The paper is devoted to investigation of the problems of geometric function

theory of a complex variable. A general problem of the description

of extremal configurations maximizing the product of the inner radii of

mutually non-overlapping domains is studied

Denega, I. V.

Targonskii, A. L.

Zhytomyr State University Library

Збiрник праць Iн-ту математики НАН України 2017, том 14, № 1, 147–155УДК 517.54I. V. Denega 1, A. L. Targonskii 21(Institute of Mathematics of the National Academy of Sciences ofUkraine, Kyiv)2(Zhytomyr Ivan Franko State University, Zhytomyr)1 iradenega@yandex.ru, 2 targonsk@zu.edu.uaSeparating transformation in aproblem on extremal decomposition ofthe complex planeDedicated to Prof. Yu. B. Zelinskii on the occasion of his 70th birthdayThe paper is devoted to investigation of the problems of geometric functi-on theory of a complex variable. A general problem of the descriptionof extremal configurations maximizing the product of the inner radii ofmutually non-overlapping domains is studied.Робота присвячена дослiдженню вiдкритих проблем геометричної тео-рiї функцiй комплексної змiнної. Зокрема, вивчається загальна пробле-ма з описання екстремальних конфiгурацiй, що мiнiмiзують добутоквнутрiшнiх радiусiв попарно неперетинних областей.Let N and R be the sets of natural and real numbers, respectively; C bethe Complex plane and let C = C∪{∞} be its one-point compactification,R+ = (0,∞). Let r(B,a) be the inner radius of the domain B ⊂ C withrespect to a point a ∈ B [1, 2]. An inner radius is a generalization of aconformal radius for multiply connected domains.Definition 1. A finite collection of arbitrary fixed domains {Bk}nk=1,n ∈ N, n > 2, such as Bk ⊂ C, Bk ∩ Bm = ∅, k 6= m, k,m = 1,n is calleda system of non-overlapping domains.Definition 2. Let n ∈ N, n > 2. A set of points An :=:={ak ∈ C : k = 1,n}is called n-radial system if: |ak| ∈ R+, k = 1,n,0 = arg a1 < arg a2 < . . . < arg an < 2pi.c© I. V. Denega, A. L. Targonskii, 2017148 I. V. Denega, A. L. TargonskiiDenote Pk(An) := {w : arg ak < argw < arg ak+1},θk := arg ak, an+1 := a1, θn+1 := 2pi, αk :=1pi argak+1ak, αn+1 := α1,k = 1, n. Let χ(t) = 12 (t+ t−1). For an arbitrary n-radial system of pointsAn = {ak}, k = 1,n, we assumeL(An) :=n∏k=1χ(∣∣∣ akak+1∣∣∣ 12αk ) |ak|.The class of n-radial systems of points for which L(An) = 1 automati-cally includes all systems of n distinct points that are located on the unitcircle.The goal of the present work is the construction of sharp upper boundsfor a functional of the formJn(γ) = [r (B0,0) r (B∞,∞)]γn∏k=1r (Bk,ak) ,where γ ∈ R+, An = {ak}nk=1 is n-radial system of points, B0, B∞,{Bk}nk=1 is system of pairwise disjoint domains, ak ∈ Bk ⊂ C, k = 0, n,∞ ∈ B∞ ⊂ C.Note that to describe the extremal configurations of the domains weuse notion of quadratic differential (see, for example, [1, P. 63–70]).The following statement holdsTheorem 1. Let n ∈ N, n > 7, 0 < γ 6 γn, γn = 0,08n2. Thenfor any n-radial system of points An = {ak}nk=1 such that L(An) = 1and any set of mutually non-overlapping domains Bk, a0 = 0 ∈ B0 ⊂ C,∞ ∈ B∞ ⊂ C, ak ∈ Bk ⊂ C, k = 1,n, the following inequality holds[r (B0,0) r (B∞,∞)]γn∏k=1r (Bk,ak) 6 [r (Λ0,0) r (Λ∞,∞)]γn∏k=1r (Λk, λk) ,where domains Λ0, Λ∞, Λk, and points 0,∞, λk, k = 1,n, are, respectively,circular domains and poles of the quadratic differentialQ(w)dw2 = −γw2n + (n2 − 2γ)wn + γw2(wn − 1)2 dw2. (1)Separating transformation in a problem on . . . 149Proof. Consider the system of functionsζ = pik(w) = −i(e−iθkw) 1αk , k = 1,n.The family of functions {pik(w)}nk=1 is called admissible for separatingtransformation of domains B0, B∞, Bk, k = 1,n, with respect to theangles {Pk}nk=1. Let Ω(1)k , k = 1,n, denote a domain of the plane Cζobtained as a result of the union of the connected component of the setpik(Bk⋂P k) containing the point pik(ak) with the own symmetric reflecti-on relative to the imaginary axis. In turn, by Ω(2)k , k = 1,n, we denote thedomain of the plane Cζ obtained as a result of the union of the connectedcomponent of the set pik(Bk+1⋂P k) containing the point pik(ak+1) withthe own symmetric reflection relative to the imaginary axis, Bn+1 := B1,pin(an+1) := pin(a1). In addition, by Ω(0)k we denote the domain of theplane Cζ obtained as a result of the union of the connected component ofthe set pik(B0⋂P k) containing the point ζ = 0 with the own symmetricreflection relative to the imaginary axis. Accordingly, by Ω(∞)k we denotethe domain of the plane Cζ obtained as a result of the union of theconnected component of the set pik(B∞⋂P k) containing the point ζ =∞with the own symmetric reflection relative to the imaginary axis. Denotepik(ak) := ω(1)k , pik(ak+1) := ω(2)k , k = 1,n. It follows from the definition ofthe function pik(w) that|pik(w)− ω(1)k | ∼1αk|ak|1αk−1 · |w − ak|, w → ak, w ∈ Pk,|pik(w)− ω(2)k | ∼1αk|ak+1|1αk−1 · |w − ak+1|, w → ak+1, w ∈ Pk,|pik(w)| ∼ |w|1αk , w → 0, w ∈ Pk,|pik(w)| ∼ |w|1αk , w →∞, w ∈ Pk.Taking into account corresponding results of papers [2, 3], we have theinequalitiesr (Bk,ak) 6 r(Ω(1)k ,ω(1)k)· r(Ω(2)k ,ω(2)k)1αk|ak|1αk−1 · 1αk−1 |ak|1αk−1−112, k = 1,n, (2)150 I. V. Denega, A. L. Targonskiir (B0,0) 6[n∏k=1rα2k(Ω(0)k ,0)] 12,r (B∞,∞) 6[n∏k=1rα2k(Ω(∞)k ,∞)] 12. (3)An equality in (2)–(3) is fully investigated in [2, Theorem 1.9]. By usingthis inequalities we have the expansionJn(γ) 6n∏k=1(r(Ω(0)k ,0)r(Ω(∞)k ,∞)) γα2k2 ×× r(Ω(1)k ,ω(1)k)· r(Ω(2)k ,ω(2)k)1αk|ak|1αk−1 · 1αk−1 |ak|1αk−1−112.Elementary calculations show thatJn(γ) 6(n∏k=1αk)n∏k=1|ak|1αk + |ak+1|1αk(|ak||ak+1|)12αk· |ak|××n∏k=1(r(Ω(0)k ,0)r(Ω(∞)k ,∞))γα2k · r(Ω(1)k ,ω(1)k)· r(Ω(2)k ,ω(2)k)(|ak|1αk + |ak+1|1αk)212.HenceJn(γ) 6(n∏k=1αk)n∏k=1(∣∣∣∣ akak+1∣∣∣∣ 12αk + ∣∣∣∣ak+1ak∣∣∣∣ 12αk)|ak|××n∏k=1(r(Ω(0)k ,0)r(Ω(∞)k ,∞))γα2k · r(Ω(1)k ,ω(1)k)· r(Ω(2)k ,ω(2)k)(|ak|1αk + |ak+1|1αk)212.Separating transformation in a problem on . . . 151Thusn∏k=1(∣∣∣∣ akak+1∣∣∣∣ 12αk + ∣∣∣∣ak+1ak∣∣∣∣ 12αk)|ak| =n∏k=1χ(∣∣∣ akak+1∣∣∣ 12αk ) |ak| = L(An),and we obtainJn(γ) 6 2n ·(n∏k=1αk)· L(An)××n∏k=1r(Ω(1)k ,ω(1)k)· r(Ω(2)k ,ω(2)k)(|ak|1αk + |ak+1|1αk)2 (r (Ω(0)k ,0) r (Ω(∞)k ,∞))γα2k12.For convenience, right side of the last inequality we multiply and divideby the value √γ and get the inequalityJn(γ) 6(2√γ)n·(n∏k=1αk√γ)· L(An)××n∏k=1r(Ω(1)k ,ω(1)k)· r(Ω(2)k ,ω(2)k)(|ak|1αk + |ak+1|1αk)2 (r (Ω(0)k ,0) r (Ω(∞)k ,∞))γα2k12,where |ak|1αk + |ak+1|1αk = |ω(2)k − ω(1)k |, k = 1,n. Equality in the lastinequality is attained if and only if the equality is attained in inequalities(2) – (3) for all k = 1,n. Each expression contained in braces in the lastinequality is value of the functionalKτ = [r (B0,0) r (B∞,∞)]τ2 · r (B1,a1) r (B2,a2)|a1 − a2|2 (4)on the system of non-overlapping domains {Ω(0)k ,Ω(1)k ,Ω(2)k ,Ω(∞)k } andcorresponding points system {0, ω(1)k , ω(2)k ,∞}, k = 1,n. Functionalevaluation (4) in the case of fixed poles was first obtained by V.N. Dubi-nin, then by G.V. Kuzmina, E. Emelyanov, A. L. Tarhonskii. Note thatin our case the points a1 and a2 of the functional (4) are not fixed and notsymmetrical. So, fractional-linear transformations can not reduce them to−1 and 1. V.N. Dubinin and G.V. Kuzmina considered the case when152 I. V. Denega, A. L. Targonskiithe points a1 and a2 are symmetrical and equal to 1 and −1, respecti-vely. Therefore decisive role is played corollary of Theorem 4.1.1 in [1,p. 169]. From here, based on invariance of the functional (4) as in theproof of Theorem 4.1.1 in [1], we have that Kτ 6 Φ(τ), τ > 0, whereΦ(τ) = τ2τ2 · |1− τ |−(1−τ)2 · (1 + τ)−(1+τ)2 . ThenJn(γ) 6(2√γ)n( n∏k=1αk√γ)[n∏k=1Φ(τk)]1/2==(2√γ)n [ n∏k=1(τ2τ2k+2k · |1− τk|−(1−τk)2 · (1 + τk)−(1+τk)2)] 12,where τk =√γ · αk, k = 1,n. LetS(x) = x2x2+2 · |1− x|−(1−x)2 · (1 + x)−(1+x)2 and Ψ(x) = ln(S(x)).S(x) is logarithmically convex function on the interval [0, x0],x0 ≈ 0,88441. Further similarly [3, 4] we consider the following extremalproblemn∏k=1S(xk) −→ max,n∑k=1xk = 2√γ, xk = αk√γ.Let X(0) ={x(0)k}nk=1be any extremal set of points in above menti-oned problem. In a similar way as in [4], we obtain the following result:if 0 < x(0)k < x(0)j , thenΨ′(x(0)k ) = Ψ′(x(0)j ), (5)where k,j = 1,n, k 6= j,Ψ′(x) = 4x ln(x)− 2(x− 1) ln |x− 1| − 2(x+ 1) ln(x+ 1) + 2x(see Fig. 1).Separating transformation in a problem on . . . 153Fig. 1: Graph of the function y = Ψ′(x)By using similar arguments as in the paper [4] and the relation (5) wewill prove that x(0)1 = x(0)2 = · · · = x(0)n . Taking into account propertiesof the function Ψ′(x) and conditions of Theorem we obtain the followingrelation: (x1 − 0,56)n + (x2 − x1) > 0 for n > 7. Therefore, we havenx1 + (x2 − x1) > 0,56n. Finally we get(n− 1)x1 + x2 > 2√γn, γn = 0,08n2, n > 7.Thus, in accordance with [4], we agree to say that the set of points{x(0)k}nk=1, n > 7, can not be extremal if x(0)n ∈ (x0, 2]. Consequently, foran extremal set{x(0)k}nk=1is only possible in the case when x(0)k ∈ (0, x0],k = 1,n, and x(0)1 = x(0)2 = . . . = x(0)n . For γ < γn, n > 7, all previ-154 I. V. Denega, A. L. Targonskiious arguments are valid. The equality case is straightforward to verify.Theorem 1 is proved.Corollary 1. Under conditions of Theorem 1 the following inequality holds[r(B0,0)r(B∞,∞)]γn∏k=1r(Bk, ak) 66(4n)n (4γn2)( 4γn2)∣∣∣ 2√γn − 1∣∣∣(2√γn −1)2 (2√γn + 1)( 2√γn +1)2n2.Equality is attained if 0,∞, ak, and B0, B∞, Bk, k = 1,n, are, respectively,poles and circular domains of the quadratic differential (1).Corollary 2. Let n ∈ N, n > 7, 0 < γ 6 γn, γn = 0,08n2. Then forany different points on the circle |ak| = 1, k = 1,n, and any pairwise non-overlapping domains Bk, a0 = 0 ∈ B0 ⊂ C, ∞ ∈ B∞ ⊂ C, ak ∈ Bk ⊂ C,k = 1,n, the following inequality holds[r (B0,0) r (B∞,∞)]γn∏k=1r (Bk,ak) 6 [r (Λ0,0) r (Λ∞,∞)]γn∏k=1r (Λk, λk) ,where domains Λ0, Λ∞, Λk, and points 0,∞, λk, k = 1,n, are, respectively,circular domains and poles of the quadratic differential (1).References[1] Bakhtin A.K., Bakhtina G.P., Zelinskii Yu. B. Topological-algebraicstructures and geometric methods in complex analysis // Proceedings ofthe Institute of Mathematics of NAS of Ukraine, 2008. — 308 p. (in Russi-an).[2] Dubinin V.N. Symmetrization method in geometric function theory ofcomplex variables // Successes Mat. Science. — 1994. — 49, No. 1 (295). —P. 3 – 76 (in Russian); Engl. transl. in: Russian Math. Surveys. — 1994. —49, № 1. — P. 1 – 79.Separating transformation in a problem on . . . 155[3] Dubinin V.N. Separating transformation of domains and problems onextremal decomposition // Notes scientific. sem. Leningr. Dep. of Math.Inst. AN USSR. — 1988. — 168. — P. 48 – 66 (in Russian); Engl. transl. in:J. Soviet Math. — 1991. — 53, No. 3. — P. 252 — 263.[4] Kovalev L.V. On the problem of extremal decomposition with free poleson a circle // Dal’nevostochnyi Mat. Sb. — 1996. — 2. — P. 96 – 98 (inRussian).[5] Lavrentev M.A. To the theory of conformal mappings // Trudy Fiz.-Mat.Inst. 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Separating transformation in a problem on extremal decomposition of the complex plane

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Separating transformation in a problem on extremal decomposition of the complex plane

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