Some inequalities for inner radii of partially overlapping domains
In this paper we consider a problem on an extremal decomposition of the complex plane in the geometric function theory.
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| author | Vyhivska, L. V. Vyhivska, L. V. |
| author_facet | Vyhivska, L. V. Vyhivska, L. V. |
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"author": "L. V. Vyhivska",
"institution": "Institute of Mathematics of NAS of Ukraine"
}
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| description | In this paper we consider a problem on an extremal decomposition of the complex plane in the geometric function theory. |
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Збiрник праць Iн-ту математики НАН України 2017, том 14, № 1, 82–89
УДК 517.54
L.V. Vyhivska
(Institute of Mathematics of the National Academy of Sciences of
Ukraine, Kyiv)
liudmylavygivska@ukr.net
Some inequalities for inner radii of
partially overlapping domains
Dedicated to Prof. Yu. B. Zelinskii on the occasion of his 70th birthday
In this paper we consider a problem on an extremal decomposition of the
complex plane in the geometric function theory.
У данiй роботi розглядається задача екстремального розбиття компле-
ксної площини у геометричнiй теорiї функцiй.
1. Denotations and definitions. Let N, R be the sets of natural
and real numbers, respectively, C be the complex plane, C = C ∪ {∞}
be its one-point compactification, and R+ = (0,∞). Let r(D,a) be an
inner radius of the domain D ⊂ C with respect to the point a ∈ D (cf.,
e.g., [1–4]). An inner radius is a generalization of a conformal radius for
multiply connected domains. An inner radius of the domainD is associated
with the generalized Green’s function gD(z,a) of the domain D by the
relations
gD(z,a) = ln
1
|z − a|
+ ln r(D,a) + o(1), z → a,
gD(z,∞) = ln |z|+ ln r(D,∞) + o(1), z →∞.
For a system of points An := {ak : a0 = 0, |ak| = 1, k = 0,n} and for
an open set D, An ⊂ D, we denote by D(ak) a connected component of
D containing ak, k = 0,n.
c© L.V. Vyhivska, 2017
Some inequalities for inner radii of partially overlapping domains 83
Denote by Pk := {w : arg ak < argw < arg ak+1}, an+1 := a1,
αk :=
1
π
arg
ak+1
ak
, αn+1 := α1, k = 1,n,
∑n
k=1 αk = 2.
We denote by
Dk(0) := D(0) ∩ P k, Dk(ak) := D(ak) ∩ P k, Dk(ak+1) := D(ak+1) ∩ P k,
for each k = 1,n, an+1 := a1.
The open set D, An ⊂ D, satisfies the non-overlapping condition with
respect to the system of points An, if the equality
[Dk(0) ∩Dk(ak)] ∪ [Dk(0) ∩Dk(ak+1)] ∪ [Dk(ak) ∩Dk(ak+1)] = ∅,
1 ≤ k ≤ n, holds for all different points ak which belong to P k.
The system of domains {Dk}nk=0 satisfies a partially overlapping condi-
tion with respect to the system of points An, if the open set D = ∪nk=0Dk
satisfies the non-overlapping condition with respect to the system An.
2. Formulation of the problem. The main goal of the work is
to obtain a sharp upper bound for the functional
Jn(γ) = rγ (D0,0)
n∏
k=1
r (Dk,ak)
where γ ∈ R+, |ak| = 1, a0 = 0, {Dk}nk=0 are the system of partially
overlapping domains such that ak ∈ Dk ⊂ C for k = 0, n. The problem
was formulated in the work [1].
3. Results and proofs. The following theorem strengthens the
main result of the work [5].
Theorem 1. Let n ∈ N, n > 4, γ ∈ (0, γn], γ4 = 4,17, γ5 = 5,71,
γ6 = 7,5, γ7 = 9,53, γ8 = 11,81, and γn = 0,1215n2 for n > 9. Then for
any different points of the unit circle |ak| = 1 such that 0 < αk < 2/
√
γ,
k = 1,n, and for any system domains Dk, ak ∈ Dk ⊂ C, k = 0,n, which
satisfy the partially overlapping condition with respect to points of the unit
circle, the following inequality holds
Jn(γ) ≤
(
4
n
)n (
4γ
n2
) γ
n(
1− γ
n2
)n+ γ
n
(
1−
√
γ
n
1 +
√
γ
n
)2
√
γ
. (1)
The equality is attained if ak and Dk, k = 0,n, are, respectively, poles
84 L.V. Vyhivska
and circular domains of the quadratic differential
Q(w)dw2 = − (n2 − γ)wn + γ
w2(wn − 1)2
dw2.
Proof. Let domains Dk, k = 0,n, satisfy conditions of Theorem, and we
have the open set D = ∪nk=0Dk. Then the inequality
r(Dk,ak) ≤ r(D,ak),
holds, and, obviously, we obtain
rγ(D0,0)
n∏
k=1
r(Dk,ak) ≤ rγ(D,0)
n∏
k=1
r(D,ak).
Further, consider the system of functions:
ζ = πk(w) = −i
(
e−iθkw
) 1
αk , k = 1,n.
The family of the functions {πk(w)}nk=1 is called admissible for the
separating transformation of the open set D, with respect to the angles
{Pk}nk=1. Let M
(1)
k , k = 1,n, denote the domain of the plane ζ, obtai-
ned as a result of the union of the connected component of the set
πk(D
⋂
P k) containing the point πk(ak) with the own symmetric reflecti-
on with respect to the imaginary axis. In turn, by M
(2)
k , k = 1,n, one
denotes the domain of the plain Cζ , which are obtained as a result of
the union of the connected component of the set πk(D
⋂
P k) containi-
ng the point πk(ak+1) with the own symmetric reflection with respect
to the imaginary axis, πn(an+1) := πn(a1). Moreover, we denote M (0)
k
as the domain of the plane Cζ , obtained as a result of the union of
the connected component of the set πk(D
⋂
P k) containing the point
ζ = 0 with the own symmetric reflection with respect to the imagi-
nary axis. Denote by πk(ak) := m
(1)
k , πk(ak+1) := m
(2)
k , k = 1,n,
πn(an+1) := m
(2)
n . From the definition of the function πk, it follows that
|πk(w)−m(1)
k | ∼
1
αk
· |w − ak|, w → ak, w ∈ Pk,
|πk(w)−m(2)
k | ∼
1
αk
· |w − ak+1|, w → ak+1, w ∈ Pk,
Some inequalities for inner radii of partially overlapping domains 85
|πk(w)| ∼ |w|
1
αk , w → 0, w ∈ Pk.
Further, using the result of the papers [1,2], we obtain the inequality
r (D, ak) ≤
[
αkαk−1r
(
M
(1)
k ,m
(1)
k
)
r
(
M
(2)
k ,m
(2)
k
)] 1
2
, k = 1,n, (2)
r (D, 0) ≤
[
n∏
k=1
rα
2
k
(
M
(0)
k ,0
)] 1
2
. (3)
From inequalities ((2)), ((3)), we obtain the inequality
Jn(γ) ≤
n∏
k=1
αk
[
n∏
k=1
rγα
2
k
(
M
(0)
k ,0
)
r
(
M
(1)
k ,m
(1)
k
)
r
(
M
(2)
k ,m
(2)
k
)] 1
2
.
Using the technique developed in [4, p. 269–274], we obtain the estimate
Jn(γ) 6
(
n∏
k=1
αk
)[
n∏
k=1
rα
2
kγ
(
G
(0)
k ,0
)
r
(
G
(1)
k ,− i
)
r
(
G
(2)
k ,i
)] 1
2
, (4)
where G(0)
k , G
(1)
k , G
(2)
k are circular domains of the quadratic differential
Q(w)dw2 =
(4− α2
kγ)w2 − α2
kγ
w2(w2 + 1)2
dw2,
such that 0 ∈ G(0)
k , −i ∈ G(1)
k , i ∈ G(2)
k . Let
S(x) = 2x
2+6 · xx
2
· (2− x)−
1
2 (2−x)
2
· (2 + x)−
1
2 (2+x)
2
, x ∈ [0,2].
Then, from inequality (4) according to [1, 2], we obtain the estimate
Jn(γ) 6 γ−n/2
(
n∏
k=1
αk
)[
n∏
k=1
S(x)
] 1
2
6 γ−n/2
[
n∏
k=1
L(x)
] 1
2
,
where
L(x) = 2x
2+6 · xx
2+2 · (2− x)−
1
2 (2−x)
2
· (2 + x)−
1
2 (2+x)
2
, x ∈ [0,2].
Consider the extremal problem
n∏
k=1
L(xk) −→ max;
n∑
k=1
xk = 2
√
γ,
86 L.V. Vyhivska
xk = αk
√
γ, 0 < xk 6 2.
Let F (x) = ln (L(x)) and X(0) =
{
x
(0)
k
}n
k=1
is any set of extremal points
of the problem which is considered above.
Repeating the arguments of [5] we obtain the statement:
if 0 < x
(0)
k < x
(0)
j < 2, k 6= j, then the following equalities hold:
F ′(x
(0)
k ) = F ′(x
(0)
j ), k,j = 1,n, k 6= j,
F ′(x) = 2x ln 2x+ (2− x) ln(2− x)−
−(2 + x) ln(2 + x) +
2
x
(see Fig. 1).
Fig. 1: A graph of the function F ′(x)
Let us verify that for the above-accepted relati-
ons the following condition is valid: x
(0)
1 =
= x
(0)
2 = . . . = x
(0)
n . Let F ′(x) = h, y0 6 h 6 1, y0 ≈ −0,17.
Consider values h:
h1 = 1, h2 = 0,95, h3 = 0,9, h4 = 0,85, · · · , h23 = −0,15, h24 = −0,17.
We need to find a solution of the equation:
F ′(x) = hk, k = 1,24. (5)
For every hk ∈ [y0, 1] the equation has two solutions:
x1(hk) ∈ (0, x0], x2(hk) ∈ (x0, 2], x0 ≈ 1,324683.
Some inequalities for inner radii of partially overlapping domains 87
Results of direct calculations are given in the following table.
k hk x1(hk) x2(hk) 3x1(hk) + x2(hk+1) 4x1(hk) + x2(hk+1)
1 1,00 0,697331 2,000000
2 0,95 0,708144 1,992640 4,084633 4,781964
3 0,90 0,719344 1,983233 4,107666 4,815810
4 0,85 0,730957 1,972549 4,130581 4,849925
5 0,80 0,743014 1,960786 4,153657 4,884614
6 0,75 0,755550 1,948028 4,177071 4,920085
7 0,70 0,768602 1,934315 4,200964 4,956513
8 0,65 0,782217 1,919654 4,225462 4,994064
9 0,60 0,796446 1,904035 4,250687 5,032904
10 0,55 0,811347 1,887429 4,276766 5,073211
11 0,50 0,826991 1,869791 4,303831 5,115178
12 0,45 0,843462 1,851059 4,332032 5,159023
13 0,40 0,860858 1,831149 4,361534 5,204996
14 0,35 0,879304 1,809955 4,392531 5,253389
15 0,30 0,898950 1,787338 4,425249 5,304553
16 0,25 0,919989 1,763115 4,459964 5,358914
17 0,20 0,942675 1,737044 4,497012 5,417001
18 0,15 0,967348 1,708794 4,536819 5,479494
19 0,10 0,994487 1,677892 4,579935 5,547283
20 0,00 1,059462 1,604865 4,588325 5,582811
21 -0,05 1,100561 1,559491 4,737878 5,797340
22 -0,10 1,152868 1,502748 4,804430 5,904991
23 -0,15 1,234855 1,416172 4,874775 6,027642
24 -0,17 1,324683 1,324683 5,029248 6,264103
Taking into consideration properties of the function F ′(x) and the
condition of Theorem, we obtain the following inequality from the table,
respectively, for n = 4,8, hk 6 h 6 hk+1, k = 1,23:
n∑
k=1
xk(h) > (n− 1)x1(hk) + x2(hk+1) > 2
√
γn,
Thus, the case
{
x
(0)
k
}n
k=1
∈ (0, x0], x0 ≈ 1,324683, n = 4,8, is possible
only for the extremal set X(0), and, therefore, x(0)1 = x
(0)
2 = · · · = x
(0)
n .
The inequality
(x1(hk)− 0,6973)n+ (x2(hk+1)− x1(hk)) > 0, n > 9,
the proof of which is based on the technique developed in [5], is true for
roots of the equation (5).
88 L.V. Vyhivska
Then
nx1(hk) + (x2(hk+1)− x1(hk)) > 0,6973n.
Solving the inequality
0,6973n > 2
√
γn,
conclude that γn = 0,1215n2, for n > 9.
Therefore, in the case n > 9, the set of the points
{
x
(0)
k
}n
k=1
can not be
the extremal, provided x(0)n ∈ (x0; 2]. Then, the case may be only for the
extremal set
{
x
(0)
k
}n
k=1
, when x
(0)
k ∈ (0, x0], k = 1,n, and x
(0)
1 = x
(0)
2 =
= . . . = x
(0)
n . For all γ < γn, n > 9, all the previous reasoning holds. The
Theorem is proved
Corollary 1. Let n ∈ N, n > 4, γ ∈ (0, γn], γ4 = 4,17, γ5 = 5,71,
γ6 = 7,5, γ7 = 9,53, γ8 = 11,81, and γn = 0,1215n2 for n > 9. Then for
any different points of the unit circle |ak| = 1 such that 0 < αk < 2/
√
γ,
k = 1,n, and for any system domains Dk, ak ∈ Dk ⊂ C, k = 0,n, which
satisfy the partially overlapping condition with respect to points of a unit
circle, the following inequality holds
rγ (D0,0)
n∏
k=1
r (Dk,ak) ≤ rγ (Λ0,0)
n∏
k=1
r (Λk,λk) ,
where λk and Λk, k = 0,n, are, respectively, poles and circular domains of
the quadratic differential
Q(w)dw2 = − (n2 − γ)wn + γ
w2(wn − 1)2
dw2.
Corollary 2 ( [5,6] ). Let n ∈ N, n > l ∈ {5,4} (l = 5 in [5], l = 4 in [6]),
γ ∈ (0, γn], γn = n. Then for different points of the unit circle |ak| = 1
such that 0 < αk < 2/
√
γ, k = 0,n, and for any non-overlapping domains
Bk, ak ∈ Bk ⊂ C, k = 1,n, a0 = 0 ∈ B0, the inequality (1) holds. The
equality is attained under the same condition as in Theorem 1.
Acknowledgments. The author is greatly indebted to
Professor Aleksander Bakhtin for reading a draft version of this
paper and providing numerous suggestions.
Some inequalities for inner radii of partially overlapping domains 89
References
[1] Dubinin V.N. Symmetrization in the geometric theory of functions of a
complex variable// Uspekhi Math. Nauk. — 1994. — 49, No. 1(295). —
P. 3 – 76 (in Russian); Engl. transl. in: Russian Mathematical Surveys. —
1994. — 49(1), No. 1. — P. 1 – 79.
[2] Dubinin V.N. The separating transformation of domains and problems on
the extremal partition // Zap. Nauchn. Semin. Leningr. Otd. Mat. Inst.
Steklova. — 1988. — 168. — P. 48 – 66 (in Russian).
[3] Dubinin V.N. Condenser capacities and symmetrization in geometric functi-
on theory. — Basel: Birkhäuser / Springer, 2014. — 344 p.
[4] Bakhtin A.K., Bakhtina G.P., Zelinskii Yu. B. Topological Algebraic
Structures and Geometric Methods in Complex Analysis. — K.: Inst. Math.
of NAS of Ukraine, 2008. — 308 p. (in Russian).
[5] Kovalev L.V. To the problem of extremal decomposition with free poles on
a circumference // Dal’nevost. Mat. Sborn. — 1996. — No. 2. — P. 96 – 98
(in Russian).
[6] Bakhtin A.K., Denega I. V. Addendum to a theorem on extremal
decomposition of the complex plane // Bulletin de la société des sci-
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LXII, No. 2. — P. 83 – 92.
|
| id | oai:trim.imath.kiev.ua:article-104 |
| institution | Transactions of Institute of Mathematics of NAS of Ukraine |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-08-04T01:00:31Z |
| publishDate | 2017 |
| publisher | Інститут математики НАН України |
| record_format | ojs |
| resource_txt_mv | trimimathkievua/ab/cd746dc563da1a3b698cec8df8c927ab.pdf |
| spelling | oai:trim.imath.kiev.ua:article-1042018-02-13T13:58:16Z Some inequalities for inner radii of partially overlapping domains Vyhivska, L. V. Vyhivska, L. V. In this paper we consider a problem on an extremal decomposition of the complex plane in the geometric function theory. Інститут математики НАН України 2017-04-25 Article Article application/pdf https://trim.imath.kiev.ua/index.php/trim/article/view/104 Transactions of Institute of Mathematics, the NAS of Ukraine; Vol. 14 No. 1 (2017): Vol. 14 No. 1 (2017): Analysis and Applications; 82-89 Сборник Трудов Института математики НАН Украины; Том 14 № 1 (2017): Том 14 № 1 (2017): Анализ и приложения; 82-89 Збірник Праць Інституту математики НАН України; Том 14 № 1 (2017): Аналіз та застосування; 82-89 3083-7529 1815-2910 en https://trim.imath.kiev.ua/index.php/trim/article/view/104/93 Авторське право (c) 2017 Праці Інституту математики НАН України |
| spellingShingle | Vyhivska, L. V. Vyhivska, L. V. Some inequalities for inner radii of partially overlapping domains |
| title | Some inequalities for inner radii of partially overlapping domains |
| title_full | Some inequalities for inner radii of partially overlapping domains |
| title_fullStr | Some inequalities for inner radii of partially overlapping domains |
| title_full_unstemmed | Some inequalities for inner radii of partially overlapping domains |
| title_short | Some inequalities for inner radii of partially overlapping domains |
| title_sort | some inequalities for inner radii of partially overlapping domains |
| url | https://trim.imath.kiev.ua/index.php/trim/article/view/104 |
| work_keys_str_mv | AT vyhivskalv someinequalitiesforinnerradiiofpartiallyoverlappingdomains AT vyhivskalv someinequalitiesforinnerradiiofpartiallyoverlappingdomains |