2005
DOI: 10.1007/s10455-005-7276-5
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Towards an Effectivisation of the Riemann Theorem

Abstract: Let Q be a connected and simply connected domain on the Riemann sphere, not coinciding with the Riemann sphere and with the whole complex plane C. Then, according to the Riemann Theorem, there exists a conformal bijection between Q and the exterior of the unit disk. In this paper we find an explicit form of this map for a broad class of domains with analytic boundaries.

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Cited by 7 publications
(8 citation statements)
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References 13 publications
(11 reference statements)
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“…with universal coefficients which will be found. For a particular function f (t 0 ) they were found in [22]. In the general case the arguments are similar.…”
Section: Taylor Expansion Of Symmetric Solutionsmentioning
confidence: 58%
See 1 more Smart Citation
“…with universal coefficients which will be found. For a particular function f (t 0 ) they were found in [22]. In the general case the arguments are similar.…”
Section: Taylor Expansion Of Symmetric Solutionsmentioning
confidence: 58%
“…r m . Our method is an extension of the method developed in [22,23] for the formal c = 1 string solution. Convergence of the Taylor series for this case was investigated in [11].…”
Section: Introductionmentioning
confidence: 99%
“…Substituting r = 1 into series (5) and comparing the result with WF (3) it is easy to find that exact solution of IDP (1-3) is equal to:…”
Section: The Internal Dirichlet Problem and Fractality Of Boundary Conditionmentioning
confidence: 99%
“…Generally speaking one may derive this mapping f (z, ε) explicitly in the framework of formalism of the harmonic moments of exterior domain C\ ε [5] but this way is too hard. On the other hand due to representation (23) of ε we can restrict ourselves by construction of approximate conformal mapping of the nearly circular domain on unit disk.…”
Section: The Internal Dirichlet Problem and Fractality Of Boundary Conditionmentioning
confidence: 99%
“…Following [15], we define the combinatorial constantsP ij Proof. Since ∆(z) = k≥1 z −k k ∂h k , the Hirota equation (26) yields…”
Section: Theh-kp Hierarchy In Terms Of ∂H Imentioning
confidence: 99%