# Geometric description of some Loewner chains with infinitely many slits

E. K. Theodosiadis  and  K. Zarvalis  Department of Mathematics, Stockholm
University, 106 91 Stockholm, Sweden<EMAIL_ADDRESS>Department of
Mathematics, Aristotle University of Thessaloniki, 54124, Thessaloniki, Greece
<EMAIL_ADDRESS>
###### Abstract.

We study the chordal Loewner equation associated with certain driving
functions that produce infinitely many slits. Specifically, for a choice of a
sequence of positive numbers $(b_{n})_{n\geq 1}$ and points of the real line
$(k_{n})_{n\geq 1}$, we explicitily solve the Loewner PDE

$\dfrac{\partial f}{\partial
t}(z,t)=-f^{\prime}(z,t)\sum_{n=1}^{+\infty}\dfrac{2b_{n}}{z-k_{n}\sqrt{1-t}}$

in $\mathbb{H}\times[0,1)$. Using techniques involving the harmonic measure,
we analyze the geometric behaviour of its solutions, as $t\rightarrow 1^{-}$.

###### Key words and phrases:

Loewner equation, spirallike functions, harmonic measure.

###### 2020 Mathematics Subject Classification:

Primary 30C45, 35C05; Secondary 30C85

##  1\. Introduction

Given an increasing family of slit domains $(H_{t})_{0\leq t<T}$ of the upper
half-plane $\mathbb{H}$, that is $H_{t}:=\mathbb{H}\setminus\gamma([0,t])$ for
a continuous curve $\gamma:[0,T]\to\overline{\mathbb{H}}$, with
$\gamma(0)\in\mathbb{R}$ and $\gamma((0,T))\subset\mathbb{H}$, there exists a
family of conformal maps
$g_{t}=g(\cdot,t):H_{t}\overset{onto}{\longrightarrow}\mathbb{H}$, normalized
in such a way that the hydrodynamic condition $g_{t}(z)-z\rightarrow 0$, as
$z\rightarrow\infty$, is satisfied. Furthermore, an application of the Schwarz
reflection principle, shows that $g_{t}$ has a Laurent expansion at infinity

$g_{t}(z)=z+\dfrac{b(t)}{z}+\cdots,$

for all $z\in\mathbb{C}$ that lie outside a disc containing
$\gamma([0,t])\cup\gamma([0,t])^{*}$, where $K^{*}$ denotes the reflection of
$K$ with respect to the real axis. The coefficient $b(t)>0$ is called the
half-plane capacity of $\gamma([0,t])$, denoted by
$\text{hcap}(\gamma([0,t]))$. Then, the chordal Loewner differential equation
reads for the initial value problem

(1.1) $\dfrac{\partial}{\partial
t}g_{t}(z)=\dfrac{b^{\prime}(t)}{g_{t}(z)-\lambda(t)},\quad g_{0}(z)=z,$

for all $z\in H_{t}$ and $0\leq t<T$, where $\lambda(t)$ is a continuous real-
valued function of $t$, given as $\lambda(t)=g_{t}^{-1}(\gamma(t))$, called
the driving function. In the literature, the half-plane capacity mostly
appears as $b(t)=2t$, which is possible by means of a time reparameterization.
See [10] for a detailed derivation of equation (1.1).

In the opposite direction, if we consider the initial value problem (1.1), we
then let $T_{z}$ be the supremum of all $t$, such that the solution to the
equation is well defined and $g_{t}(z)\in\mathbb{H}$ for all $t\leq T_{z}$.
Then, the domain $H_{t}:=\\{z\in\mathbb{H}:T_{z}>t\\}$ is simply connected,
$K_{t}:=\mathbb{H}\setminus H_{t}$ is compact and $\mathbb{H}\setminus K_{t}$
is also simply connected. Moreover, $g_{t}$ maps $H_{t}$ conformally onto
$\mathbb{H}$, satisfying the hydrodynamic condition. We refer to
$(K_{t})_{t}$, as the compact hulls generated by (1.1); see [5, Chapter 4] for
details. It is not true, in general, that an arbitrary continuous function
$\lambda(t)$ produces hulls that are curves. Several authors have studied the
relation between a driving function and the corresponding hulls, showing that
Lip-$\frac{1}{2}$ driving functions, with sufficiently small
$\text{Lip}(\frac{1}{2})$-norm, produce quasi-slit domains (see e.g. [6], [9]
and [13]).

Turning back to equation (1.1), one can consider its multiple, finite or
infinite, curve version. For instance, as we see in [14], given $n$ disjoint
Jordan curves that emanate from the real line towards the upper half-plane,
the corresponding Loewner flow is produced by the driving functions
$\lambda_{1},\dots,\lambda_{n}$. Furthermore, in [15], the authors show that
if the hulls are made up of infinitely many slits $\Gamma_{n}$ parameterized
in $[0,1]$ such that each curve $\Gamma_{j}(t)$ can be separated from the
closure of $\bigcup_{n\neq j}\Gamma_{j}(t)$ at each time $t$, by open sets,
then the Loewner equation is written as

(1.2) $\dfrac{\partial g}{\partial
t}(z,t)=\sum_{n=1}^{+\infty}\dfrac{b_{n}(t)}{g(z,t)-\lambda_{n}(t)},$

for $z\in\mathbb{H}\setminus\bigcup_{n=1}^{+\infty}\Gamma_{n}(t)$ and a.e
$0\leq t\leq 1$, where
$\sum_{n=1}^{+\infty}b_{n}(t)=\partial_{t}\textrm{hcap}(\bigcup_{n=1}^{+\infty}\Gamma_{n}(t))$.

To the authors’ best knowledge, explicit solutions to the Loewner equation,
which involve infinitely many slits, do not appear in the literature often. In
this article, our goal is to present Loewner flows driven by infinitely many
driving functions of the form $\lambda_{n}(t)=k_{n}\sqrt{1-t}$, by solving the
corresponding differential equation. Single-slit versions have been studied in
[4] and [7], where the authors begin with the driving function
$\lambda(t)=k\sqrt{1-t}$, and show that the geometry of the associated slit
varies according to $k$. In [16], the first author generalizes their result in
the multi-slit version, by considering $n$ driving functions of the form
$k_{j}\sqrt{1-t}$, to find four different geometric possibilities for the
resulting slits. More specifically, they either spiral about some point of the
upper half-plane, or intersect the real line non-tangentially (each slit by a
different angle), tangentially (all slits by the same angle, either $0$ or
$\pi$) or orthogonally (all slits by angle $\frac{\pi}{2}$). In this article,
we shall extend the preceding result in the case of infinitely many slits and
shall conclude that the same geometric possibilities occur.

To start with, we set our configuration by choosing a summable sequence
$(b_{n})_{n\geq 1}$ of positive numbers and a sequence of distinct real points
$(k_{n})_{n\geq 1}$ ordered in such a way that $\mathbb{R}$ can be written as
a countable union of bounded intervals of the form $[k_{m},k_{m^{\prime}}]$,
not containing in their interior any of the $k_{n}$’s, and unbounded intervals
of the form $(-\infty,k_{m}]$ or $[k_{m^{\prime}},+\infty)$, if such intervals
exist, again not containing any other point $k_{n}$. We formally write the
last condition for the sequence $(k_{n})_{n\geq 1}$ as

(1.3)
$\begin{split}\mathbb{R}=\overline{I_{-}}\cup\bigcup_{j=1}^{+\infty}\overline{I_{j}}\cup\overline{I_{+}},\hskip
4.2679pt\text{where $I_{j}$ are defined for all}\ j\geq 1,\text{such
that}\hskip 28.45274pt\\\ \text{there exists some}\hskip
4.2679ptj^{\prime}\neq j:I_{j}=(k_{j},k_{j^{\prime}}),\hskip 4.2679pt\text{so
that}\hskip 4.2679ptk_{n}\notin I_{j},\forall n\geq 1,\hskip 4.2679pt\hskip
24.18483pt\\\ I_{-}=(-\infty,\min_{n\geq 1}{k_{n}})\hskip
4.2679pt\text{and}\hskip 4.2679ptI_{+}=(\max_{n\geq 1}{k_{n}},+\infty),\hskip
4.2679pt\text{with}\hskip 4.2679ptk_{n}\notin I_{\pm},\forall n\geq 1.\hskip
14.22636pt\\\ \text{Furthermore, we assume that}\hskip 4.2679ptd:=\inf_{n\geq
1}\left|I_{n}\right|>0.\hskip 113.81102pt\end{split}$

Note that the left endpoint of $I_{j}$, determines its enumeration and note,
also, that depending on the choice of the sequence $(k_{n})_{n\geq 1}$, either
one of the two unbounded sets $I_{-}$ or $I_{+}$ might be empty. In
particular, if $I_{+}$ is not empty, then the sequence $(k_{n})_{n\geq 1}$ has
a maximum, say $k_{N}$. In this case, the interval $I_{N}$ as described above
ceases to exist, but this omission does not impact our study.

Then, we write equation (1.2) as

(1.4) $\dfrac{\partial g}{\partial
t}(z,t)=\sum_{n=1}^{+\infty}\dfrac{2b_{n}}{g(z,t)-k_{n}\sqrt{1-t}}$

with $g(z,0)=z$. The technique to solve this equation is straightforward and
involves the transformation $\hat{g}(z,t)=\frac{g(z,t)}{\sqrt{1-t}}$, which,
as we shall see later on, allows us to transform (1.4) into the separable
differential equation
$\frac{d\hat{g}}{P_{\mathbb{H}}(\hat{g})}=\frac{dt}{2(1-t)}$, where

$P_{\mathbb{H}}(z):=z+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{z-k_{n}}.$

It turns out that the geometric properties of $g$ depend on the nature of the
roots of the auxiliary function $P_{\mathbb{H}}$. We can already see that for
each $m\in\mathbb{N}$ and for $x\in\mathbb{R}$, $\lim_{x\rightarrow
k_{m}^{+}}P_{\mathbb{H}}(x)=+\infty$ and $\lim_{x\rightarrow
k_{m}^{-}}P_{\mathbb{H}}(x)=-\infty$, which implies that there exists some
$\lambda_{m}\in I_{m}$, real root of $P_{\mathbb{H}}$, so that
$P^{\prime}_{\mathbb{H}}(\lambda_{m})<0$. For the sake of clarity, we proceed
to the following definition.

######  Definition 1.

We characterize any root $\rho\in\mathbb{R}$ of $P_{\mathbb{H}}$, satisfying
$P_{\mathbb{H}}^{\prime}(\rho)<0$, as a standard root of $P_{\mathbb{H}}$. Via
the note above, we may see that each bounded interval $I_{m}$ contains at
least one standard root.

A basic part of our analysis is to decipher what other zeroes might exist. In
Section 3, we uncover the properties of $P_{\mathbb{H}}$, which will allow us
to determine exactly the possible additional roots, apart from the standard
roots $\lambda_{m}$ mentioned above. For example, if we consider the case when
only finitely many $b_{n}$’s are non-zero, then $P_{\mathbb{H}}$ is a rational
function and thus its roots are determined by a polynomial of finite degree.
Hence, with the use of the fundamental theorem of algebra, we can distinguish
between four cases for the roots, namely one complex root (and its conjugate),
or a multiple real root of degree 2 or 3, or distinct real roots; see [16] for
details. In the case of infinitely many driving functions, the corresponding
auxiliary function $P_{\mathbb{H}}$ is no longer rational and hence the
aforementioned analysis no longer applies. Therefore, we come up with
different strategies and we use complex analytic methods to deduce our
results. It is our intention, at first, to prove that the same four cases are
the only possibilities, as they appear in the proposition below.

######  Proposition 2.

(1) If there exists a complex root $\beta\in\mathbb{H}$ of $P_{\mathbb{H}}$,
then $\beta,\bar{\beta}$ are the unique roots in
$\mathbb{C}\setminus\mathbb{R}$, they are simple and each bounded interval
$I_{j}$ (as described in (1.3)) has exactly one root of $P_{\mathbb{H}}$,
which is the standard root $\lambda_{j}$. If there exist unbounded intervals,
they contain no real roots of $P_{\mathbb{H}}$.

(2) If some interval $I_{j}$ contains distinct simple real roots of
$P_{\mathbb{H}}$, then $I_{j}$ has exactly three distinct roots, two of which
are standard, if it is bounded, or two distinct roots, one of which is
standard, if it is unbounded. Each other interval $I_{k}$ has exactly one
root, which is the standard, if $I_{k}$ is bounded and none if $I_{k}$ is
unbounded, and $P_{\mathbb{H}}$ has no complex roots.

(3) If there exists a multiple real root $\rho_{0}$ of $P_{\mathbb{H}}$, lying
either in a bounded or in an unbounded interval (assuming that such an
unbounded interval exists), then $P_{\mathbb{H}}$ has only real roots and we
have the following cases:

  1. (a)

either $\rho_{0}$ is a double root and each bounded interval $I_{j}$ has
exactly one simple root, the standard root $\lambda_{j}$, while if there exist
unbounded intervals, they contain no roots (except possibly $\rho_{0}$),

  2. (b)

or $\rho_{0}$ is a triple root which can only lie in some bounded interval
$I_{m}$ and coincide with the standard root $\lambda_{m}$. Each other bounded
interval $I_{j}$ has exactly one simple root, the standard root $\lambda_{j}$,
while if there exist unbounded intervals, they contain no roots.

We summarize by presenting the main result of this work. But before that, let
us point out that it is useful to consider the inverse functions
$f(\cdot,t):=f_{t}=g_{t}^{-1}:\mathbb{H}\overset{onto}{\longrightarrow}H_{t}$.
It is then direct to see by equation (1.4), that $f$ satisfies the PDE in
$\mathbb{H}\times[0,1)$,

(1.5) $\dfrac{\partial f}{\partial
t}(z,t)=-f^{\prime}(z,t)\sum_{n=1}^{+\infty}\dfrac{2b_{n}}{z-k_{n}\sqrt{1-t}},$

with initial value $f(z,0)=z$. We refer to the solution of the preceding
equation as the chordal Loewner flow. Having established all possibilities for
the roots of $P_{\mathbb{H}}$ in Section 3, we then proceed to Section 4,
where we explicitly solve equation $(\ref{PDE1})$ and visualize the geometry
of the flow. In addition, we are going to describe its asymptotic behavior as
$t\rightarrow 1^{-}$. This behavior has been studied extensively in the past.
However, in this article, we provide another approach towards the aim of
tackling the problem. More specifically, we make use of a conformal invariant,
the harmonic measure. In fact, harmonic measure and its conformally invariant
nature prove to be a powerful tool for studying how a trajectory behaves
asymptotically.

We are, now, ready to state our main result.

######  Theorem 3.

Consider a summable sequence of positive numbers $(b_{n})_{n\geq 1}$ and a
sequence of distinct real point $(k_{n})_{n\geq 1}$, satisfying condition
$(\ref{initial condition})$. Then, the initial value problem $(\ref{PDE1})$
admits a unique solution in $\mathbb{H}\times[0,1)$ and we distinguish the
following cases:

  1. (1)

If $P_{\mathbb{H}}$ has a complex root $\beta\in\mathbb{H}$, then the Loewner
flow is of the form

$f(z,t)=h^{-1}\left((1-t)^{\alpha
e^{-i\psi}}h\left((1-t)^{-\frac{1}{2}}z\right)\right),$

where $h$ maps the upper half plane onto the complement of infinitely many
logarithmic spirals of angle $-\psi$, where $\alpha$ and $\psi$ depend on the
sequences $(k_{n})_{n\geq 1}$ and $(b_{n})_{n\geq 1}$.

  2. (2)

If $P_{\mathbb{H}}$ has three distinct real roots, in some interval described
in (1.3), then the Loewner flow is of the form

$f(z,t)=h^{-1}\left((1-t)^{\alpha}h\left((1-t)^{-\frac{1}{2}}z\right)\right),$

where $h$ is a Schwarz-Cristoffel map of the upper half plane, that maps
$\mathbb{H}$ onto $\mathbb{H}$ minus infinitely many line segments emanating
from the origin.

  3. (3)

If $P_{\mathbb{H}}$ has a multiple root, either double or triple, then the
Loewner flow is of the form

$f(z,t)=h^{-1}\left(\dfrac{1}{2}\log(1-t)+h\left((1-t)^{-\frac{1}{2}}z\right)\right),$

where $h$ is a univalent map of the upper half plane, that maps $\mathbb{H}$
onto:
(a) either a horizontal half-plane, minus infinitely many half-lines parallel
to $\mathbb{R}$, extending to the point at infinity from the left, if the root
is double,
(b) or the complement of infinitely many half-lines parallel to $\mathbb{R}$,
extending to the point at infinity from the left, if the root is triple.

Furthermore, for all $n\geq 1$, we have that the trajectories of the driving
functions $\hat{\gamma}^{(n)}:=\\{f(k_{n}\sqrt{1-t},t):\ t\in[0,1)\\}$, are
smooth curves of $\mathbb{H}$ starting at $k_{n}$, that spiral about the point
$\beta$ in case $(1)$, intersect $\mathbb{R}$ at one of the real roots non-
tangentially in case $(2)$ and intersect $\mathbb{R}$ at the multiple root
tangentially in case $(3a)$ or orthogonally in case $(3b)$.

Figure 1. The geometric behaviour of the tip points $f(k_{n}\sqrt{1-t},t)$ for
each case of Theorem 3.

The structure of the article is as follows: In Section 2, we collect the
preliminary results that will be needed during the course of the proofs. Then,
in Section 3, we will proceed to a complete study of the function
$P_{\mathbb{H}}$ and its nature by examining its roots. Finally, in Section 4,
we shall state and prove a series of lemmas and propositions, whose
combination will lead to Theorem 3.

##  2\. Basic tools and preliminaries

###  2.1. Theory of conformal mappings

We begin by presenting some basic definitions about spirallike domains, that
will be necessary for later. A logarithmic spiral of angle
$\psi\in(-\frac{\pi}{2},\frac{\pi}{2})$ in the complex plane, that joins the
origin with infinity and passes from some point $w_{0}\neq 0$, is defined as
the curve with parameterization $S:w=w_{0}\text{exp}(-e^{i\psi}t)$,
$-\infty\leq t\leq+\infty$.

######  Definition 4.

(1) A simply connected domain $D$, that contains the origin, is said to be
$\psi$-spirallike (with respect to $0$), if for any point $w_{0}\in D$, the
logarithmic spiral $S:w=w_{0}\exp(-e^{i\psi}t)$, $0\leq t\leq+\infty$, is
contained in $D$.

(2) A univalent function $f\in H(\mathbb{H})$, with $f(\beta)=0$ for some
$\beta\in\mathbb{H}$, is said to be $\psi$-spirallike (with respect to
$\beta$), if it maps the upper half plane onto a $\psi$-spirallike domain $D$
(with respect to 0).

For more details on spirallike functions, the interested reader may refer to
[3, §2.7]. Note that $0$-spirals are half-lines emanating from the origin and
extending to infinity. We refer to $0$-spirallike domains/functions as
starlike domains/functions. The following theorem gives an analytic
characterization of spirallike mappings.

######  Theorem 5.

Let $f\in H(\mathbb{H})$, with $f^{\prime}(\beta)\neq 0$ and $f(z)=0$ if and
only if $z=\beta$. Then, $f$ is $\psi$-spirallike, if and only if

$\text{Im}\left(e^{-i\psi}\dfrac{(z-\beta)(z-\bar{\beta})f^{\prime}(z)}{f(z)}\right)>0,$

for all $z\in\mathbb{H}$.

###### Proof.

By [3, Theorem 2.19], a function $g\in H(\mathbb{D})$, with $g^{\prime}(0)\neq
0$ and $g(z)=0$ if and only if $z=0$, is $\psi$-spirallike, if and only if

$\mathrm{Re}\left(e^{-i\psi}\dfrac{zg^{\prime}(z)}{g(z)}\right)>0,$

for all $z\in\mathbb{D}$. Considering the Möbius transform
$T(z)=\frac{z-\beta}{z-\bar{\beta}}:\mathbb{H}\rightarrow\mathbb{D}$ and
applying the preceding result to the function $g:=f\circ T^{-1}$, the desired
inequality follows. ∎

Aside from that and returning to our setting, as we mentioned in the
introduction, when $b_{n}$ is non-zero for finitely many $n\geq 1$,
$P_{\mathbb{H}}$ is a rational function. As we shall see in Section 4, in
order to solve PDE $(\ref{PDE1})$, we need to apply partial fraction
decomposition on $\frac{1}{P_{\mathbb{H}}}$, which can be done since this is a
rational function as well. However, the same method is not applicable when
$b_{n}>0$, for all $n\geq 1$. For this reason, we are going to require the
following theorem, which is an application of the residue calculus (see [8,
Section II.9]) for details.

######  Theorem 6.

[8, Theorem II.2.7] Let $f$ be an analytic function in $\mathbb{C}$, whose
only singularities are poles in the finite plane, say $(\lambda_{n})_{n\geq
1}$, and let $G_{n}$ be the principal part of $f$ at each $\lambda_{n}$,
respectively. Let $(L_{n})_{n\geq 1}$ be a sequence of contours which satisfy
the following:

  1. (i)

each contour $L_{n}$ contains finitely many poles,

  2. (ii)

$0\in L_{n}\subset\text{Int}(L_{n+1})$ for all $n\geq 1$,

  3. (iii)

$r_{n}:=\text{dist}(0,L_{n})\rightarrow+\infty$.

If $\limsup_{n\rightarrow\infty}\int_{L_{n}}|f(\zeta)|d\zeta<\infty$, then

$f(z)=\sum_{n=1}^{+\infty}G_{n}(z)$

and the convergence is uniform on compacta.

###  2.2. Harmonic Measure

During the course of several proofs, we are going to utilize one conformal
invariant, the harmonic measure. An introductory presentation of its rich
theory may be found in [12]. For the purposes of the present article, we will
only review some basic facts. Let $\Omega\subsetneq\mathbb{C}$ be a domain
with non-polar boundary. Let $E$ be a Borel subset of $\partial\Omega$. Then,
the harmonic measure of $E$ with respect to $\Omega$ is exactly the solution
of the generalized Dirichlet problem for the Laplacian in $\Omega$ with
boundary function equal to $1$ on $E$ and to $0$ on $\partial\Omega\setminus
E$. For the harmonic measure of $E$ with respect to $\Omega$ and for
$z\in\Omega$, we use the notation $\omega(z,E,\Omega)$. It is known that for a
fixed $z\in\Omega$, $\omega(z,\cdot,\Omega)$ is a Borel probability measure on
$\partial\Omega$.

As we already mentioned, the harmonic measure is conformally invariant. In
addition, it has a very useful monotonicity property. More specifically, let
$\Omega_{1}\subset\Omega_{2}\subsetneq\mathbb{C}$ be two domains with non-
polar boundaries and let $E\subset\partial\Omega_{1}\cap\partial\Omega_{2}$.
Then,

$\omega(z,E,\Omega_{1})\leq\omega(z,E,\Omega_{2}),\quad\text{for all
}z\in\Omega_{1}.$

Furthermore, later on we will need some formulas for harmonic measure in
certain particular domains (see [12, p.100]). Let $[a,b]\subset\mathbb{R}$ and
let $z\in\mathbb{H}$. Then,

$\omega(z,[a,b],\mathbb{H})=\frac{1}{\pi}\arg\left(\frac{z-b}{z-a}\right).$

As a consequence, given any curve $\gamma:[0,+\infty)\to\mathbb{H}$ satisfying
$\lim_{t\to+\infty}\gamma(t)=\infty$, it can be easily calculated that
$\lim_{t\to+\infty}\omega(\gamma(t),[a,b],\mathbb{H})=0$.

Finally, for the angular domain
$U_{\alpha,\beta}:=\\{w\in\mathbb{C}:\alpha<\arg w<\beta\\}$, where
$0\leq\alpha<\beta<2\pi$ (or $-\pi\leq\alpha<\beta<\pi$), and for $z\in
U_{\alpha,\beta}$, we have

$\omega(z,\\{w\in\mathbb{C}:\arg
w=\alpha\\},U_{\alpha,\beta})=\frac{\beta-\arg
z}{\beta-\alpha}=1-\omega(z,\\{w\in\mathbb{C}:\arg
w=\beta\\},U_{\alpha,\beta}).$

######  Remark 7.

Suppose that $E\subset\partial\mathbb{D}$, where $\mathbb{D}$ is the unit
disk, is a circular arc with endpoints $a,b$. Then, we know (see e.g. [2,
p.155]) that the level set

$D_{k}=\left\\{z\in\mathbb{D}:\omega(z,E,\mathbb{D})=k\right\\},\quad
k\in(0,1),$

is a circular arc (or a diameter in case $k=\frac{1}{2}$ and $E$ is a half-
circle) inside $\mathbb{D}$ with endpoints $a,b$ that intersects
$\partial\mathbb{D}$ with angles $k\pi$ and $(1-k)\pi$.

######  Remark 8.

An important piece of information with regard to harmonic measure is its
probabilistic interpretation. In particular, the quantity $\omega(z,E,\Omega)$
represents the probability of a Brownian motion starting at $z$ to exit the
domain $\Omega$ for the first time passing through $E$. This thought, combined
with the previous remark, justifies intuitively our usage of harmonic measure
in the study of angles.

######  Remark 9.

In many cases, instead of Borel sets, we use sets of prime ends for the
harmonic measure. This is possible through Carathéodory’s Theorem concerning
the extension of the Riemann mapping theorem to the boundary (see [11, Chapter
9] for details on prime ends and the boundary behavior of conformal mappings).
To be more exact, suppose that $\Omega\subsetneq\mathbb{C}$ is a simply
connected domain with non-polar boundary and $f:\mathbb{D}\to\Omega$ is a
corresponding Riemann mapping. Suppose that $E\subset\partial\mathbb{D}$ is
Borel. Then, this set $E$ corresponds through $f$ to a set of prime ends
$\hat{f(E)}$ of $\Omega$ (if $\Omega$ is not a Jordan domain, then there might
not be an one-to-one correspondence between $\hat{f(E)}$ and points of
$\partial\Omega$ that are limit points of $\lim\limits_{z\to\zeta}f(z)$,
$\zeta\in E$). Therefore, by the conformal invariance of the harmonic measure
and Carathéodory’s Theorem, we may write

$\omega(z,E,\mathbb{D})=\omega(f(z),\hat{f(E)},\Omega),\quad z\in\mathbb{D}.$

##  3\. Properties of the driving function

Given a summable sequence of non-negative numbers $(b_{n})_{n\geq 1}$ and a
sequence of real points $(k_{n})_{n\geq 1}$, the function

$P_{\mathbb{H}}(z):=z+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{z-k_{n}}$

is quickly verified to converge locally uniformly in
$\mathbb{C}\setminus\overline{(k_{n})_{n\geq 1}}$. As we mentioned previously,
the key role with regard to the geometry of the slits is played by the roots
of $P_{\mathbb{H}}$. If the non-zero terms of the sequence $(b_{n})_{n\geq 1}$
are finitely many, then $P_{\mathbb{H}}$ is a rational function. Therefore, we
are able to use the fundamental theorem of algebra in order to count its roots
and then distinguish the cases when it has complex roots or only real roots.
But since we do not have this theorem at our disposal in this case, we need to
come up with different techinques. We start with the following lemma.

######  Lemma 10.

Given the parameters above we have the following expressions:

(1) If $F(z,w):=\dfrac{P_{\mathbb{H}}(z)}{(z-w)(z-\bar{w})}$, then

$F(z,w)=\left(\dfrac{P_{\mathbb{H}}(w)}{z-w}-\dfrac{P_{\mathbb{H}}(\bar{w})}{z-\bar{w}}\right)\dfrac{1}{w-\bar{w}}+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{(z-k_{n})|w-k_{n}|^{2}},$

for all $z,w\in\mathbb{C}\setminus(k_{n})_{n\geq 1}$ with $z\neq w$.

(2) If
$H(z,\lambda_{1},\lambda_{2}):=\dfrac{P_{\mathbb{H}}(z)}{(z-\lambda_{1})(z-\lambda_{2})}$,
then

$H(z,\lambda_{1},\lambda_{2})=\left(\frac{P_{\mathbb{H}}(\lambda_{1})}{z-\lambda_{1}}-\frac{P_{\mathbb{H}}(\lambda_{2})}{z-\lambda_{2}}\right)\dfrac{1}{\lambda_{1}-\lambda_{2}}+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{(z-k_{n})(\lambda_{1}-k_{n})(\lambda_{2}-k_{n})},$

for all $z,\lambda_{1},\lambda_{2}\in\mathbb{C}\setminus(k_{n})_{n\geq 1}$
with $z\neq\lambda_{1},\lambda_{2}$ and $\lambda_{1}\neq\lambda_{2}$.

(3) If $G(z,\lambda):=\dfrac{P_{\mathbb{H}}(z)}{(z-\lambda)^{2}}$, then

$G(z,\lambda)=\dfrac{P_{\mathbb{H}}^{\prime}(\lambda)}{z-\lambda}+\dfrac{P_{\mathbb{H}}(\lambda)}{(z-\lambda)^{2}}+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{(z-k_{n})(\lambda-
k_{n})^{2}},$

for all $z,\lambda\in\mathbb{C}\setminus(k_{n})_{n\geq 1}$ with
$z\neq\lambda$.

###### Proof.

We commence by proving relation (2). By executing consecutive partial fraction
decompositions, we have that

$\displaystyle
H(z,\lambda_{1},\lambda_{2})=\dfrac{z}{(z-\lambda_{1})(z-\lambda_{2})}+\sum_{n=1}^{+\infty}\dfrac{\frac{4b_{n}}{z-k_{n}}}{(z-\lambda_{1})(z-\lambda_{2})}$
$\displaystyle=\dfrac{1}{\lambda_{1}-\lambda_{2}}\left(\dfrac{\lambda_{1}}{z-\lambda_{1}}-\dfrac{\lambda_{2}}{z-\lambda_{2}}+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{(z-k_{n})(z-\lambda_{1})}-\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{(z-k_{n})(z-\lambda_{2})}\right)$
$\displaystyle=\dfrac{1}{\lambda_{1}-\lambda_{2}}\left[\dfrac{\lambda_{1}}{z-\lambda_{1}}-\dfrac{\lambda_{2}}{z-\lambda_{2}}+\sum_{n=1}^{+\infty}\left(\dfrac{\frac{4b_{n}}{\lambda_{1}-k_{n}}}{z-\lambda_{1}}-\dfrac{\frac{4b_{n}}{\lambda_{1}-k_{n}}}{z-k_{n}}\right)-\sum_{n=1}^{+\infty}\left(\dfrac{\frac{4b_{n}}{\lambda_{2}-k_{n}}}{z-\lambda_{2}}-\dfrac{\frac{4b_{n}}{\lambda_{2}-k_{n}}}{z-k_{n}}\right)\right]$
$\displaystyle=\dfrac{1}{\lambda_{1}-\lambda_{2}}\left(\dfrac{\lambda_{1}+\sum_{n=1}^{+\infty}\frac{4b_{n}}{\lambda_{1}-k_{n}}}{z-\lambda_{1}}-\dfrac{\lambda_{2}+\sum_{n=1}^{+\infty}\frac{4b_{n}}{\lambda_{2}-k_{n}}}{z-\lambda_{2}}\right)+\dfrac{1}{\lambda_{1}-\lambda_{2}}\sum_{n=1}^{+\infty}\dfrac{\frac{4b_{n}}{\lambda_{2}-k_{n}}-\frac{4b_{n}}{\lambda_{1}-k_{n}}}{z-k_{n}}$

and the third statement follows. Notice that $F(z,w)=H(z,w,\bar{w})$ and
$G(z,\lambda)=H(z,\lambda,\lambda)$. Then, relations (1) and (3) follow. ∎

######  Definition 11.

An interval of the form $I_{n}=(k_{n},k_{n^{\prime}})$ that does not contain
any point of the sequence $(k_{n})_{n\geq 1}$ is called bounded interval of
$P_{\mathbb{H}}$. If there exists an interval of the form
$I_{-}=(-\infty,k_{n^{\prime}})$ or $I_{+}=(k_{n^{\prime}},+\infty)$ that does
not contain any point of the sequence $(k_{n})_{n\geq 1}$, then it is called
left or right unbounded interval of $P_{\mathbb{H}}$, respectively.

The preceding proposition is important, as it reveals the mechanism, according
to which, $P_{\mathbb{H}}$ can have a unique complex root, or a unique
multiple real root, or there exists a unique interval of $P_{\mathbb{H}}$
(bounded or unbounded) that contains three distinct real roots. Indeed, as we
shall see in a subsequent corollary, the left summands in the relations
$(1)-(3)$ above vanish, when $w,\lambda,\lambda_{j}$ are roots of
$P_{\mathbb{H}}$ and as a result the imaginary parts of $F,G$ and $H$ are
negative and positive in the upper half plane and in the lower half plane,
respectively, allowing us to deduce uniqueness.

######  Lemma 12.

A bounded interval of $P_{\mathbb{H}}$ contains either one or three roots of
$P_{\mathbb{H}}$, counting multiplicity. An unbounded interval of
$P_{\mathbb{H}}$ contains either none or two roots of $P_{\mathbb{H}}$,
counting multiplicity.

###### Proof.

In a bounded interval of $P_{\mathbb{H}}$, say $(k_{n},k_{n^{\prime}})$, we
observe that $\lim_{\mathbb{R}\ni x\to k_{n}^{+}}P_{\mathbb{H}}(x)=+\infty$
and $\lim_{\mathbb{R}\ni x\to{k_{n^{\prime}}^{-}}}P_{\mathbb{H}}(x)=-\infty$.
Thus, there exists some $\lambda_{n}\in(k_{n},k_{n^{\prime}})$, root of
$P_{\mathbb{H}}$. Because the third derivative of $P_{\mathbb{H}}$ is always
negative in $\mathbb{R}$, by Rolle’s theorem, we can have, counting
multiplicity, up to three real roots in $(k_{n},k_{n^{\prime}})$. In
particular, if $P_{\mathbb{H}}$ has exactly two distinct roots, then one of
them is necessarily a double root.

Besides, in case there exists a left unbounded interval, then we may compute
that $\lim_{x\to-\infty}P_{\mathbb{H}}(x)=-\infty$, whereas if there exists a
right unbounded interval, then $\lim_{x\to+\infty}P_{\mathbb{H}}(x)=+\infty$.
Using similar arguments as above, we may deduce the desired result. ∎

As a direct corollary we may now prove Proposition 2.

###### Proof of Proposition 2.

(1) Assume that $\beta\in\mathbb{H}$ is a complex root of $P_{\mathbb{H}}$.
Since the parameters $b_{n}$ and $k_{n}$ are real, $\bar{\beta}$ is also a
root. By Lemma 10, the imaginary part of $F(z,\beta)$ is negative for all
$z\in\mathbb{H}$ and positive for all $z\in-\mathbb{H}$. This implies that
$\beta,\bar{\beta}$ are simple roots of $P_{\mathbb{H}}$ and they are the
unique non-real roots. Finally, we have that $F^{\prime}(x,\beta)<0$, for all
$x\in\mathbb{R}\setminus(k_{n})_{n\geq 1}$, which means that $F(\cdot,\beta)$
is decreasing in each interval of $P_{\mathbb{H}}$ and the result follows.

(2) Assume that there exist two distinct roots $\lambda_{1},\lambda_{2}$ in
some interval of $P_{\mathbb{H}}$. Again, we have that
$\text{Im}(H(z,\lambda_{1},\lambda_{2}))\neq 0$, for all
$z\in\mathbb{C}\setminus\mathbb{R}$. Hence, $H$ and by extension
$P_{\mathbb{H}}$ cannot have non-real roots. Since $\lambda_{1},\lambda_{2}$
lie in the same interval, we have that
$(\lambda_{1}-k_{n})(\lambda_{2}-k_{n})>0$, for all $n\in\mathbb{N}$.
Therefore, $H^{\prime}(x,\lambda_{1},\lambda_{2})<0$, for all
$x\in\mathbb{R}\setminus(k_{n})_{n\geq 1}$ and this completes the proof.

(3) By Lemma 10, the imaginary part of $G(z,\rho_{0})$ is negative for all
$z\in\mathbb{H}$ and positive for all $z\in-\mathbb{H}$. Therefore, $G$ and
$P_{\mathbb{H}}$ have no roots in $\mathbb{C}\setminus\mathbb{R}$. As before,
we have that $G^{\prime}(x,\rho_{0})<0$ for all
$x\in\mathbb{R}\setminus(k_{n})_{n\geq 1}$, which means that $G(\cdot,\rho)$
is decreasing in each interval of $P_{\mathbb{H}}$. This monotony implies the
desired result.

∎

We recall at this point that by condition $(\ref{initial condition})$, the
terms of $(k_{n})_{n\geq 1}$ are chosen in such a way that $d:=\inf_{n\neq
m}|k_{n}-k_{m}|>0$. This allows for the very useful property that for any
$\epsilon\leq\frac{d}{4}$, $P_{\mathbb{H}}$ is convergent uniformly in the
domain $\mathbb{C}\setminus\bigcup_{n\geq 1}D(k_{n},\epsilon)$. Under this
assumption, we can prove the following result.

######  Proposition 13.

Consider a sequence of non-negative numbers $(b_{n})_{n\geq 1}$ and a sequence
of real points $(k_{n})_{n\geq 1}$ satisfying condition $(\ref{initial
condition})$ and denote by $(\lambda_{n})_{n\geq 1}$ the standard roots of
Proposition 2. We distinguish the following cases in accordance with
Proposition 2:

(1) Let $\beta\in\mathbb{H}$ be a complex root of $P_{\mathbb{H}}$ (along with
its conjugate $\bar{\beta}$) and define
$\psi:=\mathrm{Arg}(P_{\mathbb{H}}^{\prime}(\beta))$. Then,

(3.1)
$\dfrac{1}{P_{\mathbb{H}}(z)}=\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\beta)}}{z-\beta}+\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\bar{\beta})}}{z-\bar{\beta}}+\sum_{n=1}^{+\infty}\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}}{z-\lambda_{n}}.$

Furthermore, we have that $\psi\in(-\frac{\pi}{2},\frac{\pi}{2})$ and

$\sum_{n=1}^{+\infty}\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}+\dfrac{2\cos(\psi)}{|P_{\mathbb{H}}^{\prime}(\beta)|}=1.$

(2) Assume that $P_{\mathbb{H}}$ has only distinct real roots in some interval
of $P_{\mathbb{H}}$, either three in a bounded one, say
$\lambda_{j}<\rho_{1}<\rho_{2}$, or two in an unbounded, say
$\rho_{1},\rho_{2}$. Then,

(3.2)
$\dfrac{1}{P_{\mathbb{H}}(z)}=\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\rho_{1})}}{z-\rho_{1}}+\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\rho_{2})}}{z-\rho_{2}}+\sum_{n=1}^{+\infty}\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}}{z-\lambda_{n}}.$

Furthermore,

$\sum_{n=1}^{+\infty}\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}+\dfrac{1}{P_{\mathbb{H}}^{\prime}(\rho_{1})}+\dfrac{1}{P_{\mathbb{H}}^{\prime}(\rho_{2})}=1.$

(3a) Let $\rho_{0}\in\mathbb{R}$ be a real double root of $P_{\mathbb{H}}$.
Then,

(3.3)
$\dfrac{1}{P_{\mathbb{H}}(z)}=\dfrac{\frac{2}{P_{\mathbb{H}}^{(2)}(\rho_{0})}}{(z-\rho_{0})^{2}}+\dfrac{-\frac{2P_{\mathbb{H}}^{(3)}(\rho_{0})}{3(P_{\mathbb{H}}^{(2)}(\rho_{0}))^{2}}}{z-\rho_{0}}+\sum_{n=1}^{+\infty}\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}}{z-\lambda_{n}}.$

Furthermore,

$-\dfrac{2P_{\mathbb{H}}^{(3)}(\rho_{0})}{3\left(P_{\mathbb{H}}^{(2)}(\rho_{0})\right)^{2}}+\sum_{n=1}^{+\infty}\dfrac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}=1$

(3b) Let $\rho_{0}\in\mathbb{R}$ be a real triple root of $P_{\mathbb{H}}$.
Then, $\rho_{0}=\lambda_{m}$ for some $m\in\mathbb{N}$ and

(3.4)
$\dfrac{1}{P_{\mathbb{H}}(z)}=\dfrac{\frac{6}{P_{\mathbb{H}}^{(3)}(\rho_{0})}}{(z-\rho_{0})^{3}}+\dfrac{-\frac{3P_{\mathbb{H}}^{(4)}(\rho_{0})}{2\left(P_{\mathbb{H}}^{(3)}(\rho_{0})\right)^{2}}}{(z-\rho_{0})^{2}}+2\dfrac{\left(\frac{P_{\mathbb{H}}^{(4)}(\rho_{0})}{4!}\right)^{2}-\frac{P_{\mathbb{H}}^{(3)}(\rho_{0})P_{\mathbb{H}}^{(5)}(\rho_{0})}{3!5!}}{\left(\frac{P_{\mathbb{H}}^{(3)}(\rho_{0})}{3!}\right)^{3}(z-\rho_{0})}+\sum_{n\neq
m}\dfrac{\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}}{z-\lambda_{n}}.$

Furthermore,

$2\dfrac{\left(\frac{P_{\mathbb{H}}^{(4)}(\rho_{0})}{4!}\right)^{2}-\frac{P_{\mathbb{H}}^{(3)}(\rho_{0})P_{\mathbb{H}}^{(5)}(\rho_{0})}{3!5!}}{\left(\frac{P_{\mathbb{H}}^{(3)}(\rho_{0})}{3!}\right)^{3}}+\sum_{n\neq
m}\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}=1.$

###### Proof.

We first establish the representation formulas $(\ref{complex
root})-(\ref{triple root})$. Assume, without loss of generality, that the
sequence $(k_{n})_{n\geq 1}$ accumulates at both $+\infty$ and $-\infty$. So,
we can extract a subsequence $(k_{m^{1}_{n}})_{n\geq 1}$ that increases to
$+\infty$. Let $R_{n}$ be the middle points of the intervals $I_{m^{1}_{n}}$
(see condition $(\ref{initial condition})$) and observe that $(R_{n})_{n\geq
1}$ is increasing to $+\infty$. We, now, construct a sequence of rectangles
$(L_{n})_{n\geq 1}$ in the following manner: first we consider $L_{n}$ to be
the square of center $0$ with sides parallel to the axes, passing from the
point $R_{n}$. If $-R_{n}$ happens to be the middle point of an interval
$(k_{m_{n_{2}}^{2}},k_{m_{n_{1}}^{2}})$, we leave $L_{n}$ as is. If not, we
transform $L_{n}$ into a rectangle passing from the points $R_{n},\pm iR_{n}$
and $-R_{n}-\epsilon_{n}$ (or $-R_{n}+\epsilon_{n}$), where
$\epsilon_{n}<\frac{d}{2}$ is picked in such a way that $|\zeta-
k_{m}|\geq\frac{d}{2}$, for all $\zeta\in L_{n}$ and every $m\geq 1$. We then
have that $L_{n}\subset\mathrm{Int}(L_{n+1})$ and the perimeter of $L_{n}$ is
less than $8R_{n}+2\epsilon_{n}$, for all $n\geq 1$. We may assume that for
each case described above, the additional roots $\beta,\bar{\beta}$ or
$\rho_{j}$, lie in $\mathrm{Int}(L_{1})$ and
$R_{1}>\sigma:=\frac{2}{d}\sum_{n=1}^{+\infty}4b_{n}$. We then have that
$|P_{\mathbb{H}}(\zeta)|=\left|\zeta+\sum_{m=1}^{+\infty}\frac{4b_{m}}{\zeta-
k_{m}}\right|\geq R_{n}-\frac{d}{2}-\sigma$, for all $\zeta\in L_{n}$. This
implies that
$\limsup_{n\rightarrow+\infty}\int_{L_{n}}\frac{1}{|P_{\mathbb{H}}(\zeta)|}|d\zeta|<+\infty$.
For every $n\geq 1$, recall that $\lambda_{n}$ is a simple root of
$P_{\mathbb{H}}$, hence a simple pole $\frac{1}{P_{\mathbb{H}}}$ with residue
$\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}$. For the cases (1) and (2),
the roots $\beta,\bar{\beta}$ and $\rho_{j}$, $j=1,2$, are also simple poles
of $\frac{1}{P_{\mathbb{H}}}$, while in cases (3a) and (3b), the root
$\rho_{0}$ is a pole of second or third order, respectively. Therefore,
writing down the principal part of $\frac{1}{P_{\mathbb{H}}}$ around each
pole, we apply Theorem 6 and we conclude relations $(\ref{complex
root})-(\ref{triple root})$. The convergence is locally uniform.

Next, we shall establish the second part of (1). Using Lemma 10, we observe
that
$\mathrm{Im}F(\beta,\beta)=\mathrm{Im}\frac{|P_{\mathbb{H}}^{\prime}(\beta)|e^{i\psi}}{2i\mathrm{Im}\beta}<0$
and hence $\cos(\psi)>0$. To prove the last statement, we have that

$\dfrac{iy}{P_{\mathbb{H}}(iy)}=\dfrac{1}{P_{\mathbb{H}}^{\prime}(\beta)}\dfrac{iy}{iy-\beta}+\dfrac{1}{P_{\mathbb{H}}^{\prime}(\bar{\beta})}\dfrac{iy}{iy-\bar{\beta}}+\sum_{n=1}^{+\infty}\dfrac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}\dfrac{iy}{iy-\lambda_{n}},$

for all $y>0$. At first, we need to show that the
$(1/P^{\prime}_{\mathbb{H}}(\lambda_{n}))_{n\geq 1}$ is summable. Taking the
real parts in the preceding relation, we have that

$\sum_{n=1}^{+\infty}\dfrac{1}{|P_{\mathbb{H}}^{\prime}(\lambda_{n})|}\dfrac{y^{2}}{y^{2}+\lambda_{n}^{2}}=\text{Re}\left(\dfrac{1}{P_{\mathbb{H}}^{\prime}(\beta)}\dfrac{iy}{iy-\beta}+\dfrac{1}{P_{\mathbb{H}}^{\prime}(\bar{\beta})}\dfrac{iy}{iy-\bar{\beta}}-\dfrac{iy}{P_{\mathbb{H}}(iy)}\right)=:\Phi(y)>0$

for all $y\geq 0$. Aiming for a contradiction, we assume that
$\sum_{n=1}^{+\infty}\frac{1}{|P^{\prime}_{\mathbb{H}}(\lambda_{n})|}=+\infty$,
we then consider some natural $N\geq 1$, such that its $N$-th partial sum is
larger than $M:=2\max_{y\geq 0}\Phi(y)<\infty$. Then, we choose $y_{0}>0$, so
that $\frac{y_{0}^{2}}{y_{0}^{2}+\lambda_{n}^{2}}\geq\frac{1}{2}$, for all
$n=1,\dots,N$. As a result,

$\Phi(y_{0})\geq\sum_{n=1}^{N}\dfrac{1}{|P_{\mathbb{H}}^{\prime}(\lambda_{n})|}\dfrac{y_{0}^{2}}{y_{0}^{2}+\lambda_{n}^{2}}\geq\dfrac{1}{2}\sum_{n=1}^{N}\dfrac{1}{|P_{\mathbb{H}}^{\prime}(\lambda_{n})|}>\dfrac{M}{2}\geq\Phi(y_{0}).$

This contradiction shows that the series is convergent, and finally, letting
$y\rightarrow+\infty$, by the uniform convergence of the series, we deduce the
second statement of $(1)$. The other cases follow similarly, so we omit their
proof. ∎

In the last proposition, we made four different assumptions regarding the
roots of $P_{\mathbb{H}}$. Obviously, each of these cases occurs depending on
the choice of the sequences $(b_{n})_{n\geq 1}$ and $(k_{n})_{n\geq 1}$. The
following proposition states that for every such choice, one of these four
cases can turn up and they are mutually exclusive, thus, exactly one of the
four holds.

######  Proposition 14.

Given a sequence of real points $(k_{n})_{n\geq 1}$ and a summable sequence of
positive numbers $(b_{n})_{n\geq 1}$, then only one of the possible cases of
Proposition 2 can occur. Therefore, either $P_{\mathbb{H}}$ has a complex root
or one of the intervals of $P_{\mathbb{H}}$ has three roots counting
multiplicity, if the interval is bounded, or two roots counting multiplicity
if it is unbounded.

###### Proof.

Assume that $P_{\mathbb{H}}$ has no complex roots. We will prove that there
exists an interval (bounded or unbounded) that contains three or two roots
counting multiplicity. Assume on the contrary that this is not true and hence
each bounded interval $I_{n}$ contains only the standard root $\lambda_{n}$
and each unbounded (if it exists) contains no roots. Then, as in the proof of
Proposition 13, we have that
$\frac{1}{P_{\mathbb{H}}(z)}=\sum_{n=1}^{+\infty}\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}\frac{1}{z-\lambda_{n}}$,
which implies that

$1=\lim_{y\rightarrow+\infty}\frac{iy}{P_{\mathbb{H}}(iy)}=\sum_{n=1}^{+\infty}\frac{1}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}<0,$

a contradiction. Therefore, there exists some interval that contains more than
one root and by Proposition 2, this is the unique interval that will contain
three roots, if it is bounded, or two if it is unbounded, counting
multiplicity.

On the other hand, if $P_{\mathbb{H}}$ has a complex root, then by Proposition
2, this root along with its conjugate are the sole complex roots and each
bounded interval of $P_{\mathbb{H}}$ has exactly one real root, whereas if
there exists an unbounded one, it has no real roots.

∎

##  4\. The chordal Loewner flow

Recall, now, that our purpose is to solve PDE (1.5) and describe its solutions
geometrically. For this, we consider the corresponding ODE

(4.1)
$\dfrac{dw}{dt}(z,t)=\sum_{n=1}^{+\infty}\dfrac{2b_{n}}{w(z,t)-k_{n}\sqrt{1-t}}$

with initial value $w(z,0)=z$. This becomes, using the transform
$v=(1-t)^{-\frac{1}{2}}w$,

(4.2)
$\dfrac{dv}{dt}(z,t)=\dfrac{1}{2(1-t)}\left(v(z,t)+\sum_{n=1}^{+\infty}\dfrac{4b_{n}}{v(z,t)-k_{n}}\right)=\dfrac{P_{\mathbb{H}}(v(z,t))}{2(1-t)}.$

This is a separable differential equation and therefore, the initial value
problem $(\ref{PDE1})$ can be solved by integrating the equation

(4.3) $\dfrac{1}{P_{\mathbb{H}}(v)}dv=\dfrac{1}{2(1-t)}dt.$

For this reason, fix some $z_{0}\in\mathbb{H}$ and consider the function
$\phi(z):=\int_{z_{0}}^{z}\frac{d\zeta}{P_{\mathbb{H}}(\zeta)}$, which is well
defined and analytic, since $\mathbb{H}$ is simply connected. The function
$\phi$ is going to play the role of the primitive of
$\frac{1}{P_{\mathbb{H}}}$. Looking at relations $(\ref{complex
root})-(\ref{triple root})$, the following lemma will be necessary for later.

######  Lemma 15.

Let $(A_{n})_{n\geq 1}$ be an absolutely summable sequence of complex numbers
and let $(\lambda_{n})_{n\geq 1}$ be a sequence of real points. Fix some
$z_{0}\in\mathbb{H}$. Then, the function

$\psi(z):=\sum_{n=1}^{+\infty}A_{n}\log\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}$

is analytic in $\mathbb{H}$ and continuous in
$\overline{\mathbb{H}}\setminus\overline{(\lambda_{n})_{n\geq 1}}$.

###### Proof.

It suffices to prove that the series converges uniformly in a compact set
$K\subset\overline{\mathbb{H}}\setminus\overline{(\lambda_{n})_{n\geq 1}}$.
Let $R>\left|z_{0}\right|$ such that $K\subset D(0,R)$ and let
$d:=\text{dist}(K,\overline{(\lambda_{n})_{n\geq 1}})$. Clearly $d>0$. For all
those $n\geq 1$ such that $\left|\lambda_{n}\right|\leq R$ and for $z\in K$,
we have that

$\dfrac{d}{\left|z_{0}\right|+R}\leq\left|\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\leq\dfrac{\left|z\right|+\left|\lambda_{n}\right|}{\left|z_{0}-\lambda_{n}\right|}\leq\dfrac{2R}{\text{Im}z_{0}}.$

Thus, there exists some $M_{1}(K)>0$ such that
$\left|\log\left|\frac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\right|\leq
M_{1}(K)$. On the other hand, for all those $n\geq 1$ such that
$\left|\lambda_{n}\right|>R$, we have that

$\left|\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|=\left|\dfrac{z-z_{0}+z_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\leq
1+\dfrac{R+\left|z_{0}\right|}{R-\left|z_{0}\right|}=\dfrac{2R}{R-\left|z_{0}\right|}.$

Now, if $R<\left|\lambda_{n}\right|\leq 2R$, then
$\left|\frac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\geq\frac{d}{\left|z_{0}\right|+2R}$
whereas if $\left|\lambda_{n}\right|>2R$, we have that
$\left|\frac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\geq\frac{\left|\lambda_{n}\right|-R}{\left|\lambda_{n}\right|+R}>\frac{1}{3}$.
In total, we may find some $M_{2}(K)>0$, such that
$\left|\log\left|\frac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\right|\leq
M_{2}(K)$. Letting $M>\max\\{M_{1}(K),M_{2}(K)\\}$, we have that

$\displaystyle\left|\sum_{n=1}^{+\infty}A_{n}\log\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|$
$\displaystyle\leq\sum_{n=1}^{+\infty}\left|A_{n}\right|\left|\log\left|\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\right|+\sum_{n=1}^{+\infty}\left|A_{n}\right|\left|\text{arg}\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right|$
$\displaystyle\leq(M+2\pi)\sum_{n=1}^{+\infty}\left|A_{n}\right|<+\infty$

and the desired result follows. ∎

The usefulness of the preceding lemma lies in the fact that depending on the
choice of the parameters, the series
$\sum_{n=1}^{+\infty}A_{n}\log(z-\lambda_{n})$ might not be convergent. We
overcome this problem by fixing some $z_{0}$ and using the function $\psi$
above. In order to study equation (1.5), we distinguish the four possible
cases of Proposition 13. According to Proposition 14, all cases can arise and
hence we study each case separate in the upcoming subsections. We fix some
arbitrary point $z_{0}\in\mathbb{H}$, according to Lemma 15.

Roots of $P_{\mathbb{H}}$ |  Geometry of the slits
---|---
$P_{\mathbb{H}}$ has a complex root $\beta\in\mathbb{H}$. |  $\gamma_{n}$ spirals about $\beta$, $\forall n\geq 1$.
$P_{\mathbb{H}}$ has a double root $\rho_{0}\in\mathbb{R}$. |  $\gamma_{n}$ intersects $\mathbb{R}$ tangentially at $\rho_{0}$, $\forall n\geq 1$.
$P_{\mathbb{H}}$ has a triple root $\rho_{0}\in\mathbb{R}$. |  $\gamma_{n}$ intersects $\mathbb{R}$ orthogonally at $\rho_{0}$, $\forall n\geq 1$.
$P_{\mathbb{H}}$ has distinct real roots and $\exists\rho_{1}\in\mathbb{R}$, satisfying $P_{\mathbb{H}}^{\prime}(\rho_{1})>0$. |  $\gamma_{n}$ intersects $\mathbb{R}$ non-tangentially at some $\rho_{1}$, $\forall n\geq 1$.
Table 1. All options for the roots of $P_{\mathbb{H}}$ (left column) and the
corresponding behaviour of the slits (right column).

###  4.1. Spirals.

At first, assume that $P_{\mathbb{H}}$ has a complex root
$\beta\in\mathbb{H}$. Utilizing Proposition 13, equation
$(\ref{differential})$ may be written as

$\left(\dfrac{1}{v-\beta}+\dfrac{e^{2i\psi}}{v-\bar{\beta}}-\sum_{n=1}^{+\infty}\dfrac{\left|\frac{P_{\mathbb{H}}^{\prime}(\beta)}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}\right|e^{i\psi}}{v-\lambda_{n}}\right)dv=\dfrac{P_{\mathbb{H}}^{\prime}(\beta)}{2(1-t)}dt.$

Setting
$\alpha_{n}:=\left|\frac{P_{\mathbb{H}}^{\prime}(\beta)}{P_{\mathbb{H}}^{\prime}(\lambda_{n})}\right|$,
it is straightforward by Proposition 13 that the sequence
$(-a_{n}e^{i\psi})_{n\geq 1}$ is absolutely summable. Hence, through a slight
reformulation of Lemma 15, the infinite product
$\prod_{n=1}^{+\infty}(\frac{z-\lambda_{n}}{z_{0}-\lambda_{n}})^{-\alpha_{n}e^{i\psi}}$
is well defined. Therefore, the preceding equation gives the implicit solution
$h(v(z,t))=(1-t)^{-\frac{P_{\mathbb{H}}^{\prime}(\beta)}{2}}h(z)$, where

(4.4)
$h(z)=e^{\phi(z)}=\dfrac{z-\beta}{z_{0}-\beta}\left(\dfrac{z-\bar{\beta}}{z_{0}-\bar{\beta}}\right)^{e^{2i\psi}}\prod_{n=1}^{+\infty}\left(\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right)^{-\alpha_{n}e^{i\psi}}$

for all $z\in\mathbb{H}$.

######  Proposition 16.

The function $h$ defined by (4.4) is analytic and univalent in the upper half-
plane. In particular, $h$ is a $\psi$-spirallike function of $\mathbb{H}$,
with $\psi\in(-\frac{\pi}{2},\frac{\pi}{2})$.

###### Proof.

As before, by Lemma 15, the infinite product converges locally uniformly,
therefore analyticity follows directly. In general, if a sequence of analytic
functions converges locally uniformly, then so does the sequence of the
derivatives. Hence, differentiating $h$, we interchange limit and derivative
and then apply Proposition 13 to get

$\dfrac{h^{\prime}(z)}{h(z)}=\dfrac{1}{z-\beta}+\dfrac{e^{2i\psi}}{z-\bar{\beta}}+\sum_{n=1}^{+\infty}\dfrac{-\alpha_{n}e^{i\psi}}{z-\lambda_{n}}=\dfrac{P^{\prime}_{\mathbb{H}}(\beta)}{P_{\mathbb{H}}(z)}$

for all $z\in\mathbb{H}$. Using Lemma 10 and keeping in mind that
$\textrm{Im}(F(z,\beta))<0$ for all $z\in\mathbb{H}$, we have that

$\text{Im}\left(e^{-i\psi}\dfrac{(z-\beta)(z-\bar{\beta})h^{\prime}(z)}{h(z)}\right)=\text{Im}\left(\dfrac{\left|P_{\mathbb{H}}^{\prime}(\beta)\right|}{F(z,\beta)}\right)>0,$

for all $z\in\mathbb{H}$. As a consequence, according to Theorem 5, $h$ is a
$\psi$-spirallike (and by extension univalent) function of the upper half-
plane. ∎

Using the preceding proposition, the univalence of $h$ shows that

$v(z,t)=h^{-1}\left((1-t)^{-\frac{P^{\prime}_{\mathbb{H}}(\beta)}{2}}h(z)\right),$

for all $z\in\mathbb{H}$. In other words, returning back to
$w=(1-t)^{\frac{1}{2}}v$ and then to $f=w^{-1}$, we have that the PDE (1.5) is
satisfied by the function

$f(z,t)=h^{-1}\left((1-t)^{\frac{P^{\prime}_{\mathbb{H}}(\beta)}{2}}h\left((1-t)^{-\frac{1}{2}}z\right)\right),$

for all $z\in\mathbb{H}$ and $t\in[0,1)$.

######  Remark 17.

Note at this point that $f$ is independent of the choice of
$z_{0}\in\mathbb{H}$. Indeed, if we denote by
$\phi_{z_{0}}(z)=\int_{z_{0}}^{z}\frac{1}{P_{\mathbb{H}}(\zeta)}d\zeta$, then
we have that $\phi^{\prime}_{z_{0}}=\phi^{\prime}_{z^{\prime}_{0}}$, for any
other choice $z_{0}^{\prime}$. Hence, $h_{z_{0}}=ch_{z_{0}^{\prime}}$, for
some constant number $c$. However, due to the conjugation formula above, we
have that

$\displaystyle f_{z_{0}}(z,t):$
$\displaystyle=h_{z_{0}}^{-1}\left((1-t)^{\frac{P^{\prime}_{\mathbb{H}}(\beta)}{2}}h_{z_{0}}\left((1-t)^{-\frac{1}{2}}z\right)\right)$
$\displaystyle=h_{z_{0}^{\prime}}^{-1}\left((1-t)^{\frac{P^{\prime}_{\mathbb{H}}(\beta)}{2}}h_{z_{0}^{\prime}}\left((1-t)^{-\frac{1}{2}}z\right)\right)=:f_{z_{0}^{\prime}}(z,t)$

and thus $f$ is well defined.

######  Lemma 18.

The function $h$ defined by $(\ref{koenigs1})$ maps the upper half-plane onto
the complement of infinitely many logarithmic spirals joining the tip points
$h(k_{n})$ to the point at infinity. Moreover, $h(x)\rightarrow\infty$, as
$x\rightarrow\infty$.

###### Proof.

By Proposition 16, it suffices to find the image of the real line under $h$.
By $(\ref{koenigs1})$, using the elementary trigonometric identities
$e^{2i\psi}+1=2\cos(\psi)e^{i\psi}$ and $e^{2i\psi}-1=2i\sin(\psi)e^{i\psi}$,
a series of straightforward computations shows that for each $m\geq 1$,
$h(x)=h(k_{m})C\exp{(e^{i\psi}S_{m}(x)-ie^{i\psi}\sum_{n=1}^{+\infty}a_{n}\mathrm{Arg}(\frac{x-\lambda_{n}}{k_{m}-\lambda_{n}}))}$,
where

$S_{m}(x)=2\log\left|\dfrac{x-\beta}{k_{m}-\beta}\right|\cos(\psi)+2\mathrm{Arg}\left(\dfrac{x-\beta}{k_{m}-\beta}\right)\sin(\psi)-\sum_{n=1}^{+\infty}a_{n}\log\left|\dfrac{x-\lambda_{n}}{k_{m}-\lambda_{n}}\right|$

and $C\in\mathbb{C}$ is some absolute constant. Assume, for the sake of
simplicity, that $C=1$ (see Remark 17). Now, let
$\lambda_{m^{*}}\in\mathbb{R}$ be the largest $\lambda_{j}$, so that
$k_{m}\in(\lambda_{m^{*}},\lambda_{m})$. Restricting this interval, we see
that $h(x)=h(k_{m})\exp(e^{i\psi}S_{m}(x))$, which implies the first part of
the lemma.

For the second part, again by straightforward calculations, we have that

$\displaystyle\log\left|h(x)\right|$
$\displaystyle=\cos(\psi)\left(2\log\left|x-\beta\right|\cos(\psi)-\sum_{n=1}^{+\infty}a_{n}\log\left|\dfrac{x-\lambda_{n}}{z_{0}-\lambda_{n}}\right|\right)$
$\displaystyle+2\cos(\psi)\sin(\psi)\mathrm{Arg}(x-\beta)+\sin(\psi)\sum_{n=1}^{+\infty}a_{n}\mathrm{Arg}\left(\dfrac{x-\lambda_{n}}{z_{0}-\lambda_{n}}\right).$

Recall that the only possible accumulation points of the sequence
$(\lambda_{n})_{n\geq 1}$ are $\pm\infty$. Let us assume, without loss of
generality, that it accumulates at both $+\infty$ and $-\infty$. We will show
that $\left|h(x)\right|\rightarrow\infty$, as $x\rightarrow+\infty$. Consider
large enough $x>0$, say $x>\mathrm{Re}\beta+1$. By the first part of
Proposition 13 and the fact the $\cos(\psi)>0$, we get that

$\displaystyle\log\left|h(x)\right|$
$\displaystyle\approx\log\left|x-\beta\right|\left|P_{\mathbb{H}}^{\prime}(\beta)\right|+\sum_{n=1}^{+\infty}a_{n}\log\left|\dfrac{x-\beta}{x-\lambda_{n}}(z_{0}-\lambda_{n})\right|$
$\displaystyle\geq\log\left|x-\beta\right|\left|P_{\mathbb{H}}^{\prime}(\beta)\right|+\sum_{n=1}^{+\infty}a_{n}\log\left|\dfrac{x-\mathrm{Re}\beta}{x-\lambda_{n}}(z_{0}-\lambda_{n})\right|,$

where the symbol $\approx$ denotes comparability. We will prove that the
preceding sum is bounded below, therefore the right-hand side tends to
infinity. To see this, we split the sum to those indices such that
$\lambda_{n}>\mathrm{Re}\beta+1$ and the rest. It is easy to see that for
those indices, we have that
$\left|\frac{x-\mathrm{Re}\beta}{x-\lambda_{n}}\right|\geq\min\\{1,\frac{1}{\lambda_{n}-\mathrm{Re}\beta-1}\\}=\frac{1}{\lambda_{n}-\mathrm{Re}\beta-1}$,
for large $n$, since $(\lambda_{n})_{n\geq 1}$ accumulate at $\infty$. This
implies that

$\sum_{\lambda_{n}\geq\mathrm{Re}\beta+1}a_{n}\log\left|\dfrac{x-\mathrm{Re}\beta}{x-\lambda_{n}}(z_{0}-\lambda_{n})\right|\geq\sum_{\lambda_{n}\geq\mathrm{Re}\beta+1}a_{n}\log\left|\dfrac{z_{0}-\lambda_{n}}{\mathrm{Re}\beta+1-\lambda_{n}}\right|.$

The case for the indices so that $\lambda_{n}<0$ is easier and one can see
that $\log|\frac{x-\mathrm{Re}\beta}{x-\lambda_{n}}|$ is positive. Combining
everything together, we deduce that $\left|h(x)\right|\rightarrow\infty$, as
$x\rightarrow+\infty$ and similarly when $x$ tends to $-\infty$, the result
follows. ∎

Figure 2. The image of $h$, when $P_{\mathbb{H}}$ has a complex root.

By Lemma 18 we see that the image of $h$ is the complement of infinitely many
spirals with tip points $h(k_{n})$, so that $h(h_{n})\rightarrow\infty$, as in
Figure 2. One can also show that there exists a spirallike sector of angle
$\psi$ and amplitude $\frac{1}{\cos(\psi)}\sum_{n=1}^{+\infty}a_{n}\pi$, so
that it does not contain any spirals $\Gamma(s)=h(k_{n})e^{e^{i\psi}s}$,
$s\in\mathbb{R}$ (see Figure 2). Such a sector is defined as the set
$\mathrm{Spir}[\psi,\alpha,\theta_{0}]=\\{e^{e^{i\psi}t+i\theta}:t\in\mathbb{R},\theta\in(\theta_{0}-\alpha,\theta_{0}+\alpha)\\}$
(see [1], p. 385). Note that in order to determine the amplitude of the
sector, this is done by computing the amplitude of the points of intersection
of the boundary spirals $S_{1}$, $S_{2}$ of the sector, with the unit circle.
Thus, if $\Gamma(s_{0,n})\in\partial\mathbb{D}$, where
$s_{0,n}=-\frac{1}{\cos(\psi)}\log\left|h(k_{n})\right|$, we then let
$\zeta_{1}:=\lim_{k_{n}\rightarrow+\infty}\Gamma(s_{0,n})$ and
$\zeta_{2}:=\lim_{k_{n}\rightarrow-\infty}\Gamma(s_{0,n})$ to be the points
$S_{1}\cap\partial\mathbb{D}$ and $S_{1}\cap\partial\mathbb{D}$ respectively.
It is, now, a matter of straightforward calculations to determine the points
$\zeta_{j}\in\partial\mathbb{D}$, thus the amplitude of the sector.

###  4.2. Non-tangential intersections.

Assume that $P_{\mathbb{H}}$ has two distinct real roots $\rho_{1},\rho_{2}$
in some interval of $P_{\mathbb{H}}$. Relabeling, if necessary, we assume that
$\lambda_{m}<\rho_{1}<\rho_{2}$ if $\rho_{1},\rho_{2}$ lie in a bounded
interval $I_{m}$, $\rho_{1}<\rho_{2}$ if they lie in the left unbounded
interval, or $\rho_{2}<\rho_{1}$ if they lie in the right unbounded interval.
This way, we ensure that
$P^{\prime}_{\mathbb{H}}(\rho_{1})>0>P^{\prime}_{\mathbb{H}}(\rho_{2})$. For
the rest we will cover the first two cases where $\rho_{1}<\rho_{2}$. Then,
the Möbius transform $T(z)=\frac{z-\rho_{2}}{z-\rho_{1}}$ maps the upper half-
plane onto itself. We shall use this fact shortly. For the case where we have
a right unbounded interval, thus $\rho_{2}<\rho_{1}$, the results of this
section follow similarly by considering $-T$. We know that
$P_{\mathbb{H}}^{\prime}(\lambda_{n})<0$ for all $n\geq 1$. We then work as
follows. With the help of Proposition 13, the ODE (4.3) is written as

$\left(\dfrac{-1}{v-\rho_{1}}+\dfrac{\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{\left|P^{\prime}_{\mathbb{H}}(\rho_{2})\right|}}{v-\rho_{2}}+\sum_{n=1}^{+\infty}\dfrac{\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{\left|P^{\prime}_{\mathbb{H}}(\lambda_{n})\right|}}{v-\lambda_{n}}\right)dv=\dfrac{-P^{\prime}_{\mathbb{H}}(\rho_{1})}{2(1-t)}dt.$

Hence, considering after integration the function

(4.5)
$h(z)=\dfrac{(z-\rho_{2})^{b}}{z-\rho_{1}}\prod_{n=1}^{+\infty}\left(\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}\right)^{a_{n}}$

for all $z\in\mathbb{H}$, where
$b:=\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{\left|P^{\prime}_{\mathbb{H}}(\rho_{2})\right|}>0$
and
$a_{n}:=\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{\left|P^{\prime}_{\mathbb{H}}(\lambda_{n})\right|}>0$,
$n\geq 1$, we deduce the implicit solution
$h(v(z,t))=(1-t)^{\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{2}}h(z)$.

######  Proposition 19.

The function $h$ defined by $(\ref{koenigs2})$ is analytic and univalent in
the upper half-plane.

###### Proof.

By Lemma 15, $h$ is well defined and analytic in the upper half-plane. To
prove that it is univalent, at first we observe that
$\frac{h^{\prime}(z)}{h(z)}=\frac{-P^{\prime}_{\mathbb{H}}(\rho_{1})}{P_{\mathbb{H}}(z)}$.
Consider, now, the Möbius transform
$T(z)=\frac{z-\rho_{2}}{z-\rho_{1}}:\mathbb{H}\rightarrow\mathbb{H}$. Taking
into account the elementary identity
$(T^{-1}(z)-\rho_{1})(T^{-1}(z)-\rho_{2})=(\rho_{2}-\rho_{1})z(T^{-1})^{\prime}(z)$,
we then deduce that

$\dfrac{(h\circ T^{-1})^{\prime}(z)}{h\circ
T^{-1}(z)}=\dfrac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{\rho_{1}-\rho_{2}}\dfrac{(T^{-1}(z)-\rho_{1})(T^{-1}(z)-\rho_{2})}{zP_{\mathbb{H}}(T^{-1}(z))}=\dfrac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{\rho_{1}-\rho_{2}}\dfrac{1}{zH(T^{-1}(z),\rho_{1},\rho_{2})}$

for all $z\in\mathbb{H}$, where $H$ is given by Lemma 10. Next, we shall show
that the above quotient has constant sign in the upper half-plane. It suffices
to show that $\text{Im}(T(z)H(z,\rho_{1},\rho_{2}))<0$ for all
$z\in\mathbb{H}$. But we observe, again by Lemma 10, that
$T(z)H(z,\rho_{1},\rho_{2})=G(z,\rho_{1})$. As a result, since $\rho_{1}$ is a
root and since $P^{\prime}_{\mathbb{H}}(\rho_{1})>0$, we have that
$\text{Im}G(z,\rho_{1})<0$. Thus, the imaginary part of the logarithmic
derivative of $h\circ T^{-1}$ has constant sign, so it is univalent and this
concludes the proof. ∎

######  Lemma 20.

The function $h$ maps the upper half-plane onto
$\mathbb{H}\setminus\bigcup_{n=1}^{\infty}[0,h(k_{n})]$, where
$(0,h(k_{n})]\subset\mathbb{H}$ is a line segment emanating from the origin to
the tip point $h(k_{n})$. Moreover, we have that $h(k_{n})\rightarrow 0$, as
$n\rightarrow+\infty$.

###### Proof.

First of all, we observe that the sequences $(\lambda_{n})_{n\geq 1}$ and
$(k_{n})_{n\geq 1}$ do not have any accumulation points in $\mathbb{R}$. So,
there exists some $\epsilon>0$ sufficiently small such that
$\lambda_{n},k_{n}\notin[\rho_{1}-\epsilon,\rho_{1}+\epsilon]$, for all
$n\in\mathbb{N}$. On the contrary, both $(\lambda_{n})_{n\geq 1}$ and
$(k_{n})_{n\geq 1}$ accumulate at $-\infty$ and/or $+\infty$. Therefore, we
are interested in the limit $\lim\limits_{\mathbb{R}\ni x\to\infty}h(x)$. To
calculate it, we consider the function
$g(x)=h(T^{-1}(x))=h(\frac{x\rho_{1}-\rho_{2}}{x-1})$,
$x\in\mathbb{R}\setminus\\{1\\}$. Obviously, $\lim\limits_{x\to
1}g(x)=\lim\limits_{x\to\pm\infty}h(x)$. After several simple computations, we
find that

$g(x)=x^{b}\frac{(x-1)^{1-b}}{(\rho_{1}-\rho_{2})^{1-b}}\prod\limits_{n=1}^{+\infty}\left(\frac{\rho_{1}-\lambda_{n}}{z_{0}-\lambda_{n}}\frac{x-T(\lambda_{n})}{x-1}\right)^{a_{n}}.$

Observe that the sequence $(T(\lambda_{n}))_{n\geq 1}$ is a bounded sequence
of real points and beacuse the series
$\sum_{n=1}^{+\infty}a_{n}\log\left|\frac{\rho_{1}-\lambda_{n}}{z_{o}-\lambda_{n}}\right|$
converges by Lemma 15, taking the absolute value, we write

$\displaystyle\log\left|g(x)\right|$
$\displaystyle=(b-1)\log\left|\rho_{1}-\rho_{2}\right|+\sum_{n=1}^{+\infty}a_{n}\log\left|\frac{\rho_{1}-\lambda_{n}}{z_{o}-\lambda_{n}}\right|+b\log\left|x\right|$
$\displaystyle+(1-b-\sum_{n=1}^{+\infty}a_{n})\log\left|x-1\right|+\sum_{n=1}^{+\infty}a_{n}\log\left|x-T(\lambda_{n})\right|.$

Now, since $T(\lambda_{n})\rightarrow 1$, there exists some $N>1$, such that
$\left|T(\lambda_{n})-1\right|<\frac{1}{2}$, for all $n\geq N$. So, for all
$x\in(\frac{1}{2},\frac{3}{2})$, we have that
$\log\left|x-T(\lambda_{n})\right|$ is negative for all $n\geq N$. Note that
$1-b-\sum_{n=1}^{+\infty}a_{n}=P_{\mathbb{H}}^{\prime}(\rho_{1})>0$, due to
Proposition 13. Gathering everything together, we obtain

$\displaystyle\log\left|g(x)\right|$
$\displaystyle=C+b\log\left|x\right|+\sum_{n=1}^{N-1}a_{n}\log\left|x-T(\lambda_{n})\right|$
$\displaystyle+P_{\mathbb{H}}^{\prime}(\rho_{1})\log\left|x-1\right|+\sum_{n=N}^{+\infty}a_{n}\log\left|x-T(\lambda_{n})\right|\rightarrow-\infty,$

as $x\to 1$. As a result, $\lim\limits_{x\to 1}g(x)=0$, which leads to
$\lim\limits_{x\to+\infty}h(x)=\lim\limits_{x\to-\infty}h(x)=0$. This implies
that $\lim_{n\to+\infty}h(k_{n})=0$, as well, and we have the desired result.

On another note, let $M>0$. Then, there exists $\delta>0$ such that
$|h(x)|<\delta$, for all $x\in(-\infty,-M]\cup[M,+\infty)$. In addition, due
to compactness and the fact that $\rho_{1}$ is the only singularity of $h$ in
$\mathbb{R}$, there also exists some $\delta^{\prime}$ such that
$|h(x)|<\delta^{\prime}$, for all
$x\in[-M,M]\setminus[\rho_{1}-\epsilon,\rho_{1}+\epsilon]$. Picking
$R=\max\\{\delta,\delta^{\prime}\\}$, we have that $|z|<R$, for all
$z\in\bigcup_{n=1}^{+\infty}[0,h(k_{n})]$.

Now, regarding the image of $h$, analyzing $g(x)$, we find that

$\mathrm{Arg}(g(x))=C^{\prime}+b\mathrm{Arg}(x)+P_{\mathbb{H}}^{\prime}(\rho_{1})\mathrm{Arg}(x-1)+\sum_{n=1}^{+\infty}a_{n}\mathrm{Arg}(x-T(\lambda_{n}))$

for some constant $C^{\prime}$. As we saw in Remark 17, we may assume that
$C^{\prime}=0$. Because $T(x)=\frac{x-\rho_{2}}{x-\rho_{1}}$ and because there
exists no $\lambda_{n}\in(\rho_{1},\rho_{2})$, we have that
$T(\lambda_{n})>0$, for all $n\geq 1$ ($T$ maps the line segment
$(\rho_{1},\rho_{2})$ onto the negative line). Therefore, we can see that for
$x<0$,

$\mathrm{Arg}(g(x))=b\pi+P^{\prime}_{\mathbb{H}}(\rho_{1})\pi+\sum_{n=1}^{+\infty}a_{n}\pi=\pi
P^{\prime}_{\mathbb{H}}(\rho_{1})\left(1-\frac{1}{P^{\prime}_{\mathbb{H}}(\rho_{2})}-\frac{1}{P^{\prime}_{\mathbb{H}}(\lambda_{n})}\right)=\pi,$

through Proposition 13. Thus, $g$ maps the negative line onto itself. To see
the image of the positive line, we consider the angles

$\theta_{1}:=P_{\mathbb{H}}^{\prime}(\rho_{1})\pi+\sum_{n:T(\lambda_{n})>1}a_{n}\pi\quad\text{and}\quad\theta_{2}:=\sum_{n:T(\lambda_{n})>1}a_{n}\pi.$

Figure 3. The image of $h$, when $P_{\mathbb{H}}$ has distinct real roots.
There exists an angular sector of amplitude
$P_{\mathbb{H}}^{\prime}(\rho_{1})\pi$, that contains no slits.

We then deduce, similarly as before, that $g$ maps $[0,1]$ onto the union of
line segments of the form $[0,h(k_{n})]$, for all $n\geq 1$, so that
$T(\lambda_{n})<1$. Assume without loss of generality that $T(\lambda_{n})<1$
for infinitely many indices. Otherwise, we only have finitely many such
segments, so we comment no further. Those segments, accumulate to the half-
line $L_{\theta_{1}}:=\\{xe^{i\theta_{1}}:x\geq 0\\}$ and by the first part of
the proof, the tip points of the segments accumulate at the origin. Similarly,
the image of the line $[1,+\infty]$ is the union of the line segments of the
form $[0,h(k_{n})]$, for all $n\geq 1$, so that $T(\lambda_{n})>1$, and the
half-line $[0,+\infty]$; recall that $(T(\lambda_{n}))_{n\geq 1}$ is bounded,
so picking $n_{0}\geq 1$ such that $T(\lambda_{n_{0}})=\max T(\lambda_{n})$,
then $g([T(\lambda_{n_{0}}),+\infty])=[0,+\infty]$. Again, those segments
accumulate to the half-line $L_{\theta_{2}}:=\\{xe^{i\theta_{2}}:x\geq 0\\}$
with the tip points accumulating at zero (see Figure 3). ∎

The Loewner flow in this case is given by the conjugation formula

(4.6)
$f(z,t)=h^{-1}\left(\left(1-t\right)^{-\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{2}}h\left(\left(1-t\right)^{-\frac{1}{2}}z\right)\right)$

for all $z\in\mathbb{H}$ and $0\leq t<1$. In order to study the geometry of
the slits as $t\rightarrow 1^{-}$, we follow the formula above and thus, we
extend the tip points $h(k_{n})$ to infinity by applying the mapping $z\mapsto
e^{\frac{P^{\prime}_{\mathbb{H}}(\rho_{1})}{2(1-t)}}z$ and then we consider
their preimage of $\\{xh(k_{n}):x\geq 1\\}$ under $h$. The following
proposition shows that the trajectory of $f(k_{n}\sqrt{1-t},t)$ collides at
the point $\rho_{1}\in\mathbb{R}$ as $t\rightarrow 1^{-}$, non-tangentially.

######  Proposition 21.

For each $n\in\mathbb{N}$, the trace
$\hat{\gamma}_{n}:=\\{f(k_{n}\sqrt{1-t},t):[0,1)\\}$ is a smooth curve
intersecting the real line non-tangentially at the root $\rho_{1}$.

###### Proof.

Fix $n\in\mathbb{N}$ and assume without loss of generality that
$k_{n}>\rho_{1}$. Consider the curve $\gamma_{n}:[0,1)\to\mathbb{H}$ with
$\gamma_{n}(t)=f(k_{n}\sqrt{1-t},t)$. Surely $\lim\limits_{t\to
1}\gamma_{n}(t)=\rho_{1}$. We only need to deal with the angle of the
convergence and we are going to do this through the use of harmonic measure.
By (4.5), $h$ maps the upper half-plane $\mathbb{H}$ conformally onto the
simply connected domain
$\Omega:=\mathbb{H}\setminus\bigcup\limits_{j=1}^{+\infty}[0,h(k_{j})]$. In
addition, the image of $\gamma_{n}$ through $h$ is the ray $\\{re^{i\arg
h(k_{n})}:r>|h(k_{n})|\\}$ and
$\lim\limits_{t\to+\infty}h(\gamma_{n}(t))=\infty$. It is easy to see that
$h\circ\gamma_{n}$ separates the prime ends of $\Omega$ into two connected
components. The one consists of the prime ends corresponding to $(-\infty,0]$,
to $\bigcup\limits_{j\in\mathbb{N}}\\{[0,h(k_{j})]:\arg h(k_{j})>\arg
h(k_{n})\\}$ and the prime ends corresponding to $[0,h(k_{n})]$ defined by
crosscuts with arguments larger than $\arg h(k_{n})$. The other is the
complement. We denote them by $\partial\Omega^{+}$ and $\partial\Omega^{-}$,
respectively (see Figure 4).

Our first objective is to prove that $\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\bigcup\limits_{j=1}^{+\infty}[0,h(k_{j})],\Omega)=0$.
Indeed, by the previous lemma, there exists some $R>0$ such that $|z|<R$, for
all $z\in\bigcup\limits_{j=1}^{+\infty}[0,h(k_{j})]$. Therefore, through a new
conformal mapping $g$ of $\Omega$ onto $\mathbb{H}$ which fixes $\infty$ and
sends $h(k_{n})$ to $0$, we may map a subset of
$\bigcup\limits_{j=1}^{+\infty}[0,h(k_{j})]$ onto the closed segment
$[a,b]\subset\mathbb{R}$. Then, by conformal invariance,

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\bigcup\limits_{j=1}^{+\infty}[0,h(k_{j})],\Omega)\leq\lim\limits_{t\to
1}\omega(g(h(\gamma_{n}(t))),[a,b],\mathbb{H})=0,$

in view of Subsection 2.1. Combining this with the domain monotonicity
property, we have

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)=\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),(-\infty,0],\Omega)\leq\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),(-\infty,0],\mathbb{H}).$

In a similar fashion,

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega)\leq\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),[0,+\infty),\mathbb{H}).$

Figure 4. The domain $\Omega$ and its prime ends

However, in the last two relations, both the left-hand and right-hand sides
add up to 1. This implies that the equalities prevail. Consequently,

$\displaystyle\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega)$ $\displaystyle=$
$\displaystyle\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),[0,+\infty),\mathbb{H})$ $\displaystyle=$
$\displaystyle\lim\limits_{r\to+\infty}\omega(re^{i\arg
h(k_{n})},\\{w\in\mathbb{C}:\arg w=0\\},U_{0,\pi})$ $\displaystyle=$
$\displaystyle\lim\limits_{r\to+\infty}\frac{\pi-\arg h(k_{n})}{\pi}.$

Using a Riemann mapping of $\Omega$ onto the unit disk, the conformal
invariance of the harmonic measure and the preservation of the orientation,
Remark 7 reveals that the image of $h\circ\gamma_{n}$ inside the unit disk
converges to some point of the unit circle by angle $\pi(1-\frac{\pi-\arg
h(k_{n})}{\pi})=\arg h(k_{n})$. Finally, through a Möbius transformation which
preserves angles in the whole complex plane, we return to our initial setting,
to see that $\gamma_{n}$ intersects the real line at $\rho_{1}$ by angle
$\pi-\arg h(k_{n})\in(0,\pi)$, thus non-tangentially. Remember that
$h(k_{n})\in\mathbb{H}$ or equivalently $0<\arg h(k_{n})<\pi$. The difference
between the angles in the unit disk and the upper half-plane derives from the
fact that in the unit disk we use to count the angle with the help of the
tangent, while in the upper half-plane the angle is usually counted by
starting from the “positive” semi-axis. ∎

###  4.3. Tangential intersections.

Assume, now, that $P_{\mathbb{H}}$ has a double root $\rho_{0}\in\mathbb{R}$.
Fix some $z_{0}\in\mathbb{H}$ and consider the function
$h(z):=\int_{z_{0}}^{z}\frac{d\zeta}{P_{\mathbb{H}}(\zeta)}$, which is well
defined and analytic, since $\mathbb{H}$ is simply connected. Therefore, by
$(\ref{double root})$ and due to the uniform convergence on compacta, we have
that

(4.7)
$h(z)=\sum_{n=1}^{+\infty}A_{n}\log\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}+\left(1-\sum_{n=1}^{+\infty}A_{n}\right)\log(z-\rho_{0})+\dfrac{B}{z-\rho_{0}}+\text{const.}$

for all $z\in\mathbb{H}$ where the parameters $A_{n}<0$ and $B>0$ are given by
$(\ref{double root})$ and the implicit solution
$h(v(z,t))=-\frac{1}{2}\log(1-t)+h(z)$. The following proposition shows that
$h$ is a well defined analytic and univalent function of the upper half-plane.

######  Proposition 22.

Let $(A_{n})_{n\geq 1}$ be a summable sequence of negative numbers,
$(\lambda_{n})_{n\geq 1}$ be a sequence of real points and let also
$\rho_{0}\in\mathbb{R}$, $B\in\mathbb{R}$ and $C>0$. Fix some
$z_{0}\in\mathbb{H}$. Then, the function

$\phi(z)=\sum_{n=1}^{+\infty}A_{n}\log\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}+(1-\sum_{n=1}^{+\infty}A_{n})\log(z-\rho_{0})+\dfrac{B}{z-\rho_{0}}+\dfrac{C}{(z-\rho_{0})^{2}}$

is analytic and univalent in $\mathbb{H}$.

###### Proof.

Analyticity follows directly from Lemma 15. To prove that $h$ is univalent, we
precompose with the Möbius transform $T(z)=\frac{1}{\rho_{0}-z}$. Note that
$T:\mathbb{H}\rightarrow\mathbb{H}$. By differentiating, we get that

$(\phi\circ
T^{-1})^{\prime}(z)=\sum_{n=1}^{+\infty}\dfrac{A_{n}}{z-T(\lambda_{n})}-\dfrac{1}{z}-B+2Cz$

which has positive imaginary part for all $z\in\mathbb{H}$. Hence, $\phi\circ
T^{-1}$ and by extension $\phi$ are univalent in $\mathbb{H}$, since the upper
half-plane is a convex domain. ∎

######  Lemma 23.

Assume that $h$ is given by $(\ref{koenigs3})$. Then, it maps the upper half-
plane onto a horizontal half-plane, minus infinitely many horizontal half-
lines that extend to infinity from the left. In addition,
$\mathrm{Re}h(x)\rightarrow+\infty$, as $x\rightarrow\pm\infty$.

###### Proof.

We argue as in Lemma 20. Again, the only accumulation points of the sequences
$(\lambda_{n})_{n\geq 1}$ and $(k_{n})_{n\geq 1}$ are $\pm\infty$.We, now,
consider the function $g(x)=h(T^{-1}(x))=h(\frac{\rho_{0}x-1}{x})$,
$x\in\mathbb{R}\setminus\\{0\\}$. It is easy to see that $\lim\limits_{x\to
0}g(x)=\lim\limits_{x\to\pm\infty}h(x)$. After a series of calculations, we
see that

$g(x)=\sum\limits_{n=1}^{+\infty}A_{n}\log\left(\frac{\rho_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}\frac{x-T(\lambda_{n})}{x}\right)+\left(1-\sum\limits_{n=1}^{+\infty}A_{n}\right)\log\left(-\frac{1}{x}\right)-Bx+\text{const.},$

where $B>0$ and $A_{n}<0$, for every $n\in\mathbb{N}$. Focusing on the real
parts, we are led to

$\displaystyle\text{Re}g(x)$
$\displaystyle=\sum\limits_{n=1}^{+\infty}A_{n}\log\left|\frac{\rho_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}\frac{x-T(\lambda_{n})}{x}\right|-\left(1-\sum\limits_{n=1}^{+\infty}A_{n}\right)\log|x|-Bx+\text{const.}$
$\displaystyle=\sum\limits_{n=1}^{+\infty}A_{n}\log\left|\frac{\rho_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}(x-T(\lambda_{n}))\right|-\log|x|-Bx+\text{const.}$

Figure 5. The image of $h$, when
$k_{n_{0}}<\lambda_{n_{0}}<\rho_{0}<k_{n_{0}^{\prime}}$, for some $n_{0}\geq
1$.

We focus on the infinite sum. Since the only possible limit points of
$(\lambda_{n})_{n\geq 1}$ are $\pm\infty$ and since $(T(\lambda_{n}))_{n\geq
1}$ is bounded and converges to $0$ as $n\to+\infty$, we have

$\sum\limits_{n=1}^{+\infty}A_{n}\log\left|\frac{\rho_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}(x-T(\lambda_{n}))\right|\approx\sum\limits_{n=1}^{N}A_{n}\log\left|\frac{\rho_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}(x-T(\lambda_{n}))\right|+\sum\limits_{n=N+1}^{+\infty}A_{n}\log|x|,$

for some large enough $N\in\mathbb{N}$, where $\approx$ denotes comparability
for large $n$. As a result,

$\mathrm{Re}g(x)\approx\sum\limits_{n=1}^{N}A_{n}\log\left|\frac{\rho_{0}-\lambda_{n}}{z_{0}-\lambda_{n}}(x-T(\lambda_{n}))\right|-\left(1-\sum\limits_{n=N+1}^{+\infty}A_{n}\right)\log|x|-Bx+\text{const.}$

Keeping in mind that the sequence $(A_{n})_{n\geq 1}$ is summable with
negative terms, we quickly see that $\lim\limits_{x\to
0}\mathrm{Re}g(x)=+\infty$. Also, taking into account that
$\mathrm{Re}h(\lambda_{n})=+\infty$, we deduce that
$\lim\limits_{x\to\pm\infty}\mathrm{Re}h(x)=+\infty$.

Finally, to see the image of $h$, we find $h(\mathbb{R})$. Without loss of
generality, we assume that $(k_{n})_{n\geq 1}$ accumulates at both $\pm\infty$
(and hence so does the sequence $(\lambda_{n})_{n\geq 1}$) and let also
$k_{n_{0}}<\lambda_{n_{0}}<\rho_{0}<k_{n_{o}^{\prime}}$. Now, it is direct to
see that

(4.8)
$\mathrm{Im}h(x)=\sum_{n=1}^{+\infty}A_{n}\mathrm{Arg}(x-\lambda_{n})+(1-\sum_{n=1}^{+\infty}A_{n})\mathrm{Arg}(x-\rho_{0})+\text{const}.$

Again, for simplicity we take the constant to be zero. For any $m\geq 1$, so
that $\lambda_{m}<\lambda_{n_{0}}$, it is
$\mathrm{Im}h(x)=(1-\sum_{\lambda_{n}\leq\lambda_{m}}A_{n})\pi$, for all
$x\in(\lambda_{m},\lambda_{m^{\prime}})$. Similarly for
$\lambda_{m}>\rho_{0}$, we get that for all
$x\in(\lambda_{m},\lambda_{m^{\prime}})$,
$\mathrm{Im}h(x)=\sum_{\lambda_{n}>\lambda_{m}}A_{n}\pi$. To conclude, we have
that $\mathrm{Im}h(x)=(1-\sum_{\lambda_{n}\leq\rho_{0}}A_{n})\pi$, in
$(\lambda_{n_{0}},\rho_{0})$ and because $\mathrm{Re}h(x)\rightarrow-\infty$,
as $x\rightarrow\rho_{0}-$, we deduce that $h$ maps the interval
$(\lambda_{n_{0}},\rho_{0})$ onto the line
$\mathbb{R}+i\sum_{\lambda_{n}\leq\rho_{0}}A_{n}\pi$, as in Figure 5. ∎

By the proof of the lemma above, we see that as
$\lambda_{m}\rightarrow-\infty$, then
$\mathrm{Im}h(k_{m})=(1-\sum_{\lambda_{n}\leq\lambda_{m}}A_{n})\pi\rightarrow\pi$
and as $\lambda_{m}\rightarrow+\infty$, then
$\mathrm{Im}h(k_{m})=\sum_{\lambda_{n}>\lambda_{m}}A_{n}\pi\rightarrow 0$.
Also, $\mathrm{Re}h(k_{m})\rightarrow+\infty$ and hence, we see that there is
a horizontal strip of amplitude $\pi$ that contains no slits, whereas the tip
points of the half-lines accumulate at the boundary of this strip from above
and below, while also “disappearing” to the right as we see in Figure 5.

Following similar steps as before, the Loewner flow in this case is given by
the conjugation formula

$f(z,t)=h^{-1}\left(\frac{1}{2}\log(1-t)+h\left((1-t)^{-\frac{1}{2}}z\right)\right).$

Once again we enquire about the convergence of the corresponding trajectories.

######  Proposition 24.

For each $n\in\mathbb{N}$, the trace
$\hat{\gamma}_{n}:=\\{f(k_{n}\sqrt{1-t},t):t\in[0,1)\\}$ is a smooth curve
intersecting the real line tangentially at the root $\rho_{0}$. In particular,
either all curves converge to $\rho_{0}$ by angle $0$ or all of them converge
by angle $\pi$.

###### Proof.

Again, we will utilize the harmonic measure. Towards this goal, fix
$n\in\mathbb{N}$ and consider the curve $\gamma_{n}:[0,1)\to\mathbb{H}$ with
$\gamma_{n}(t)=f(k_{n}\sqrt{1-t},t)$. As before, $\lim_{t\to
1^{-}}\gamma_{n}(t)=\rho_{1}$. This time, $h$ maps the upper half-plane
$\mathbb{H}$ onto a horizontal half-plane minus a sequence of horizontal half-
lines stretching to $\infty$ in the positive direction. To be more formal, set
$L_{j}=\\{z\in\mathbb{C}:\text{Re}z\geq\text{Re}h(k_{j}),\text{Im}z=\text{Im}h(k_{j})\\}$.
In our first case, we assume that there exists some $a>\text{Im}h(k_{j})$, for
all $j\in\mathbb{N}$ such that
$\Omega:=h(\mathbb{H})=\\{z\in\mathbb{C}:\text{Im}z<a\\}\setminus\bigcup_{j=1}^{+\infty}L_{j}$.
By the previous lemma, we may see that there exists $R\in\mathbb{R}$ such that
$\text{Re}h(k_{j})>R$, for all $j\in\mathbb{N}$. Arguing as before, the curve
$\gamma_{n}$ separates the prime ends of $\Omega$ into two connected
components $\partial\Omega^{+}$ and $\partial\Omega^{-}$, where
$\partial\Omega^{+}$ consists of the prime ends corresponding to the
horizontal line $L:=\\{z\in\mathbb{C}:\text{Im}z=a\\}$, the prime ends
corresponding to
$\bigcup_{j\in\mathbb{N}}\\{L_{j}:\text{Im}h(k_{j})>\text{Im}h(k_{n})\\}$ and
the prime ends corresponding to the half-line $L_{n}$ defined by crosscuts
with imaginary parts larger than $\text{Im}h(k_{n})$. Naturally,
$\partial\Omega^{-}$ consists of all the remaining prime ends. Our objective
is to prove that $\lim_{t\to
1^{-}}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)=1$. An important note
towards this direction is the fact that
$h\circ\gamma_{n}([0,1))=\\{z\in\mathbb{C}:\text{Re}z<\text{Re}h(k_{n}),\text{Im}z=\text{Im}h(k_{n})\\}$,
with $\lim_{t\to 1^{-}}\text{Re}h(\gamma_{n}(t))=-\infty$.

Figure 6. The sets of Proposition 24

The existence of the real number $R$ that bounds the real parts of all half-
lines, allows us to proceed to the following construction: we may find some
point $\zeta\in L$ and a ray $A$ emanating from $\zeta$ such that
$A\setminus\\{\zeta\\}\subset\Omega$ (an obvious example is a point $\zeta$
that rests sufficiently to the left and the half-line $A$ that is
perpendicular to $L$ at $\zeta$). Set $L^{+}=\\{z\in
L:\text{Re}z\leq\text{Re}\zeta\\}$ and $L^{-}=L\setminus L^{+}$. Then, the
angular simply connected domain $U$ bounded by $L^{+}$ and $A$ is contained
inside $\Omega$ (for the whole construction see Figure 6). Moreover, it is
easy to see that $U$ eventually contains the curve $h\circ\gamma_{n}$ or
equivalently there exists some $t_{0}\in[0,1)$ such that $h(\gamma_{n}(t))\in
U$, for all $t\in[t_{0},1)$. Since $\partial\Omega^{+}\supset L^{+}$ and
$\Omega\supset U$, we are led to

$\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)\geq\omega(h(\gamma_{n}(t)),L^{+},\Omega)\geq\omega(\gamma_{n}(t),L^{+},U),$

for all $t\in[t_{0},1)$. By the construction of the ray $A$, there exists some
$\beta\in(\pi,2\pi)$ such that

$\displaystyle\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)$ $\displaystyle\geq$
$\displaystyle\lim\limits_{t\to 1}\omega(h(\gamma_{n}(t)),L^{+},U)$
$\displaystyle\geq$ $\displaystyle\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t))-\zeta,\\{w\in\mathbb{C}:\arg
w=\pi\\},U_{\pi,\beta})$ $\displaystyle=$ $\displaystyle\lim\limits_{t\to
1}\frac{\beta-\arg(h(\gamma_{n}(t))-\zeta)}{\beta-\pi}$ $\displaystyle=$
$\displaystyle 1,$

since $\lim_{t\to 1}\arg h(\gamma_{n}(t))-\zeta=\lim_{t\to 1}\arg
h(\gamma_{n}(t))=\pi$. However, the harmonic measure has $1$ as an upper bound
and as a result

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)=1=\frac{\pi}{\pi}.$

Using again a Riemann map of $\Omega$ onto the unit disk, Remark 7 and a
Möbius transformation to return to $\mathbb{H}$, we see that each $\gamma_{n}$
coverges to $\rho_{1}$ by angle $0$ and thus, tangentially. Finally, if at the
start, our domain was of the form
$\Omega:=\\{z\in\mathbb{C}:\text{Im}z>a\\}\setminus\bigcup_{j=1}^{+\infty}L_{j}$,
then each curve $\gamma_{n}$ would converge to $\rho_{1}$ by angle $\pi$,
again tangentially. ∎

###  4.4. Orthogonal intersections.

In the final case, we assume that $P_{\mathbb{H}}$ has a triple root
$\rho_{0}\in\mathbb{R}$, which means according to the preliminary analysis,
that $\rho_{0}$ is a double root coinciding with some $\lambda_{n_{0}}$, hence
triple. Integrating in $(\ref{differential})$ and using $(\ref{double root})$,
we deduce that $h(v(z,t))=-\frac{1}{2}\log(1-t)+h(z)$, where

(4.9) $h(z)=\sum_{n\neq
n_{0}}A_{n}\log\dfrac{z-\lambda_{n}}{z_{0}-\lambda_{n}}+(1-\sum_{n\neq
n_{0}}\log(z-\rho_{0})+\dfrac{B}{z-\rho_{0}}+\dfrac{C}{(z-\rho_{0})^{2}}.$

where the parameters $A_{n}<0$, $B$ and $C>0$ are given by $(\ref{triple
root})$. By Lemma 15 and Proposition 22, $h$ is analytic and univalent in the
upper half-plane.

######  Lemma 25.

Assume that $h$ is given by $(\ref{koenigs4})$. Then it maps the upper half-
plane onto the complement of inifinitely many horizontal half-lines and we
have that:

  1. (1)

$\mathrm{Re}h(x)\rightarrow+\infty$, as $x\rightarrow\pm\infty$ and

  2. (2)

there exists some $Q>0$, so that $|\mathrm{Im}h(k_{n})|<Q$, for all
$n\in\mathbb{N}$.

Figure 7. The image of $h$, when $\rho_{0}$ is a triple root.

###### Proof.

The proof of the lemma follows an identical procedure to the lemma of the
preceding section, albeit with minor modifications. For the sake of brevity,
we skip this proof, but we only note some adjustments. We write, without loss
of generality, $k_{n_{0}}<\lambda_{n_{0}}=\rho_{0}<k_{n_{0}^{\prime}}$.
Through the fact that $\mathrm{Re}h(x)$ equals

$\sum_{n\neq
n_{0}}A_{n}\log\left|\frac{x-\lambda_{n}}{z_{0}-\lambda_{n}}\right|+\dfrac{(1-\sum_{n\neq
n_{0}}A_{n})\log\left|x-\rho_{0}\right|(x-\rho_{0})^{2}+B(x-\rho_{0})+C}{(x-\rho_{0})^{2}}$

we see that $\mathrm{Re}h(x)\rightarrow+\infty$, as $x\rightarrow\rho_{0}$,
while by $(\ref{image of h3})$ we see that $h((k_{n_{0}},\rho_{0}))$ and
$h((\rho_{0},k_{n_{0}^{\prime}}))$ are the half-lines lying above and below
all other half-lines, respectively, as we see in Figure 7. ∎

Again, we see that the strip $\\{0<\mathrm{Im}z<\pi\\}$ contains no half-
lines, while the tip points accumulate on the boundary of the strip and the
point at infinity towards the right. In addition, the Loewner flow is the same
as in the previous case. Finally, the following proposition allows us to see
that the trajectories of $f(k_{n}\sqrt{1-t},t)$ collide to the real line,
orthogonally at the point $\rho_{0}$, for all $n\geq 1$.

######  Proposition 26.

For each $n\in\mathbb{N}$, the trace
$\hat{\gamma}_{n}:=\\{f(k_{n}\sqrt{1-t},t):t\in[0,1)\\}$ is a smooth curve
intersecting the real line orthogonally at the root $\rho_{0}$ (i.e. with
angle $\frac{\pi}{2}$).

###### Proof.

Fix $n\in\mathbb{N}$ and set
$L_{j}=\\{z\in\mathbb{C}:\text{Re}z\geq\text{Re}h(k_{j}),\text{Im}z=\text{Im}h(k_{j})\\}$.
This time, (4.9) dictates that
$\Omega:=h(\mathbb{H})=\mathbb{C}\setminus\bigcup_{j=1}^{+\infty}L_{j}$. In a
similar fashion as in the previous cases, we denote by $\partial\Omega^{+}$
the prime ends of $\Omega$ corresponding to
$\bigcup_{j\in\mathbb{N}}\\{L_{j}:\text{Im}h(k_{j})>\text{Im}h(k_{n})\\}$
along with the prime ends corresponding to the half-line $L_{n}$ and defined
by crosscuts with imaginary parts larger than $\text{Im}h(k_{n})$. Once more,
we denote by $\partial\Omega^{+}$ the set of the remaining prime ends. Recall
that
$h\circ\gamma_{n}([0,1))=\\{z\in\mathbb{C}:\text{Re}z<\text{Re}h(k_{n}),\text{Im}z=\text{Im}h(k_{n})\\}$
and $\lim_{t\to 1^{-}}\text{Re}h(\gamma_{n}(t))=-\infty$. Our aim is to prove
that

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)=\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega)=\frac{1}{2}.$

By the previous lemma, we see that there exist $n_{1},n_{2}\in\mathbb{N}$ such
that $\text{Im}h(k_{n_{1}})\leq\text{Im}h(k_{j})\leq\text{Im}h(k_{n_{2}})$,
for all $j\in\mathbb{N}$. So, denote by $L^{+}\subset\partial\Omega^{+}$ the
prime ends corresponding to the half-line $L_{n_{2}}$ and defined by crosscuts
with imaginary parts larger than $\text{Im}h(k_{n_{2}})$. On the other side.
denote by $L^{-}\subset\partial\Omega^{-}$ the prime ends corresponding to
$L_{n_{1}}$ and defined by crosscuts with imaginary parts smaller than
$\text{Im}h(k_{n_{1}})$. As in the tangential case, there exists some
$R\in\mathbb{R}$ such that $\text{Re}h(k_{j})\geq R$, for all
$j\in\mathbb{N}$. Therefore, through the conformal invariance of the harmonic
measure and a mapping of $\Omega$ onto $\mathbb{H}$ that maps $h(k_{n})$ to
$0$ and fixes infinity, we may see that

$\lim\limits_{t\to 1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+}\setminus
L^{+},\Omega)=\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-}\setminus L^{-},\Omega)=0.$

Figure 8. The sets of Proposition 26

We will use two auxiliary “Koebe-like” domains to reach the desired
conclusions. We set $\Omega_{1}=\mathbb{C}\setminus L_{n_{1}}$ and
$\Omega_{2}=\mathbb{C}\setminus L_{n_{2}}$. Trivially, both of them are
supersets of $\Omega$. In particular, in the sense of prime ends,
$L^{-}\subset\partial\Omega\cap\partial\Omega_{1}$ and
$L^{+}\subset\partial\Omega\cap\partial\Omega_{2}$ (see Figure 8). By the
domain monotonicity property of harmonic measure,

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),L^{-},\Omega_{1})\geq\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),L^{-},\Omega)=\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega).$

Through a chain of conformal mappings, we are going to estimate the harmonic
measure with regard to the domain $\Omega_{1}$. As a matter of fact,

$\displaystyle\lim\limits_{t\to 1}\omega(h(\gamma_{n}(t)),L^{-},\Omega_{1})$
$\displaystyle=$ $\displaystyle\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t))-h(k_{n_{1}}),L^{-}-h(k_{n_{1}}),\Omega_{1}-h(k_{n_{1}}))$
$\displaystyle=$ $\displaystyle\lim\limits_{t\to
1}\omega(-h(\gamma_{n}(t))+h(k_{n_{1}}),-L^{-}+h(k_{n_{1}}),\mathbb{C}\setminus(-\infty,0])$
$\displaystyle=$ $\displaystyle\lim\limits_{t\to
1}\omega(\sqrt{-h(\gamma_{n}(t))+h(k_{n_{1}})},\\{z\in\mathbb{C}:\arg
z=-\frac{\pi}{2}\\},-i\mathbb{H})$ $\displaystyle=$
$\displaystyle\lim\limits_{t\to
1}\frac{\arg\sqrt{-h(\gamma_{n}(t))+h(k_{n_{1}})}+\frac{\pi}{2}}{\pi}$
$\displaystyle=$ $\displaystyle\frac{1}{2}.$

Combining with the inequality above, we get $\lim_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega)\leq\frac{1}{2}$.
Following an almost identical procedure, but this time with the Koebe domain
$\Omega_{2}$, we may prove that $\lim_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)\leq\frac{1}{2}$ as well.
However, since $\partial\Omega=\partial\Omega^{+}\cup\partial\Omega^{-}$, it
is obvious that

$\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega)+\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)=1,$

for all $t\in[0,1)$. So, it follows that

$\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{-},\Omega)=\lim\limits_{t\to
1}\omega(h(\gamma_{n}(t)),\partial\Omega^{+},\Omega)=\frac{1}{2}.$

Lastly, Remark 7 once again implies the desired result. ∎

To conclude, the combination of all the lemmas and propositions of this
section provides the proof of Theorem 3.

## Acknowledgements

We would like to thank Alan Sola for his advice and careful consideration
during the preparation of the current work. We also thank Dimitrios Betsakos
for the helpful conversations.

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