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

Alexander Kuznetsov

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 w1,w2 Ω with fixed hyperbolical distance between them are estimated. As a consequence a nontrivial lower estimate for I(Ω) 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.

### 1. Introduction

Let Ω be a simply connected domain on the complex plane and w Ω. Let ρΩ (w) be the conformal radius of Ω at the point w, and dΩ (w) be the Euclidean distance from the point w to the boundary Ω of the domain Ω.

Ic(Ω) = ΩρΩ2(x + iy)dxdy is the conformal moment of Ω, and I(Ω) = ΩdΩ2(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 rigidity P (Ω) of the domain Ω by simple geometric characteristics of the domain [23]. As a solution the following inequalities

 I(∂Ω) ≤ Ic(Ω) ≤ P(Ω) ≤ 4Ic(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.[45]). The first inequality in (1) is not sharp when Ic (Ω) and I(Ω) are finite. F.G. Avkhadiev set a problem to find lower and upper sharp bounds for the ratio

 I(Ω) = Ic(Ω) I(∂Ω), (2)

when Ic(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 w1,w2 Ω, and let a holomorphic univalent function f(z) maps the unit disk U = {z : z < 1} onto Ω so that f(0) = w1 and f(r) = w2, where r (0, 1) is a given constant. The normalization of f means, that the hyperbolical distance between w1 and w2 is constant. Let us consider the functional

 G(f,α) = ∣f′(0)∣α + ∣f′(r)∣α(1 −∣r∣2)α dΩ(f(0))α + dΩ(f(r))α ,α > 0, (3)

on the class S0 of univalent holomorphic functions f(z),z U.

Obviously G(f,α) = G(β + γf,α),γ0. It is possible to consider the functional G(f,α) as a function of Ω and two points w1,w2 Ω with fixed hyperbolical distance between them. Namely,

G(f,α) = ρΩ(w1)α + ρ Ω(w2)α dΩ(w1)α + dΩ(w2)α.

Let rα be the unique solution of the equation

1 4r (1 r)2 + (1 r)2α (1 + r)2α + (1 + r)2α (1 r)2α min γ,θe1 2reiγ r2e2iγ (1 + reiγ)2 + + 2r2 1 r2 1 reiγ 1 + reiγ + 4reiθ 1 r2 1 reiγ 1 + reiγ 4r (1 r)2 = 0,

on the interval (0, 1). Then we have

Theorem 1. If r < rα, then every function f(z) minimizing the functional G(f,α) on the class S0 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

 F(f,r,c) = ∣f′(0)∣4c + ∣f′(r)∣4(1 − r2)4 df(U)(f(0))2∣f′(0)∣2c + df(U)(f(r))2∣f′(r)∣2(1 − r2)2, c > 0, 0 < r < 1, (4)

on class S0. We have

Proposition 1. If r < r0, then every function f(z) minimizing the functional F(f,r,c) on the class S0 maps U onto an arc biangle, bounded by circle arcs centered at the points f(0) and f(r).

The value of r0 will be given in the proof of Proposition 1.

By fθ,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(fθ,d,r,c) = sup fS0F(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 Ic(Ω) and I(Ω) are finite, the estimate

I(Ω) min r1fθ,d1(d)r0{F(fθ,d,fθ,d1(d), 1 1 r2),F(fθ,d,fθ,d1(d), 1 r2)} = = 1.031021...

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) = z (1z)2, kɛ (z) = k1((1 ɛ)k(z)) and kɛ γ (z) = eiγk ɛ(eiγz). For ɛ small enough we have:

 kɛγ(z) = z − ɛz1 − zeiγ 1 + zeiγ + O(ɛ2). (5)

The function kɛγ(z) maps U onto U minus radial slit with endpoint at eiγ. Note that the slit length tends to 0 as ɛ 0.

Assume the opposite. Then there are two open disks D1 ,D2 U, U D1 D2, with centers at points 0 and f(r) and radii d(0) and d(f(r)) respectively. Therefore, there are γ and ɛ0 > 0 such that D1,D2 fɛ(U), where fɛ(z) = f(kɛγ(z)), 0 ɛ ɛ 0.

By construction d(fɛ(0)) = d(f(0)) = d(0), and fɛ(0) = (1 ɛ)f(0) = (1 ɛ). Moreover taking into account (5), we have

 fɛ(z) = f(z) − ɛf′(z)z1 − zeiγ 1 + zeiγ + O(ɛ2). (6)

Using (6) and Dm fɛ(U),m = 1, 2 we get

 d(fɛ(r)) ≥ d(f(r)) − ɛf′(r)r1 − reiγ 1 + reiγ + O(ɛ2) ≥ ≥ d(f(r))1 − ɛ ∣f′(r)∣ d(f(r))r1 + r 1 − r + O(ɛ2). (7)

Note that

f(r) d(r) = (1 r2)f(r) (1 r2)d(f(r)),

and f(r)(1 r2) is the conformal radius of the domain. Thus f (r)d(r) 4 1r2. Therefore

d(fɛ(r)) d(f(r))1 ɛ4r 1 + r (1 r2)(1 r) + O(ɛ2) = d(f(r))1 ɛ 4r (1 r)2 + O(ɛ2).

Let p(r) = 4r (1r)2.

Using (6) we have

fɛ(z) = f(z) ɛzf(z)1 zeiγ 1 + zeiγ + O(ɛ2).

So

fɛ(z) = f(z)1 ɛe1 2zeiγ z2e2iγ (1 + zeiγ)2 + zf(z) f(z) 1 zeiγ 1 + zeiγ + O(ɛ2).

Let q(r) = e12reiγr2e2iγ (1+reiγ)2 + rf(r) f(r) 1reiγ 1+reiγ. Taking into account (7), we obtain

 G(fɛ,α) ≤ ∣f′(0)∣α(1 − αɛ) + ∣f′(r)∣α(1 −∣r∣2)α(1 − q(r)αɛ) d(f(0))α + d(f(r))α(1 − p(r)αɛ) + O(ɛ2). (8)

Let

A = f(0)α,

B = (1 r2)αf(r)α,

C = d(f(0))α,

D = d(f(r))α.

Then inequality (8) takes the form

G(fɛ,α) A(1 ɛα) + B(1 q(r)αɛ) C + D(1 p(r)αɛ) + O(ɛ2),

or

G(fɛ,α) A + B C + D (C + D)(A + Bq(r)) (A + B)Dp(r) (C + D)2 αɛ + O(ɛ2).

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) + C D + B A(q(r) − p(r)) > 0. (9)

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

C D d(f(0)) d(f(0)) + f(r)α.

Since f S,

C D 1 1 + 4f(r)α.

Using the growth theorem in the class S, we have

C D (1 r)2 (1 r)2 + 4rα = (1 r)2α (1 + r)2α.

By the distortion theorem for the class S

(1 r)2α (1 + r)2α A B (1 + r)2α (1 r)2α.

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

1 p(r) + (1 r)2α (1 + r)2α + (1 r)2α (1 + r)2α(q(r) p(r)) > 0

and

1 p(r) + (1 r)2α (1 + r)2α + (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])

 rf′′(r) f′(r) − 2r2 1 − r2 ≤ 4r 1 − r2 (10)

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

 M1(r) = 1 − 4r (1 − r)2 + (1 − r)2α (1 + r)2α+ + (1 − r)2α (1 + r)2α min γ,θ∈ℝR(r,γ,θ) − 4r (1 − r)2 > 0, (11)
 M2(r) = 1 − 4r (1 − r)2 + (1 − r)2α (1 + r)2α+ + (1 + r)2α (1 − r)2α min γ,θ∈ℝR(r,γ,θ) − 4r (1 − r)2 > 0, (12)

where

R(r,γ,θ) = e1 2reiγ r2e2iγ (1 + reiγ)2 + 2r2 1 r2 1 reiγ 1 + reiγ + 4reiθ 1 r2 1 reiγ 1 + reiγ

imply (9).

Consider the functions

P(r) = 1 4r (1 r)2

and

Q(r) = min γ,θR(r,γ,θ) 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 22 = 0.1715..., and Q(r) equals zero at the point r = 0.086427.... Therefore we have that the roots of equations M1(r) = 0 and M2(r) = 0 lie in the interval (r , 1). It is easy to see that, for r (r, 1), M2(r) > M1(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 M2(r) is decreasing on this interval too. Thus, we conclude that if r < rα, where rα is the single root of the equation

1 4r (1 r)2 + (1 r)2α (1 + r)2α + (1 + r)2α (1 r)2α min γ,θR(r,γ,θ) 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 rα > r.

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

A = cf(0)4,

B = f(r)4(1 r2)4,

C = d(f(0))2f(0)2c,

D = d(g(r))2f(r)2(1 r2)2,

we have

F(fɛ,c,r) A + B C + D ɛ4(A + q(r)B)(C + D) 2(A + B)(C + (p + q)D) (C + D)2 + O(ɛ2),

where fɛ(z) = f(kɛγ(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

 q(r) > p(r), p(r) + q(r) < 2, q(r) > 0.5 (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

 minγ,θ∈ℝR(r,γ,θ) − 4r (1 − r)2 > 0, 4r (1 − r)2 + max γ,θ∈ℝR(r,γ,θ) < 2, minγ,θ∈ℝR(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 r01 = 0.086427..., the second at r02 = 0.071796..., the third at r03 = 0.071796.... thus we conclude that if

r < r0 = max{r01,r02,r03},

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 U1 , U2 and construct a map ξ(z) : U1 U2 such that Jacobian of ξ(z) equals the unity almost everywhere and the inequality

s(ξ(z),z) = z ξ(z) 1 z¯ξ(z) < r0,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 s1 , s2,s3,s4 be the intersection of the square s with the first, the second, the third and the fourth quarters of the complex plane. Let c1 , c3 be the sectors of U , cutting off the sides of the square s, parallel to the axis OX, and by c2,c4 sectors, cutting off the sides of the square s, paralleling to the axis OY.

Let s1,s2,c1,c2 U1, s3 , s4,c2,c4 U2 and

ξ(z) = z r 2 i r 2,z s1 z + r 2 i r 2,z s2 eiπ 2 z, z c1,c3.

Let us find the lower and upper sharp bounds of s(ξ(z),z), for z c1,c3. It is easy to see that

s(ξ(z),z) = z1 i 1 + z2i = 2z 1 + z4.

The function r 1+r4 increases on the interval (0, 1) and so we have

2r 4 + r4 s(ξ(z),z) 2r 1 + r4,z c1,c3.

By construction, s(ξ(z),z) < r0a.e., so

r = 1 1 r0 4 r0 = 0.0507682....

Let us find the lower and upper sharp bounds of s(ξ(z),z), for z s1. Let z = x + iy, then we have

s(ξ(z),z) = r 1 (x + iy)x r 2 iy r 2.

So

s(ξ(z),z)2 = r2 r2(xy)2 2 + 1 x2 y2 + r(x+y) 2 2.

Let us consider the function

q(x,y) = 1 x2 y2 + r(x + y) 2 .

When z s1, the function q(x,y) is nonnegative, and attains its minimal value at points r 2, i r 2, 0, r 2 + i r 2. Thus the function r2 2 (x y)2 is minimal at x = y, then

r2(x y)2 2 + 1 x2 y2 + r(x + y) 2 2

attains minimum at the point 0, so we have s(ξ(z),z) r, for z s1.

Let find the upper sharp bound for the function

p(x,y) = r2(x y)2 2 + 1 x2 y2 + r(x + y) 2 2.

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

r 22 + i r 22,

r4 r2 22 + ir + 4 r2 22 ,

r + 4 r2 22 + ir4 r2 22 ,

r4 + r2 22 + ir4 + r2 22 ,

r + 4 + r2 22 + ir + 4 + r2 22 .

Note that only the first point lies in s1 provided r = r 011 1 r 04 = 0.0507....

The second partial derivative at the point r 22 + i r 22 equals

pxx r 22, r 22 = pyy r 22, r 22 = (16 + 24r2 + r4) 16 ,

pxy r 22, r 22 = (4 r2)2 16 .

So we have

pxxpyy pxy2 = r2(4 + r2)2 4 > 0.

Therefore the function p(x,y) may attain its maximum only in this point or on the boundary of the square s1 or at the point r 22 + i r 22. Let us show that p(x,y) attains maximum at the point r 22 + i r 22. Simple calculations show that

px r 2,y = 2r1 r 2 y2 < 0,y 0, r 2

px(0,y) = 2r(1 y2) > 0,y 0, r 2

pyx, r 2 = 2r1 r 2 x2 < 0,x 0, r 2

py(x, 0) = 2r(1 x2) > 0.x 0, r 2

Thus we have

4r 4 + r2 s(ξ(z),z) 2r 1 + r4 = r0,z s1,s2,c1,c3.

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 π n and translating into each other after rotation on angle π n. Let us find n such that

s(ξ(z),z) = r1 eiπ n 1 r2eiπ n r0,

or

cos π n 2r2 r 02(1 + r4) 2r2(1 r02) .

Thus

n A(r,r0) = π arccos 2r2r02(1+r4) 2r2(1r02) .

Let

n = [A(r,r0)] + 1.

It is easy to check

s(z,ξ(z)) = r1 eiπ n 1 r2eiπ n = r 2 2 cos π [A(r,r0)]+1 1 + r4 2r2 cos π [A(r,r0)]+1 r 2 2 cos π A(r,r0)+1 1 + r4 2r2 cos π A(r,r0)+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

r1 = 4r 4 + r2 = 0.0507355... s(ξ(z),z) r0 = 0.0717968....,z U1.

By construction we have that the measure of the set {z : s(z,ξ(z)) = r0} is zero.

Let us find the lower and upper sharp bounds for

 c(z) = 1 −∣z∣2 1 −∣ξ(z)∣2 (16)

in z U1. By construction we have c(z) = 1, if z c1,c2 or z > r. Let us consider the case of z s1. Then it is easy to see that (16) attains minimum at the point 2 2 r + i2 2 r and maximum at the point 0. So we have that (16) lies between 1 r2 and (1 r2)1.

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

Ic(Ω) = Uf(z)4(1 z2)2dxdy = U1f(z)4(1 z2)2dxdy + U2f(z)4(1 z2)2dxdy = U1f(z)4(1 z2)2 + f(ξ(z))4(1 ξ(z)2)2dxdy.

Similarly,

I(Ω) = Uf(z)2d(f(z))2dxdy = U1f(z)2d(f(z))2 + f(ξ(z))2d(f(ξ(z)))2dxdy.

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

 ∣f′(z)∣4(1 −∣z∣2)2 + ∣f′(ξ(z))∣4(1 −∣ξ(z)∣2)2 ∣f′(z)∣2d(f(z))2 + ∣f′(ξ(z))∣2d(f(ξ(z)))2 . (17)

Making change of variable ψ(w) = wz 1z¯w, we have that (17) is equal to

 ∣g′(0)∣4c + ∣g′(r)∣4(1 − r2)4 d(0)2∣g′(0)∣2c + d(g(r))2∣g′(r)∣2(1 − r2)2, (18)

where g(w) = f ψ(w), r = s(ξ(z),z) and c = (1z2)2 (1ξ(z)2)2. Using

r1 < r < r0,(1 r2)2 (1 ξ(z)2)2 (1 z2)2 1 (1 r2)2,

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

I(Ω) min r1fθ,d1(d)r0{F(fθ,d,fθ,d1(d), 1 1 r2),F(fθ,d,fθ,d1(d), 1r2)}.

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)

[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