Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 16, 2004, 57 – 69

© A. Kuznetsov

ABSTRACT. In this paper we consider a lower estimate for the ratio $I\left(\Omega \right)$ of the conformal moment of a simple connected domain $\Omega$ in the complex plane to the moment of inertia of this domain about its boundary. Related functionals depending on a simple connected domain $\Omega$ and two points $w_1,w_2\in \Omega$ with fixed hyperbolical distance between them are estimated. As a consequence a nontrivial lower estimate for $I\left(\Omega \right)$ is obtained.

1. Introduction

Let $\Omega$ be a simply connected domain on the complex plane $\mathbb{ℂ}$ and $w\in \Omega$. Let ${\rho }_{\Omega }\left(w\right)$ be the conformal radius of $\Omega$ at the point $w$, and $d_{\Omega }\left(w\right)$ be the Euclidean distance from the point $w$ to the boundary $\partial \Omega$ of the domain $\Omega$.

$I_c\left(\Omega \right)={\int }_{\Omega }{\rho }_{\Omega }^2\left(x+iy\right)dxdy$ is the conformal moment of $\Omega$, and $I\left(\partial \Omega \right)={\int }_{\Omega }{d}_{\Omega }^2\left(x+iy\right)dxdy$ is the moment of inertia of $\Omega$ about $\partial \Omega$. This functionals were introduced by F.G. Avkhadiev [1] for solution of the classical St. Venant problem of finding two-side estimates for the torsional rigidity $P\left(\Omega \right)$ of the domain $\Omega$ by simple geometric characteristics of the domain [2, 3]. As a solution the following inequalities

were obtained. The first and the last inequalities in (1) is the corollary of well-known inequalities for the ratio of the conformal radius of a domain at a point and the distance from this point to the domain boundary (see for e.g.[4, 5]). The first inequality in (1) is not sharp when $I_c\left(\Omega \right)$ and $I\left(\partial \Omega \right)$ are finite. F.G. Avkhadiev set a problem to find lower and upper sharp bounds for the ratio

when $I_c\left(\Omega \right)$ and $I\left(\partial \Omega \right)$ are finite. Note that $I\left(\Omega \right)$ is invariant under linear transforms of $\Omega$.

The first and the last inequalities in (1) imply that the region of values $I\left(\Omega \right)$ is a subset of $\left[1,16\right]$. In this paper we give a better lower estimate for $I\left(\Omega \right)$.

Let $w_1,w_2\in \Omega$, and let a holomorphic univalent function $f\left(z\right)$ maps the unit disk $\mathbb{U}=\left\{z:\mid z\mid <1\right\}$ onto $\Omega$ so that $f\left(0\right)=w_1$ and $f\left(r\right)=w_2$, where $r\in \left(0,1\right)$ is a given constant. The normalization of $f$ means, that the hyperbolical distance between $w_1$ and $w_2$ is constant. Let us consider the functional

on the class $S^0$ of univalent holomorphic functions $f\left(z\right),z\in \mathbb{U}$.

Obviously $G\left(f,\alpha \right)=G\left(\beta +\gamma f,\alpha \right),\gamma \ne 0$. It is possible to consider the functional $G\left(f,\alpha \right)$ as a function of $\Omega$ and two points $w_1,w_2\in \Omega$ with fixed hyperbolical distance between them. Namely,

Let $r_{\alpha }$ be the unique solution of the equation

on the interval $\left(0,1\right)$. Then we have

To improve the lower estimate of $I\left(\Omega \right)$ we need to find the sharp bound of the functional

on class $S^0$. We have

The value of $r_0$ will be given in the proof of Proposition 1.

By $f_{\theta ,d}$ denote a function mapping $\mathbb{U}$ onto the arc biangle satisfying the following requirement: one of bounding circles has center at the origin and unit radius, the other bounding circle has center at the point $d>0$ and $\theta$ is the argument of the intersection point of this circle lying in the upper half-plane. Taking into account that the functional $F\left(f,r,c\right)$ is invariant under linear transforms of the complex plane, we conclude that there are $d>0,\theta \in \left[0,\pi \right]$ such that $F\left(f_{\theta ,d},r,c\right)=\underset{f\in S^0}{sup}F\left(f,r,c\right)$.

Now we can give a lower estimate for $I\left(\Omega \right)$ in terms of the functional $F\left(f,r,c\right)$.

The values of $r^{\ast },r_1$ will be given in the proof of Theorem 2.

2. Proof of theorem 1 and proposition 1

Proof of Theorem 1. It is possible to assume that $f\in S$, where $S$ is the class of all holomorphic univalent functions $f\left(z\right)$ in $\mathbb{U}$ normalized by $f\left(0\right)=f^{\prime }\left(0\right)-1=0$. Let $k\left(z\right)=\frac{z}{{\left(1-z\right)}^2}$, $k_{\varepsilon }\left(z\right)=k^{-1}\left(\left(1-\varepsilon \right)k\left(z\right)\right)$ and $k_{\varepsilon }^{\gamma }\left(z\right)=e^{-i\gamma }{k}_{\varepsilon }\left(e^{i\gamma }z\right)$. For $\varepsilon$ small enough we have:

The function $k_{\varepsilon }^{\gamma }\left(z\right)$ maps $\mathbb{U}$ onto $\mathbb{U}$ minus radial slit with endpoint at $-e^{-i\gamma }$. Note that the slit length tends to $0$ as $\varepsilon \to 0$.

Assume the opposite. Then there are two open disks $D_1,D_2\subset \mathbb{U},\mathbb{U}⁄\subset D_1\cup D_2$, with centers at points $0$ and $f\left(r\right)$ and radii $d\left(0\right)$ and $d\left(f\left(r\right)\right)$ respectively. Therefore, there are $\gamma \in \mathbb{ℝ}$ and ${\varepsilon }_0>0$ such that $D_1,D_2\subset f_{\varepsilon }\left(\mathbb{U}\right)$, where $f_{\varepsilon }\left(z\right)=f\left(k_{\varepsilon }^{\gamma }\left(z\right)\right),0\le \varepsilon \le {\varepsilon }_0$.

By construction $d\left(f_{\varepsilon }\left(0\right)\right)=d\left(f\left(0\right)\right)=d\left(0\right)$, and $\mid f_{\varepsilon }^{\prime }\left(0\right)\mid =\left(1-\varepsilon \right)\mid f^{\prime }\left(0\right)\mid =\left(1-\varepsilon \right)$. Moreover taking into account (5), we have

Using (6) and $D_m\subset f_{\varepsilon }\left(\mathbb{U}\right),m=1,2$ we get

Note that

and $\mid f^{\prime }\left(r\right)\mid \left(1-r^2\right)$ is the conformal radius of the domain. Thus $\mid f^{\prime }\left(r\right)\mid ∕d\left(r\right)\le \frac{4}{1-r^2}$. Therefore

Let $p\left(r\right)=\frac{4r}{{\left(1-r\right)}^2}$.

Using (6) we have

So

Let $q\left(r\right)=\Re e\left[\right.\frac{1-2r{e}^{i\gamma }-r^2{e}^{2i\gamma }}{{\left(1+r{e}^{i\gamma }\right)}^2}+\frac{r{f}^{\prime \prime }\left(r\right)}{f^{\prime }\left(r\right)}\frac{1-r{e}^{i\gamma }}{1+r{e}^{i\gamma }}\left]\right.$. Taking into account (7), we obtain

Let

Then inequality (8) takes the form

or

Therefore, in order to prove that every function $f\left(z\right)$ minimizing the functional $G\left(f,\alpha \right)$ on class $S$ map $\mathbb{U}$ onto an arc biangle, it is sufficient to show that

or

Let us find a lower bound of $\frac{C}{D}$. Using the obvious inequality $d\left(f\left(r\right)\right)\le d\left(f\left(0\right)\right)+\mid f\left(r\right)\mid$, we have

Since $f\in S$,

Using the growth theorem in the class $S$, we have

By the distortion theorem for the class $S$

Tacking into account that the left part of (9) is monotonic on $\frac{B}{A}$, we conclude that the following inequalities

and

imply inequality (9). Substituting into this inequality $p,q$, using the estimate for function $f\in S$ (see foe e.g. [6, p. 32])

and the maximum principle for harmonic functions we have, that inequalities

where

imply (9).

Consider the functions

and

They are monotonic on the interval $\left(0,1\right)$. So they may have only one zero on this interval. Computations show, that $P\left(r\right)$ equals zero at the point $3-2\sqrt{2}=0.1715...$, and $Q\left(r\right)$ equals zero at the point $r^{\prime }=0.086427...$. Therefore we have that the roots of equations $M_1\left(r\right)=0$ and $M_2\left(r\right)=0$ lie in the interval $\left(r^{\prime },1\right)$. It is easy to see that, for $r\in \left(r^{\prime },1\right)$, $M_2\left(r\right)>M_1\left(r\right)$ and so if inequality (12) holds, then inequality (11) holds too. Using the inequality $Q\left(r\right)<0$ and the fact that $Q\left(r\right)$ is decreasing on the interval $\left(r^{\prime },1\right)$ we have, that the function $M_2\left(r\right)$ is decreasing on this interval too. Thus, we conclude that if $r<r_{\alpha }$, where $r_{\alpha }$ is the single root of the equation

on the interval $\left(0,1\right)$, then every function minimizing functional (3) maps $\mathbb{U}$ onto an arc biangle, bounded by two circle arcs centered at the points $f\left(0\right)$ and $f\left(r\right)$. Note that $r_{\alpha }>r^{\prime }$.

Proof of Proposition 1. Using the same argument as in the proof of theorem 1 and following notation

we have

where $f_{\varepsilon }\left(z\right)=f\left(k_{\varepsilon }^{\gamma }\left(z\right)\right)$. Thus the inequality

or

implies that any function minimizing functional (4), maps $\mathbb{U}$ onto an arc biangle. Using inequalities $A,B,C,D>0$ we have that inequalities

guarantee inequality (13). Substituting in (14) the functions $p\left(r\right)$ and $q\left(r\right)$, using estimate (10) and the maximum principle for harmonic functions, we have that the following inequalities

imply (13).

Functions in the left-side of (15) are monotonic on the interval (0,1). Computations show that, the first was violated at $r_{01}=0.086427...$, the second at $r_{02}=0.071796...$, the third at $r_{03}=0.071796...$. thus we conclude that if

then every function every minimizing functional (4) maps $\mathbb{U}$ onto an arc biangle, bounded by circle arcs centered at the points $f\left(0\right)$ and $f\left(r\right)$.

3. Proof of Theorem 2

Proof of Theorem 2. First we divide $\mathbb{U}$ into two subsets ${\mathbb{U}}_1,{\mathbb{U}}_2$ and construct a map $\xi \left(z\right):{\mathbb{U}}_1\to {\mathbb{U}}_2$ such that Jacobian of $\xi \left(z\right)$ equals the unity almost everywhere and the inequality

holds.

First, we consider a disk ${\mathbb{U}}^{\ast }$ of radius $r^{\ast }$ centered at the origin.

Let us consider the square $s$ inscribed in ${\mathbb{U}}^{\ast }$ with sides parallel to the axes of coordinates. Let $s_1,s_2,s_3,s_4$ be the intersection of the square $s$ with the first, the second, the third and the fourth quarters of the complex plane. Let $c_1,c_3$ be the sectors of ${\mathbb{U}}^{\ast }$, cutting off the sides of the square $s$, parallel to the axis OX, and by $c_2,c_4$ sectors, cutting off the sides of the square $s$, paralleling to the axis OY.

Let $s_1,s_2,c_1,c_2\subset {\mathbb{U}}_1$, $s_3,s_4,c_2,c_4\subset {\mathbb{U}}_2$ and

Let us find the lower and upper sharp bounds of $s\left(\xi \left(z\right),z\right)$, for $z\in c_1,c_3$. It is easy to see that

The function $\frac{r}{\sqrt{1+r^4}}$ increases on the interval $\left(0,1\right)$ and so we have

By construction, $s\left(\xi \left(z\right),z\right)<r_0\text{a.e.}$, so

Let us find the lower and upper sharp bounds of $s\left(\xi \left(z\right),z\right)$, for $z\in s_1$. Let $z=x+iy$, then we have

So

Let us consider the function

When $z\in s_1$, the function $q\left(x,y\right)$ is nonnegative, and attains its minimal value at points $\frac{r^{\ast }}{\sqrt{2}}$, $i\frac{r^{\ast }}{\sqrt{2}}$, $0$, $\frac{r^{\ast }}{\sqrt{2}}+i\frac{r^{\ast }}{\sqrt{2}}$. Thus the function $\frac{r^{\ast 2}}{2}{\left(x-y\right)}^2$ is minimal at $x=y$, then

attains minimum at the point $0$, so we have $s\left(\xi \left(z\right),z\right)\le r^{\ast }$, for $z\in s_1$.

Let find the upper sharp bound for the function

Calculations show that the stationary points of $p\left(x,y\right)$ are the following

Note that only the first point lies in $s_1$ provided $r^{\ast }=r_0^{-1}\sqrt{1-\sqrt{1-r_0^4}}=0.0507...$.

The second partial derivative at the point $\frac{r^{\ast }}{2\sqrt{2}}+i\frac{r^{\ast }}{2\sqrt{2}}$ equals

So we have

Therefore the function $p\left(x,y\right)$ may attain its maximum only in this point or on the boundary of the square $s_1$ or at the point $\frac{r^{\ast }}{2\sqrt{2}}+i\frac{r^{\ast }}{2\sqrt{2}}$. Let us show that $p\left(x,y\right)$ attains maximum at the point $\frac{r^{\ast }}{2\sqrt{2}}+i\frac{r^{\ast }}{2\sqrt{2}}$. Simple calculations show that

Thus we have

Let us consider the ring $1>\mid z\mid >r^{\ast }$. Divide the circle $\mid z\mid =r$, $1>r>r^{\ast }$ into two disjoint subsets, consisting of pairwise disjoint arc, contracting angle $\frac{\pi }{n}$ and translating into each other after rotation on angle $\frac{\pi }{n}$. Let us find $n$ such that

or