Alexander Kuznetsov ON A PROBLEM OF AVHADIEV
(submitted by F. Avhadiev)
ABSTRACT. In this paper we consider a lower estimate for the ratioI(Ω)of the conformal moment of a simple connected domainΩin thecomplex plane to the moment of inertia of this domain about itsboundary. Related functionals depending on a simple connected domainΩand two pointsw1,w2∈Ωwith fixedhyperbolical distance between them are estimated. As a consequence a nontrivial lowerestimate for I(Ω)is obtained.
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, Grant04-01-00083 and the Program ”Russian Universities”, Grant ur.04.01.040.
1. Introduction
Let Ω
be a simply connected domain on the complex plane
ℂ 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 [2, 3]. 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.[4, 5]). 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
on the class S0 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 w1,w2∈Ω
with fixed hyperbolical distance between them. Namely,
Theorem 1.If r<rα,then every function f(z)minimizing the functional G(f,α)on the class S0maps Uonto 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
Proposition 1.If r<r0,then every function f(z)minimizing the functional F(f,r,c)on the class S0maps Uonto 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)= supf∈S0F(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 domainΩ, forwhich Ic(Ω)and I(∂Ω)are finite, the estimate
The values of r∗,r1
will be given in the proof of Theorem 2.
2. Proof of theorem 1 and proposition 1
Proof of Theorem1. 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(1−z)2,
kɛ(z)=k−1((1−ɛ)k(z)) and
kɛγ(z)=e−iγkɛ(eiγz). For
ɛ small
enough we have:
kɛγ(z)=z−ɛz1−zeiγ1+zeiγ+O(ɛ2).
(5)
The function kɛγ(z)
maps U onto
U minus radial slit with
endpoint at −e−iγ. 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∈ℝ
and ɛ0>0 such
that D1,D2⊂fɛ(U),
where fɛ(z)=f(kɛγ(z)),0≤ɛ≤ɛ0.
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
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)+CD+BA(q(r)−p(r))>0.
(9)
Let us find a lower bound of CD.
Using the obvious inequality d(f(r))≤d(f(0))+∣f(r)∣,
we have
CD≥d(f(0))d(f(0))+∣f(r)∣α.
Since f∈S,
CD≥11+4∣f(r)∣α.
Using the growth theorem in the class
S, we
have
CD≥(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≤AB≤(1+r)2α(1−r)2α.
Tacking into account that the left part of (9) is monotonic on
BA, we
conclude that the following inequalities
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
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 Proposition1. Using the same argument as in the proof of
theorem 1 and following notation
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
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 Theorem2. 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−ir∗2,z∈s1z+r∗2−ir∗2,z∈s2eiπ2z,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)=∣z∣∣1−i∣∣1+∣z∣2i∣=2∣z∣1+∣z∣4.
The function r1+r4 increases
on the interval (0,1)
and so we have
2r∗4+r∗4≤s(ξ(z),z)≤2r∗1+r∗4,z∈c1,c3.
By construction, s(ξ(z),z)<r0a.e.,
so
r∗=1−1−r04r0=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=r∗2r∗2(x−y)22+1−x2−y2+r∗(x+y)22.
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,
ir∗2,
0,
r∗2+ir∗2. Thus the
function r∗22(x−y)2 is
minimal at x=y,
then
r∗2(x−y)22+1−x2−y2+r∗(x+y)22
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)=r∗2(x−y)22+1−x2−y2+r∗(x+y)22.
Calculations show that the stationary points of
p(x,y) are
the following
r∗22+ir∗22,
r∗−4−r∗222+ir∗+4−r∗222,
r∗+4−r∗222+ir∗−4−r∗222,
r∗−4+r∗222+ir∗−4+r∗222,
r∗+4+r∗222+ir∗+4+r∗222.
Note that only the first point lies in
s1 provided
r∗=r0−11−1−r04=0.0507....
The second partial derivative at the point
r∗22+ir∗22
equals
pxxr∗22,r∗22=pyyr∗22,r∗22=(16+24r∗2+r∗4)16,
pxyr∗22,r∗22=(4−r∗2)216.
So we have
pxxpyy−pxy2=r∗2(4+r∗2)24>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+ir∗22. Let us show that
p(x,y) attains maximum
at the point r∗22+ir∗22.
Simple calculations show that
pxr∗2,y=−2r∗1−r∗2−y2<0,y∈0,r∗2
px(0,y)=2r∗(1−y2)>0,y∈0,r∗2
pyx,r∗2=−2r∗1−r∗2−x2<0,x∈0,r∗2
py(x,0)=2r∗(1−x2)>0.x∈0,r∗2
Thus we have
4r∗4+r∗2≤s(ξ(z),z)≤2r∗1+r∗4=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
πn
and translating into each other after rotation on angle
πn. Let us
find n
such that
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∣21−∣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
22r+i22r 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