2016
DOI: 10.5186/aasfm.2016.4142
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On tangential slit solution of the Loewner equation

Abstract: Abstract. For a non-tangential slit γ(t), the behavior of the driving function λ(t) near zero in the Loewner equation is well understood; for tangential slit, the situation is less clear. In this paper, we investigate the tangential slit Γ p , p > 0, where Γ is a circular arc tangent at 0; Γ p has order p+1 p near zero. Our main result is to give the exact expression of λ(t), and its Hölder exponent near 0 in terms of p, which has a natural connection with the known results. We also extend this to a general ty… Show more

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Cited by 11 publications
(7 citation statements)
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References 26 publications
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“…Then for t small enough, the hull K t driven by λ is contained in the region {x + iy : 0 ≤ x, 0 < y < 26 a x 2−2r }. This proposition may be of independent interest, as it provides a converse of sorts to recent work by Lau and Wu [15]. In particular, Lau and Wu show that if the curve leaves the real line tangentially by lying in the domain {x + iy : 0 < x, ax r < y < bx r } for r > 1, then we have some control on the associated driving function, namely lim sup t→0 t −1/(r+1) |λ(t)| < ∞.…”
Section: Introductionmentioning
confidence: 56%
“…Then for t small enough, the hull K t driven by λ is contained in the region {x + iy : 0 ≤ x, 0 < y < 26 a x 2−2r }. This proposition may be of independent interest, as it provides a converse of sorts to recent work by Lau and Wu [15]. In particular, Lau and Wu show that if the curve leaves the real line tangentially by lying in the domain {x + iy : 0 < x, ax r < y < bx r } for r > 1, then we have some control on the associated driving function, namely lim sup t→0 t −1/(r+1) |λ(t)| < ∞.…”
Section: Introductionmentioning
confidence: 56%
“…We refer to [6] where the problem was solved for the circular arc in H tangential to R at 0. This result was generalized in [7] for powers of this arc and in [8] for tangential curves close to this arc. It was proved in [6] that the tangential circular arc of radius 1 and centered at i is generated by the driving function λ(t) = 3α(t) + 2 −α(t)π where α = α(t) is an algebraic function satisfying the equation α(3α + 4 √ −απ) = −6t, t ≥ 0.…”
Section: Discussionmentioning
confidence: 90%
“…Similarly to (7) derive an implicit representation for z(t) in Case (iii). Integrate the differential equation for (z 4 (t)) ′ to get the needed equation…”
Section: Case (I)mentioning
confidence: 99%
“…2. Ðàçëîaeåíèå óïðàâëÿþùåé ôóíêöèè λ(t) â ñëó÷àå êàñàòåëüíîãî ðàçðåçà  [27] ïîëó÷åí ñëåäóþùèé ðåçóëüòàò. Ïóñòü îòîáðàaeåíèå g, g : D → Π + , ãîëîìîðôíî è îäíîëèñòíî â îäíîñâÿçíîé îáëàñòè D = Π + \Γ s , Γ s ⊂ Π + ∪ {0}, ïðåäñòàâëÿþùåé ñîáîé âåðõíþþ ïîëóïëîñêîñòü ñ ðàçðåçîì âäîëü êðèâîé…”
Section: îïðåäåëåíèå ïàðàìåòðîâ ñåìåéñòâà îòîáðàAeåíèéunclassified