Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 16, 2004, 57 – 69
© A. Kuznetsov
ON A PROBLEM OF AVHADIEV
(submitted by F. Avhadiev)
ABSTRACT. In this paper we consider a lower estimate for the ratio of the conformal moment of a simple connected domain in the complex plane to the moment of inertia of this domain about its boundary. Related functionals depending on a simple connected domain and two points with fixed hyperbolical distance between them are estimated. As a consequence a nontrivial lower estimate for is obtained.
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.
Let be a simply connected domain on the complex plane and . Let be the conformal radius of at the point , and be the Euclidean distance from the point to the boundary of the domain .
is the conformal moment of , and is the moment of inertia of about . This functionals were introduced by F.G. Avkhadiev  for solution of the classical St. Venant problem of finding two-side estimates for the torsional rigidity of the domain 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 and are finite. F.G. Avkhadiev set a problem to find lower and upper sharp bounds for the ratio
when and are finite. Note that is invariant under linear transforms of .
The first and the last inequalities in (1) imply that the region of values is a subset of . In this paper we give a better lower estimate for .
Let , and let a holomorphic univalent function maps the unit disk onto so that and , where is a given constant. The normalization of means, that the hyperbolical distance between and is constant. Let us consider the functional
on the class of univalent holomorphic functions .
Obviously . It is possible to consider the functional as a function of and two points with fixed hyperbolical distance between them. Namely,
Let be the unique solution of the equation
on the interval . Then we have
To improve the lower estimate of we need to find the sharp bound of the functional
on class . We have
The value of will be given in the proof of Proposition 1.
By denote a function mapping 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 and is the argument of the intersection point of this circle lying in the upper half-plane. Taking into account that the functional is invariant under linear transforms of the complex plane, we conclude that there are such that .
Now we can give a lower estimate for in terms of the functional .
The values of will be given in the proof of Theorem 2.
Proof of Theorem 1. It is possible to assume that , where is the class of all holomorphic univalent functions in normalized by . Let , and . For small enough we have:
The function maps onto minus radial slit with endpoint at . Note that the slit length tends to as .
Assume the opposite. Then there are two open disks , with centers at points and and radii and respectively. Therefore, there are and such that , where .
By construction , and . Moreover taking into account (5), we have
Using (6) and we get
and is the conformal radius of the domain. Thus . Therefore
Using (6) we have
Let . Taking into account (7), we obtain
Then inequality (8) takes the form
Therefore, in order to prove that every function minimizing the functional on class map onto an arc biangle, it is sufficient to show that
Let us find a lower bound of . Using the obvious inequality , we have
Using the growth theorem in the class , we have
By the distortion theorem for the class
Tacking into account that the left part of (9) is monotonic on , we conclude that the following inequalities
imply inequality (9). Substituting into this inequality , using the estimate for function (see foe e.g. [6, p. 32])
and the maximum principle for harmonic functions we have, that inequalities
Consider the functions
They are monotonic on the interval . So they may have only one zero on this interval. Computations show, that equals zero at the point , and equals zero at the point . Therefore we have that the roots of equations and lie in the interval . It is easy to see that, for , and so if inequality (12) holds, then inequality (11) holds too. Using the inequality and the fact that is decreasing on the interval we have, that the function is decreasing on this interval too. Thus, we conclude that if , where is the single root of the equation
on the interval , then every function minimizing functional (3) maps onto an arc biangle, bounded by two circle arcs centered at the points and . Note that .
Proof of Proposition 1. Using the same argument as in the proof of theorem 1 and following notation
where . Thus the inequality
implies that any function minimizing functional (4), maps onto an arc biangle. Using inequalities we have that inequalities
guarantee inequality (13). Substituting in (14) the functions and , using estimate (10) and the maximum principle for harmonic functions, we have that the following inequalities
Functions in the left-side of (15) are monotonic on the interval (0,1). Computations show that, the first was violated at , the second at , the third at . thus we conclude that if
then every function every minimizing functional (4) maps onto an arc biangle, bounded by circle arcs centered at the points and .
Proof of Theorem 2. First we divide into two subsets and construct a map such that Jacobian of equals the unity almost everywhere and the inequality
First, we consider a disk of radius centered at the origin.
Let us consider the square inscribed in with sides parallel to the axes of coordinates. Let be the intersection of the square with the first, the second, the third and the fourth quarters of the complex plane. Let be the sectors of , cutting off the sides of the square , parallel to the axis OX, and by sectors, cutting off the sides of the square , paralleling to the axis OY.
Let , and
Let us find the lower and upper sharp bounds of , for . It is easy to see that
The function increases on the interval and so we have
By construction, , so
Let us find the lower and upper sharp bounds of , for . Let , then we have
Let us consider the function
When , the function is nonnegative, and attains its minimal value at points , , , . Thus the function is minimal at , then
attains minimum at the point , so we have , for .
Let find the upper sharp bound for the function
Calculations show that the stationary points of are the following
Note that only the first point lies in provided .
The second partial derivative at the point equals
So we have
Therefore the function may attain its maximum only in this point or on the boundary of the square or at the point . Let us show that attains maximum at the point . Simple calculations show that
Thus we have
Let us consider the ring . Divide the circle , into two disjoint subsets, consisting of pairwise disjoint arc, contracting angle and translating into each other after rotation on angle . Let us find such that
It is easy to check
Calculations show that it attains minimum on the set at the point , and minimal value equals .
Thus, we have
By construction we have that the measure of the set is zero.
Let us find the lower and upper sharp bounds for
in . By construction we have , if or . Let us consider the case of . Then it is easy to see that (16) attains minimum at the point and maximum at the point . So we have that (16) lies between and .
Now we can give an improved lower estimate of (2). Let map onto , then we have
So we have that functional (2) is not greater then
Making change of variable , we have that (17) is equal to
where , and . Using
proposition 1 and that (18) is monotonic on , we have
Numerical calculations show that the right hand side of above inequality is not less then .
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