CC BY
63Probl. Anal. Issues Anal. Vol. 8 (26), No 1, 2019, pp. 17-31 17
DOI: 10.15393/j3.art.2019.5452
UDC 517.54
A. K. Bakhtin, I. V. Denega
SHARP ESTIMATES OF PRODUCTS OF INNER RADII OF NON-OVERLAPPING DOMAINS IN THE COMPLEX PLANE
Abstract. In the paper we study a generalization of the extremal problem of geometric theory of functions of a complex variable on non-overlapping domains with free poles: Fix any y e R+ and find the maximum (and describe all extremals) of the functional
n
[r (Bo, 0) r (Bx, to)]7 n r (Bk,ak), k=l
where n e N, n ^ 2, a0 = 0, |ak| = 1, B0, Bn, {Bk}n=l is a system of mutually non-overlapping domains, ak e Bk c C, k = 0, n, ^ e c C, (r(B,a) is an inner radius of the domain B c C at a e B). Instead of the classical condition that the poles are on the unit circle, we require that the system of free poles is an n-radial system of points normalized by some "control" functional. A partial solution of this problem is obtained. Key words: inner radius of a domain, non-overlapping domains, radial system of points, separating transformation, quadratic differential, Green's function
2010 Mathematical Subject Classification: 30C75
Let N, R be the sets of natural and real numbers, respectively, C be the complex plane, C = CU{to} be a one-point compactification, and R+ = (0, to). Let x(t) = 1 (t + t-1), t e R+, be the Zhukovskii function. Let r(B,a) be an inner radius of the domain B c C relative to the point a e B.
The system of points An := {ak e C,k = 1, n}, n e N, n ^ 2 is called n-radial, if |ak| e R+ for k = 1, n and 0 = arg al < ... < arg an < 2n.
Denote
Pk = Pk(An) := {w : arg ak < arg w < arg ak+i}, an+i := ai, (g) Petrozavodsk State University, 2019
1
1 «fc+i j - \—^
ak :=-arg —, a„+i := ai, k = 1,n, = 2.
k=1
n
ak
For any n-radial system of points An = {ak}, k = 1,n, we introduce the "control" functional
L(An) := n
X
k=1
ak
ak+i
|ak |.
The class of n-radial systems of points for which L(An) = 1 contains automatically all systems of n different points of the unit circle. Consider the following extremal problem.
Problem 1. For any fixed value of y G R+, find the maximum of the functional
n
Jn(7) = [r (Bo, 0) r (B^, n r (Bk, «k) , (1)
k=1
where n G N, n ^ 2, a0 = 0, An = {ak}n=1 are n-radial systems of points, such that L (An) = 1, B0, B^,, {Bk}n=1 is a system of mutually non-overlapping domains, ak G Bk C C, k = 0,n, ^ G B^ C C; also, describe all extremals.
This problem belongs to the class of extremal problems with free poles. Problems of this type have been studied in many papers (see, for example, [1-16]). For y = 1 and n ^ 2, an estimate of the functional Jn(Y) for the system of non-overlapping domains was found in the paper [6, p. 59]. Kuz'mina [15, p. 267] strengthened this result for simply connected don > 2.
mains and showed that the estimate is correct for y G ^0, n^
Note that for n =2 the Kuz'mina's estimate of the functional (1) coincides with the Dubinin's estimate. Some partial cases of the above-posed problem were considered in [2,3,5]. Let
S(x) = x2x2+2 ■ |1 - x|-(1-x)2 ■ (1 + x)-(1+x)2 and tf(x) = ln(S(x)).
2
tf'(x) = 4xln(x) - 2(x - 1) ln |x - 1| - 2(x + 1)ln(x + 1) + - (see Fig. 4).
x
The function S(x) is logarithmically convex on the interval [0,x0], xo w 0.88441. Let tf'(x) = t, yo ^ t < 0, yo ~ -1.06. The equation (x) = tk has two solutions x1(t) G (0,x0] and x2(t) G (x0, ro].
1
2a
k
y 0,5
Figure 4: The function plot y = ^'(x)
Let = min ((n — 1)xi(i) + x2(t)) the following proposition is true.
/Y o
tni
then 7n = -f
Then
Y0
n
-2M . Then, for
Theorem 1. Let n e N, n ^ 2, y e (0, yOO] any n-radial system of points An = {ak }n=l such that L (An) = 1, and any
Bo
Bk
system of mutually non-overlapping domains Bo, a0 = 0 e B0 c C, to e c C, ak e Bk c C, k = 1, n, the following inequality holds:
[r (Bo, 0) r (B^, to)]7 n r (Bk,ak) ^ [r (Ao, 0) r (A^,to)]7 n r (Ak, Ak)
k=i
k=1
where the domains A0, An, Ak, and the points 0, to, Ak, k = 1,n, are, respectively, circular domains and poles of the quadratic differential
Q(w)dw2 = —
Yw
2n
+ (n2 — 2y)wn + y , 2 — dw .
w2(wn — 1)2
(3)
Proof. Let Z = nk (w) denote a univalent branch of the multivalent analy-. _
tic function — i (e-iargakw) , k = 1,n, that maps Pk onto the right halfplane Re Z > 0 conformally in the one-sheet way. Consider the system of
. __(1) _
functions Z = nk(w) = —i (e-iargakw)ak, k = 1,n. Let Qk', k = 1,n, denote a domain of the plane Cz, obtained as a result of the union of the
2
2
connected component of the set nk(Bk P| Pk), containing the point nk(ak), with its symmetric reflection with respect to the imaginary axis. In turn, by k = I , n, we denote the domain of the plane Cz, obtained as a result of the union of the connected component of the set nk (Bk+1 P| Pk), containing the point nk (ak+1), with its symmetric reflection with respect to the imaginary axis, Bn+1 := B1, nn(an+1) := nn(a1). In addition, Q^ denotes a domain of the plane Cz obtained as a result of the union of the connected component of the set nk(B0 H Pfc), containing the point Z = 0, with its symmetric reflection with respect to the imaginary axis. Similarly, Q^ denotes a domain of the plane Cz obtained as a result of the union of the connected component of the set nk (B^, P| Pk), containing the point Z = to, with its symmetric reflection with respect to the imaginary axis.
It is clear that nk(ak) := wj^, nk(ak+1) := ^k2), k = 1, n, nn(an+1) := The definition of the functions nk yields
(2) L.-
____(2)
1
|nk(w) - |afcI"k 1 ■ |w - ak|, w ^ ak, w G Pk
ak 1
|nk(w) - w(2)| ~ — |ak+i|1 ■ |w - ak+11, W ^ ak+1, w G Pk, ak
|nk(w)| ~ |w| ak , w — 0, w G Pk,
i _
|nk(w)| ~ |w| ak , w —^ TO, w G Pk.
Using the corresponding results for the separating transformation [6,7], we get the inequalities
r(Bk ,ak) ^
r ■ r to2), ^
"k-1, wk-1
-1 1
«fc-1
|ak I'
1
r (Bo, 0) ^
r (Bœ, to) ^
H r"k ("i0), 0
,k=1
,k=1
Ilrak to
(6)
The conditions of realization of the sign of equality in inequalities (4)-(6) are described in [7, p. 29]. On the basis of those relations, we
2
1
1
fc-1
2
get the inequality
IT A IT |akIak + |ak+i|ak | | Jn(Y) ^ M -IT- ■ |ak|x
k=i k=i (|ak || ak+i |)2ak
x
II r K" , 0 r «k^, to
Y«k r Ki),4i)) ■ r («k^k
)(2) , .(2)
k=i ^|ak |ak + |ak+i|ak
Further, from the last relation we have
1 \ 2
Jn(Y) ^ n n
x
k=i
, k=i / k=i
n(r («M r («?»,to
ak 2«k + ak+i
ak+i ak
2 r f«(i) Yak ' I «k
1
2«k
|ak |x
kk
J_ J_\2
|ak|ak + |ak+i |ak
where |^ki)| = |akIak, |^k2)I = |ak+i|ak, ^ - ^k^ = KIak + |ak+i| Taking into account the fact that
i ak
n
k=i
1
ak
ak+i
l
2«k
+
ak+i
ak
l
2«k
|ak1 = n X k=i
ak
ak+i
l
2«k
|ak | = L (An)
we obtain the following inequality
n
Jn(Y) ^ 2n ■ m ■L (An)
x
x
n
k=i
k=i
r I «k0), 0) r («^, TO
r ( «M") ■ r («kVf' <■
J_ J_\2
|ak |ak + |ak+i|ak
Equality in the last inequality is achieved when equality is realized in the inequalities (4)-(6) for all k = 1,n. Based on the last relation, Theorem 4.1.1 in [1], Corollary 4.1.3 in [1], and the invariance of the functional
r (Bi,ai) r (B3, a3)\Y /r (B2,a2) r (B4,a4)
|ai - a3|2
|a2 — a4|2
1
1
2
2
we have
H akVy) ■ L (An)
x
k=1
X
n
k=1
r m k0), 0) r m
oo
\ Y«k r ( nk
(1) ~(1)
W
■ r(n k2),Wk2)
|«k|ak + |ak+11
2
ak
where the domains nk0), nk^, nk1), nk2) and points 0, to, wk1), wk2), are, respectively, the circular domains and the poles of the quadratic differential
2
Q(z)dz2 = --
z4 + 2(1--2 )z2 + 1
!ak z 2(z2 + 1)2
dz2
Each term in the braces of the last inequality is a value of the functional
r2 r (B1,«1) r (B2,a2)
K = [r (B0, 0) r (B^, to)]1
|a1 - a2|2
7)
on the system of nonoverlapping domains {nk0), nk1), nk2), nk^}, and the corresponding system of points {0, wk1), wk2), to} (k = 1,n).
An estimate of the functional (7) in the case of fixed poles was first obtained in [6], and then in the papers [9,15]. On the basis of Lemma 4.1.2 [1], we get the estimate
KT ^ $(t), t ^ 0, where $(t) = t2t2 ■ |1 - t|-(1-t)2 ■ (1 + t)-(1+t)2. Then
•My) < "
n
n ak VY)
, k=1
n $(Tk)
k=1
1/2
(Vy)
n (t^2 -|1 - Tk|-(1-r')2 ■ (1+ Tk)-(1+")2) ,k=1
where Tk = V! ■ ak, k = 1,n.
Consider the function S(x) = x2x2+2- x|-(1-x) ■ (1 + x)-(1+x) . The function S(x) is logarithmically convex on the interval [0,x0], x0 w 0.88441. Now we consider an extremal problem
S(xk)
max,
X]xk = 2 Vy, xk = ak VY.
k=1
k=1
n
2
2
Let X(0) = |xk0)} be an arbitrary extremal point of the problem. The following result holds (obtained similarly [12]):
(40)) = *' (40)) =... = n*n0)), (9)
where ^'(x) = 4x ln(x) — 2(x — 1) ln |x — 11 — 2(x + 1) ln(x + 1) + X (see Fig. 4). X
Further it will be necessary for us to show that the following condition holds:
zi0) = z20) = ■■■ = xW for all y G (0,Yn].
Let ^'(z) = t, y0 ^ t < 0, y0 ~ —1.06. We find a solution of equation ^'(z) = tk, k = 1, 53. Since Vtk G [y0, 0), it follows that the equation has two solutions zi(t) G (0,x0], x2(t) G (z0, to].
Consider the following values of t: t1 = —0.02, t2 = —0.04, t3 = —0.06, t4 = —0.08, ■ ■ ■, t52 = —1.04, t53 = y0. Direct calculations are presented in Table 2.
Consider the case n = 2. From the analysis of the tabular data for n = 2, we get that the minimum of the sum zi(tk) + x2(tk+i) is achieved for the interval [—0.62; —0.64] and is equal to 1.709336 (see Table 3). The relation zi(t) + x2(t) = 2-^y holds for each y G (0;0.73]. Let y = 0.73; then the value 2-^y is less than the minimum 1.709336. Thus, for n = 2 and y G (0; 0.73], we obtain that x2 does not belong to (x0, to), that is zi and x2 belong to the interval (0,x0] and zi = x2. From inequalities (8) and (9) for n = 2, we have
4 „/ VY
J2(Y) ^ y ■ s 2
Y V 2
For n = 3, the minimum of the value 2zi(tk) + z2(tk+i) on the whole graph is achieved on the interval [—0.48; —0.50] and is equal to 2.381211 (see Table 4). Similarly, 2zi(t) + x2(t) = 2-^Y. Let y = 1.41; then 2^Y = 2.3748. Thus, for y G (0; 1.41] the situation z2 G (x0, to) is not possible. In this way, we obtain zi,x2,x3 G (0,z0] and Xi — Z2 — X3.
Then, taking into account the inequalities (8) and (9) for n = 3, we have
,,) < (£)" is(^V3/2
Similarly, the situation holds for all y G (0,Yn], n = 4, 5, 6.
k tk Xl(tk) X2 (tk ) k tk Xl(tk) X2 (tk )
0 0 0.581421 TO 27 -0.54 0.671495 1.047944
1 -0.02 0.584192 2.607677 28 -0.56 0.675680 1.041549
2 -0.04 0.586996 2.095431 29 -0.58 0.679954 1.035639
3 -0.06 0.589833 1.849825 30 -0.6 0.684325 1.030184
4 -0.08 0.592706 1.696659 31 -0.62 0.688797 1.025157
5 -0.1 0.595614 1.588941 32 -0.64 0.693377 1.020539
6 -0.12 0.598559 1.507710 33 -0.66 0.698072 1.016313
7 -0.14 0.601542 1.443586 34 -0.68 0.702890 1.012468
8 -0.16 0.604564 1.391304 35 -0.7 0.707842 1.008999
9 -0.18 0.607626 1.347643 36 -0.72 0.712936 1.005911
10 -0.2 0.610729 1.310499 37 -0.74 0.718185 1.003228
11 -0.22 0.613876 1.278433 38 -0.76 0.723604 1.001015
12 -0.24 0.617066 1.250421 39 -0.78 0.729208 0.999457
13 -0.26 0.620302 1.225709 40 -0.8 0.735017 0.997390
14 -0.28 0.623585 1.203729 41 -0.82 0.741053 0.994797
15 -0.3 0.626917 1.184045 42 -0.84 0.747345 0.991762
16 -0.32 0.630299 1.166313 43 -0.86 0.753926 0.988295
17 -0.34 0.633734 1.150260 44 -0.88 0.760838 0.984381
18 -0.36 0.637223 1.135664 45 -0.9 0.768138 0.979982
19 -0.38 0.640770 1.122345 46 -0.92 0.775896 0.975038
20 -0.4 0.644375 1.110153 47 -0.94 0.784212 0.969461
21 -0.42 0.648041 1.098962 48 -0.96 0.793228 0.963114
22 -0.44 0.651772 1.088668 49 -0.98 0.803162 0.955787
23 -0.46 0.655569 1.079182 50 -1 0.814378 0.947120
24 -0.48 0.659437 1.070427 51 -1.02 0.827585 0.936407
25 -0.5 0.663378 1.062338 52 -1.04 0.844608 0.921828
26 -0.52 0.667396 1.054860 53 -1.06 0.884406 0.884406
Table 2: Two solutions of the equation ^'(x) = tk, k = 1,53
From Table 2, for an arbitrary n ^ 7, the following inequality holds:
(n - 1)x1(tk) + x2(tk+1) > nx1(tk) + (x2(tk+1) - x1(tk)) > 0.58n,
since x1(tk) ^ 0.5830 and x2(tk+1) - x1(tk) ^ 0. Using the condition
(n - 1)x1(t) + x2(t) = 2V!n,
we assume that 2VYn = 0.58n. Thus, Yn = 0.084n2, that is, when y g (0; 0.084n2] then the sum (n - 1)x1(t) + x2(t) is less than 0.58n. Thus, for n ^ 7 and y G (0,!n], we obtain
./„(!) S (A) "
s( 2V!
n
n/2
k tk Xl(tk ) + X2(tk+l) k tk Xl(tk ) + X2(tk + l)
0 0 27 -0,54 1,715340
1 -0,02 3,189098 28 -0,56 1,713044
2 -0,04 2,679623 29 -0,58 1,711318
3 -0,06 2,436820 30 -0,6 1,710138
4 -0,08 2,286492 31 -0,62 1,709482
5 -0,1 2,181647 32 -0,64 1,709336
6 -0,12 2,103324 33 -0,66 1,709690
7 -0,14 2,042145 34 -0,68 1,710540
8 -0,16 1,992846 35 -0,7 1,711889
9 -0,18 1,952207 36 -0,72 1,713753
10 -0,2 1,918125 37 -0,74 1,716163
11 -0,22 1,889163 38 -0,76 1,719200
12 -0,24 1,864297 39 -0,78 1,723061
13 -0,26 1,842775 40 -0,8 1,726598
14 -0,28 1,824031 41 -0,82 1,729814
15 -0,3 1,807630 42 -0,84 1,732815
16 -0,32 1,793230 43 -0,86 1,735640
17 -0,34 1,780559 44 -0,88 1,738307
18 -0,36 1,769398 45 -0,9 1,740820
19 -0,38 1,759569 46 -0,92 1,743176
20 -0,4 1,750923 47 -0,94 1,745356
21 -0,42 1,743337 48 -0,96 1,747326
22 -0,44 1,736709 49 -0,98 1,749015
23 -0,46 1,730953 50 -1 1,750281
24 -0,48 1,725996 51 -1,02 1,750785
25 -0,5 1,721775 52 -1,04 1,749413
26 -0,52 1,718238 53 -1,06 1,729015
Table 3: Minimum of the sum xi(tk) + £2(^+1), k = 1,53
The equality case is straightforward to verify. Theorem 1 is proved. □
From Theorem 1, we obtain the following results.
Corollary 1. Let n G N, n ^ 2, y G (0,Yn], Y2 = 0.7304, y3 = 1.4175, Y4 = 2.2983, y5 = 3.3683, Ye = 4.6244, and Yn = 0.084n2, n ^ 7. Then for any n-radial system of points An = {ak }n=i such that L (An) = = 1, and any system of mutually non-overlapping domains B0, B^, Bk, a0 = 0 G B0 C C, tog B^ C C, ak G Bk C C, k = 1, n, the inequality (2) holds. Equality is attained in the same case as in Theorem 1.
Corollary 2. Under the conditions of Theorem 1, the following inequa-
lity holds:
[r(B0, 0)r(B^, to)]7n r(Bk,«k)
k=1
^ n
2y
4y ^ «
2
11 _ 4Y
1 .,2
4y I n + 2
n-2V!
n +
2^7
'10)
Equality in this inequality is achieved when 0, to, ak and B0, B^, Bk, k = 1,n, are, respectively, poles and circular domains of the quadratic differential (3).
n
k tk 2xi (tk) + X2(tk+i) k tk 2xi (tk) + X2(tk+i)
0 0 27 -0.54 2.382735
1 -0.02 3.770519 28 -0.56 2.384539
2 -0.04 3.263814 29 -0.58 2.386998
3 -0.06 3.023816 30 -0.6 2.390093
4 -0.08 2.876325 31 -0.62 2.393807
5 -0.1 2.774353 32 -0.64 2.398133
6 -0.12 2.698938 33 -0.66 2.403067
7 -0.14 2.640704 34 -0.68 2.408612
8 -0.16 2.594388 35 -0.7 2.414780
9 -0.18 2.556771 36 -0.72 2.421594
10 -0.2 2.525751 37 -0.74 2.429099
11 -0.22 2.499892 38 -0.76 2.437386
12 -0.24 2.478172 39 -0.78 2.446665
13 -0.26 2.459841 40 -0.8 2.455806
14 -0.28 2.444333 41 -0.82 2.464831
15 -0.3 2.431215 42 -0.84 2.473869
16 -0.32 2.420146 43 -0.86 2.482985
17 -0.34 2.410858 44 -0.88 2.492232
18 -0.36 2.403133 45 -0.9 2.501659
19 -0.38 2.396792 46 -0.92 2.511314
20 -0.4 2.391692 47 -0.94 2.521252
21 -0.42 2.387712 48 -0.96 2.531538
22 -0.44 2.384750 49 -0.98 2.542243
23 -0.46 2.382725 50 -1 2.553443
24 -0.48 2.381565 51 -1.02 2.565162
25 -0.5 2.381211 52 -1.04 2.576998
26 -0.52 2.381615 53 -1.06 2.573623
Table 4: Minimum of the sum 2x1(tk) + x2(tk+1), k = 1,53
Corollary 3. Let n G N, n ^ 2, y G (0,Yn], Y2 = 0.7304, ys = 1.4175, Y4 = 2.2983, Y5 = 3.3683, yg = 4.6244, and Yn = 0.084 n2, n ^ 7. Then, for any other points of the unit circle |w| = 1 and any set of mutually
non-overlapping domains B0, B^, Bk, a0 = 0 G B0 C C, tog B^ C C, ak G Bk C C, k = 1,n, the inequality (2) holds. Equality is attained in the same case as in Theorem 1.
If we consider a sufficiently strict restriction on the distribution of the angles ak, k = 1,n, then we can get a stronger result. Let -0 ~ 0.884414 be a root of the equation
-2 1
ln^ = -2. (11)
1 — -2 -2
Then the following proposition is true.
Theorem 2. Let n G N, n ^ 2, y G (0,Yn), Yn = i-0n2. Then for any n-radial system of points An = {ak}n=i such that L (An) = 1, 0 < ak ^ -0/^Y, where -0 is a root of equation (11), k = 1, n, and for any collection of pairwise nonoverlapping domains B0, B^, Bk, a0 = 0 G B0 C C, tog B^ C C, ak G Bk C C, k = 1,n, the inequality (10) holds. Equality is attained in the same case as in Corollary 2.
Proof. The proof of Theorem 2 practically repeats the proof of Theorem 1, only the logarithmic convexity of the function S(z) on the segment (0,y0] and relation below are used in the final stage of the proof. The reation is
n
1 n / y^ zk"
n £ ln S (Xk) ^ ln s
k=i \
It is equivalent to
ln(n S (Xk )j < ln( sQ
Equality in this inequality is attained if
2^7
Tl = T2 = ... = Tn = -,
n
i.e., if ak = n, k = 1,n. In this case, relation (7) yields
Jn(7) ^ J0(y) = ( n
n
(r ^ 0) r to)) n2----2-
n
where D0, D^,, D1 and D2 are the circular domains of the quadratic differential
^ + 2 (* - 2^-2 + ^ = - "--- d-2. (12)
From whence, we have, eventually,
J-(Y)*( ty
Using a specific formula for S (x), we get the basic inequality of Theorem 2. Changing the variable in (12) by the formula - = -iwn, we get the quadratic differential (3). The sign of equality in inequality (10) is verified directly. Theorem 2 is proved. □
Corollary 1. Let n G N, n ^ 2, y G (0,Yn], Yn = 0.19n2. Then for any n-radial system of points An = {ak}n=1 such that L (An) = 1, 0 < ak ^ y0/VY, y0 ~ 0.88441, k = 1,n, and any set of mutually non-overlapping domains B0, B^, Bk, a0 = 0 G B0 C C, tog B^ C C, ak G Bk C C, k = 1,n, the inequality (2) holds. Equality is attained in the same case as in Theorem 1.
Consider the following problem, which was formulated as an open problem in the case y =1 in the paper by Dubinin [7].
Problem 2. Find, for any fixed value of y G (0,n], the maximum of the functional
n
rY(B0, 0) n r(Bk,ak), k=1
where B0, B1, B2,..., Bn, n ^ 2, is any system of pairwise non-overlapping domains in C, where the domains B1,..., Bn have symmetry with respect to the unit circle, a0 = 0, |ak| = 1, k = 1,n, ak G Bk C C, k = 0,n; describe all extremals of the functional.
This problem was solved for y =1 and n ^ 2 by Kovalev [13,14]. The following theorem substantially complements the results of the papers [4,13,14]. We obtain the following results assuming that B0 C U (here U denotes the unit circle).
Theorem 3. Let n G N, n ^ 2, y G (0,Yn), Yn = 2y0n2. Then, for any n-radial system of points An = {ak}n=1, such that |ak | = 1,
n
2
0 < ak ^ yo/VY, where y0 is a root of equation (11), k = 1,n, and any set of mutually non-overlapping domains B0, Bk, a0 = 0 G B0 C U, ak G Bk C C, k = 1, n, where the domains Bk have symmetry with respect to the unit circle |w| = 1 for all k = 1, n, the following inequality holds:
rY (B0, 0)J] r (Bfc, ak) ^ rY (A0, 0) r (Afc, Ak). (13)
fc=i fc=i
Equality in (13) is attained when 0, Ak and A0, Ak, k = 1,n, are, respectively, the poles and the circular domains of the quadratic differential
\ j 2 Yw2n + 2(n2 - y)wn + y 2
Q(w)dw2 = — ^-^—-^ dw2. (14)
w2(wn — 1)2
Proof. Note (see [6, p.59]) that if the domains Bk have symmetry with respect to the unit circle |w| = 1 for all k = 1,n, and the domain B0 C
n
U, then the extremal problem for the functional rY (B0, 0) r (Bk, ak)
fc=i
can be reduced, by easy transformations, to the study of the functional
n
rY/2(B0, 0) r7/2(B^, to) r(Bk, ak). Thus, using this property and proofs
fc=i
of Theorem 1 and Theorem 2, we obtain the result of Theorem 3. □
Using Corollary 3 and proofs of Theorem 3 and Theorem 1, it is not difficult to obtain the following result.
Theorem 4. Let n G N, 7 G (0,7n], 72 = 1.4608, 73 = 2.8350, 74 = 4.5966, 75 = 6.7366, 76 = 9.2488, 7n = 0.168 n2, n ^ 7. Then, for any other points of the unit circle |w| = 1 and any system of mutually non-overlapping domains B0, Bk, a0 = 0 G B0 C U, ak G Bk C C, k = 1, n, where the domains Bk have symmetry with respect to the unit circle |w| = 1 for all k = 1,n, the following inequality holds:
rY (B0, 0)n r (Bfc,ak) ^ -
4
v-7 h — 12+Y
k— 1 /v-) 2
Y Y2Y
n — v/2y
n + v/2y
Equality in the inequality is achieved when ak and Bk, k = 0, n, are, respectively, poles and circular domains of the quadratic differential (14).
n
n
References
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Received September 19, 2018. In revised form, September 21, 2018. Accepted December 28, 2018. Published online January 11, 2019.
Institute of Mathematics of the National Academy of Sciences of Ukraine Department of complex analysis and potential theory 01004 Ukraine, Kiev-4, 3, Tereschenkivska st. A. K. Bakhtin
E-mail: abahtin@imath.kiev.ua I. V. Denega
E-mail: iradenega@gmail.com
