# On the solution of a conformal mapping problem by means of Weierstrass
functions

Smirnov Matvey  119991 Russia, Moscow GSP-1, ul. Gubkina 8, Institute for
Numerical Mathematics, Russian Academy of Sciences<EMAIL_ADDRESS>
###### Abstract.

The conformal mapping problem for the section of a channel filled with porous
material under a rectangular dam onto the upper half-plane is considered.
Similar problems arise in computing of fluid flow in hydraulic structures. As
a solution method, the representation of Christoffel–Schwartz elliptic
integral in terms of Weierstrass functions is used. The calculation is based
on Taylor series for the sigma function, the coefficients of which are
determined recursively. A simple formula for a conformal mapping is obtained,
which depends on four parameters and uses the sigma function. A numerical
experiment was carried out for a specific area. The degeneration of the
region, which consists in the dam width tending to zero, is considered, and it
is shown that the resulting formula has a limit that implements the solution
of the limiting problem. A refined proof of Weierstrass recursive formula for
the coefficients of Taylor series of the sigma function is presented.

Keywords. Conformal mappings, Christoffel-Schwartz integral, elliptic
funtions, Weierstrass sigma function, degeneration of Weierstrass functions.

##  1\. Introduction

A region $\Omega\subset\mathbb{C}$, the boundary of which is a polygonal curve
with angles multiple of $\pi/2$, is considered in this article. This region
models the shape of a channel under a dam. The calculation of fluid flow in
such channel boils down to a conformal mapping problem of $\Omega$ onto the
upper half-plane. The solution of such problems is given by Christoffel-
Schwartz integral (see, e.g., [15] or [10]), which in this case is naturally
defined on an elliptic Riemann surface. In this paper the simple formula,
which writes integral in terms of Weierstrass sigma function (see e.g., [1] or
[9]), was found. This approach allows to avoid numerical integration and the
mapping parameters can be found from a simple nonlinear system of equations;
therefore, the computation is significantly simplified.

Similar problems have been considered in [4], [6], [7], [5], and [3], where
Christoffel-Schwartz integral was efficiently represented by theta functions
(see, e.g., [11]). In the paper [3] the application of Lauricella functions to
these problems was studied, and also different approaches were compared.

The main advantage of Weierstrass functions over theta functions is that they
have limiting values when the surface degenerates. The paper analyzes behavior
of the constructed conformal mapping under the condition that the dam width
tends to zero. It turns out that conformal mappings have a limit, which is a
solution to the limiting problem. Thus, it is shown that the solution is
stable under the considered degeneration.

The described property of the Weierstrass sigma function loses its value if
the standard method is used for calculations, which expresses the sigma
function in terms of the theta function (since it does not withstand
degeneration). Thus, it is necessary to use an independent method for
calculating sigma functions. In this paper, we use the expression for the
coefficients of its expansion into Taylor series obtained by Weierstrass (see
[17]). Since the proof presented there is apparently not complete (at one
point, the analyticity of the sigma function in three variables in a
neighborhood of zero is used, which is not obvious), we give a more detailed
proof in the appendix. The above formula, however, is not sufficient for the
final numerical solution, since Taylor series are not suitable for
calculations with large arguments (namely, such a need arises with
degeneration). Thus, the problem of constructing an efficient computational
method for the sigma function independent of theta functions still remains
unsolved. If such a method is available, it will be possible to construct
formulas that are stable under various degenerations and use them in
calculations. The results of this work illustrate the need in such methods.

Problems in which hyperelliptic Riemann surfaces of higher genus arise can
also be solved using the theory of sigma functions developed by Klein and
Baker in [14] and [2] respectively (more detailed exposition can be found in
[16]). There is hope that it will be possible to prove the stability of
formulas expressing the solution of the above problems in terms of high-order
sigma functions. Thus, the construction of Weierstrass-type recurrent formulas
(which are known for genus 1 and 2; see [16]) and calculation methods for
sigma functions can be extremely useful in applied problems.

##  2\. The statement and the origin of the problem

Consider region $\Omega$ in the complex plane pictured on Figure 1.

$w_{4}$$w_{3}$$w_{2}$$w_{1}$$\delta$$h$$h^{-}$$h^{+}$ Figure 1. Region
$\Omega$.

It is bounded from below by a line and from above by a polygonal curve with
four vertices $w_{1},w_{2},w_{3},w_{4}$ (it is convenient to think that this
region also has two vertices at $\pm\infty$). Let the line, which bounds the
region from below be parallel to the real axis, and let the vertex $w_{4}$ be
at the origin. Then, this region is determined by four real parameters
$h^{-},h^{+},h,\delta$, where $h$ is equal to the length of segment
$[w_{1},w_{2}]$, $\delta$ is the length of $[w_{2},w_{3}]$, and $h^{-}$ and
$h^{+}$ are equal to the distance from the line that bounds the region from
below to $w_{4}$ and $w_{1}$ respectively. These parameters are positive and
satisfy inequalities $h^{-}-h^{+}+h>0$, which corresponds to positivity of the
length of $[w_{3},w_{4}]$, and $h<h^{+}$. The region is determined uniquely by
these parameters.

Regions similar to $\Omega$ arise in problems connected with the computation
of fluid flow through the porous material under a dam. Since the flow is
continuous and satisfies the Darcy’s law, the pressure $p$ is a harmonic
function in $\Omega$. Assuming that segments $[w_{1},w_{2}]$, $[w_{2},w_{3}]$,
$[w_{3},w_{4}]$, and channel’s bottom are impenetrable, we obtain natural
boundary conditions: the normal derivative $\partial p/\partial n$ vanishes on
impenetrable segments of the boundary, while on the remaining segments (i.e.
on the half-lines starting from $w_{1}$ and $w_{4}$) $p$ is locally constant.

Consider a real-valued function $q$ in the region $\Omega$ such that $f=p+iq$
is holomorphic (such function exists because $\Omega$ is simply connected).
The normal derivative of $p$ vanishing condition is easily equivalent to
constancy of $q$ on the corresponding boundary segment. It follows, that, if
$f$ is a function that conformally maps $\Omega$ onto a rectangle in a way,
such that $w_{1}$, $w_{4}$, and vertices at infinity are mapped to the
vertices of a rectangle, then $p=\operatorname{Re}f$ is a solution to the
original problem. $q=\operatorname{Im}f$ is called the current function. Its
level lines are the streamlines of the fluid under the dam. It is clear, that
it is enough to solve the problem of conformal mapping of $\Omega$ onto upper
half-plane $\mathbb{C}_{+}$ in order to solve the specified problem.

In what follows, the conformal mapping problem will be solved explicitly using
the tools of Weierstrass elliptic functions. Below we show the calculation of
streamlines in $\Omega$ obtained using the method constructed in this work.

Figure 2. Streamlines in the region $\Omega$.

##  3\. The solution of the conformal mapping problem

###  3.1. The general form of the solution and parameter determination

Since $\Omega$ is simply connected, there is a conformal mapping
$W:\mathbb{C}_{+}\rightarrow\Omega$, where
$\mathbb{C}_{+}=\\{z\in\mathbb{C}:\operatorname{Im}z>0\\}$ is the upper half-
plane (see, e.g., [8] or [15]). Using, if necessary, a suitable automorphism
of $\mathbb{C}_{+}$, one can make it such that the point $w_{4}$ is the
preimage under $W$ (more precisely, under its continuation to the boundary) of
$\infty$. Then, by Christoffel-Schwartz theorem (see [10]), there exist
$x^{-}<x^{+}<x_{1}<x_{2}<x_{3}\in\mathbb{R}$ and $C\in\mathbb{C}$ such that

(3.1)
$dW=\phi=C\frac{\sqrt{(x-x_{2})(x-x_{3})}}{(x-x^{-})(x-x^{+})\sqrt{x-x_{1}}}dx.$

######  Remark 3.1.

Here $x_{i}$ is the preimage of $w_{i}$ under $W$, and $x^{-}$ and $x^{+}$ are
the points on the boundary of the upper half-plane in which $W$ goes to
infinity (preimages of the vertices at infinity).

The differential form $\phi$ can be considered on the hyperelliptic Riemann
surface $V$ of genus $1$, defined by equation
$y^{2}=F(x)=4(x-x_{1})(x-x_{2})(x-x_{3})$. Using the shift of the upper half-
plane we can set $x_{1}+x_{2}+x_{3}=0$ without loss of generality. Thus,
$F(x)=4x^{3}-g_{2}x-g_{3}$ for some real $g_{2},g_{3}$ (that are determined by
$x_{1},x_{2},x_{3}$). We can rewrite $\phi$ on this surface in the form

(3.2) $\phi=2C\frac{(x-x_{2})(x-x_{3})}{y(x-x^{-})(x-x^{+})}dx.$

Let us fix the branch of $\sqrt{F(x)}$ in the region that is made from
$\mathbb{C}$ by throwing off segment $[x_{1},x_{2}]$ and half-line
$[x_{3},\infty]$. Let this branch have positive values as the argument tends
to half-line $(x_{3},\infty)$ from the upper half-plane. Recalling that $dx/y$
is a holomorphic (everywhere non-zero) form on $V$, we obtain that $\phi$ has
two zeros of multiplicity $2$ at $(x_{2},0)$ and $(x_{3},0)$ and also four
simple poles at $(x^{-},\pm\sqrt{F(x^{-})})$ and $(x^{+},\pm\sqrt{F(x^{+})})$.
Note that the residues of this form at these poles are equal to $\pm
h^{-}/\pi$ and $\mp h^{+}/\pi$ respectively.

Now we shall use Abel map (see, e.g., [12]) which identifies $V$ with
$\operatorname{Jac}(V)$ (as usual, we set the infinity as the initial point
and $dx/y$ as the basis of holomorphic forms). Let us introduce the half-
periods

$\omega=\int_{x_{1}}^{x_{2}}\frac{dx}{y},\;\;\omega^{\prime}=-\int_{x_{2}}^{x_{3}}\frac{dx}{y},$

and quantities $\eta=\zeta(\omega)$ and
$\eta^{\prime}=\zeta(\omega^{\prime})$, where $\zeta$ is the Weierstrass zeta
function (see [1]). It is easy to see that $\omega,\eta\in\mathbb{R}$ and
$\omega^{\prime},\eta^{\prime}\in i\mathbb{R}$. The set of points
$(x,\sqrt{F(x)})$, where $x\in\mathbb{C}_{+}$, is mapped by this map onto the
rectangle with vertices $0,\omega^{\prime},\omega^{\prime}-\omega,-\omega$.
Let us denote the images of the points $(x^{-},\sqrt{F(x^{-})})$ and
$(x^{+},\sqrt{F(x^{+})})$ by $z^{-}$ and $z^{+}$ respectively (see Figure 3,
where the preimages of points are indicated in the brackets).

$0\;(\infty)$$\omega^{\prime}\;(x_{1})$$\omega^{\prime}-\omega\;(x_{2})$$-\omega\;(x_{3})$$z^{-}$$z^{+}$
Figure 3. Image of the upper half-plane under the Abel map.

The images of $(x^{-},-\sqrt{F(x^{-})})$ and $(x^{+},-\sqrt{F(x^{+})})$ in
this case are equal to $-z^{-}$ and $-z^{+}$. Consider the differential form
$\psi$ on the torus that corresponds to $\phi$ under this identification of
$V$ with $\operatorname{Jac}(V)$. This form has $4$ simple poles in the points
$\pm z^{-}$ and $\pm z^{+}$ and its residues are equal to $\pm h^{-}/\pi$ and
$\mp h^{+}/\pi$ respectively.

Now we use the method of representing elliptic functions by Weierstrass
functions that is described in [1]. Consider meromorphic function

(3.3)
$g(z)=\frac{h^{-}}{\pi}(\zeta(z-z^{-})-\zeta(z+z^{-}))-\frac{h^{+}}{\pi}(\zeta(z-z^{+})-\zeta(z+z^{+})).$

Using the quasiperiodic properties of $\zeta$ (see [1]) it is easy to conclude
that $g$ is elliptic. Form $g(z)dz$ has the same simple poles as $\psi$ with
the same residues. Therefore, $\psi-g(z)dz$ is a holomorphic form on the
torus. Since the space of holomorphic $1$-forms on the torus is one-
dimensional it follows that $\psi-g(z)dz=Ddz$, where $D$ is a constant (note
that $D\in i\mathbb{R}$).

Now we return to the map $W$. It is clear that

$W(x)=-\int_{x}^{\infty}\phi.$

In view of that, let

(3.4) $Q(z)=\int_{0}^{z}\psi.$

Obviously, $W(x)$ is equal to $Q(z)$, where $z$ is the image of
$(x,\sqrt{F(x)})$ under the Abel map. Thus, $Q$ conformally maps the rectangle
with vertices $0,\omega^{\prime},\omega^{\prime}-\omega,-\omega$ onto
$\Omega$, and $\omega^{\prime}$ is mapped to $w_{1}$, $\omega^{\prime}-\omega$
is mapped to $w_{2}$, and $-\omega$ to $w_{3}$ (also $0$ is mapped to
$w_{4}$). Now we can derive the system of equations from the previously
obtained relations:

(3.5) $g(-\omega)+D=0,\;\;\;g(\omega^{\prime}-\omega)+D=0,$ (3.6)
$Q(\omega^{\prime}-\omega)-Q(\omega^{\prime})=-ih,\;\;\;Q(-\omega)-Q(\omega^{\prime}-\omega)=-\delta.$

######  Remark 3.2.

The first pair of equations follows from the fact that $\phi$ has zeros in the
points $(x_{2},0)$ and $(x_{3},0)$, and the second pair is a consequence of
relations $w_{3}-w_{2}=-\delta$, $w_{2}-w_{1}=-ih$.

It remains to derive a reasonable formula for $Q$. Recall that $\zeta$ is a
logarithmic derivative of $\sigma$. It easily follows that

(3.7)
$Q(z)=Dz+\frac{h^{-}}{\pi}\ln\left(\frac{\sigma(z-z^{-})}{\sigma(z+z^{-})}\right)-\frac{h^{+}}{\pi}\ln\left(\frac{\sigma(z-z^{+})}{\sigma(z+z^{+})}\right)-i(h^{-}-h^{+}),$

where $\ln$ denotes the branch of logarithm in the plane cut by negative
imaginary half-line such that $\ln(1)=0$. Substituting in (3.5) the formula
for $g$ from (3.3) and using quasiperiodicity of $\sigma$ (see, e.g., [1]), we
obtain the system of equations:

(3.8) $\begin{dcases}-D\omega-\frac{2h^{+}}{\pi}\eta
z^{+}+\frac{2h^{-}}{\pi}\eta z^{-}=-ih,\\\
-D\omega^{\prime}-\frac{2h^{+}}{\pi}\eta^{\prime}z^{+}+\frac{2h^{-}}{\pi}\eta^{\prime}z^{-}=-\delta,\\\
D+\frac{h^{-}}{\pi}(\zeta(\omega-z^{-})-\zeta(\omega+z^{-}))-\frac{h^{+}}{\pi}(\zeta(\omega-z^{+})-\zeta(\omega+z^{+}))=0,\\\
D+\frac{h^{-}}{\pi}(\zeta(\omega^{\prime}+\omega-z^{-})-\zeta(\omega^{\prime}+\omega+z^{-}))\\\
\qquad-\frac{h^{+}}{\pi}(\zeta(\omega^{\prime}+\omega-z^{+})-\zeta(\omega^{\prime}+\omega+z^{+}))=0.\end{dcases}$

In this system of equations there are five variables (since the quantities
$\omega,\omega^{\prime},\eta,\eta^{\prime}$ are determined by $g_{2}$ and
$g_{3}$) $g_{2},g_{3},D,z^{+},z^{-}$ (the first two are real and the other are
imaginary) and four equations (3.8) (the first, third and fourth equations are
imaginary and the second one is real). Thus, it is natural to consider a
single parameter family of curves that necessarily contains a suitable one,
i.e. to consider functions $g_{2}=g_{2}(\gamma)$ and $g_{3}=g_{3}(\gamma)$ and
use the system (3.8) to determine the parameters $\gamma,D,z^{+},z^{-}$.

In what follows we shall use the family of curves that is defined by the roots
of the polynomial $F$: $x_{1}=\gamma-1/2$, $x_{2}=-2\gamma$,
$x_{3}=\gamma+1/2$, $\gamma\in(-1/6,1/6)$ (more detailed analysis of this
family is given during the study of the degeneration $\delta\rightarrow 0$).
This family corresponds to the normalization condition $x_{3}-x_{1}=1$ in
addition to the already given relation $x_{1}+x_{2}+x_{3}=0$.

(a) The rectangle $P$ and contours in it.

(b) The image of the rectangle and the contours.

(c) The behaviour near $[w_{2},w_{3}]$.

Figure 4. The conformal mapping $Q$.

###  3.2. On the numerical implementation

It was decided to use for numerical implementation the explicit computation of
the sigma function depending on parameters $g_{2},g_{3}$ through its Taylor
series (see [17] or Theorem A.4). It is clear that for the effective solution
of the system (3.8) it is necessary to compute all the quantities in it and
their derivatives with respect to parameters. In the end it reduces to the
computation of $\omega$ and $\omega^{\prime}$ and their derivatives with
respect to $g_{2}$ and $g_{3}$ and, also, $\zeta$ and its derivatives with
respect to $z,g_{2},g_{3}$. Since

$\zeta=\frac{1}{\sigma}\frac{\partial\sigma}{\partial z},$

the problem of computation of $\zeta$ and its derivatives can be solved
easily. In order to compute $\omega$ we note that $\sigma$ has zeros exactly
in the points of the lattice
$\\{2m\omega+2n\omega^{\prime}:n,m\in\mathbb{Z}\\}$ and these zeros are
simple. An effective way to localize a simple zero $z_{0}$ of a holomorphic
function $f$ is to compute integral of $zf^{\prime}(z)/2\pi if(z)$ on a
contour enclosing $z_{0}$. To find a suitable contour it is possible to apply
a variant of binary search using that $\omega\geq\pi/2$. Using the specified
method either directly, or for an approximate calculation of zero and
subsequent application of equation solving methods, it is easy to construct an
effective and precise algorithm of computation of $\omega$ (and
$\omega^{\prime}$). In order to compute their derivatives it is possible to
differentiate the integral of $z\sigma^{\prime}(z)/\sigma(z)$ by $g_{2}$ or
$g_{3}$, and to compute it explicitly by determining the residue in the zero
of $\sigma$. Thus, the solution of the system (3.8) can be completely reduced
to the computation of the sigma function and its derivatives with respect to
$z$, $g_{2}$, and $g_{3}$.

We demonstrate the solution of a specific problem by this method. Let
$h^{+}=\pi$, $h^{-}=\pi+0.5$, $h=0.5$, $\delta=0.2$. We shall search for the
solution in the one parameter family of curves defined by $x_{1}=\gamma-1/2$,
$x_{2}=-2\gamma$, $x_{3}=\gamma+1/2$, $\gamma\in(-1/6,1/6)$. The solution of
the system (3.8) is

$(\gamma,D,z^{+},z^{-})=(0.1051616134,0.0203152915i,1.3043479103i,0.7195735824i).$

Given this $\gamma$ we obtain $\omega=1.6518996331$,
$\omega^{\prime}=2.2939120295i$. On Figure 4 the image of the rectangle $P$
with vertices $0,\omega^{\prime},\omega^{\prime}-\omega,-\omega$ under the map
$Q$ is shown.

##  4\. Stability of the solution under the degeneration of the region

Here we shall consider a problem of conformal mapping of the upper half-plane
onto the region $\widetilde{\Omega}$ that comes from $\Omega$ with
degeneration $\delta\rightarrow 0$ (see Figure 5) and analyse behaviour of the
solution under the condition that no other degeneration is happening (i.e.
quantities $h^{-},h^{+},h,h^{-}+h-h^{+},h^{+}-h$ have positive limits).

$w_{4}$$w_{2}=w_{3}$$w_{1}$$h$$h^{-}$$h^{+}$ Figure 5. Region
$\widetilde{\Omega}$.

$\widetilde{\Omega}$ is determined by three parameters $h,h^{+}$ and $h^{-}$.
A conformal mapping of the upper half-plane onto $\widetilde{\Omega}$ can be
found by the analogous method (using Christoffel-Schwartz theorem). In this
case, since the corresponding Riemann surface has genus $0$, the solution can
be expressed in elementary functions. Another method (that is considered here)
is to apply formula (3.7), using the fact that $\sigma$ is defined for $g_{2}$
and $g_{3}$ such that $F(x)=4x^{3}-g_{2}x-g_{3}$ has multiple roots. It is
natural to suppose that the solution can be found by taking the limit under
gluing the roots, that are mapped to $w_{2}$ and $w_{3}$. Together with that
the stability of the solution under $\delta\rightarrow 0$ will be proved.

###  4.1. Gluing of the roots

Again consider a family of curves depending on $\gamma\in(-1/6,1/6)$ that is
given by $F_{\gamma}(x)=4(x-x_{1}(\gamma))(x-x_{2}(\gamma))(x-x_{3}(\gamma))$,
where $x_{1}(\gamma)=\gamma-1/2$, $x_{2}(\gamma)=-2\gamma$,
$x_{3}(\gamma)=\gamma+1/2$. Under $\gamma\rightarrow-1/6$ the roots $x_{2}$
and $x_{3}$ glue. The limiting values of $g_{2}$ and $g_{3}$ are $4/3$ and
$-8/27$ respectively. For each $\gamma$ we define quantities $\omega(\gamma)$,
$\omega^{\prime}(\gamma)$, $\eta(\gamma)$, $\eta^{\prime}(\gamma)$. In what
follows we shall omit dependence on $\gamma$.

######  Lemma 4.1.

Under $\gamma\rightarrow-1/6$ we have

(4.1)
$\omega,\eta\rightarrow\infty,\;\;\omega^{\prime}\rightarrow\frac{i\pi}{2},\;\;\eta^{\prime}\rightarrow-\frac{i\pi}{6},\;\;\frac{\eta}{\omega}\rightarrow-\frac{1}{3}.$

Moreover,

(4.2)
$\begin{gathered}\sigma(z,\frac{4}{3},-\frac{8}{27})=e^{-\frac{z^{2}}{6}}\sinh(z),\;\;\zeta(z,\frac{4}{3},-\frac{8}{27})=\coth(z)-\frac{z}{3},\\\
\wp(z,\frac{4}{3},-\frac{8}{27})=\frac{1}{\sinh^{2}(z)}+\frac{1}{3}.\end{gathered}$

Finally, there exists $\varepsilon>0$ such that for $\gamma+1/6<\varepsilon$
the estimation

(4.3) $-c_{1}\ln(\gamma+1/6)\leq\omega(\gamma)\leq-c_{2}\ln(\gamma+1/6)$

holds, where $0<c_{1}<c_{2}$.

###### Proof.

(4.1) easily follows from integral representations

$\omega(\gamma)=\frac{1}{2}\int_{\gamma-\frac{1}{2}}^{-2\gamma}\frac{dx}{\sqrt{(x-\gamma-1/2)(x-\gamma+1/2)(x+2\gamma)}},$
$\omega^{\prime}(\gamma)=\frac{1}{2}\int_{-2\gamma}^{\gamma+\frac{1}{2}}\frac{dx}{\sqrt{-(x-\gamma-1/2)(x-\gamma+1/2)(x+2\gamma)}},$
$\eta(\gamma)=-\frac{1}{2}\int_{\gamma-\frac{1}{2}}^{-2\gamma}\frac{xdx}{\sqrt{(x-\gamma-1/2)(x-\gamma+1/2)(x+2\gamma)}},$
$\eta^{\prime}(\gamma)=-\frac{1}{2}\int_{-2\gamma}^{\gamma+\frac{1}{2}}\frac{xdx}{\sqrt{-(x-\gamma-1/2)(x-\gamma+1/2)(x+2\gamma)}}.$

To derive (4.2) one can pass to the limit $\gamma\rightarrow-1/6$ using the
formula that represents $\sigma$ as the infinite product (see [1]). We obtain

$\sigma(z,\frac{4}{3},-\frac{8}{27})=z\prod_{n\neq
0}\left(1-\frac{z}{in\pi}\right)e^{\frac{z}{in\pi}-\frac{z^{2}}{2n^{2}\pi^{2}}}.$

Using classical identities

$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6},\;\;\prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)=\frac{\sin(x)}{x},$

we derive the first equation of (4.2). The rest follow from equalities
$\zeta(z)=\sigma^{\prime}(z)/\sigma(z)$, $\wp(z)=-\zeta^{\prime}(z)$.

Now we estimate the growth of $\omega(\gamma)$. Consider equality

$\omega(\gamma)=\frac{1}{2}\int_{0}^{1/2-3\gamma}\frac{dt}{\sqrt{t(1-t)(1/2-3\gamma-t)}}.$

Note that for small $\gamma$ integral on the segment $[0,1/2]$ is bounded and,
therefore,

$\omega(\gamma)\sim\frac{1}{2}\int_{1/2}^{1/2-3\gamma}\frac{dt}{\sqrt{t(1-t)(1/2-3\gamma-t)}}.$

The estimation of the integral in rhs is quite simple since $1/\sqrt{t}$ is
bounded from below and above by positive constants and the remaining integral
can be calculated explicitly. ∎

Since $\omega^{\prime}\rightarrow i\pi/2$ and $\omega\rightarrow\infty$, it is
natural to suppose that (3.7) can give a formula for a conformal mapping of
the half-strip

$S=\\{z\in\mathbb{C}:\operatorname{Re}z<0,\operatorname{Im}z\in(0,\pi/2)\\}$

onto $\widetilde{\Omega}$. Let

(4.4)
$\widetilde{Q}(z)=Dz+\frac{h^{-}}{\pi}\ln\left(\frac{\sigma(z-z^{-})}{\sigma(z+z^{-})}\right)-\frac{h^{+}}{\pi}\ln\left(\frac{\sigma(z-z^{+})}{\sigma(z+z^{+})}\right)-i(h^{-}-h^{+}),$

where $\sigma$ is taken at values $g_{2}=4/3$, $g_{3}=-8/27$ and
$D\in\mathbb{C}$, $z^{-},z^{+}\in(0,i\pi/2)$ are the parameters. Substituting
(4.2) into (4.4), we obtain

(4.5)
$\begin{multlined}\widetilde{Q}(z)=z\left(D+\frac{2h^{-}z^{-}}{3\pi}-\frac{2h^{+}z^{+}}{3\pi}\right)\\\
+\frac{h^{-}}{\pi}\ln\frac{\sinh(z-z^{-})}{\sinh(z+z^{-})}-\frac{h^{+}}{\pi}\ln\frac{\sinh(z-z^{+})}{\sinh(z+z^{+})}-i(h^{-}-h^{+}).\end{multlined}\widetilde{Q}(z)=z\left(D+\frac{2h^{-}z^{-}}{3\pi}-\frac{2h^{+}z^{+}}{3\pi}\right)\\\
+\frac{h^{-}}{\pi}\ln\frac{\sinh(z-z^{-})}{\sinh(z+z^{-})}-\frac{h^{+}}{\pi}\ln\frac{\sinh(z-z^{+})}{\sinh(z+z^{+})}-i(h^{-}-h^{+}).$

It is easy to check that if $\widetilde{Q}$ has a non-zero linear term it has
no limit under $\operatorname{Re}z\rightarrow-\infty$. If

$D+\frac{2h^{-}z^{-}}{3\pi}-\frac{2h^{+}z^{+}}{3\pi}=0,$

its limit is equal to $2(h^{-}z^{-}-h^{+}z^{+})/\pi-i(h^{-}-h^{+})$. Thus, if
$\widetilde{Q}$ conformally maps $S$ onto $\widetilde{\Omega}$, then the
conditions

(4.6) $\begin{dcases}D+\frac{2h^{-}z^{-}}{3\pi}-\frac{2h^{+}z^{+}}{3\pi}=0,\\\
h^{-}z^{-}-h^{+}z^{+}=-\frac{ih\pi}{2}\end{dcases}$

hold. Obviously, these conditions are sufficient under the additional
assumption that the derivative of $\widetilde{Q}$ is not vanishing on $S$ and
its boundary. Using tedious but elementary calculations it can be shown that
this condition is equivalent to

(4.7) $h^{-}\sinh(2z^{-})=h^{+}\sinh(2z^{+}).$

Thus, (4.6) and (4.7) determine the parameters $D,z^{-},z^{+}$, at which
$\widetilde{Q}$ is the desired conformal mapping.

We show that the obtained formula reduces to Christoffel-Schwartz integral in
the upper half-plane under the variable change $x=\wp(z)=1/\sinh^{2}(z)+1/3$.
Changing the variable in (4.4) and using that the linear term vanishes we
obtain the following formula for the conformal mapping of $\mathbb{C}_{+}$
onto $\widetilde{\Omega}$:

(4.8)
$\widetilde{W}(x)=\frac{h^{-}}{\pi}\ln\left(\frac{\sqrt{x^{-}+2/3}-\sqrt{x+2/3}}{\sqrt{x^{-}+2/3}+\sqrt{x+2/3}}\right)+\frac{h^{+}}{\pi}\ln\left(\frac{\sqrt{x^{+}+2/3}-\sqrt{x+2/3}}{\sqrt{x^{+}+2/3}+\sqrt{x+2/3}}\right).$

The equations on the parameters $z^{-}$ and $z^{+}$ can be rewritten in the
form

(4.9)
$\begin{dcases}\frac{h^{-}\sqrt{x^{-}+1}}{x^{-}}=\frac{h^{+}\sqrt{x^{+}+1}}{x^{+}},\\\
h^{-}\ln\left(\frac{1+\sqrt{x^{-}+4/3}}{\sqrt{x^{-}+1/3}}\right)+h^{+}\ln\left(\frac{1+\sqrt{x^{+}+4/3}}{\sqrt{x^{+}+1/3}}\right)=-\frac{ih\pi}{2}.\end{dcases}$

Differentiating $\widetilde{W}$ and using the first equation from (4.9) we
obtain

$d\widetilde{W}=\frac{h^{-}\sqrt{x^{-}+1}-h^{+}\sqrt{x^{+}+1}}{\pi}\frac{(x-1/3)dx}{\sqrt{x+2/3}(x-x^{-})(x-x^{+})}.$

Therefore we found the exact form of the constant in Christoffel-Schwartz
integral. The remaining parameters $x^{-}$ and $x^{+}$ can be found from the
system of equations (4.9).

###  4.2. Passing to the limit

Consider a sequence of regions determined by the parameters
$(h^{-}_{n},h^{+}_{n},h_{n},\delta_{n})$. Assume that they have limits
$(h^{-}_{\lim},h^{+}_{\lim},h_{\lim},0)$. Also we assume that
$h^{+}_{\lim}-h_{\lim}>0$ and $h^{-}_{\lim}-h^{+}_{\lim}+h_{\lim}>0$. We shall
prove that the parameters $(D_{n},\gamma_{n},z^{-}_{n},z^{+}_{n})$, given by
the solution of the system (3.8) for the corresponding regions, have limits
$(D_{\lim},-1/6,z^{-}_{\lim},z^{+}_{\lim})$, and, moreover, parameters
$(D_{\lim},z^{-}_{\lim},z^{+}_{\lim})$ satisfy (4.6) and (4.7). Thus, in view
of the fact that the Weierstrass sigma function is entire, it follows that the
constructed solution is stable.

The following proof is rather long and technical and, therefore, we shall omit
most of the calculations.

In the following estimations we shall also use parameters
$x^{-}_{n},x^{+}_{n},x_{1}^{(n)},x_{2}^{(n)},x_{3}^{(n)},C_{n}$ of the map
$W_{n}$.

######  Lemma 4.2.

Assume that $\gamma_{n}\rightarrow-1/6$ and
$\delta_{n}\omega(\gamma_{n})\rightarrow 0$. Then the indicated convergence
holds.

###### Proof.

Note that $\gamma_{n}\rightarrow-1/6$ implies that sequences $z^{-}_{n}$ and
$z^{+}_{n}$ are bounded. The first equation in (3.8) then implies that $D_{n}$
is also bounded. Passing to the subsequences we can assume that all these
sequences are convergent (if we succeed to prove that the limits satisfy (4.6)
and (4.7), then uniqueness of the solution implies that all the subsequences
converge to the same limit, and, therefore, the initial sequences converge).
The first equation in (4.6) is obtained by passing to the limit in the second
equation in (3.8) in view of Lemma 4.1. Multiplying the first equation in
(3.8) by $\omega^{\prime}$ and the second one by $\omega$, subtracting and
passing to limit leads to the second equation in (4.6) (the term
$\delta\omega$ by assumption tends to zero).

Now we derive (4.7) for the limits of the sequences. Recall the following
notation of the theory of elliptic functions (see [1]):

$\zeta_{2}(z)=\zeta(z+\omega)-\eta,$
$\zeta_{3}(z)=\zeta(z+\omega+\omega^{\prime})+\eta+\eta^{\prime}.$

These functions are connected to $\sigma_{2},\sigma_{3}$:

$\zeta_{k}=\frac{1}{\sigma_{k}}\frac{d\sigma_{k}}{dz}=\frac{d\ln\sigma_{k}}{dz}.$

Finally, $\sigma_{k}=\sigma\sqrt{\wp-x_{k}}$. The last two equations in (3.8)
can be rewritten as

(4.10)
$D+\frac{h^{-}}{\pi}(\zeta_{k}(-z^{-})-\zeta(z^{-}))-\frac{h^{+}}{\pi}(\zeta_{k}(-z^{+})-\zeta(z^{+}))=0,\;\;k=2,3.$

Since

$\begin{multlined}\zeta_{2}(z)-\zeta_{3}(z)=\frac{\sigma^{\prime}(z)\sqrt{\wp-
x_{2}}+\wp^{\prime}(z)\sigma(z)(\wp-x_{2})^{-1/2}}{\sigma(z)\sqrt{\wp-
x_{2}}}\\\ -\frac{\sigma^{\prime}(z)\sqrt{\wp-
x_{3}}+\wp^{\prime}(z)\sigma(z)(\wp-x_{3})^{-1/2}}{\sigma(z)\sqrt{\wp-
x_{3}}},\end{multlined}\zeta_{2}(z)-\zeta_{3}(z)=\frac{\sigma^{\prime}(z)\sqrt{\wp-
x_{2}}+\wp^{\prime}(z)\sigma(z)(\wp-x_{2})^{-1/2}}{\sigma(z)\sqrt{\wp-
x_{2}}}\\\ -\frac{\sigma^{\prime}(z)\sqrt{\wp-
x_{3}}+\wp^{\prime}(z)\sigma(z)(\wp-x_{3})^{-1/2}}{\sigma(z)\sqrt{\wp-
x_{3}}},$

it follows that

(4.11) $\zeta_{2}(z)-\zeta_{3}(z)=\frac{\wp^{\prime}(z)(x_{2}-x_{3})}{(\wp-
x_{2})(\wp-x_{3})}.$

Equation (4.7) is derived by passing to the limit (using Lemma 4.1) from
equations (4.10) (from which the constant $D$ can be eliminated) and
substitution the formula for $\zeta_{2}-\zeta_{3}$ from (4.11). ∎

######  Lemma 4.3.

Inequalities $|C_{n}|\geq a_{1}$, $|C_{n}|\leq
a_{2}\sqrt{x_{3}^{(n)}-x^{-}_{n}}$ hold for some positive constants
$a_{1},a_{2}$. Moreover, the sequence $x^{+}_{n}$ is bounded from below.

###### Proof.

The estimations for $C^{(n)}$ easily follow from

$|C_{n}|\bigintss_{x_{3}^{(n)}}^{+\infty}\frac{\sqrt{\left(x-x_{2}^{(n)}\right)\left(x-x_{3}^{(n)}\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{-}_{n})(x-x^{+}_{n})}=h^{-}_{n}-h^{+}_{n}+h_{n}.$

To prove the boundedness from below for $x^{+}_{n}$ it suffices to consider
equality

$|C_{n}|\bigintss_{x_{1}^{(n)}}^{x_{2}^{(n)}}\frac{\sqrt{\left(x_{2}^{(n)}-x\right)\left(x_{3}^{(n)}-x\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{-}_{n})(x-x^{+}_{n})}=h_{n}.$

∎

######  Lemma 4.4.

Assume that sequence $x^{-}_{n}$ is bounded from below. Then there exist
constants $0<b_{1}<b_{2}$ such that
$b_{1}\sqrt{\delta_{n}}\leq|x_{2}^{(n)}-x_{3}^{(n)}|\leq
b_{2}\sqrt{\delta_{n}}$.

###### Proof.

It follows from easy estimation for the integral in equality

$|C_{n}|\bigintss_{x_{2}^{(n)}}^{x_{3}^{(n)}}\frac{\sqrt{\left(x-x_{2}^{(n)}\right)\left(x_{3}^{(n)}-x\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{-}_{n})(x-x^{+}_{n})}=\delta_{n}.$

∎

The foregoing Lemmas imply that it remains to prove that sequence $x^{-}_{n}$
is bounded from below (in view of the asymptotics (4.3)). Assume that this is
not true. Passing to the subsequences we can assume that
$x^{-}_{n}\rightarrow-\infty$ and $x_{1}^{(n)},x_{2}^{(n)},x_{3}^{(n)}$,
$x^{+}_{n}$ are convergent.

######  Lemma 4.5.

In the foregoing assumptions statements $x_{3}^{(n)}-x_{2}^{(n)}\rightarrow
0$, $x_{1}^{(n)}-x^{+}_{n}\rightarrow 0$ hold.

###### Proof.

Equality

$|C_{n}|\frac{\sqrt{\left(x_{2}^{(n)}-x^{+}_{n}\right)\left(x_{3}^{(n)}-x^{+}_{n}\right)}}{\sqrt{x_{1}^{(n)}-x^{+}_{n}}(x^{+}_{n}-x^{-}_{n})}=\frac{h^{+}_{n}}{\pi}$

implies $x_{1}^{(n)}-x^{+}_{n}\rightarrow 0$.

To prove that $x_{3}^{(n)}-x_{2}^{(n)}\rightarrow 0$ we return to parameters
$(D_{n},\gamma_{n},z^{-}_{n},z^{+}_{n})$. Assume that
$x_{2}^{(n)}-x_{1}^{(n)}\rightarrow 0$. Then
$\omega^{\prime}(\gamma_{n}),\eta^{\prime}(\gamma_{n})\rightarrow\infty$, and
$\omega$ and $\eta$ have finite limits. Moreover, Legendre identity (see,
e.g., [1] or [9]) implies

$\lim_{n\rightarrow\infty}\frac{\omega^{\prime}(\gamma_{n})}{\eta^{\prime}(\gamma_{n})}=\lim_{n\rightarrow\infty}\frac{\omega(\gamma_{n})}{\eta(\gamma_{n})}.$

The first two equations from (3.8) imply

$\lim_{n\rightarrow\infty}\left(\frac{2h^{+}_{n}}{\pi}z^{+}_{n}\left(\frac{\omega^{\prime}}{\eta^{\prime}}-\frac{\omega}{\eta}\right)-i\frac{h_{n}}{\omega}\right)=0.$

Therefore,

$\lim_{n\rightarrow\infty}\left(\frac{h^{+}_{n}z^{+}_{n}}{\omega^{\prime}}-h_{n}\right)=0,$

and, passing to the limit, we obtain $h_{\lim}\geq h^{+}_{\lim}$. This
contradicts the assumptions made.

Now assume that $x_{2}^{(n)}-x_{1}^{(n)}\nrightarrow 0$
and$x_{3}^{(n)}-x_{2}^{(n)}\nrightarrow 0$. Then both periods $\omega$ and
$\omega^{\prime}$ have finite limits. In this case $z^{-}_{n}\rightarrow 0$,
$z^{+}_{n}\rightarrow\omega^{\prime}(\gamma_{\lim})$. It is obvious that
$D_{n}$ also is convergent and, passing to the limit in the second equation in
(3.8), we obtain

$-D_{\lim}\omega^{\prime}-\frac{2h^{+}_{\lim}\omega^{\prime}\eta^{\prime}}{\pi}=0.$

Substituting into the first equation we get

$\frac{2h^{+}_{\lim}}{\pi}\omega\eta^{\prime}-\frac{2h^{+}_{\lim}}{\pi}\omega^{\prime}\eta=-ih_{\lim},$

implying $h^{+}_{\lim}=h_{\lim}$. ∎

Now we have enough preparation to deduce a contradiction from
$x^{-}_{n}\rightarrow-\infty$. In order to do this we shall analyse
asymptotics of some sequences (in what follows the equivalence of sequences
means that the quotient of them tends to $1$).

Equality

$|C_{n}|\frac{\sqrt{\left(x_{2}^{(n)}-x^{-}_{n}\right)\left(x_{3}^{(n)}-x^{-}_{n}\right)}}{\sqrt{x_{1}^{(n)}-x^{-}_{n}}(x^{+}_{n}-x^{-}_{n})}=\frac{h^{-}_{n}}{\pi}$

implies that

(4.12) $|C_{n}|\sim\frac{h^{-}_{n}}{\pi}\sqrt{|x^{-}_{n}|}.$

On the other hand

$|C_{n}|\frac{\sqrt{\left(x_{2}^{(n)}-x^{+}_{n}\right)\left(x_{3}^{(n)}-x^{+}_{n}\right)}}{\sqrt{x_{1}^{(n)}-x^{+}_{n}}(x^{+}_{n}-x^{-}_{n})}=\frac{h^{+}_{n}}{\pi},$

and, therefore,

(4.13)
$\sqrt{x_{1}^{(n)}-x^{+}_{n}}\sim\frac{h^{-}_{n}}{h^{+}_{n}}\frac{1}{\sqrt{|x^{-}_{n}|}}.$

Now consider equality

$|C_{n}|\bigintss_{x_{1}^{(n)}}^{x_{2}^{(n)}}\frac{\sqrt{\left(x_{2}^{(n)}-x\right)\left(x_{3}^{(n)}-x\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{-}_{n})(x-x^{+}_{n})}=h_{n}.$

Using (4.12), it is easy to show that the sequence in lhs is equivalent to
sequence

$\frac{h^{-}_{n}}{\pi\sqrt{|x^{-}_{n}|}}\bigintss_{x_{1}^{(n)}}^{x_{2}^{(n)}}\frac{\sqrt{\left(x_{2}^{(n)}-x\right)\left(x_{3}^{(n)}-x\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{+}_{n})}.$

Now, changing the variable, we obtain

$\bigintss_{x_{1}^{(n)}}^{x_{2}^{(n)}}\frac{\sqrt{\left(x_{2}^{(n)}-x\right)\left(x_{3}^{(n)}-x\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{+}_{n})}=\bigintss_{0}^{x_{2}^{(n)}-x_{1}^{(n)}}\frac{\sqrt{\left(x_{2}^{(n)}-x_{1}^{(n)}-x\right)(1-x)}dx}{\sqrt{x}(x+x_{1}^{(n)}-x^{+}_{n})}.$

It appears that the asymptotics of the last integral is independent of the
convergence rate of $|x_{2}^{(n)}-x_{1}^{(n)}|\rightarrow 1$. Namely, for all
sequences $\alpha_{n}\rightarrow 1$ and $a_{n}\rightarrow 0$ the equivalence

$\int_{0}^{\alpha_{n}}\frac{\sqrt{(\alpha_{n}-x)(1-x)}dx}{\sqrt{x}(x+a_{n})}\sim\int_{0}^{1}\frac{(1-x)dx}{\sqrt{x}(x+a_{n})}\sim\frac{\pi}{\sqrt{a_{n}}}$

holds. Finally, in view of (4.13), we obtain

$h_{n}=|C_{n}|\bigintss_{x_{1}^{(n)}}^{x_{2}^{(n)}}\frac{\sqrt{\left(x_{2}^{(n)}-x\right)\left(x_{3}^{(n)}-x\right)}dx}{\sqrt{x-x_{1}^{(n)}}(x-x^{-}_{n})(x-x^{+}_{n})}\sim\frac{h^{-}_{n}}{\pi\sqrt{|x^{-}_{n}|}}\frac{\pi}{\sqrt{x_{1}^{(n)}-x^{+}_{n}}}\sim
h^{+}_{n}.$

It contradicts the assumption $h_{\lim}<h^{+}_{\lim}$.

##  5\. Conclusion

The simple expression through the Weierstrass sigma function for a conformal
mapping of a polygonal region $\Omega$ was obtained. For the specific example
the numerical experiment was carried out. The behaviour under degeneration was
analyzed and it was shown that the formula is stable and converges to the
solution of the limiting problem.

The future research direction can be connected either with the construction
and analysis of the solutions to similar problems corresponding, for example,
to Riemann surfaces of genus $2$, or with the development of the sigma
function theory: construction of the recurrent formulas for higher genus and
elaboration of computational methods independent of the theta function theory.

##  6\. Acknowledgements

The author expresses his gratitude to A. Bogatyrev and O. Grigoriev for posing
the problem and useful discussions, and also to K. Malkov for help in the
computer implementation of the calculations. The author also thanks the Center
for Continuing Professional Education “Sirius University” for the invitation
to the educational module “Computational Technologies, Multidimensional Data
Analysis, and Modelling”, during which some of the results of this work were
obtained.

##  Appendix A On the coefficients of the Weierstrass sigma function Taylor
series

Here we prove that the sigma function is an entire function of three variables
and derive a recurrent formula for its Taylor series coefficients, that was
originally established by Weierstrass in [17]. The proof given there has a gap
connected to the analyticity of the sigma function in a neighbourhood of zero.
Perhaps, this fact can be proved by an independent argument but, since
Weierstrass does not give any references (and omits this issue completely), we
decided to provide a complete proof here.

The homogeneity condition

$\sigma(\frac{z}{\lambda},\lambda^{4}g_{2},\lambda^{6}g_{3})=\frac{1}{\lambda}\sigma(z,g_{2},g_{3})$

easily implies the following differential equation for the $\sigma$ function:

(A.1) $z\frac{\partial\sigma}{\partial z}-4g_{2}\frac{\partial\sigma}{\partial
g_{2}}-6g_{3}\frac{\partial\sigma}{\partial g_{3}}-\sigma=0.$

Further, using the definition of the $\sigma$ function and the standard
differential equation for the $\wp$ function, one can derive an equation (for
a proof see [13])

(A.2) $\frac{\partial^{2}\sigma}{\partial
z^{2}}-12g_{3}\frac{\partial\sigma}{\partial
g_{2}}-\frac{2}{3}g_{2}^{2}\frac{\partial\sigma}{\partial
g_{3}}+\frac{1}{12}g_{2}z^{2}\sigma=0.$

Let $f$ be an entire function of three variables $(z,g_{2},g_{3})$ satisfying
(A.1) and (A.2). We derive a relation between the Taylor series coefficients
$f_{mnk}$ of $f$:

$f=\sum_{m,n,k=0}^{\infty}f_{mnk}g_{2}^{m}g_{3}^{n}z^{k}.$

(A.1) implies that $f_{mnk}=0$, if $k\neq 4m+6n+1$. Therefore, $f$ can be
written in the form

$f=\sum_{m,n=0}^{\infty}a_{mn}g_{2}^{m}g_{3}^{n}z^{4m+6n+1}.$

Now, substituting this expression of $f$ into (A.2), we obtain the equality

(A.3)
$a_{mn}=\frac{12(m+1)a_{m+1,n-1}+\frac{2}{3}(n+1)a_{m-2,n+1}-\frac{1}{12}a_{m-1,n}}{(4m+6n+1)(4m+6n)},$

in which for convenience $a_{mn}$ is defined by zero when $m$ or $n$ is
negative. It is easy to see that (A.3) uniquely determines sequence $a_{mn}$
for given $a_{00}$. To prove this let us introduce an order relation on pairs
of nonnegative integers $(m,n)$: $(m,n)\leq(m^{\prime},n^{\prime})$, if
$m+n<m^{\prime}+n^{\prime}$ or if $m+n=m^{\prime}+n^{\prime}$ and $n\leq
n^{\prime}$. It is easy to see that we defined a well-order on
$\mathbb{Z}_{+}\times\mathbb{Z}_{+}$, and in (A.3) indices of the terms
$a_{mn}$ in rhs are strictly less then $(m,n)$. Thus, it is proved that (A.3)
determines $a_{mn}$ recursively for given $a_{00}$.

If the sigma function was an entire function of three variables, or, at least,
holomorphic in some neighbourhood of zero, then the recurrence relation (A.3)
for its Taylor series coefficients would be proved. The difficulty is that the
domain of $\sigma$ is the set
$\\{(z,g_{2},g_{3})\in\mathbb{C}^{3}:g_{2}^{3}-27g_{3}^{2}\neq 0\\}$. The
following considerations prove the entirety of $\sigma$ and the recurrence
relation (A.3).

######  Remark A.1.

It is known (see, e.g., [1] or [9]) that condition $g_{2}^{3}-27g_{3}^{2}\neq
0$ is equivalent to simplicity of the roots of polynomial
$4x^{3}-g_{2}x-g_{3}$.

######  Lemma A.2.

Let $a_{mn}$ satisfy the recurrence relation (A.3). Then for all
$q>(28+\sqrt{811})/36\approx 1.569$ there exists $C>0$ such that

(A.4) $|a_{mn}|\leq C\frac{q^{2m+3n}}{(2m+3n)!}.$

###### Proof.

Substituting in (A.3) this estimation and it is easy to show that for the
existence of a constant, it suffices that the inequality

$\frac{6(m+1)}{4m+6n+1}\frac{q^{2m+3n-1}}{(2m+3n)!}+\frac{(n+1)q^{2m+3n-1}}{6(4m+6n+1)(2m+3n)!}+\frac{q^{2m+3n-2}}{48(2m+3n)!}\leq\frac{q^{2m+3n}}{(2m+3n)!}$

holds starting from some index $(m,n)$ (in terms of the foregoing ordering).
For this, in turn, it suffices to satisfy the inequality

$\frac{1}{48}+q\left(\frac{3}{2}+\frac{1}{18}\right)<q^{2}.$

Solving the quadratic equation, we obtain the required statement. ∎

Lemma A.2 allows to define an entire function

$h(z,g_{2},g_{3})=\sum_{m,n=0}^{\infty}a_{mn}g_{2}^{m}g_{3}^{n}z^{4m+6n+1},$

where $a_{mn}$ are determined by recurrence relation (A.3) and initial
condition $a_{00}=1$. We shall prove that $h\equiv\sigma$ for $(g_{2},g_{3})$
such that $g_{2}^{3}-27g_{3}^{2}\neq 0$.

######  Lemma A.3.

Let $f$ be a holomorphic function of variables $(z,g_{2},g_{3})$, defined on a
set $\mathbb{C}\times U$, where $U\subset\mathbb{C}^{2}$ is open, satisfying
equation (A.2). Assume that $f$ is odd in variable $z$. Then $f$ can be
represented by series

(A.5) $f(z,g_{2},g_{3})=\sum_{n=0}^{\infty}c_{n}(g_{2},g_{3})z^{2n+1},$

and in $U$ the recurrence relation

(A.6) $(2n+3)(2n+2)c_{n+1}-12g_{3}\frac{\partial c_{n}}{\partial
g_{2}}-\frac{2}{3}g_{2}^{2}\frac{\partial c_{n}}{\partial
g_{3}}+\frac{1}{12}g_{2}c_{n-1},$

holds, where $n\geq 0$ (for $n=0$ we set $c_{n-1}=0$).

###### Proof.

Indeed the representability of $f$ by the series follows from entirety of $f$
by $z$. Its coefficients $c_{n}(g_{2},g_{3})$ are given by

$c_{n}(g_{2},g_{3})=\frac{1}{(2n+1)!}\frac{\partial^{2n+1}f}{\partial
z^{2n+1}}|_{z=0}.$

It is easy to see that the series (A.5) can be differentiated term-by-term,
and therefore we can substitute it in (A.2). Collecting the coefficient at
$z^{2n+1}$, we obtain (A.6). ∎

Recurrence relation (A.6) can be used to prove, that $\sigma$ and $h$ coincide
on the domain of the $\sigma$ function. Indeed, if the first terms of their
expansions coincide, then these function coincide (note that they are both odd
in $z$). Indeed, $\partial\sigma/\partial z|_{z=0}\equiv\partial h/\partial
z|_{z=0}\equiv 1$. Thus, $h$ is the analytic continuation of the $\sigma$
function to an entire function of variables $(z,g_{2},g_{3})$. This completes
the proof of the following theorem.

######  Theorem A.4 (Weierstrass).

The $\sigma$ function is entire and for all $(z,g_{2},g_{3})\in\mathbb{C}^{3}$
equality

(A.7)
$\sigma(z,g_{2},g_{3})=\sum_{m,n=0}^{\infty}a_{mn}g_{2}^{m}g_{3}^{n}z^{4m+6n+1}$

holds, where coefficients $a_{mn}$ are determined by recurrence relation (A.3)
and initial condition $a_{00}=1$.

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