since $0\in\mathscr{S}_{1}^{+},\mathscr{S}_{2}^{+}$ (see Lemma 5), then
$\psi_{1}$ and $\psi_{2}$ are obviously analytic around $0$. However, in the
case $p_{1,1}=0$ and $p_{1,0}=p_{0,1}$, because
$0\in\mathscr{S}_{1},\mathscr{S}_{2}$, we only define the limit values
$\psi_{1}^{+}(0)$ and $\psi_{2}^{+}(0)$. And in the case $p_{1,1}=0$ and
$p_{0,1}>p_{1,0}$, $0\in\mathscr{S}_{1}^{+}$ but
$0\notin\mathscr{S}_{2}^{+}\cup\mathscr{S}_{2}$, $\psi_{1}$ is analytic around
$0$ but $\psi_{2}(0)$ has not been defined yet.

The following lemma presents analytic continuations of $\psi_{1}$ and
$\psi_{2}$ around $0$.

######  Proposition 18.

In the case $p_{1,1}=0$ and $p_{1,0}=p_{0,1}$, $\psi_{1}$ and $\psi_{2}$ can
be extended analytically around $0$, such that
$\psi_{1}(0)=\psi_{2}(0)\in\mathscr{L}\cap\mathbb{R}$ and
$\psi_{1}^{\prime}(0),\psi_{2}^{\prime}(0)\neq 0$.

In the case $p_{1,1}=0$ and $p_{0,1}>p_{1,0}$, $\psi_{2}$ can be extended
analytically to $0$ along the real line, such that
$\psi_{1}(0)=\psi_{2}(0)\in\mathscr{L}^{+}\cap\mathbb{R}$ and
$\psi_{2}^{\prime}(0)\neq 0$.

###### Proof.

Consider first the case $p_{1,1}=0$ and $p_{1,0}=p_{0,1}$. Let $V$ be an open
and small enough neighborhood of $0$ such that all the assertions in Lemma 9
hold true for $Y(x)$ on $V$. We then introduce the function

$\omega(x):=\begin{cases}\psi_{1}(x),&x\in V\cap\mathscr{S}_{1}^{+},\\\
\psi_{2}(Y(x)),&x\in V\cap\mathscr{S}_{1}^{-}.\end{cases}$

Since $\psi_{1}$ is well defined in $\mathscr{S}_{1}^{+}$, $\psi_{2}$ is well
defined in $\mathscr{S}_{2}^{+}$, and $Y(V\cap\mathscr{S}_{1}^{-})$ is a
subset of $\mathscr{S}_{2}^{+}$ (Lemma 9), then $\omega(x)$ is also well
defined. Further, $\omega(x)$ is sectionally analytic on
$V\cap\mathscr{S}_{1}^{+}$, $V\cap\mathscr{S}_{2}^{+}$ and continuous on $V$
(since $\psi^{+}_{1}(x)=\psi_{2}^{+}(Y(x))$ for all $x\in
V\cap\mathscr{S}_{1}$), then by Morera’s theorem, $\omega(x)$ is analytic on
$V$. In other words, $\psi_{2}(Y(x))$ is an analytic continuation of
$\psi_{1}(x)$ around $0$. Since $\mathscr{S}_{1}$ and $\mathscr{L}$ are smooth
respectively at $0$ and $\psi_{1}(0)$, then $\psi_{1}^{\prime}(0)\neq 0$ by
[34, Thm. 3.9]. We have analogous arguments for the analyticity of $\psi_{2}$
around $0$.

We move to the case $p_{1,1}=0$ and $p_{0,1}>p_{1,0}$, where
$0\in\mathscr{S}_{1}^{+},\mathscr{S}_{2}^{-}$. Let $y_{0}$ denote the
intersection of $\mathbb{R}$ and $\mathscr{S}_{2}\setminus\\{1\\}$. By the
same arguments as in the previous case, $\psi_{1}(X(y))$ is an analytic
continuation of $\psi_{2}(y)$ around $y_{0}$. Since the Taylor expansion of
$X(y)$ at $y_{0}$ possesses all real coefficients, then the extension of
$X(y)$ along the segment $[0,y_{0}]$ is also real and cannot escape
$\mathscr{S}_{1}^{+}$ (because $|X(y)|<|y|$ for all $y\in(0,y_{0})$ close
enough to $y_{0}$ by Lemma 10). Thus, $\psi_{1}(X(y))$ is well defined on
$[0,y_{0}]$ and forms an analytic continuation of $\psi_{2}(y)$. Hence,
$\psi_{2}(0)=\psi_{1}(X(0))=\psi_{1}(0)\in\mathscr{L}^{+}$ and
$\psi_{2}^{\prime}(0)=\psi_{1}^{\prime}(X(0))X^{\prime}(0)\neq 0$. ∎

##  4\. Proof of the main theorems

In this section, we will outline the proof of the main theorems, but not
present them in detail since the strategy is similar as in the symmetric case
of [24]. From the functional equation (1.9) and the conformal mappings in
Section 3, one can construct a boundary value problem (Lemma 19) whose
polynomial solutions form the class of harmonic functions
$(h_{n})_{n\in\mathbb{N}}$ in Theorem 2. Such harmonic functions are shown to
satisfy the following features: first, at any point $(i,j)\in\mathbb{N}^{2}$,
for all $n$ large enough, $h_{n}(i,j)=0$ (Lemma 20); second,
$\\{h_{n}(\cdot,1)\\}_{n\geqslant 1}$ forms a basis of the space
$\\{h(\cdot,1):h\text{ is harmonic}\\}$ (Lemma 21). Consequently, any infinite
linear combination $\sum_{n\geqslant 1}a_{n}h_{n}$ with
$\\{a_{n}\\}_{n\geqslant 1}\subset\mathbb{R}$, when evaluated at any point
$(i,j)\in\mathbb{N}^{2}$, possesses a finite value. On the other hand, any
discrete harmonic function can be expressed uniquely as an infinite sum
$\sum_{n\geqslant 1}a_{n}h_{n}$ with $\\{a_{n}\\}_{n\geqslant
1}\subset\mathbb{R}$ (Theorem 3).

We first set

(4.1)
$F(t):=\begin{cases}K{\cdot}H(\psi_{1}^{-1}(t),0)-\frac{1}{2}K{\cdot}H(0,0),&\text{if
}t\in\mathscr{L}^{+},\\\
-K{\cdot}H(0,\psi_{2}^{-1}(t))+\frac{1}{2}K{\cdot}H(0,0),&\text{if
}t\in\mathscr{L}^{-}.\\\ \end{cases}$

$F^{+}(t)$$F^{-}(t)$ Figure 8. Description of the BVP in Lemma 19: If $F$ is
analytic on $\mathscr{L}^{+}$ (blue domain) and $\mathscr{L}^{-}$ (red
domain), and continuous on $\mathscr{L}$ (i.e., $F^{+}(t)=F^{-}(t)$ for all
$t\in\mathscr{L}$), then $F$ is analytic on $\mathbb{C}$.

The following lemma is a direct consequence of the functional equation (1.9).

######  Lemma 19.

Assume (H1)–(H6), (K1)–(K5), and that $H(x,0)$ and $H(0,y)$ have radii of
convergence equal to or greater than $1$. Then $F$ defined in (4.1) satisfies
the following BVP (see Figure 8):

  1. (i)

$F$ is analytic on $\mathscr{L}^{+}$ and admits a continuous extension $F^{+}$
to $\mathscr{L}$;

  2. (ii)

$F$ is analytic on $\mathscr{L}^{-}$ and admits a continuous extension $F^{-}$
to $\mathscr{L}$;

  3. (iii)

For all $t\in\mathscr{L}$,

(4.2) $F^{+}(t)-F^{-}(t)=0.$

Consequently, if $F$ is bounded at infinity, then $F$ is the zero function and
so is the associated harmonic function. On the other hand, if $F$ has a pole
of order $n>0$ at infinity, then $F$ is a polynomial of degree $n$ satisfying:

(4.3) $F(\psi_{1}(0))=-F(\psi_{2}(0))=\frac{K\cdot
H(0,0)}{2}=-\frac{1}{2}p_{1,1}h(1,1).$

###### Proof.

The general solutions of the BVP are in fact entire functions, and under the
conditions on the growth at infinity, one can infer these solutions as
constant functions or polynomials (see more details in [24, Lem. 9 and Cor.
10]). If $F$ is a constant function (resp. a polynomial), evaluating $F$ at
$\psi_{1}(0)$ and $\psi_{2}(0)$ yields that $F$ is the zero function (resp.
$F$ satisfies Eq. (4.3)). ∎

###### Proof of Theorem 2.

As one can easily verify, the polynomials introduced in (1.14) are solutions
of the BVP in Lemma 19 and form a basis of $\mathbb{R}[X]$. Consequently, the
functions $H_{n}(x,y)$ defined in (1.15) satisfy the functional equation
(1.9). The remaining work is to prove that these $H_{n}(x,y)$ are bivariate
power series around $(0,0)$, which always holds true in the case
$p_{1,1}\not=0$ (because $K(0,0)\not=0$ in this case). In the case
$p_{1,1}=0$, since $\psi_{1}(0)=\psi_{2}(0)$ by Prop. 18, then $H_{n}(x,y)$
can be rewritten as

$H_{n}(x,y)=\frac{\psi_{1}(x)-\psi_{2}(y)}{K(x,y)}\sum_{k=0}^{n-1}(\psi_{1}(x)-\psi_{1}(0))^{k}(\psi_{2}(y)-\psi_{1}(0))^{n-1-k}.$

Since $p_{0,1}$ and $p_{1,0}$ cannot simultaneously vanish, we can assume
further $0\not=p_{0,1}\geqslant p_{1,0}$. Hence, on a neighborhood of $(0,0)$,
$\partial_{x}K(x,y)\not=0$, which implies that $X(y)$ is the unique solution
of $K(x,y)$ (as a function of $x$) around $0$ by the implicit function
theorem. Moreover, since $\partial_{x}(\psi_{1}(x)-\psi_{2}(y))\not=0$ on a
neighborhood of $(0,0)$, and $\psi_{1}(X(y))=\psi_{2}(y)$ around $0$ (recall
from the proof of Prop. 18 that $\psi_{1}(X(y))$ is an analytic continuation
of $\psi_{2}(y)$ around $0$), then $X(y)$ is the unique solution of
$\big{(}\psi_{1}(x)-\psi_{2}(y)\big{)}$ (as functions of $x$) around $0$ by
the implicit function theorem. Thanks to the Weierstrass preparation theorem
for analytic functions in several variables (see [23, Chap. 2, Sec. B, Thm.
2]), we can write:

$\frac{\psi_{1}(x)-\psi_{2}(y)}{K(x,y)}=\frac{u(x,y)(x-X(y))}{v(x,y)(x-X(y))}=\frac{u(x,y)}{v(x,y)},$

where $u(x,y)$ and $v(x,y)$ are analytic around $(0,0)$ and not vanishing at
$(0,0)$. Thus, $(\psi_{1}(x)-\psi_{2}(y))/K(x,y)$ is analytic around $(0,0)$,
and so is $H_{n}(x,y)$.

We move to the properties of $H_{n}(x,y)$. Since $K(x,y)$ has no solution in
$\mathscr{S}_{1}^{+}\times\mathscr{S}_{2}^{+}$ (Prop. 1(ii)), then
$H_{n}(x,y)$ is analytic in this domain. And lastly, one can compare the
Laplace transform of $h_{n,\sigma}(x,y)$, which can be explicitly computed,
with the Laplace transform of $h_{n}(x,y)$, which can be approximated through
$H_{n}(x,y)$ (see [24, Sec. 4.2, Prop. 11]). ∎

Although the proof of Theorem 2 shares the same framework as that of [24, Thm.
1] (the symmetric case), there are some major differences that we want to
emphasize:

  * •

In the symmetric case of [24], the conformal mappings $\psi_{1}$ and
$\psi_{2}$ can always be constructed. Further, the shift function $\alpha$ is
the identity, we thus do not need the step of solving the conformal welding
problem with quasisymmetric shift (Subsection 3.2). In the non-symmetric case,
we have to restrict the analysis under Assumptions (K1)–(K5) to ensure the
existence of such mappings;

  * •

As presented above, the proof is mostly based on the behaviors of $\psi_{1}$,
$\psi_{2}$, $K(x,y)$ around $(0,0)$. In the case $p_{1,1}=0$,
$p_{0,1}>p_{1,0}$, we have $0\in\mathscr{S}_{2}^{-}$, which does not appear in
the symmetric case. We therefore need analytic continuation arguments (Prop.
1(i), Prop. 18) to show that $\psi_{2}(z)$ is well defined and analytic around
$0$, and $K(x,y)$ has no solution in
$\mathscr{S}_{1}^{+}\times\mathscr{S}_{2}^{+}$ (Prop. 1(ii)).

In order to prove Theorem 3, we recall from [24] the following two lemmas
presenting important properties of the harmonic functions
$\\{h_{n}\\}_{n\geqslant 1}$ in Theorem 2.

######  Lemma 20.

The harmonic function $h_{n}$ defined in Theorem 2 satisfies the following
assertions:

  * •

In the case $p_{1,1}=0$,

    1. (i)

For all $i,j\geqslant 1$ such that $i+j\leqslant n$, we have $h_{n}(i,j)=0$;

    2. (ii)

For all $i,j\geqslant 1$ such that $i+j=n+1$, we have $h_{n}(i,j)\not=0$.

  * •

In the case $p_{1,1}\not=0$,

    1. (i)

For all $1\leqslant i,j\leqslant\lfloor n/2\rfloor$, we have $h_{n}(i,j)=0$;

    2. (ii)

$h_{n}(\lfloor n/2+1\rfloor,1),h_{n}(1,\lfloor n/2+1\rfloor)\neq 0$.

######  Lemma 21.

For any discrete harmonic function $h(i,j)$, we have:

  * •

In the case $p_{1,1}=0$, there exist unique sequences $\\{a_{n}\\}_{n\geqslant
1},\\{b_{n}\\}_{n\geqslant 1}\subset\mathbb{R}$ such that:

$h(i,1)=\sum_{n\geqslant 1}a_{n}h_{n}(i,1)\quad\text{and}\quad
h(1,i)=\sum_{n\geqslant 1}b_{n}h_{n}(1,i),\quad\text{for all }i\in\mathbb{N};$

  * •

In the case $p_{1,1}\not=0$, there exists a unique sequence
$\\{a_{n}\\}_{n\geqslant 1}\subset\mathbb{R}$ such that:

$h(i,1)=\sum_{n\geqslant 1}a_{n}h_{n}(i,1)\quad\text{and}\quad
h(1,i)=\sum_{n\geqslant 1}a_{n}h_{n}(1,i),\quad\text{for all }i\in\mathbb{N},$

where $\\{h_{n}\\}_{n\geqslant 1}$ are defined in Theorem 2.

By Lemma 20, the linear systems of equations in Lemma 21 are solvable and have
unique solutions. We do not mention its detailed proof since it is similar as
in [24, Lem. 14, 15].

###### Proof of Theorem 3.

Observe first that any infinite sum $\sum_{n\geqslant 1}a_{n}h_{n}$, with
$\\{a_{n}\\}_{n\geqslant 1}\subset\mathbb{R}$, evaluated at any point
$(i,j)\in\mathbb{N}^{2}$, has a finite value, since $h_{n}(i,j)=0$ for all
$n\geqslant i+j$ by Lemma 20. Thus, the mapping $\Phi$ in Theorem 3 is well
defined. The injectivity of $\Phi$ is implied by Lemma 21.

Now let $h(i,j)$ be any harmonic function and $H(x,y)$ be its generating
function. To show the surjectivity of $\Phi$, we prove that there exists a
sequence $\\{a_{n}\\}_{n\geqslant 1}\subset\mathbb{R}$ such that
$h(i,j)=\sum_{n\geqslant 1}a_{n}h_{n}(i,j)$ for all $(i,j)\in\mathbb{N}^{2}$.

In the case $p_{1,1}\not=0$, let $\\{a_{n}\\}_{n\geqslant 1}$ be the sequence
satisfying Lemma 21 and $\widetilde{H}(x,y)$ be the generating function of
$\sum_{n\geqslant 1}a_{n}h_{n}(i,j)$. We recall that $H(x,y)$ and
$\widetilde{H}(x,y)$ are determined by Eq. (1.9):

$\displaystyle H(x,y)=\frac{K\cdot H(x,0)+K\cdot H(0,y)-K\cdot
H(0,0)}{K(x,y)},$
$\displaystyle\widetilde{H}(x,y)=\frac{K\cdot\widetilde{H}(x,0)+K\cdot\widetilde{H}(0,y)-K\cdot\widetilde{H}(0,0)}{K(x,y)}.$

Moreover, the quotients are well defined since $K(0,0)\not=0$. Lemma 21
implies that $H(x,0)$ and $\widetilde{H}(x,0)$ coincide and so do $H(0,y)$ and
$\widetilde{H}(0,y)$. Thus, $H(x,y)$ and $\widetilde{H}(x,y)$ also coincide
and $h\equiv\sum_{n\geqslant 1}a_{n}h_{n}$.

We move to the case $p_{1,1}=0$, and further assume $0\not=p_{0,1}\geqslant
p_{1,0}$. Then the function $X(y)$ is well defined around $0$ (Prop. 1(i)) and
by the Weierstrass preparation theorem in the ring of formal power series
$\mathbb{R}[[x,y]]$, one can write:

$K(x,y)=u(x,y)(x-X(y)),$

where $u$ is invertible in $\mathbb{R}[[x,y]]$. By Lemma 21, there exists a
unique sequence $\\{a_{n}\\}_{n\geqslant 1}$ such that
$h(i,1)=\sum_{n\geqslant 1}a_{n}h_{n}(i,1)$ for all $i\in\mathbb{N}$. Let
$\widetilde{H}(x,y)$ denote the generating function of $\sum_{n\geqslant
1}a_{n}h_{n}(i,1)$, we then have $H(x,0)=\widetilde{H}(x,0)$. Consider the
division of $K\cdot H(x,0)$ (resp. $K\cdot\widetilde{H}(x,0)$) by $(x-X(y))$.
Applying the Weierstrass division theorem to $K(x,y)$ in the ring
$\mathbb{R}[[x,y]]$, there exists a unique formal series $G\in\mathbb{R}[[y]]$
(resp. $\widetilde{G}\in\mathbb{R}[[y]]$) such that $K\cdot H(x,0)+G(y)$
(resp. $K\cdot\widetilde{H}(x,0)+\widetilde{G}(y)$) is divisible by $x-X(y)$.
Since $H(x,0)$ and $\widetilde{H}(x,0)$ coincide, then so do $G(y)$ and
$\widetilde{G}(y)$. And thus, the quotients of $K\cdot H(x,0)+G(y)$ and
$K\cdot\widetilde{H}(x,0)+\widetilde{G}(y)$ by $K(x,y)$, which are
respectively $H(x,y)$ and $\widetilde{H}(x,y)$, also coincide. This shows the
surjectivity of $\Phi$ in the case $p_{1,1}=0$. ∎

##  5\. Small jump random walks

In this section, we take a closer look to models associated with small jump
random walks, namely, $p_{i,j}=0$ if $|i|\geqslant 2$ or $|j|\geqslant 2$. Our
goal is to derive explicit expressions of the conformal mappings $\psi_{1}$
and $\psi_{2}$ in Theorem 2. Such models have been carefully studied in [18,
Sec. 6.5], where the kernel’s zero set can be parametrized through a rational
uniformization. We first recall some facts about the uniformization, then
construct conformal mappings. At the end of the section, we also give a
concrete example where the shift function and all the conformal mappings in
Section 3 can be explicitly expressed.

$x^{-1}(s)$$y^{-1}(s)$$z^{3}+z^{-3}$ Figure 9. The model
$p_{1,0}=p_{0,-1}=p_{-1,1}=1/3$ (also known as the tandem walk, due to its
links with queuing theory): the inverses $x^{-1}(s)$ and $y^{-1}(s)$ map
respectively and conformally $\mathscr{S}_{1}^{+}\setminus[x_{1},1]$ (red
domain in the top left figure) and $\mathscr{S}_{2}^{+}\setminus[x_{1},1]$
(blue domain in the bottom left figure) onto lower (in red) and upper (in
blue) domains in the cone; the mapping $\omega(z):=z^{3}+z^{-3}$ maps the cone
onto the plane $\mathbb{C}$ cut along some segments; $\psi_{1}(z)$ and
$\psi_{2}(z)$ in Theorem 2 can be chosen as $\psi_{1}=\omega\circ x^{-1}$ and
$\psi_{2}=\omega\circ y^{-1}$.

Consider the problem under Assumptions (H1)–(H4) with the additional
hypothesis $p_{i,j}=0$ if $|i|\geqslant 2$ or $|j|\geqslant 2$. Notice that
the problem of any model $\\{p_{k,\ell}\\}_{k,\ell}$ with $p_{0,0}\not=0$ is
equivalent to the problem of the model $\\{p^{\prime}_{k,\ell}\\}_{k,\ell}$
with $p^{\prime}_{0,0}=0$ and $p^{\prime}_{k,\ell}=p_{k,\ell}/(1-p_{0,0})$ for
all $k,\ell$. Then without loss of generality, we assume further $p_{0,0}=0$.
We can write the kernel under the form

$\displaystyle K(x,y)=a(x)y^{2}+b(x)y+c(x),$

where

$\displaystyle a(x)=-(p_{-1,-1}x^{2}+p_{0,-1}x+p_{1,-1}),$ $\displaystyle
b(x)=-(p_{-1,0}x^{2}-x+p_{1,0}),$ $\displaystyle
c(x)=-(p_{-1,1}x^{2}+p_{0,1}x+p_{1,1}).$

The corresponding discriminant $d(x)=b(x)^{2}-4a(x)c(x)$ is a polynomial of
order $3$ or $4$, and in either case, $d(x)$ always has one root
$x_{1}\in[-1,1)$, a double root $1$. If $d(x)$ has order $4$, then the
remaining root, denoted as $x_{4}$, is in $(1,\infty)\cup(-\infty,-1]$. If
$d(x)$ has order $3$, we denote conventionally $x_{4}=\infty$. We also obtain
$y_{1}\in[-1,1)$ and $y_{4}\in(-\infty,-1]\cup(1,\infty]$ similarly.

Put

$\displaystyle
s_{0}=\frac{2-(x_{1}+x_{4})+2\sqrt{(1-x_{1})(1-x_{4})}}{x_{4}-x_{1}},$
$\displaystyle
s_{1}=\frac{x_{1}+x_{4}-2x_{1}x_{4}+2\sqrt{x_{1}x_{4}(1-x_{1})(1-x_{4})}}{x_{4}-x_{1}},$
$\displaystyle
s_{2}=\frac{2-(y_{1}+y_{4})+2\sqrt{(1-y_{1})(1-y_{4})}}{y_{4}-y_{1}},$
$\displaystyle
s_{3}=\frac{y_{1}+y_{4}-2y_{1}y_{4}+2\sqrt{y_{1}y_{4}(1-y_{1})(1-y_{4})}}{y_{4}-y_{1}},$
$\displaystyle\rho=e^{-i\theta},$

where $\theta:=(\theta_{1}+\theta_{2})/2$ and $\theta_{1}$, $\theta_{2}$ are
defined in (2.6).

We recall an important result about the rational uniformization. We refer to
[19, Sec. 2.3] for the proof.

######  Lemma 22.

Assuming (H1)–(H4) and that $p_{i,j}=0$ if $|i|\geqslant 2$ or $|j|\geqslant
2$, one has

$\\{(x,y)\in(\mathbb{C}\cup\\{\infty\\})^{2}:K(x,y)=0\\}=\\{(x(s),y(s)):s\in\mathbb{C}\cup\\{\infty\\}\\},$

where

$x(s)=\frac{(s-s_{1})(s-\frac{1}{s_{1}})}{(s-s_{0})(s-\frac{1}{s_{0}})}\quad\text{and}\quad
y(s)=\frac{(\rho s-s_{3})(\rho s-\frac{1}{s_{3}})}{(\rho s-s_{2})(\rho
s-\frac{1}{s_{2}})}.$

Moreover, the above rational functions admit the involutions $x(s)=x(1/s)$ and
$y(s)=y(1/(\rho^{2}s))$.

Put $\mathscr{P}:=x^{-1}(\mathscr{S}_{1})\cap y^{-1}(\mathscr{S}_{2})$.

######  Lemma 23.

Under Assumptions (H1)–(H4) and (K1)–(K5), $\mathscr{P}$ is a non-self-
intersecting curve with two endpoints $0$ and infinity, lying in the cone
$\mathscr{E}:=\\{re^{i\phi}:r\geqslant 0,\phi\in[0,\theta]\\}$. Consequently,
$\mathscr{P}$ divides $\mathscr{E}$ into two domains, denoted as
$\mathscr{P}^{+}$ and $\mathscr{P}^{-}$ (see Figure 9), such that:

  1. (i)

$x(s)$ maps conformally $\mathscr{P}^{+}$ onto
$\mathscr{S}_{1}^{+}\setminus[x_{1},1]$;

  2. (ii)

$y(s)$ maps conformally $\mathscr{P}^{-}$ onto
$\mathscr{S}_{2}^{+}\setminus[y_{1},1]$.

###### Proof.

We first rewrite $x(s)$ and $y(s)$ as

$x(s)=1+\frac{\left(s_{0}+\frac{1}{s_{0}}\right)-\left(s_{1}+\frac{1}{s_{1}}\right)}{\left(s+\frac{1}{s}\right)-\left(s_{0}+\frac{1}{s_{0}}\right)}\quad\text{and}\quad
y(s)=1+\frac{\left(s_{2}+\frac{1}{s_{2}}\right)-\left(s_{3}+\frac{1}{s_{3}}\right)}{\left(\rho
s+\frac{1}{\rho s}\right)-\left(s_{2}+\frac{1}{s_{2}}\right)}.$

It is worth mentioning that $\\{s_{i}+1/s_{i}\\}_{0\leqslant i\leqslant 3}$
are real, and $x(s)$ and $y(s)$ are composed by Möbius transformations (which
have the form $(az+b)/(cz+d)$ for $a,b,c,d\in\mathbb{C}$) and Joukowsky
transformation (which has the form $z+1/z$).

Let us specify the image of $\mathscr{E}$ under the maps $x(s)$ and $y(s)$. It
is easily seen that

(5.1) $s_{0}+\frac{1}{s_{0}}=\frac{4-2(x_{1}+x_{4})}{x_{4}-x_{1}}<2\leqslant
s+\frac{1}{s}$

for all $s\in[0,\infty]$, and $x^{\prime}(s)$ does not change the sign on
$(0,1)$ and $(1,\infty)$. Moreover, $x(0)=x(\infty)=1$ and $x(1)=x_{0}$.
Hence, $x(s)$ is a one-to-one mapping from $[0,1]$ onto $[x_{1},1]$, and from
$[1,\infty]$ onto $[x_{1},1]$. Moreover, since $x^{\prime}(s)$ never vanishes
on $e^{i\theta}\mathbb{R}_{+}$, then $x(s)$ is a one-to-one mapping from
$e^{i\theta}\mathbb{R}_{+}$ onto a Jordan curve $\mathscr{J}_{1}$ passing
through $1$, since $x(0)=x(\infty)=1$. Similarly, $y(s)$ is a one-to-one
mapping from $e^{i\theta}[0,1]$ and $e^{i\theta}[1,\infty]$ onto $[y_{1},1]$,
and $\mathbb{R}_{+}$ onto a Jordan curve $\mathscr{J}_{2}$ passing through
$1$. This implies that $x(s)$ (resp. $y(s)$) maps conformally $\mathscr{E}$
onto $\mathscr{J}_{1}^{+}\setminus[x_{1},1]$ (resp.
$\mathscr{J}_{2}^{+}\setminus[y_{1},1]$).

We now prove that
$\mathscr{S}_{1}\subset\mathscr{J}_{1}^{+}\setminus[x_{1},1]$ and
$\mathscr{S}_{2}\subset\mathscr{J}_{2}^{+}\setminus[y_{1},1]$. We first show
that $(x_{1},1)\subset\mathscr{S}_{1}^{+}$ . Reasoning by contradiction, if
$\mathscr{S}_{1}$ cuts $(x_{1},1)$ at any point, then there exists
$s^{\prime}\in(0,\infty)\setminus\\{1\\}$ such that $x(s^{\prime})$ is the
intersection of $\mathscr{S}_{1}$ and $(x_{1},1)$, and $y(s^{\prime})$ is the
intersection of $\mathscr{S}_{2}$ and $\mathbb{R}$. In other words,
$y(s^{\prime})$ is real, which cannot not hold, since $\left(\rho
s+\frac{1}{\rho s}\right)$ is not real for any
$s\in(0,\infty)\setminus\\{1\\}$. Hence,
$(x_{1},1)\subset\mathscr{S}_{1}^{+}$.

Similarly, $(y_{1},1)\subset\mathscr{S}_{2}^{+}$. By Prop. 1(ii), we then know
that $\mathscr{J}_{1}$ and $\mathscr{J}_{2}$ are respectively outside of
$\mathscr{S}_{1}^{+}$ and $\mathscr{S}_{2}^{+}$. Hence,
$\mathscr{S}_{1}\subset\mathscr{J}_{1}^{+}\setminus[x_{1},1]$ and
$\mathscr{S}_{2}\subset\mathscr{J}_{2}^{+}\setminus[y_{1},1]$.

Now consider the set $x^{-1}(\mathscr{S}_{1})$. By the position of
$\mathscr{S}_{1}$ and the involution of $x(s)$ in Lemma 22, we know that
$x^{-1}(\mathscr{S}_{1})$ (resp. $y^{-1}(\mathscr{S}_{2})$) is a union of two
curves, both of which have two endpoints $0$ and infinity, but one is in
$\mathscr{E}$ and the other one is in $e^{-i\theta}\mathscr{E}$ (resp.
$e^{i\theta}\mathscr{E}$). This implies that $\mathscr{P}\subset\mathscr{E}$.
The other statement in the lemma also follows. ∎

Now put

$\omega(z):=z^{\pi/\theta}+z^{-\pi/\theta}.$

It is seen that $\omega(z)$ is the composition of $z\mapsto z+1/z$ and
$z\mapsto z^{\pi/\theta}$. Recall $z\mapsto z^{\pi/\theta}$ maps the cone
$\mathscr{E}$ onto the upper half plane $\mathscr{H}^{+}$. The mapping
$z\mapsto z+1/z$ is the Joukowsky transform, which maps conformally the upper
unit disk onto the upper half plane $\mathscr{H}^{+}$, with the following one-
to-one correspondence on the boundary:

  * •

$[-1,1]$ corresponds to $[-\infty,-2]\cup[2,\infty]$;

  * •

The upper semicircle $\\{e^{i\phi}:\phi\in[0,\pi]\\}$ corresponds to $[-2,2]$.

$z\mapsto z+1/z$ also maps conformally the set
$\\{re^{i\phi}:r>1,\phi\in(0,\pi)\\}$ onto the lower half plane
$\mathscr{H}^{-}$, with the following one-to-one correspondence on the
boundary:

  * •

$[-\infty,-1]\cup[1,\infty]$ corresponds to $[-\infty,-2]\cup[2,\infty]$;

  * •

The upper semicircle $\\{e^{i\phi}:\phi\in[0,\pi]\\}$ corresponds to $[-2,2]$.

Hence, $z\mapsto z+1/z$ maps conformally the upper half plane onto the plan
cut $\mathbb{C}\setminus((-\infty,-2]\cup[2,\infty))$. Thus, $\omega(z)$ maps
conformally $\mathscr{E}$ onto
$\mathbb{C}\setminus((-\infty,-2]\cup[2,\infty))$.

Let $\mathscr{L}$ denote $\omega(\mathscr{P})$, which is an infinite curve,
and let $\mathscr{L}^{+}$, $\mathscr{L}^{-}$ respectively denote
$\omega(\mathscr{P}^{+})$, $\omega(\mathscr{P}^{-})$. One can verify that
$\omega\circ x^{-1}$ (resp. $\omega\circ y^{-1}$) maps conformally
$\mathscr{S}_{1}^{+}$ (resp. $\mathscr{S}_{2}$) onto $\mathscr{L}^{+}$ (resp.
$\mathscr{L}^{-}$). Now we put

$\displaystyle\psi_{1}(z)=\omega\circ
x^{-1}(z)=2T_{\pi/\theta}\left(\frac{(s_{0}+1/s_{0})z-(s_{1}+1/s_{1})}{2(z-1)}\right),$
$\displaystyle\psi_{2}(z)=\omega\circ
y^{-1}(z)=-2T_{\pi/\theta}\left(\frac{(s_{2}+1/s_{2})z-(s_{3}+1/s_{3})}{2(z-1)}\right),$

where $T_{n}(z)$ is a generalization of Chebyshev polynomial of the first kind
to non-integer order $n$, and is defined by:

$T_{\pi/\theta}(z)=\frac{1}{2}\left((z+\sqrt{z^{2}-1})^{\pi/\theta}+(z-\sqrt{z^{2}-1})^{\pi/\theta}\right),\quad
z\in\mathbb{C}\setminus(-\infty,-1),$

(see Figure 9). It can be verified that the mappings $\psi_{1}$ and $\psi_{2}$
satisfy Lemma 17 and thus can be chosen to construct harmonic functions in
Theorem 2.

### Weighted simple random walk

We now give a concrete example where all the conformal mappings in Section 3
can be explicitly expressed.

Consider the random walk with the transition probabilities:

$p_{0,1}=p_{0,-1}=\frac{3}{8},\,p_{1,0}=p_{-1,0}=\frac{1}{8}.$

The kernel then takes the form:

$K(x,y)=xy-\frac{1}{8}\big{(}3x+y+3xy^{2}+x^{2}y\big{)}.$

By solving the equation $K(\eta s,\eta s^{-1})=0$, one obtains:

$\displaystyle\mathscr{S}_{1}=\big{\\{}\frac{4s+{i}\sqrt{3}(s^{2}-1)}{s^{2}+3}s:s={e}^{{i}t},t\in[0,\pi)\big{\\}},$
$\displaystyle\mathscr{S}_{2}=\big{\\{}\frac{4s+{i}\sqrt{3}(s^{2}-1)}{s^{2}+3}s^{-1}:s={e}^{{i}t},t\in[0,\pi)\big{\\}}.$

Using the uniformization constructed above, we have:

$x(s)=\frac{(s-e^{i\pi/6})(s-e^{-i\pi/6})}{(s-e^{i5\pi/6})(s-e^{-i5\pi/6})}\quad\text{and}\quad
y(s)=\frac{(e^{-i\pi/3}s-e^{i\pi/3})(e^{-i\pi/3}s-e^{-i\pi/3})}{(e^{-i\pi/3}s-e^{i2\pi/3})(e^{-i\pi/3}s-e^{-i2\pi/3})}.$

In particular, $x(s)$ maps conformally the cone
$\\{re^{i\phi}:r>0,\phi\in(0,\pi/3)\\}$ onto
$\mathscr{S}_{1}^{+}\setminus[7-4\sqrt{3},1]$, and $y(s)$ maps conformally the
cone $\\{re^{i\phi}:r>0,\phi\in(\pi/3,\pi/2)\\}$ onto
$\mathscr{S}_{2}^{+}\setminus[1/3,1]$. We consider some conformal mappings
between half-planes, disks and cones:

$\displaystyle\mu:\\{a+bi:a>0\\}\to\mathscr{C}^{+},\quad
z\mapsto\frac{z-1}{z+1},$
$\displaystyle\pi_{1}:\\{re^{i\phi}:r>0,\phi\in(0,\pi/3)\\}\to\\{a+bi:a>0\\},\quad
z\mapsto z^{3/2}+z^{-3/2},$
$\displaystyle\pi_{2}:\\{re^{i\phi}:r>0,\phi\in(\pi/3,\pi/2)\\}\to\\{a+bi:a>0\\},\quad
z\mapsto-(ze^{-i\pi/6})^{3}-(ze^{-i\pi/6})^{-3}.$

Now put

$\displaystyle\pi_{1}(z)=\mu\circ\omega_{1}\circ
x^{-1}(z)=\frac{2T_{3/2}(-\frac{\sqrt{3}}{2}\frac{z+1}{z-1})-1}{2T_{3/2}(-\frac{\sqrt{3}}{2}\frac{z+1}{z-1})+1},$
$\displaystyle\pi_{2}(z)=\mu\circ\omega_{2}\circ
y^{-1}(z)=\frac{2T_{3}(-\frac{1}{2}\frac{z+1}{z-1})-1}{2T_{3}(-\frac{1}{2}\frac{z+1}{z-1})+1}.$

It can be verified that $\pi_{1}$ and $\pi_{2}$ are conformal mappings
respectively from $\mathscr{S}_{1}^{+}$ and $\mathscr{S}_{2}^{+}$ onto
$\mathscr{C}^{+}$, and satisfy Lemma 11. Accordingly, the shift function
$\alpha:\mathbb{R}\to\mathbb{R}$ in (3.1) has the form

$\alpha(z)=\text{sign}(z)\sqrt{\sqrt{z^{2}+4}-2},$

and $\chi_{1}$ and $\chi_{2}$ in Lemma 16 admit the expressions

$\chi_{1}(z)=\frac{2T_{4/3}(-iz/2)-1}{2T_{4/3}(-iz/2)+1}\quad\text{and}\quad\chi_{2}(z)=\frac{-2T_{2/3}(iz/2)-1}{-2T_{2/3}(iz/2)+1}.$

The conformal mappings $\psi_{1}$ and $\psi_{2}$ in Lemma 17 then follow:

$\psi_{1}(z)=2T_{2}(-\frac{\sqrt{3}}{2}\frac{z+1}{z-1})=\frac{z^{2}+10z+1}{(z-1)^{2}}\quad\text{and}\quad\psi_{2}(z)=-2T_{2}(-\frac{1}{2}\frac{z+1}{z-1})=\frac{z^{2}-6z+1}{(z-1)^{2}}.$

With the family of polynomials $\\{P_{n}(x)\\}_{n\geqslant
1}=\\{(x-1)^{n}\\}_{n\geqslant 1}$, the harmonic functions $h_{n}(i,j)$ and
theirs generating functions $H_{n}(x,y)$ in Theorem 2 can be explicitly
expressed, for example, we have:

$\displaystyle H_{1}(x,y)=\frac{-32}{(x-1)^{2}(y-1)^{2}}=-32\sum_{i,j\geqslant
1}ijx^{i-1}y^{j-1},$ $\displaystyle
H_{2}(x,y)=\frac{128(x^{2}y-3xy^{2}+4xy-3x+y)}{(x-1)^{4}(y-1)^{4}}=128\sum_{i,j\geqslant
1}\frac{ij(-3i^{2}+2j^{2}+2)}{6}x^{i-1}y^{j-1},$ $\displaystyle...$

We finally remark that $-h_{1}(i,j)/32=ij$ is the unique positive harmonic
function (up to multiplicative factors) for this example.

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