# Genus three embedded doubly periodic minimal surfaces with parallel ends

Peter Connor  Department of Mathematical Sciences
Indiana University South Bend
South Bend
IN 46634
USA<EMAIL_ADDRESS>,  Shoichi Fujimori  Department of Mathematics
Hiroshima University
Higashihiroshima
Hiroshima 739-8526
Japan<EMAIL_ADDRESS>and  Phillip Marmorino  Department of
Mathematics
University of Notre Dame
Notre Dame
IN 46556
USA<EMAIL_ADDRESS>
(Date: May 22, 2021)

###### Abstract.

We construct a one-parameter family of embedded doubly periodic minimal
surfaces of genus three with four parallel ends. The Weierstrass data for each
surface of the family are given and the two dimensional period problem is
solved.

###### Key words and phrases:

minimal surface, doubly periodic, embedding

###### 2020 Mathematics Subject Classification:

Primary 53A10; Secondary 53C42, 49Q05

The first author was supported in part by a Faculty Research Grant from
Indiana University South Bend.

The second author was supported in part by Grant-in-Aid for Scientific
Research (C) No. 21K03226 from Japan Society for the Promotion of Science.

The third author was supported in part by a SMART Summer Fellowship from
Indiana University South Bend.

##  1\. Introduction

A minimal surface $M$ in Euclidean space is called doubly periodic if it is
invariant under two linearly independent translations, which can be assumed to
be horizontal. The first example of a doubly periodic minimal surface was
discovered by Scherk [8].

Let $L$ be the two-dimensional lattice generated by the maximal group of these
translations. Meeks and Rosenberg [6] showed that if $M/L$ is complete,
properly embedded, and of finite topology then the quotient has a finite
number of annular top and bottom ends, each of which is asymptotic to flat
annuli. This type of end is referred to as a Scherk end. The top and bottom
ends are either parallel or not. In the parallel case, Meeks and Rosenberg
proved that the number of top and bottom ends are equal to the same even
number.

Karcher [4] and Meeks-Rosenberg [6] constructed a three-parameter family of
genus one examples with parallel ends. Wei [10] added a handle to one of the
Karcher and Meeks-Roseberg examples to produce a one-parameter family of genus
two examples. Connor and Weber [1] proved that, for each genus $g$, there is a
three-parameter family of genus $g$ examples with parallel ends that have a
foliation of $\mathbb{R}^{3}$ by vertical planes as a limit. The Connor-Weber
examples are only known to exist in a small neighborhood of the limit
foliation. The Karcher and Meeks-Rosenberg genus one examples and the Wei
genus two example are all members of the Connor-Weber examples.

The goal of this paper is to construct a one-parameter family $M_{\lambda}$ of
embedded doubly periodic minimal surfaces with parallel ends with genus three
in the quotient. The parameter $\lambda$ ranges over the open interval
$(0,1)$. See Figure 1.1. These surfaces are constructed using the Weierstrass
Representation by adding a handle to Wei’s genus two surfaces along a
horizontal symmetry plane.

######  Theorem 1.1.

There exists a one-parameter family $M_{\lambda}$ of properly embedded doubly
periodic minimal surfaces with parallel ends such that $M_{\lambda}/L$ has
genus three and four parallel ends.

This theorem is proven in Sections 2 through 4. The data of the surface
$M_{\lambda}$ are given in Section 2, Equations (2.1), (2.7), and (2.8) with
three parameters $\lambda$, $\lambda_{1}$, and $\lambda_{2}$. Then in Section
3, Proposition 3.1, we prove that for any $\lambda$ there exist $\lambda_{1}$
and $\lambda_{2}$ so that the surface $M_{\lambda}$ is well-defined as a
doubly periodic minimal surface. The embeddedness of $M_{\lambda}$ is proven
in Section 4, Proposition 4.1.

Figure 1.1. $M_{\lambda}$ with $\lambda=0.4$ (left), $\lambda=0.9$ (center),
and $\lambda=0.99$ (right).

Connor and Weber showed the existence of these surfaces in a small
neighborhood of their limit as a foliation of $\mathbb{R}^{3}$ by vertical
planes. This result significantly expands the known parameters for which these
surfaces exist. As $\lambda\rightarrow 0$, $M_{\lambda}$ limits as a foliation
of vertical planes, the Connor-Weber limit. As $\lambda\rightarrow 1$,
$M_{\lambda}$ limits as two copies of a genus one Scherk surface glued
together end-to-end.

Weber [9] constructed a two-parameter family of triply periodic minimal
surfaces of genus 5, which is intuitively considered as the result of adding
two handles to a family of Schwarz CLP surfaces. See Figure 1.2. The surfaces
$M_{\lambda}$ can be considered as limiting surfaces of these triply periodic
minimal surfaces.

Figure 1.2. Left: Triply periodic minimal surface of genus 3, called Schwarz
CLP. Center: Weber’s triply periodic minimal surface of genus 5. Right: A
limit of the Weber’s surface.

##  2\. Weierstrass Representation

We use the Weierstrass Representation of minimal surfaces. Let $M$ be a
minimal surface and $R$ the underlying Riemann surface of $M$. Then $M$ can be
expressed by $X:R\to\mathbb{R}^{3}$,

(2.1) $X(z)=\operatorname{Re}\int_{z_{0}}^{z}(\phi_{1},\phi_{2},\phi_{3}),$

where

$(\phi_{1},\phi_{2},\phi_{3})=\left(\frac{1}{2}\left(\frac{1}{G}-G\right)dh,\frac{i}{2}\left(\frac{1}{G}+G\right)dh,dh\right),$

$z_{0},z\in R$, $G$ a meromorphic function called the Gauss map, and $dh$ a
holomorphic one-form called the height differential. This is the Weierstrass
Representation. See for example [5] for details. We call $(R,G,dh)$ the
Weierstrass data of the minimal surface $M$.

Conversely, given a Riemann surface $R$ which is biholomorphic to a compact
Riemann surface $\overline{R}$ with finitely many points removed, a
meromorphic function $G$ and a holomorphic one-form $dh$ on $R$, the triple
$(R,G,dh)$ is the Weierstrass representation for a doubly periodic minimal
surface with horizontal periods $\bm{v}_{1}$ and $\bm{v}_{2}$ and vertical
Scherk ends if the following conditions are met.

  1. (1)

The zeros of $dh$ are the zeros and poles of $G$ on $R$ with the same
multiplicity.

  2. (2)

$G$ and $dh$ extends meromorphically to $\overline{R}$, and the height
differential $dh$ has poles of order one and $G$ has finite value at the ends
$\overline{R}-R$.

  3. (3)

For each closed curve $\gamma$ on $R$,

$\operatorname{Re}\int_{\gamma}(\phi_{1},\phi_{2},\phi_{3})=(0,0,0)\pmod{\bm{v}_{1},\bm{v}_{2}}.$

Of course, $X$ depends on the path of integration. Condition (3) ensures that
$X$ is well-defined and is called the period problem.

######  Remark 2.1.

Denoting the universal cover of $R$ by $\widetilde{R}$, the minimal surface
$X^{*}:\widetilde{M}\to\mathbb{R}^{3}$ with the Weierstrass data $(R,G,idh)$
is called the conjugate surface to $M$, and is denoted by $M^{*}$. It is known
that any curve of $R$ which is mapped by $X$ to a nonstraight planar geodesic
of $M$ is mapped by $X^{*}$ to a straight line in $M^{*}$. Furthermore, since
the Gauss map $G$ and the first fundamental form $(|G|^{-1}+|G|)^{2}|dh|^{2}$
are the same for both $M$ and $M^{*}$, it follows that the planar geodesic in
$M$ will lie in a plane perpendicular to the corresponding line in $M^{*}$ and
that the planar geodesic in $M$ will have the same length as the line in
$M^{*}$.

To give the Weierstrass data $(R,G,dh)$ of our minimal surfaces, we need some
preparations. Throughout this paper, we assume $\lambda$, $\lambda_{1}$, and
$\lambda_{2}$ are real constants such that

$0<\lambda<1<\lambda_{1}<\lambda_{2}.$

We set

$\displaystyle g_{1}(z)$
$\displaystyle=(z-\lambda)(z+1)(z-\lambda_{1})(z+\lambda_{2}),$ $\displaystyle
g_{2}(z)$ $\displaystyle=(z+\lambda)(z-1)(z+\lambda_{1})(z-\lambda_{2}).$

Then

$g_{1}(z)+g_{2}(z)=2(z^{2}-a^{2})(z^{2}-b^{2}),$

where

$\displaystyle
a=\sqrt{\dfrac{\alpha-\sqrt{\alpha^{2}-4\lambda\lambda_{1}\lambda_{2}}}{2}},\qquad
b=\sqrt{\dfrac{\alpha+\sqrt{\alpha^{2}-4\lambda\lambda_{1}\lambda_{2}}}{2}},$
$\displaystyle\alpha=\lambda+\lambda_{1}-\lambda_{2}-\lambda\lambda_{1}+\lambda\lambda_{2}+\lambda_{1}\lambda_{2}.$

It is easy to verify that $\alpha$ satisfies

$2\lambda<\alpha<2\lambda_{2}^{2}.$

Also, $a$ and $b$ satisfy the following lemma.

######  Lemma 2.2.

Both $a$ and $b$ as above are real and satisfy

$\lambda<a<1\qquad\text{and}\qquad\lambda_{1}<b<\lambda_{2}.$

###### Proof.

By direct computations, we have

(2.2)
$\displaystyle(\alpha^{2}-4\lambda\lambda_{1}\lambda_{2})-(\alpha-2\lambda)^{2}$
$\displaystyle=-4\lambda(1-\lambda)(\lambda_{2}-\lambda_{1})<0,$ (2.3)
$\displaystyle(\alpha^{2}-4\lambda\lambda_{1}\lambda_{2})-(\alpha-2)^{2}$
$\displaystyle=4(1-\lambda)(\lambda_{1}-1)(1+\lambda_{2})>0,$ (2.4)
$\displaystyle(\alpha^{2}-4\lambda\lambda_{1}\lambda_{2})-(\alpha-2\lambda_{1}^{2})^{2}$
$\displaystyle=4\lambda_{1}(\lambda_{1}-1)(\lambda+\lambda_{1})(\lambda_{2}-\lambda_{1})>0,$
(2.5)
$\displaystyle(\alpha^{2}-4\lambda\lambda_{1}\lambda_{2})-(\alpha-2\lambda_{2}^{2})^{2}$
$\displaystyle=-4\lambda_{2}(\lambda_{2}-\lambda)(1+\lambda_{2})(\lambda_{2}-\lambda_{1})<0.$

By (2.3) or (2.4), we see $\alpha^{2}-4\lambda\lambda_{1}\lambda_{2}>0$ and
hence both $a$ and $b$ are positive real. Since $\alpha>2\lambda$, (2.2)
yields

$\sqrt{\alpha^{2}-4\lambda\lambda_{1}\lambda_{2}}<\alpha-2\lambda.$

This shows $\sqrt{\lambda}<a$. By (2.3), we have

$\sqrt{\alpha^{2}-4\lambda\lambda_{1}\lambda_{2}}>\big{|}\alpha-2\big{|}.$

Thus

$a^{2}=\frac{\alpha-\sqrt{\alpha^{2}-4\lambda\lambda_{1}\lambda_{2}}}{2}<\frac{\alpha-\big{|}\alpha-2\big{|}}{2}=\begin{cases}1&(\alpha-2\geq
0)\\\ \alpha-1&(\alpha-2<0).\\\ \end{cases}$

This implies $a<1$. Similarly, (2.4) and (2.5) imply $\lambda_{1}<b$ and
$b<\lambda_{2}$, respectively.

Therefore we have

(2.6) $0<\lambda<\sqrt{\lambda}<a<1<\lambda_{1}<b<\lambda_{2}.$

∎

We consider the following compact Riemann surface of genus 3

$\overline{R}=\\{(z,w)\in(\mathbb{C}\cup\\{\infty\\})^{2}\;;\;g_{2}(z)w^{2}=g_{1}(z)\\}$

and define

(2.7) $R=\overline{R}\setminus\\{(a,i),(-a,i),(a,-i),(-a,-i)\\}.$

Define the Gauss map and height differential by

(2.8) $G=w,\qquad dh=\frac{dz}{z^{2}-a^{2}}.$

We denote by $M_{\lambda}$ the minimal surfaces with the Weierstrass data
(2.7) and (2.8). Note that Weierstrass Representation conditions (1) and (2)
are satisfied by $G$ and $dh$. In Section 3 we prove that for each
$\lambda\in(0,1)$, there exist $\lambda_{1}$ and $\lambda_{2}$ so that the
period problems are solved. This is why we denote the subscript $\lambda$ of
$M_{\lambda}$.

The following lemma is obvious.

######  Lemma 2.3.

The Riemann surface $\overline{R}$ has automorphisms

$\tau_{1}:(z,w)\mapsto(\overline{z},-\overline{w}),\qquad\tau_{2}:(z,w)\mapsto(\overline{z},\overline{w}),\qquad\tau_{3}:(z,w)\mapsto(-\overline{z},\overline{w}^{-1}),$

with

$\tau_{1}^{*}(\phi_{1},\phi_{1},\phi_{3})=(-\overline{\phi_{1}},\overline{\phi_{2}},\overline{\phi_{3}}),\quad\tau_{2}^{*}(\phi_{1},\phi_{1},\phi_{3})=(\overline{\phi_{1}},-\overline{\phi_{2}},\overline{\phi_{3}}),\quad\tau_{3}^{*}(\phi_{1},\phi_{1},\phi_{3})=(\overline{\phi_{1}},\overline{\phi_{2}},-\overline{\phi_{3}}).$

##  3\. Period Problem

In this section we prove the following proposition.

######  Proposition 3.1.

For each $\lambda\in(0,1)$, there exist $\lambda_{1}$, $\lambda_{2}$
($1<\lambda_{1}<\lambda_{2}$) and $\mathbb{T}^{2}$ such that (2.1), considered
as a map into $\mathbb{T}^{2}\times\mathbb{R}$, with the Weierstrass data
(2.7) and (2.8) is single-valued on $R$.

Since $\phi_{3}$ can be written as

$\phi_{3}=d\left(\frac{1}{2a}\log\frac{z-a}{z+a}\right),$

we see that

$\operatorname{Re}\int_{\gamma}\phi_{3}=0$

for any loop $\gamma$ on $R$.

Now we consider the periods of $\phi_{1}$ and $\phi_{2}$ for loops around the
ends. Note that $\phi_{2}$ is holomorphic on $\overline{R}$. In fact, we have

$\phi_{2}=\frac{i}{2}\frac{g_{1}(z)+g_{2}(z)}{g_{2}(z)w(z^{2}-a^{2})}dz=\frac{i(z^{2}-b^{2})}{g_{2}(z)w}dz,$

and hence $\phi_{2}$ has no residues at the ends. That is, $\phi_{2}$ has no
periods at ends. We also note that the residues of $\phi_{1}$ at ends are
either $i/(2a)$ or $-i/(2a)$. Therefore, for any loop $\gamma$ around the
ends, we have

$\operatorname{Re}\int_{\gamma}(\phi_{1},\phi_{2},\phi_{3})\in\\{n\bm{v}_{1}:n\in\mathbb{Z}\\},\qquad\bm{v}_{1}=(\pi/a,0,0).$

Next we consider the periods of $\phi_{1}$ and $\phi_{2}$ for the homology
basis of $\overline{R}$. We fix the $z$-plane in $\overline{R}$ with
$w(\infty)=+1$. Let $\gamma_{j}$ ($j=1,2,3$) be loops in the $z$-plane winding
once around $[\lambda,1]$, $[1,\lambda_{1}]$, $[\lambda_{1},\lambda_{2}]$,
respectively, counterclockwise, and we denote by the same notations the lifts
of $\gamma_{j}$ to $\overline{R}$. Then
$\\{\gamma_{j},\tau_{2}\circ\gamma_{j}:j=1,2,3\\}$ gives a homology basis of
$\overline{R}$, where $\tau_{2}$ is defined in Lemma 2.3. By Lemma 2.3, it
suffices consider the following six periods

$\operatorname{Re}\int_{\gamma_{j}}\phi_{k},\qquad(j=1,2,3,\;\;k=1,2).$

By the symmetry arguments as in [10, Proof of Lemma 3.2], we have

$\operatorname{Re}\int_{\gamma_{1}}\phi_{1}=\operatorname{Re}\int_{\gamma_{2}}\phi_{2}=\operatorname{Re}\int_{\gamma_{3}}\phi_{1}=0.$

So all we need to consider are the periods of

$\operatorname{Re}\int_{\gamma_{1}}\phi_{2},\qquad\operatorname{Re}\int_{\gamma_{2}}\phi_{1},\qquad\operatorname{Re}\int_{\gamma_{3}}\phi_{2}.$

By a direct computation, we have

$v_{2}:=\operatorname{Re}\int_{\gamma_{1}}\phi_{2}=2\int_{\lambda}^{1}\frac{b^{2}-t^{2}}{\sqrt{(t^{2}-\lambda^{2})(1-t^{2})(\lambda_{1}^{2}-t^{2})(\lambda_{2}^{2}-t^{2})}}dt>0.$

Thus we have the period $\bm{v}_{2}=(0,v_{2},0)$ in the direction parallel to
the $x_{2}$-axis. We set

(3.1)
$\mathbb{T}^{2}=\mathbb{R}^{2}/\Lambda,\quad\Lambda=\\{n_{1}\bm{v}_{1}+n_{2}\bm{v}_{2}:n_{1},n_{2}\in\mathbb{Z}\\}.$

If we have

(3.2)
$\operatorname{Re}\int_{\gamma_{2}}\phi_{1}=0\quad\text{and}\quad\operatorname{Re}\int_{\gamma_{3}}\phi_{2}=0,$

then we can conclude that $X$ is well-defined in
$\mathbb{T}^{2}\times\mathbb{R}$. It is easy to verify that the left equation
in (3.2) holds if and only if

(3.3) $\int_{1}^{\lambda_{1}}\phi_{1}(t)=0,$

where

$\phi_{1}(t)=\frac{t\big{(}(1-\lambda-\lambda_{1}+\lambda_{2})t^{2}+\lambda\lambda_{1}-\lambda\lambda_{2}-\lambda_{1}\lambda_{2}+\lambda\lambda_{1}\lambda_{2}\big{)}}{(t^{2}-a^{2})\sqrt{(t^{2}-\lambda^{2})(t^{2}-1)(\lambda_{1}^{2}-t^{2})(\lambda_{2}^{2}-t^{2})}}dt,$

and the right equation in (3.2) holds if and only if

(3.4) $\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)=0,$

where

$\phi_{2}(t)=\frac{t^{2}-b^{2}}{\sqrt{(t^{2}-\lambda^{2})(t^{2}-1)(t^{2}-\lambda_{1}^{2})(\lambda_{2}^{2}-t^{2})}}dt.$

We fix $\lambda\in(0,1)$ and set

(3.5)
$\xi_{1}=\xi_{1}(\lambda_{1},\lambda_{2})=\int_{1}^{\lambda_{1}}\phi_{1}(t),\qquad\xi_{2}=\xi_{2}(\lambda_{1},\lambda_{2})=\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t).$

We prove the following:

######  Proposition 3.2.

For any $\lambda\in(0,1)$, there exist $\lambda_{1}$ and $\lambda_{2}$ with
$1<\lambda_{1}<\lambda_{2}$ so that (3.3) and (3.4) hold.

To prove this proposition, we prepare five lemmas.

######  Lemma 3.3.

Suppose that $\lambda\lambda_{2}>1$. Then $\xi_{1}(\lambda_{1},\lambda_{2})>0$
for any $\lambda$ and $\lambda_{2}$ as long as $\lambda_{1}$ is sufficiently
close to 1.

###### Proof.

We fix $\lambda$ and $\lambda_{2}$ so that $\lambda\lambda_{2}>1$. We set
$t(s)=\sqrt{(\lambda_{1}^{2}-1)s+1}$. Then

$\int_{1}^{\lambda_{1}}\phi_{1}(t)=\int_{0}^{1}\tilde{\phi}_{1}(s),$

where

$\tilde{\phi}_{1}(s)=\frac{(1-\lambda-\lambda_{1}+\lambda_{2})t(s)^{2}+\lambda\lambda_{1}-\lambda\lambda_{2}-\lambda_{1}\lambda_{2}+\lambda\lambda_{1}\lambda_{2}}{(t(s)^{2}-a^{2})\sqrt{(t(s)^{2}-\lambda^{2})(t(s)^{2}-1)(\lambda_{1}^{2}-t(s)^{2})(\lambda_{2}^{2}-t(s)^{2})}}\frac{\lambda_{1}^{2}-1}{2}ds.$

Since $\lambda\lambda_{2}>1$ and $\alpha\to 1+\lambda\lambda_{2}$ as
$\lambda_{1}\to 1$, we see that

$a^{2}\to\frac{1+\lambda\lambda_{2}-\big{|}\lambda\lambda_{2}-1\big{|}}{2}=1$

as $\lambda_{1}\to 1$. Then a straightforward calculation yields that

$\lim_{\lambda_{1}\to
1}\tilde{\phi}_{1}(s)=\frac{\lambda\lambda_{2}-1}{2\sqrt{(1-\lambda^{2})(\lambda_{2}^{2}-1)}}\frac{2(\lambda_{2}-\lambda)s-(1-\lambda)(1+\lambda_{2})}{2(\lambda\lambda_{2}-1)s+(1-\lambda)(1+\lambda_{2})}\frac{1}{\sqrt{s(1-s)}}ds$

and one can explicitly integrate the above function as follows.

$\int_{0}^{1}\lim_{\lambda_{1}\to
1}\tilde{\phi}_{1}(s)=\frac{\pi}{2}\left(\frac{\lambda_{2}-\lambda}{\sqrt{(1-\lambda^{2})(\lambda_{2}^{2}-1)}}-1\right).$

Hence the lemma holds if and only if

$\lambda_{2}-\lambda>\sqrt{(1-\lambda^{2})(\lambda_{2}^{2}-1)}$

holds, and this inequality is equivalent to

$(\lambda\lambda_{2}-1)^{2}>0,$

and this inequality holds by the assumption. This proves the lemma. ∎

######  Lemma 3.4.

$\xi_{1}(\lambda_{1},\lambda_{2})<0$ for any $\lambda$ as long as both
$\lambda_{1}$ and $\lambda_{2}$ are sufficiently large.

###### Proof.

We fix $\lambda$ and set $\lambda_{2}=\lambda_{1}+\mu$ for an arbitrary
positive real $\mu$. Then

$\alpha=\lambda-\mu+\lambda\mu+\lambda_{1}^{2}+\lambda_{1}\mu$

and

$\frac{\alpha}{\lambda_{1}^{2}}\to 1$

as $\lambda_{1}\to\infty$. We also have

$\displaystyle a^{2}$
$\displaystyle=\frac{\alpha-\sqrt{\alpha^{2}-4\lambda\lambda_{1}(\lambda_{1}+\mu)}}{2}=\frac{2\lambda\lambda_{1}(\lambda_{1}+\mu)}{\alpha+\sqrt{\alpha^{2}+4\lambda\lambda_{1}(\lambda_{1}+\mu)}}$
$\displaystyle=\frac{2\lambda\left(1+\frac{\mu}{\lambda_{1}}\right)}{\frac{\alpha}{\lambda_{1}^{2}}+\sqrt{\left(\frac{\alpha}{\lambda_{1}^{2}}\right)^{2}+\frac{4\lambda}{\lambda_{1}^{2}}\left(1+\frac{\mu}{\lambda_{1}}\right)}}\to\lambda$

as $\lambda_{1}\to\infty$. We set

$\gamma=\sqrt{\frac{\lambda\mu+(1-\lambda)\mu\lambda_{1}+(1-\lambda)\lambda_{1}^{2}}{1-\lambda+\mu}}\in(1,\lambda_{1}).$

Then we see that

(3.6)
$\gamma\to\infty,\qquad\frac{\gamma}{\lambda_{1}}\to\sqrt{\frac{1-\lambda}{1-\lambda+\mu}}$

as $\lambda_{1}\to\infty$. Now we set $u=1/t$. Then

$\displaystyle\frac{1}{1-\lambda+\mu}\int_{1}^{\lambda_{1}}\phi_{1}(t)$
$\displaystyle=\int_{1}^{\lambda_{1}}\frac{t(t^{2}-\gamma^{2})}{(t^{2}-a^{2})\sqrt{(t^{2}-\lambda^{2})(t^{2}-1)(\lambda_{1}^{2}-t^{2})((\lambda_{1}+\mu)^{2}-t^{2})}}dt$
$\displaystyle=\int_{1/\lambda_{1}}^{1}\frac{u(1-\gamma^{2}u^{2})}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du$
(3.7)
$\displaystyle=\int_{1/\lambda_{1}}^{1/\gamma}\frac{u(1-\gamma^{2}u^{2})}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du$
(3.8)
$\displaystyle\quad-\int_{1/\gamma}^{1}\frac{u(\gamma^{2}u^{2}-1)}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du.$

We see that $\eqref{eq:34-1}\to 0$ as $\lambda_{1}\to\infty$. In fact,

$\displaystyle\eqref{eq:34-1}$
$\displaystyle<\frac{\frac{1}{\gamma}\left(1-\frac{\gamma^{2}}{\lambda_{1}^{2}}\right)}{\left(1-\frac{a^{2}}{\gamma^{2}}\right)\sqrt{\left(1-\frac{\lambda^{2}}{\gamma^{2}}\right)\left(1-\frac{1}{\gamma^{2}}\right)\left(\frac{\lambda_{1}}{\lambda_{1}}+1\right)\left(\frac{(\lambda_{1}+\mu)^{2}}{\lambda_{1}^{2}}-1\right)}}\int_{1/\lambda_{1}}^{1/\gamma}\frac{du}{\sqrt{\lambda_{1}u-1}}$
$\displaystyle=\frac{1}{\sqrt{\gamma^{2}\left(\frac{(\lambda_{1}+\mu)^{2}}{\lambda_{1}^{2}}-1\right)}}\frac{1-\frac{\gamma^{2}}{\lambda_{1}^{2}}}{\left(1-\frac{a^{2}}{\gamma^{2}}\right)\sqrt{2\left(1-\frac{\lambda^{2}}{\gamma^{2}}\right)\left(1-\frac{1}{\gamma^{2}}\right)}}\int_{1/\lambda_{1}}^{1/\gamma}\frac{du}{\sqrt{\lambda_{1}u-1}},$

and by (3.6), we have

$\gamma^{2}\left(\frac{(\lambda_{1}+\mu)^{2}}{\lambda_{1}^{2}}-1\right)=\frac{\gamma^{2}}{\lambda_{1}^{2}}\left(2\lambda_{1}+\mu\right)\mu\to\infty$

and also

$\int_{1/\lambda_{1}}^{1/\gamma}\frac{du}{\sqrt{\lambda_{1}u-1}}=\frac{2}{\lambda_{1}}\sqrt{\frac{\lambda_{1}}{\gamma}-1}\to
0$

as $\lambda_{1}\to\infty$. Thus $\eqref{eq:34-1}\to 0$ as
$\lambda_{1}\to\infty$. Next we estimate (3.8). We are considering
$\lambda_{1}$ to be large, and so there exists $\epsilon_{1}\in(0,1-1/\gamma)$
which does not depend on $\lambda_{1}$. We fix this $\epsilon_{1}$ and we
split the integral as follows.

$\displaystyle\int_{1/\gamma}^{1}\frac{u(\gamma^{2}u^{2}-1)}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du$
(3.9)
$\displaystyle=\int_{1/\gamma}^{(1/\gamma)+\epsilon_{1}}\frac{u(\gamma^{2}u^{2}-1)}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du$
(3.10)
$\displaystyle\quad+\int_{(1/\gamma)+\epsilon_{1}}^{1}\frac{u(\gamma^{2}u^{2}-1)}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du.$

By the mean value theorem for integrals, there exists
$u_{0}\in(1/\gamma,(1/\gamma)+\epsilon_{1})$ such that

$\displaystyle\eqref{eq:34-3}$
$\displaystyle=\epsilon_{1}\frac{u_{0}(\gamma^{2}u_{0}^{2}-1)}{(1-a^{2}u_{0}^{2})\sqrt{(1-\lambda^{2}u_{0}^{2})(1-u_{0}^{2})(\lambda_{1}^{2}u_{0}^{2}-1)((\lambda_{1}+\mu)^{2}u_{0}^{2}-1)}}$
$\displaystyle=\epsilon_{1}\frac{\gamma^{2}}{\lambda_{1}^{2}}\frac{u_{0}\left(u_{0}^{2}-\frac{1}{\gamma^{2}}\right)}{(1-a^{2}u_{0}^{2})\sqrt{(1-\lambda^{2}u_{0}^{2})(1-u_{0}^{2})\left(u_{0}^{2}-\frac{1}{\lambda_{1}^{2}}\right)\left(\left(1+\frac{\mu}{\lambda_{1}}\right)^{2}u_{0}^{2}-\frac{1}{\lambda_{1}^{2}}\right)}}$

which cannot be negative as $\lambda_{1}\to\infty$. On the other hand,

$\displaystyle\eqref{eq:34-4}$
$\displaystyle>\frac{\left(\frac{1}{\gamma}+\epsilon_{1}\right)\left(\gamma^{2}\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}-1\right)}{\left(1-a^{2}\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}\right)\sqrt{\left(1-\lambda^{2}\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}\right)\left(1-\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}\right)(\lambda_{1}^{2}-1)((\lambda_{1}+\mu)^{2}-1)}}$
$\displaystyle=\frac{\gamma^{2}}{\lambda_{1}^{2}}\frac{\left(\frac{1}{\gamma}+\epsilon_{1}\right)\left(\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}-\frac{1}{\gamma^{2}}\right)}{\left(1-a^{2}\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}\right)\sqrt{\left(1-\lambda^{2}\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}\right)\left(1-\left(\frac{1}{\gamma}+\epsilon_{1}\right)^{2}\right)\left(1-\frac{1}{\lambda_{1}^{2}}\right)\left(\left(1+\frac{\mu}{\lambda_{1}}\right)^{2}-\frac{1}{\lambda_{1}^{2}}\right)}}$
$\displaystyle\to\frac{1-\lambda}{1-\lambda+\mu}\frac{\epsilon_{1}^{3}}{(1-\lambda\epsilon_{1}^{2})\sqrt{(1-\lambda^{2}\epsilon_{1}^{2})(1-\epsilon_{1}^{2})}}>0$

as $\lambda_{1}\to\infty$. Thus

$\lim_{\lambda_{1}\to\infty}\int_{1/\gamma}^{1}\frac{u(\gamma^{2}u^{2}-1)}{(1-a^{2}u^{2})\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(\lambda_{1}^{2}u^{2}-1)((\lambda_{1}+\mu)^{2}u^{2}-1)}}du>0.$

Therefore

$\lim_{\lambda_{1}\to\infty}\frac{1}{1-\lambda+\mu}\int_{1}^{\lambda_{1}}\phi_{1}(t)dt=0-(\text{positive})<0.$

This proves the lemma. ∎

######  Lemma 3.5.

Suppose that $\lambda\lambda_{2}\leq 1$. Then
$\xi_{2}(\lambda_{1},\lambda_{2})>0$ for any $\lambda$ and $\lambda_{2}$ as
long as $\lambda_{1}$ is sufficiently close to 1.

###### Proof.

We fix $\lambda$ and $\lambda_{2}$ so that $\lambda\lambda_{2}\leq 1$. Since
$\alpha\to 1+\lambda\lambda_{2}$ as $\lambda_{1}\to 1$, we see that

$\lim_{\lambda_{1}\to
1}b^{2}=\frac{1+\lambda\lambda_{2}+\big{|}1-\lambda\lambda_{2}\big{|}}{2}=1.$

We set $t=t(s)=(\lambda_{2}-\lambda_{1})s+\lambda_{1}$. Then
$\lambda<1<t(s)<\lambda_{1}<\lambda_{2}$ for $s\in(0,1)$ and hence

$\begin{split}\lim_{\lambda_{1}\to
1}\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)&=\lim_{\lambda_{1}\to
1}\int_{0}^{1}\frac{(t(s)^{2}-b^{2})(\lambda_{2}-\lambda_{1})ds}{\sqrt{(t(s)^{2}-\lambda^{2})(t(s)^{2}-1)(t(s)^{2}-\lambda_{1}^{2})(\lambda_{2}^{2}-t(s)^{2})}}\\\
&=\int_{0}^{1}\frac{(t(s)^{2}-1)(\lambda_{2}-1)ds}{\sqrt{(t(s)^{2}-\lambda^{2})(t(s)^{2}-1)^{2}(\lambda_{2}^{2}-t(s)^{2})}}\\\
&=\int_{0}^{1}\frac{(\lambda_{2}-1)ds}{\sqrt{(t(s)^{2}-\lambda^{2})(\lambda_{2}^{2}-t(s)^{2})}}>0.\end{split}$

This proves the lemma. ∎

######  Lemma 3.6.

$\xi_{2}(\lambda_{1},\lambda_{2})>0$ for any $\lambda$ and $\lambda_{1}$ as
long as $\lambda_{2}$ is sufficiently close to $\lambda_{1}$.

###### Proof.

Fix $\lambda$ and $\lambda_{1}$. We first show that

(3.11)
$\lim_{\lambda_{2}\to\lambda_{1}}\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)=0.$

To do this, we set

$\phi_{21}(t)=\frac{t^{2}-b^{2}}{\sqrt{(t^{2}-\lambda^{2})(t^{2}-1)}},\qquad\phi_{22}(t)=\frac{1}{\sqrt{(t^{2}-\lambda_{1}^{2})(\lambda_{2}^{2}-t^{2})}}.$

Then $\phi_{2}(t)=\phi_{21}(t)\phi_{22}(t)dt$. By the generalized mean value
theorem for integrals, there exists $t_{0}\in(\lambda_{1},\lambda_{2})$ such
that

$\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)=\phi_{21}(t_{0})\int_{\lambda_{1}}^{\lambda_{2}}\phi_{22}(t)dt.$

Since $0<\lambda_{1}<\lambda_{2}$, It is known that

$\int_{\lambda_{1}}^{\lambda_{2}}\phi_{22}(t)dt=\int_{\lambda_{1}}^{\lambda_{2}}\frac{dt}{\sqrt{(t^{2}-\lambda_{1}^{2})(\lambda_{2}^{2}-t^{2})}}=\frac{1}{\lambda_{2}}\mathbb{K}\left(1-\frac{\lambda_{1}^{2}}{\lambda_{2}^{2}}\right),$

where $\mathbb{K}$ is the complete elliptic integral of the first kind. Hence

$\lim_{\lambda_{2}\to\lambda_{1}}\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)=\frac{1}{\lambda_{1}}\mathbb{K}(0)=\frac{\pi}{2\lambda_{1}}>0.$

On the other hand, since $\alpha\to\lambda+\lambda_{1}^{2}$ as
$\lambda_{2}\to\lambda_{1}$, we see that $b^{2}\to\lambda_{1}^{2}$ as
$\lambda_{2}\to\lambda_{1}$. Thus

$\lim_{\lambda_{2}\to\lambda_{1}}\phi_{21}(t_{0})=0$

and hence (3.11) holds. We next show that

(3.12)
$\lim_{\lambda_{2}\to\lambda_{1}}\frac{\partial}{\partial\lambda_{2}}\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)>0.$

We set $t=t(s)=(\lambda_{2}-\lambda_{1})s+\lambda_{1}$. Then

$\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)=\int_{0}^{1}\tilde{\phi}_{2}(s),$

where

$\tilde{\phi}_{2}(s)=\frac{t(s)^{2}-b^{2}}{\sqrt{(t(s)^{2}-\lambda^{2})(\lambda_{2}^{2}-t(s)^{2})}}\frac{ds}{\sqrt{s((\lambda_{2}-\lambda_{1})s+2\lambda_{1})(-(\lambda_{2}-\lambda_{1})s^{2}-2\lambda_{1}s+\lambda_{1}+\lambda_{2})}}.$

A straightforward but extremely long calculation yields that

$\lim_{\lambda_{2}\to\lambda_{1}}\frac{\partial}{\partial\lambda_{2}}\tilde{\phi}_{2}=\frac{2(\lambda_{1}^{2}-\lambda)s-(\lambda+\lambda_{1})(\lambda_{1}-1)}{2(\lambda_{1}^{2}-\lambda)\sqrt{(\lambda_{1}^{2}-\lambda^{2})(\lambda_{1}^{2}-1)}\sqrt{s(1-s)}}ds.$

Therefore

$\displaystyle\lim_{\lambda_{2}\to\lambda_{1}}\frac{\partial}{\partial\lambda_{2}}\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)$
$\displaystyle=\int_{0}^{1}\frac{2(\lambda_{1}^{2}-\lambda)s-(\lambda+\lambda_{1})(\lambda_{1}-1)}{2(\lambda_{1}^{2}-\lambda)\sqrt{(\lambda_{1}^{2}-\lambda^{2})(\lambda_{1}^{2}-1)}\sqrt{s(1-s)}}ds$
$\displaystyle=\frac{(1-\lambda)\lambda_{1}\pi}{2(\lambda_{1}^{2}-\lambda)\sqrt{(\lambda_{1}^{2}-\lambda^{2})(\lambda_{1}^{2}-1)}}>0,$

and hence (3.12) is proven. (3.11) and (3.12) proves the lemma. ∎

######  Lemma 3.7.

$\xi_{2}(\lambda_{1},\lambda_{2})<0$ for any $\lambda$ and $\lambda_{1}$ as
long as $\lambda_{2}$ is sufficiently large.

###### Proof.

Fix $\lambda$ and $\lambda_{1}$. We first remark that

(3.13) $\frac{b^{2}}{\lambda_{2}}\to\lambda+\lambda_{1}-1,\qquad b\to\infty$

as $\lambda_{2}\to\infty$. Now we set $u=1/t$. Then

$\displaystyle\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)$
$\displaystyle=\int_{1/\lambda_{2}}^{1/\lambda_{1}}\frac{1-b^{2}u^{2}}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du$
(3.14)
$\displaystyle=\int_{1/\lambda_{2}}^{1/b}\frac{1-b^{2}u^{2}}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du$
(3.15)
$\displaystyle\quad-\int_{1/b}^{1/\lambda_{1}}\frac{b^{2}u^{2}-1}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du.$

We see that $\eqref{eq:36-1}\to 0$ as $\lambda_{2}\to\infty$. In fact,

$\eqref{eq:36-1}<\frac{1-\frac{b^{2}}{\lambda_{2}^{2}}}{\sqrt{\left(1-\frac{\lambda^{2}}{b^{2}}\right)\left(1-\frac{1}{b^{2}}\right)\left(1-\frac{\lambda_{1}^{2}}{b^{2}}\right)\left(\frac{\lambda_{2}}{\lambda_{2}}+1\right)}}\int_{1/\lambda_{2}}^{1/b}\frac{du}{\sqrt{\lambda_{2}u-1}}$

and by (3.13), we have

$\int_{1/\lambda_{2}}^{1/b}\frac{du}{\sqrt{\lambda_{2}u-1}}=\frac{2}{\lambda_{2}}\sqrt{\frac{\lambda_{2}}{b}-1}=2\sqrt{\frac{1}{b\lambda_{2}}-\frac{1}{\lambda_{2}^{2}}}\to
0$

as $\lambda_{2}\to\infty$. Thus $\eqref{eq:36-1}\to 0$ as
$\lambda_{2}\to\infty$.

Next we estimate (3.15). We are considering $\lambda_{2}$ to be large, and so
there exists $\epsilon_{2}\in(0,(1/\lambda_{1})-(1/b))$ which does not depend
on $\lambda_{2}$. We fix this $\epsilon_{2}$ and we split the integral as
follows.

$\displaystyle\int_{1/b}^{1/\lambda_{1}}\frac{b^{2}u^{2}-1}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du$
(3.16)
$\displaystyle=\int_{1/b}^{(1/b)+\epsilon_{2}}\frac{b^{2}u^{2}-1}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du$
(3.17)
$\displaystyle\quad+\int_{(1/b)+\epsilon_{2}}^{1/\lambda_{1}}\frac{b^{2}u^{2}-1}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du.$

By the mean value theorem for integrals, there exists
$u_{1}\in(1/b,(1/b)+\epsilon_{2})$ such that

$\displaystyle\eqref{eq:36-3}$
$\displaystyle=\epsilon_{2}\frac{b^{2}u_{1}^{2}-1}{\sqrt{(1-\lambda^{2}u_{1}^{2})(1-u_{1}^{2})(1-\lambda_{1}^{2}u_{1}^{2})(\lambda_{2}^{2}u_{1}^{2}-1)}}$
$\displaystyle=\epsilon_{2}\frac{\frac{b^{2}}{\lambda_{2}}u_{1}^{2}-\frac{1}{\lambda_{2}}}{\sqrt{(1-\lambda^{2}u_{1}^{2})(1-u_{1}^{2})(1-\lambda_{1}^{2}u_{1}^{2})\left(u_{1}^{2}-\frac{1}{\lambda_{2}^{2}}\right)}}$

which converges to a non-negative number as $\lambda_{2}\to\infty$. On the
other hand,

$\eqref{eq:36-4}>\frac{1}{\sqrt{\left(1-\lambda^{2}\left(\frac{1}{b}+\epsilon_{2}\right)^{2}\right)\left(1-\left(\frac{1}{b}+\epsilon_{2}\right)^{2}\right)\left(1-\lambda_{1}^{2}\left(\frac{1}{b}+\epsilon_{2}\right)^{2}\right)}}\int_{(1/b)+\epsilon_{2}}^{1/\lambda_{1}}\frac{b^{2}u^{2}-1}{\sqrt{\lambda_{2}^{2}u^{2}-1}}du.$

By the mean value theorem for integrals, there exists
$u_{2}\in((1/b)+\epsilon_{2},1/\lambda_{1})$ such that

$\int_{(1/b)+\epsilon_{2}}^{1/\lambda_{1}}\frac{b^{2}u^{2}-1}{\sqrt{\lambda_{2}^{2}u^{2}-1}}du=\left(\frac{1}{\lambda_{1}}-\left(\frac{1}{b}+\epsilon_{2}\right)\right)\frac{b^{2}u_{2}^{2}-1}{\sqrt{\lambda_{2}^{2}u_{2}^{2}-1}}=\left(\frac{1}{\lambda_{1}}-\frac{1}{b}-\epsilon_{2}\right)\frac{\frac{b^{2}}{\lambda_{2}}u_{2}-\frac{1}{\lambda_{2}u_{2}}}{\sqrt{1-\frac{1}{\lambda_{2}^{2}u_{2}^{2}}}}.$

Note that $u_{2}\geq\epsilon_{2}$ as $\lambda_{2}\to\infty$ because
$u_{2}>(1/b)+\epsilon_{2}$. Thus

$\lim_{\lambda_{2}\to\infty}\int_{(1/b)+\epsilon_{2}}^{1/\lambda_{1}}\frac{b^{2}u^{2}-1}{\sqrt{\lambda_{2}^{2}u^{2}-1}}du=\left(\frac{1}{\lambda_{1}}-\epsilon_{2}\right)(\lambda+\lambda_{1}-1)u_{2}>0$

and hence

$\displaystyle\int_{1/b}^{1/\lambda_{1}}$
$\displaystyle\frac{b^{2}u^{2}-1}{\sqrt{(1-\lambda^{2}u^{2})(1-u^{2})(1-\lambda_{1}^{2}u^{2})(\lambda_{2}^{2}u^{2}-1)}}du$
$\displaystyle\to\frac{1}{\sqrt{\left(1-\lambda^{2}\epsilon_{2}^{2}\right)\left(1-\epsilon_{2}^{2}\right)\left(1-\lambda_{1}^{2}\epsilon_{2}^{2}\right)}}\left(\frac{1}{\lambda_{1}}-\epsilon_{2}\right)(\lambda+\lambda_{1}-1)u_{2}>0$

as $\lambda_{2}\to\infty$. Therefore

$\lim_{\lambda_{2}\to\infty}\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)=0-(\text{positive})<0.$

This completes the proof. ∎

###### Proof of Proposition 3.2.

Choose any $\lambda\in(0,1)$. We compile our results for the sign of

$(\xi_{1},\xi_{2})=\left(\int_{1}^{\lambda_{1}}\phi_{1}(t),\int_{\lambda_{1}}^{\lambda_{2}}\phi_{2}(t)\right)$

into a parallelogram on which $1<\lambda_{1}<\lambda_{2}$. In the
$(\lambda_{1},\lambda_{2})$-plane, the vertices of this parallelogram are at
$(\lambda_{1},\lambda_{2})=(1+\epsilon,1+2\epsilon),(N,N+\epsilon),(N,2N),$
and $(1+\epsilon,1+N+\epsilon)$. We label these points respectively as $A$,
$B$, $C$, and $D$. We also label the point $(1+\epsilon,1/\lambda)$ as $E$.
Here $\epsilon$ very small and $N$ very large are chosen so that $\xi_{1}$ is
positive on $\overline{DE}$ (Lemma 3.3) and negative on $\overline{BC}$ (Lemma
3.4) and additionally $\xi_{2}$ is positive on $\overline{AE}$ (Lemma 3.5) and
on $\overline{AB}$ (Lemma 3.6) and negative on $\overline{CD}$ (Lemma 3.7).
See Figure 3.1.

Figure 3.1. Signs of $\xi_{1}$ and $\xi_{2}$.

Since $\xi_{1}$ is continuous on the parallelogram and $\xi_{1}>0$ on
$\overline{DE}$ and $\xi_{1}<0$ on $\overline{BC}$, there exists a curve

$c_{1}=\\{(\lambda_{1},\lambda_{2}):\xi_{1}(\lambda_{1},\lambda_{2})=0\\}$

in the parallelogram so that an endpoint of $c_{1}$ lies on $\overline{CD}$
and the other endpoint lies on $\overline{EA}\cup\overline{AB}$. Similarly
there exists a curve

$c_{2}=\\{(\lambda_{1},\lambda_{2}):\xi_{2}(\lambda_{1},\lambda_{2})=0\\}$

in the parallelogram so that an endpoint of $c_{2}$ lies on $\overline{BC}$
and the other endpoint lies on $\overline{DE}$. $c_{1}$ divides the
parallelogram into two regions, one includes $\overline{BC}$ and the other
includes $\overline{DE}$. On the other hand, the endpoints of $c_{2}$ lie on
$\overline{BC}$ and $\overline{DE}$. Therefore $c_{2}$ must intersect to
$c_{1}$. The intersection $(\lambda_{1},\lambda_{2})$ satisfies

$\xi_{1}(\lambda_{1},\lambda_{2})=\xi_{2}(\lambda_{1},\lambda_{2})=0.$

Thus the minimal surface exists for that $\lambda$, and because $\lambda$ was
arbitrary, the minimal surface exists for all $\lambda\in(0,1)$. ∎

Summing up, we have proven Proposition 3.1.

##  4\. Embeddedness

In this section we prove the following proposition.

######  Proposition 4.1.

For each $\lambda\in(0,1)$ and corresponding $\lambda_{1}$ and $\lambda_{2}$,
the surface $M_{\lambda}$ given by Theorem 1.1 is embedded.

To prove this, we first recall three facts, which are central to the arguments
in this section.

######  Fact 4.2 (Schwarz Reflection Principle [5, Theorem 1.5.1]).

Suppose a minimal surface $M$ contains in its boundary a curve $C$ that is
either a straight line or a nonstraight planar geodesic. Then $M$ can be
extend smoothly across $C$ by respectively $180^{\circ}$-rotation about $C$ or
reflection through the plane containing $C$.

######  Fact 4.3 (Krust [5, Theorem 2.4.1], see also [2, Theorem 3.1]).

If an embedded minimal surface $X:\mathbb{D}\to\mathbb{R}^{3}$,
$\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ can be written as a graph over a
convex domain in a plane, then the conjugate surface
$X^{*}:\mathbb{D}\to\mathbb{R}^{3}$ is also a graph over a domain in the same
plane.

######  Fact 4.4 (Jenkins and Serrin [3], see also [7, Theorem 3.3, Remark
3.4]).

Let $D$ be a bounded convex domain such that $\partial D$ contains two sets of
finite numbers of open straight segments $\\{A_{i}\\}$, $\\{B_{j}\\}$ with the
property that no two segments $A_{i}$ and no two segments $B_{j}$ have a
common endpoint. Let the remaining portion of $\partial D$ consist of a finite
number of open arcs $\\{C_{k}\\}$, and of endpoints of $\\{A_{i}\\}$,
$\\{B_{j}\\}$, and $\\{C_{k}\\}$. Let $\mathcal{P}$ denote a simple closed
polygon whose vertices are chosen from among the endpoints of the $A_{i}$ and
$B_{j}$. Let

$\displaystyle\alpha$
$\displaystyle=\sum_{A_{i}\subset\mathcal{P}}\mathrm{Length}\,A_{i},$
$\displaystyle\beta$
$\displaystyle=\sum_{B_{j}\subset\mathcal{P}}\mathrm{Length}\,B_{j},$
$\displaystyle\gamma$ $\displaystyle=\mbox{\rm length of perimeter of
$\mathcal{P}$}.$

Then if $\\{C_{k}\\}\neq\emptyset$, there exists a solution of the minimal
surface equation in $D$ which assumes the value $+\infty$ on each $A_{i}$, and
$-\infty$ on each $B_{j}$, and any assigned bounded continuous data on each
open arc $C_{k}$ if and only if

(4.1) $2\alpha<\gamma\quad\text{and}\quad 2\beta<\gamma$

for each polygon $\mathcal{P}$ chosen as above. Moreover, the solution is
unique when it exists.

###### Proof of Proposition 4.1.

The surface $M_{\lambda}$ has symmetry planes parallel to the three coordinate
planes. Thus, $M_{\lambda}$ is made up of eight congruent pieces. Hence, we
only need to show that one of these eight pieces is embedded. Let $R_{1}$ be
the portion of the domain $R$ restricted to the first quadrant of the
$z$-plane with $w(\infty)=+1$. We set $\Omega_{1}:=X(R_{1})$. See Figure 4.1,
left.

Figure 4.1. Fundamental piece $\Omega_{1}$ (left) and its conjugate
$\Omega_{1}^{*}$ (right)

We will show that $\Omega_{1}$ is a graph over a domain in the
$(x_{2},x_{3})$-plane. By Lemma 2.3 and Fact 4.2, the boundary of $\Omega_{1}$
consists of seven planar symmetry curves:

  1. (1)

$s_{1}$ is the real interval $[\lambda,a)$ and $S_{1}:=X(s_{1})$ lies in a
plane parallel to the $(x_{2},x_{3})$-plane;

  2. (2)

$s_{2}$ is the real interval $[0,\lambda]$ and $S_{2}:=X(s_{2})$ lies in a
plane parallel to the $(x_{1},x_{3})$-plane;

  3. (3)

$s_{3}$ is the positive imaginary axis and $S_{3}:=X(s_{3})$ lies in a plane
parallel to the $(x_{1},x_{2})$-plane;

  4. (4)

$s_{4}$ is the real interval $[\lambda_{2},\infty)$ and $S_{4}:=X(s_{4})$ lies
in a plane parallel to the $(x_{1},x_{3})$-plane different than the plane
containing $S_{2}$;

  5. (5)

$s_{5}$ is the real interval $[\lambda_{1},\lambda_{2}]$ and $S_{5}:=X(s_{5})$
lies in a plane parallel to the $(x_{2},x_{3})$-plane different than the plane
containing $S_{1}$;

  6. (6)

$s_{6}$ is the real interval $[1,\lambda_{1}]$ and $S_{6}:=X(s_{6})$ lies in
plane containing $S_{4}$;

  7. (7)

$s_{7}$ is the real interval $(a,1]$ and $S_{7}:=X(s_{7})$ lies in plane
containing $S_{5}$.

Take the conjugate $\Omega_{1}^{*}:=X^{*}(R_{1})$ of $\Omega_{1}$. See Figure
4.1, right. By Fact 4.3, it suffices to prove that $\Omega_{1}^{*}$ is a graph
over a domain in the $(x_{2},x_{3})$-plane. By Remark 2.1, the boundary of
$\Omega_{1}^{*}$ consists of seven straight lines:

  1. (1)

$S_{1}^{*}:=X^{*}(s_{1})$ is a straight line parallel to the $x_{1}$-axis;

  2. (2)

$S_{2}^{*}:=X^{*}(s_{2})$ is a straight segment parallel to the $x_{2}$-axis;

  3. (3)

$S_{3}^{*}:=X^{*}(s_{3})$ is a straight segment parallel to the $x_{3}$-axis;

  4. (4)

$S_{4}^{*}:=X^{*}(s_{4})$ is a straight segment parallel to the $x_{2}$-axis;

  5. (5)

$S_{5}^{*}:=X^{*}(s_{5})$ is a straight segment parallel to the $x_{1}$-axis;

  6. (6)

$S_{6}^{*}:=X^{*}(s_{6})$ is a straight segment parallel to the $x_{2}$-axis;

  7. (7)

$S_{7}^{*}:=X^{*}(s_{7})$ is a straight line parallel to the $x_{1}$-axis.

Since $G(a)=i$, two straight lines $S_{1}^{*}$ and $S_{7}^{*}$ lie in a plane
parallel to the $(x_{1},x_{3})$-plane. After a suitable translation, we may
assume that the $(x_{2},x_{3})$-plane contains $S_{2}^{*}$, $S_{3}^{*}$, and
$S_{4}^{*}$. Now we extend $\Omega_{1}^{*}$ by $180^{\circ}$-rotation with
respect to the straight segment $S_{3}^{*}$. See Figure 4.2, left.

Figure 4.2. $\Omega^{*}_{2}$ (left), and their projections onto
$x_{2}x_{3}$-plane (right)

We denote by $\Omega^{*}_{2}$ the resulting surface, which is two copies of
$\Omega_{1}^{*}$. We also denote by $\hat{S}_{2}^{*}$, $\hat{S}_{4}^{*}$,
$\hat{S}_{6}^{*}$ the copies of $S_{2}^{*}$, $S_{4}^{*}$, $S_{6}^{*}$,
respectively. Let $\pi:\mathbb{R}^{3}\to\mathbb{R}^{2}$ be the projection from
$\mathbb{R}^{3}$ into the $(x_{2},x_{3})$-plane, that is,
$\pi(x_{1},x_{2},x_{3})=(x_{2},x_{3})$. Let $C_{k}$ ($k=1,2,3,4$) be straight
segments in $(x_{2},x_{3})$-plane as

$C_{1}=\pi(\hat{S}_{6}^{*}),\quad C_{2}=S_{4}^{*}\cup\hat{S}_{4}^{*},\quad
C_{3}=\pi(S_{6}^{*}),\quad C_{4}=S_{2}^{*}\cup\hat{S}_{2}^{*}.$

We also label $A_{1}$ (resp. $B_{1}$) the straight segment parallel to the
$x_{3}$-axis connecting $C_{3}$ (resp. $C_{1}$) and $C_{4}$. See Figure 4.2,
right. Let $D$ be the rectangular domain in the $(x_{2},x_{3})$-plane such
that the boundary $\partial D$ consists of $A_{1}$, $B_{1}$, $C_{k}$
($k=1,2,3,4$), and their endpoints. Then by Fact 4.4, $\Omega^{*}_{2}$ is a
graph over $D$ (we remark that in this setting $\mathcal{P}$ is a triangle
containing either $A_{1}$ and $C_{4}$ or $B_{1}$ and $C_{4}$, and it is
obvious that the condition (4.1) holds for both cases). Moreover, the
uniqueness of Fact 4.4 yields that $\Omega_{1}^{*}$ is also a graph over a
domain in the $(x_{2},x_{3})$-plane. This completes the proof. ∎

## Acknowledgements

The authors thank Professors Wayne Rossman and Kenji Takeda for fruitful
discussions.

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