Lobachevskii Journal of Mathematics

http://ljm.ksu.ru

Vol. 16, 2004, 57 – 69 \begin{matrix}c \\ ◯ \end{matrix} A. Kuznetsov

Alexander Kuznetsov

ON A PROBLEM OF AVHADIEV

(submitted by F. Avhadiev)

Abstract. In this paper we consider a lower estimate for the ratio I(Ω) of the conformal moment of a simple connected domain Ω in the complex plane to the moment of inertia of this domain about its boundary. Related functionals depending on a simple connected domain Ω and two points w_{1}, w_{2} ∈ Ω with fixed hyperbolical distance between them are estimated. As a consequence a nontrivial lower estimate for I(Ω) is obtained.

1. Introduction

Let Ω be a simply connected domain on the complex plane ℂ and w ∈ Ω. Let ρ_{Ω}(w) be the conformal radius of Ω at the point w, and d_{Ω}(w) be the Euclidean distance from the point w to the boundary ∂Ω of the domain Ω.

I_{c}(Ω) = ∫_{Ω} ρ_{Ω}^{2}(x + iy)dxdy is the conformal moment of Ω, and I(∂Ω) = ∫_{Ω} d_{Ω}^{2}(x + iy)dxdy is the moment of inertia of Ω about ∂Ω. 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

2000 Mathematical Subject Classification. 30A10,30C75. Key words and phrases. Moment of inertia of domain about its boundary,

conformal moment of domain, univalent functions.

Partially supported by Russian Foundation for Basic Research, Grant 04-01-00083 and the Program ”Russian Universities”, Grant ur.04.01.040.

rigidity P (Ω) of the domain Ω by simple geometric characteristics of the domain [2, 3]. As a solution the following inequalities

I(∂Ω) ⩽ I_{c}(Ω) ⩽ P (Ω) ⩽ 4I_{c}(Ω) ⩽ 64I(∂Ω)  (1)

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}(Ω) and I(∂Ω) are finite. F.G. Avkhadiev set a problem to find lower and upper sharp bounds for the ratio

I(Ω) = \frac{I_{c}(Ω)}{I(∂Ω)},  (2)

when I_{c}(Ω) and I(∂Ω) are finite. Note that I(Ω) is invariant under linear transforms of Ω.

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

Let w_{1}, w_{2} ∈ Ω, and let a holomorphic univalent function f(z) maps the unit disk 𝕌 = {z : |z| < 1} onto Ω so that f(0) = w_{1} and f(r) = w_{2}, where r ∈ (0, 1) 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

G(f, α) = \frac{|f′(0)|^{α} + |f′(r)|^{α}(1 − |r|^{2})^{α}}{d_{Ω}(f(0))^{α} + d_{Ω}(f(r))^{α}} , α > 0,  (3)

on the class S^{0} of univalent holomorphic functions f(z), z ∈ 𝕌.

Obviously G(f, α) = G(β + γf, α), γ ≠ 0. It is possible to consider the functional G(f, α) as a function of Ω and two points w_{1}, w_{2} ∈ Ω with fixed hyperbolical distance between them. Namely,

G(f, α) = \frac{ρ_{Ω}(w_{1})^{α} + ρ_{Ω}(w_{2})^{α}}{d_{Ω}(w_{1})^{α} + d_{Ω}(w_{2})^{α}} .

Let r_{α} be the unique solution of the equation

\begin{matrix}1 − \frac{4r}{(1 − r)^{2}} + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} + \frac{(1 + r)^{2α}}{(1 − r)^{2α}} ( min_{γ,θ∈ℝ}ℜe[\frac{1 − 2re^{iγ} − r^{2}e^{2iγ}}{(1 + re^{iγ})^{2}} + \\ +\frac{2r^{2}}{1 − r^{2}} \frac{1 − re^{iγ}}{1 + re^{iγ}} + \frac{4re^{iθ}}{1 − r^{2}} \frac{1 − re^{iγ}}{1 + re^{iγ}} ] − \frac{4r}{(1 − r)^{2}} ) = 0, \end{matrix}

on the interval (0, 1). Then we have Theorem 1. If r < r_{α}, then every function f(z) minimizing the functional G(f, α) on the class S^{0} maps 𝕌 onto an arc biangle bounded by circle arcs centered at the points f(0) and f(r).

To improve the lower estimate of I(Ω) we need to find the sharp bound of the functional

\begin{matrix}F (f, r, c) = \frac{|f′(0)|^{4}c + |f′(r)|^{4}(1 − r^{2})^{4}}{d_{f(𝕌)}(f(0))^{2}|f′(0)|^{2}c + d_{f(𝕌)}(f(r))^{2}|f′(r)|^{2}(1 − r^{2})^{2}} , \\ c > 0, 0 < r < 1, \end{matrix} (4)

on class S^{0}. We have

Proposition 1. If r < r_{0}, then every function f(z) minimizing the functional F (f, r, c) on the class S^{0} maps 𝕌 onto an arc biangle, bounded by circle arcs centered at the points f(0) and f(r).

The value of r_{0} will be given in the proof of Proposition 1.

By f_{θ,d} denote a function mapping 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 θ is the argument of the intersection point of this circle lying in the upper half-plane. Taking into account that the functional F (f, r, c) is invariant under linear transforms of the complex plane, we conclude that there are d > 0, θ ∈ [0, π] such that F (f_{θ,d}, r, c) = sup_{f∈S^{0}} F (f, r, c).

Now we can give a lower estimate for I(Ω) in terms of the functional F (f, r, c).

Theorem 2. For any simply connected domain Ω, for which I_{c}(Ω) and I(∂Ω) are finite, the estimate

\begin{matrix}I(Ω) ⩾ min_{r_{1}⩽f_{θ,d}^{−1}(d)⩽r_{0}}\{F (f_{θ,d}, f_{θ,d}^{−1}(d), \frac{1}{1 − r^{∗2}} ), F (f_{θ,d}, f_{θ,d}^{−1}(d), 1 − r^{∗2})\} = \\ = 1.031021... \end{matrix}

holds.

The values of r^{∗}, 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 ∈ S, where S is the class of all holomorphic univalent functions f(z) in 𝕌 normalized by f(0) = f′(0) − 1 = 0. Let k(z) = \frac{z}{(1−z)^{2}} , k_{ε}(z) = k^{−1}((1 − ε)k(z)) and k_{ε}^{γ}(z) = e^{−iγ}k_{ε}(e^{iγ}z). For ε small enough we have:

k_{ε}^{γ}(z) = z − εz \frac{1 − ze^{iγ}}{1 + ze^{iγ}} + O(ε^{2}).  (5)

The function k_{ε}^{γ}(z) maps 𝕌 onto 𝕌 minus radial slit with endpoint at −e^{−iγ}. Note that the slit length tends to 0 as ε → 0.

Assume the opposite. Then there are two open disks D_{1}, D_{2} ⊂ 𝕌, 𝕌 ⊄ D_{1} ∪ D_{2}, with centers at points 0 and f(r) and radii d(0) and d(f(r)) respectively. Therefore, there are γ ∈ ℝ and ε_{0} > 0 such that D_{1}, D_{2} ⊂ f_{ε}(𝕌), where f_{ε}(z) = f(k_{ε}^{γ}(z)), 0 ⩽ ε ⩽ ε_{0}.

By construction d(f_{ε}(0)) = d(f(0)) = d(0), and |f′_{ε}(0)| = (1−ε)|f′(0)| = (1 − ε). Moreover taking into account (5), we have

f_{ε}(z) = f(z) − εf′(z)z \frac{1 − ze^{iγ}}{1 + ze^{iγ}} + O(ε^{2}).  (6)

Using (6) and D_{m} ⊂ f_{ε}(𝕌), m = 1, 2 we get

\begin{matrix}d(f_{ε}(r)) ⩾ d(f(r)) − ε|f′(r)r\frac{1 − re^{iγ}}{1 + re^{iγ}} | + O(ε^{2}) ⩾ \\ ⩾ d(f(r))(1 − ε\frac{|f′(r)|}{d(f(r))}r\frac{1 + r}{1 − r}) + O(ε^{2}). \end{matrix}  (7)

Note that

\frac{|f′(r)|}{d(r)} = \frac{(1 − r^{2})|f′(r)|}{(1 − r^{2})d(f(r))},

and |f′(r)|(1−r^{2}) is the conformal radius of the domain. Thus |f′(r)|/d(r) ⩽ \frac{4}{1−r^{2}} . Therefore

\begin{matrix}d(f_{ε}(r)) ⩾ d(f(r))(1 − ε4r\frac{1 + r}{(1 − r^{2})(1 − r)}) + O(ε^{2}) = \\ d(f(r))(1 − ε\frac{4r}{(1 − r)^{2}} ) + O(ε^{2}). \end{matrix}

Let p(r) = \frac{4r}{(1−r)^{2}} .

Using (6) we have

f′_{ε}(z) = f′(z) − ε(zf′(z)\frac{1 − ze^{iγ}}{1 + ze^{iγ}} )′ + O(ε^{2}).

So

|f′_{ε}(z)| = |f′(z)|(1 − εℜe[\frac{1 − 2ze^{iγ} − z^{2}e^{2iγ}}{(1 + ze^{iγ})^{2}} + \frac{zf′′(z)}{f′(z)} \frac{1 − ze^{iγ}}{1 + ze^{iγ}} ]) + O(ε^{2}).

Let q(r) = ℜe[\frac{1−2re^{iγ}−r^{2}e^{2iγ}}{(1+re^{iγ})^{2}} + \frac{rf′′(r)}{f′(r)} \frac{1−re^{iγ}}{1+re^{iγ}} ]. Taking into account (7), we obtain

G(f_{ε}, α) ⩽ \frac{|f′(0)|^{α}(1 − αε) + |f′(r)|^{α}(1 − |r|^{2})^{α}(1 − q(r)αε)}{d(f(0))^{α} + d(f(r))^{α}(1 − p(r)αε)} + O(ε^{2}).

(8)

Let

A = |f′(0)|^{α},
B = (1 − r^{2})^{α}|f′(r)|^{α},
C = d(f(0))^{α},
D = d(f(r))^{α}.

Then inequality (8) takes the form

G(f_{ε}, α) ⩽ \frac{A(1 − εα) + B(1 − q(r)αε)}{C + D(1 − p(r)αε)} + O(ε^{2}),

or

G(f_{ε}, α) ⩽ \frac{A + B}{C + D} − \frac{(C + D)(A + Bq(r)) − (A + B)Dp(r)}{(C + D)^{2}}αε + O(ε^{2}).

Therefore, in order to prove that every function f(z) minimizing the functional G(f, α) on class S map 𝕌 onto an arc biangle, it is sufficient to show that

(C + D)(A + Bq(r)) − (A + B)Dp(r) > 0,

or

1 − p(r) + \frac{C}{D} + \frac{B}{A} (q(r) − p(r)) > 0.  (9)

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

\frac{C}{D} ⩾ (\frac{d(f(0))}{d(f(0)) + |f(r)|})^{α}.

Since f ∈ S,

\frac{C}{D} ⩾ (\frac{1}{1 + 4|f(r)|})^{α}.

Using the growth theorem in the class S, we have

\frac{C}{D} ⩾ (\frac{(1 − r)^{2}}{(1 − r)^{2} + 4r})^{α} = \frac{(1 − r)^{2α}}{(1 + r)^{2α}} .

By the distortion theorem for the class S

\frac{(1 − r)^{2α}}{(1 + r)^{2α}} ⩽ \frac{A}{B} ⩽ \frac{(1 + r)^{2α}}{(1 − r)^{2α}} .

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

1 − p(r) + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} (q(r) − p(r)) > 0

and

1 − p(r) + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} + \frac{(1 + r)^{2α}}{(1 − r)^{2α}} (q(r) − p(r)) > 0,

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

|\frac{rf′′(r)}{f′(r)} − \frac{2r^{2}}{1 − r^{2}} | ⩽ \frac{4r}{1 − r^{2}}  (10)

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

\begin{matrix}M_{1}(r) = 1 − \frac{4r}{(1 − r)^{2}} + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} + \\ +\frac{(1 − r)^{2α}}{(1 + r)^{2α}} ( min_{γ,θ∈ℝ}R(r, γ, θ) − \frac{4r}{(1 − r)^{2}} ) > 0, \end{matrix}  (11)

\begin{matrix}M_{2}(r) = 1 − \frac{4r}{(1 − r)^{2}} + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} + \\ +\frac{(1 + r)^{2α}}{(1 − r)^{2α}} ( min_{γ,θ∈ℝ}R(r, γ, θ) − \frac{4r}{(1 − r)^{2}} ) > 0, \end{matrix}  (12)

where

R(r, γ, θ) = ℜe[\frac{1 − 2re^{iγ} − r^{2}e^{2iγ}}{(1 + re^{iγ})^{2}} + \frac{2r^{2}}{1 − r^{2}} \frac{1 − re^{iγ}}{1 + re^{iγ}} + \frac{4re^{iθ}}{1 − r^{2}} \frac{1 − re^{iγ}}{1 + re^{iγ}} ]

imply (9).

Consider the functions

P (r) = 1 − \frac{4r}{(1 − r)^{2}}

and

Q(r) = min_{γ,θ∈ℝ}R(r, γ, θ) − \frac{4r}{(1 − r)^{2}} .

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

1 − \frac{4r}{(1 − r)^{2}} + \frac{(1 − r)^{2α}}{(1 + r)^{2α}} + \frac{(1 + r)^{2α}}{(1 − r)^{2α}} ( min_{γ,θ∈ℝ}R(r, γ, θ) − \frac{4r}{(1 − r)^{2}} ) = 0

on the interval (0, 1), then every function minimizing functional (3) maps 𝕌 onto an arc biangle, bounded by two circle arcs centered at the points f(0) and f(r). Note that r_{α} > r′.

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

A = c|f′(0)|^{4},
B = |f′(r)|^{4}(1 − r^{2})^{4},
C = d(f(0))^{2}|f′(0)|^{2}c,
D = d(g(r))^{2}|f′(r)|^{2}(1 − r^{2})^{2},

we have

\begin{matrix}F (f_{ε}, c, r) ⩽ \frac{A + B}{C + D}− \\ −ε\frac{4(A + q(r)B)(C + D) − 2(A + B)(C + (p + q)D)}{(C + D)^{2}} + O(ε^{2}), \end{matrix}

where f_{ε}(z) = f(k_{ε}^{γ}(z)). Thus the inequality

4(A + q(r)B)(C + D) − 2(A + B)(C + (p(r) + q(r))D) > 0

or

A(C +D(2−p(r)−q(r)))+B(D(q(r)−p(r))+C(−1+2q(r)))) > 0 (13)

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

\begin{matrix}q(r) > p(r), \\ p(r) + q(r) < 2, \\ q(r) > 0.5 \end{matrix}  (14)

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

min_{γ,θ∈ℝ}R(r, γ, θ) − \frac{4r}{(1 − r)^{2}} > 0,

\frac{4r}{(1 − r)^{2}} + max_{γ,θ∈ℝ}R(r, γ, θ) < 2,

min_{γ,θ∈ℝ}R(r, γ, θ) > 0.5

(15)

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

r < r_{0} = max{r_{01}, r_{02}, r_{03}},

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

3. Proof of Theorem 2

Proof of Theorem 2. First we divide 𝕌 into two subsets 𝕌_{1}, 𝕌_{2} and construct a map ξ(z) : 𝕌_{1} → 𝕌_{2} such that Jacobian of ξ(z) equals the unity almost everywhere and the inequality

s(ξ(z), z) = |\frac{z − ξ(z)}{1 − \overline{z}ξ(z)}| < r_{0}, a.e.

holds.

First, we consider a disk 𝕌^{∗} of radius r^{∗} centered at the origin.

Let us consider the square s inscribed in 𝕌^{∗} 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 𝕌^{∗}, 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} ⊂ 𝕌_{1}, s_{3}, s_{4}, c_{2}, c_{4} ⊂ 𝕌_{2} and

ξ(z) = \begin{cases}z − \frac{r^{∗}}{\sqrt{2}} − i\frac{r^{∗}}{\sqrt{2}}, z ∈ s_{1} \\ z + \frac{r^{∗}}{\sqrt{2}} − i\frac{r^{∗}}{\sqrt{2}}, z ∈ s_{2} \\ e^{i \frac{π}{2}} z, z ∈ c_{1}, c_{3}. \end{cases}

Let us find the lower and upper sharp bounds of s(ξ(z), z), for z ∈ c_{1}, c_{3}. It is easy to see that

s(ξ(z), z) = \frac{|z||1 − i|}{|1 + |z|^{2}i|} = \frac{\sqrt{2}|z|}{\sqrt{1 + |z|^{4}}} .

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

\frac{2r^{∗}}{\sqrt{4 + r^{∗4}}} ⩽ s(ξ(z), z) ⩽ \frac{\sqrt{2}r^{∗}}{\sqrt{1 + r^{∗4}}} , z ∈ c_{1}, c_{3}.

By construction, s(ξ(z), z) < r_{0} a.e., so

r^{∗} = \frac{\sqrt{1 − \sqrt{1 − r_{0}^{4}}}}{r_{0}} = 0.0507682....

Let us find the lower and upper sharp bounds of s(ξ(z), z), for z ∈ s_{1}. Let z = x + iy, then we have

s(ξ(z), z) = \frac{r^{∗}}{|1 − (x + iy)(x − \frac{r^{∗}}{\sqrt{2}} − i(y − \frac{r^{∗}}{\sqrt{2}}))|}.

So

s(ξ(z), z)^{2} = \frac{r^{∗2}}{\frac{r^{∗2}(x−y)^{2}}{2} + (1 − x^{2} − y^{2} + \frac{r^{∗}(x+y)}{\sqrt{2}})^{2}} .

Let us consider the function

q(x, y) = 1 − x^{2} − y^{2} + \frac{r^{∗}(x + y)}{\sqrt{2}}.

When z ∈ s_{1}, the function q(x, y) is nonnegative, and attains its minimal value at points \frac{r^{∗}}{\sqrt{2}}, i\frac{r^{∗}}{\sqrt{2}}, 0, \frac{r^{∗}}{\sqrt{2}} + i\frac{r^{∗}}{\sqrt{2}}. Thus the function \frac{r^{∗2}}{2} (x − y)^{2} is minimal at x = y, then

\frac{r^{∗2}(x − y)^{2}}{2} + (1 − x^{2} − y^{2} + \frac{r^{∗}(x + y)}{\sqrt{2}})^{2}

attains minimum at the point 0, so we have s(ξ(z), z) ⩽ r^{∗}, for z ∈ s_{1}. Let find the upper sharp bound for the function

p(x, y) = \frac{r^{∗2}(x − y)^{2}}{2} + (1 − x^{2} − y^{2} + \frac{r^{∗}(x + y)}{\sqrt{2}})^{2}.

Calculations show that the stationary points of p(x, y) are the following

\frac{r^{∗}}{2\sqrt{2}} + i\frac{r^{∗}}{2\sqrt{2}},

\frac{r^{∗} − \sqrt{4 − r^{∗2}}}{2\sqrt{2}} + i\frac{r^{∗} + \sqrt{4 − r^{∗2}}}{2\sqrt{2}} ,

\frac{r^{∗} + \sqrt{4 − r^{∗2}}}{2\sqrt{2}} + i\frac{r^{∗} − \sqrt{4 − r^{∗2}}}{2\sqrt{2}} ,

\frac{r^{∗} − \sqrt{4 + r^{∗2}}}{2\sqrt{2}} + i\frac{r^{∗} − \sqrt{4 + r^{∗2}}}{2\sqrt{2}} ,

\frac{r^{∗} + \sqrt{4 + r^{∗2}}}{2\sqrt{2}} + i\frac{r^{∗} + \sqrt{4 + r^{∗2}}}{2\sqrt{2}} .

Note that only the first point lies in s_{1} provided r^{∗} = r_{0}^{−1}\sqrt{1 − \sqrt{1 − r_{0}^{4}}} = 0.0507....

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

p_{xx}(\frac{r^{∗}}{2\sqrt{2}}, \frac{r^{∗}}{2\sqrt{2}}) = p_{yy}(\frac{r^{∗}}{2\sqrt{2}}, \frac{r^{∗}}{2\sqrt{2}}) = \frac{(16 + 24r^{∗2} + r^{∗4})}{16},

p_{xy}(\frac{r^{∗}}{2\sqrt{2}}, \frac{r^{∗}}{2\sqrt{2}}) = \frac{(4 − r^{∗2})^{2}}{16} .

So we have

p_{xx}p_{yy} − p_{xy}^{2} = \frac{r^{∗2}(4 + r^{∗2})^{2}}{4} > 0.

Therefore the function p(x, y) may attain its maximum only in this point or on the boundary of the square s_{1} or at the point \frac{r^{∗}}{2\sqrt{2}} + i\frac{r^{∗}}{2\sqrt{2}}. Let us show that p(x, y) attains maximum at the point \frac{r^{∗}}{2\sqrt{2}} + i\frac{r^{∗}}{2\sqrt{2}}. Simple calculations show that

p_{x}(\frac{r^{∗}}{\sqrt{2}}, y) = −\sqrt{2}r^{∗}(1 − (\frac{r^{∗}}{\sqrt{2}} − y)^{2}) < 0, y ∈ [0, \frac{r^{∗}}{\sqrt{2}}]

p_{x}(0, y) = \sqrt{2}r^{∗}(1 − y^{2}) > 0, y ∈ [0, \frac{r^{∗}}{\sqrt{2}}]

p_{y}(x, \frac{r^{∗}}{\sqrt{2}}) = −\sqrt{2}r^{∗}(1 − (\frac{r^{∗}}{\sqrt{2}} − x)^{2}) < 0, x ∈ [0, \frac{r^{∗}}{\sqrt{2}}]

p_{y}(x, 0) = \sqrt{2}r^{∗}(1 − x^{2}) > 0. x ∈ [0, \frac{r^{∗}}{\sqrt{2}}]

Thus we have

\frac{4r^{∗}}{4 + r^{∗2}} ⩽ s(ξ(z), z) ⩽ \frac{\sqrt{2}r^{∗}}{\sqrt{1 + r^{∗4}}} = r_{0}, z ∈ s_{1}, s_{2}, c_{1}, c_{3}.

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

s(ξ(z), z) = \frac{r|1 − e^{i \frac{π}{n}} |}{|1 − r^{2}e^{i \frac{π}{n}} |} ⩽ r_{0},

or

cos \frac{π}{n} ⩾ \frac{2r^{2} − r_{0}^{2}(1 + r^{4})}{2r^{2}(1 − r_{0}^{2})}.

Thus

n ⩾ A(r, r_{0}) = \frac{π}{arccos \frac{2r^{2}−r_{0}^{2}(1+r^{4})}{2r^{2}(1−r_{0}^{2})}} .

Let

n = [A(r, r_{0})] + 1.

It is easy to check

s(z, ξ(z)) = \frac{r|1 − e^{i \frac{π}{n}} |}{|1 − r^{2}e^{i \frac{π}{n}} |} = r\sqrt{\frac{2 − 2 cos \frac{π}{[A(r,r_{0})]+1}}{1 + r^{4} − 2r^{2} cos \frac{π}{[A(r,r_{0})]+1}}} ⩾

r\sqrt{\frac{2 − 2 cos \frac{π}{A(r,r_{0})+1}}{1 + r^{4} − 2r^{2} cos \frac{π}{A(r,r_{0})+1}}} .

Calculations show that it attains minimum on the set [r^{∗}, 1) at the point r = r^{∗}, and minimal value equals 0.0508335....

Thus, we have

r_{1} = \frac{4r^{∗}}{4 + r^{∗2}} = 0.0507355... ⩽ s(ξ(z), z) ⩽ r_{0} = 0.0717968...., z ∈ 𝕌_{1}.

By construction we have that the measure of the set {z : s(z, ξ(z)) = r_{0}} is zero.

Let us find the lower and upper sharp bounds for

c(z) = \frac{1 − |z|^{2}}{1 − |ξ(z)|^{2}}  (16)

in z ∈ 𝕌_{1}. By construction we have c(z) = 1, if z ∈ c_{1}, c_{2} or |z| > r^{∗}. Let us consider the case of z ∈ s_{1}. Then it is easy to see that (16) attains minimum at the point \frac{\sqrt{2}}{2}r + i\frac{\sqrt{2}}{2}r and maximum at the point 0. So we have that (16) lies between 1 − r^{∗2} and (1 − r^{∗2})^{−1}.

Now we can give an improved lower estimate of (2). Let f(z) map 𝕌 onto Ω, then we have

I_{c}(Ω) = ∫_{𝕌} [|f′(z)|^{4}(1 − |z|^{2})^{2}]dxdy =
∫_{𝕌_{1}} [|f′(z)|^{4}(1 − |z|^{2})^{2}]dxdy + ∫_{𝕌_{2}} [|f′(z)|^{4}(1 − |z|^{2})^{2}]dxdy =

∫_{𝕌_{1}} [|f′(z)|^{4}(1 − |z|^{2})^{2} + |f′(ξ(z))|^{4}(1 − |ξ(z)|^{2})^{2}]dxdy.

Similarly,

\begin{matrix}I(∂Ω) = ∫_{𝕌} [|f′(z)|^{2}d(f(z))^{2}]dxdy = \\ ∫_{𝕌_{1}} [|f′(z)|^{2}d(f(z))^{2} + |f′(ξ(z))|^{2}d(f(ξ(z)))^{2}]dxdy. \end{matrix}

So we have that functional (2) is not greater then

\frac{|f′(z)|^{4}(1 − |z|^{2})^{2} + |f′(ξ(z))|^{4}(1 − |ξ(z)|^{2})^{2}}{|f′(z)|^{2}d(f(z))^{2} + |f′(ξ(z))|^{2}d(f(ξ(z)))^{2}} .  (17)

Making change of variable ψ(w) = \frac{w−z}{1−\overline{z}w}, we have that (17) is equal to

\frac{|g′(0)|^{4}c + |g′(r)|^{4}(1 − r^{2})^{4}}{d(0)^{2}|g′(0)|^{2}c + d(g(r))^{2}|g′(r)|^{2}(1 − r^{2})^{2}} ,  (18)

where g(w) = f ∘ ψ(w), r = s(ξ(z), z) and c = \frac{(1−|z|^{2})^{2}}{(1−|ξ(z)|^{2})^{2}} . Using

r_{1} < r < r_{0}, (1 − r^{∗2})^{2} ⩽ \frac{(1 − |ξ(z)|^{2})^{2}}{(1 − |z|^{2})^{2}} ⩽ \frac{1}{(1 − r^{∗2})^{2}} ,

proposition 1 and that (18) is monotonic on c, we have

I(Ω) ⩾ min_{r_{1}⩽f_{θ,d}^{−1}(d)⩽r_{0}}\{F (f_{θ,d}, f_{θ,d}^{−1}(d), \frac{1}{1 − r^{∗2}} ), F (f_{θ,d}, f_{θ,d}^{−1}(d), 1 − r^{∗2})\}.

Numerical calculations show that the right hand side of above inequality is not less then 1.031021....

References

[1] Avhadiev F.G. Solution of generalized St Venant problem // Matem. Sborn. v.189 (1998) N 12 p.3-12 (in Russian)

[2] Saint-Venant B. Memoir about torsion of prisms. Memoir about winding of prisms. M.:GIFML,1961. (in Russian)

[3] Timoshenko S.P. History of science of strength of materials. M.:GITTL, 1957. (in Russian)

[4] Goluzin G. M. Geometric theory of functions of a complex variable. M. : Nauka, 1966. (in Russian)

ON A PROBLEM OF AVHADIEV 69

[5] Nevanlinna R.Uniformization. Springer-Verlag, 1967.

[6] Duren P.L. Univalent functions. Springer-Verlag, 1983.

Department of Mathematical Analysis, Mathematical Department, Saratov State University, ul. Astrakhanskaya, 83, Saratov:410012, RUSSIA E-mail address: KuznetsovAA@pisem.net

Received July 6, 2004