tailieunhanh - On non-homogeneous Riemann boundary value problem

In this paper we consider non-homogeneous Riemann boundary value problem with unbounded oscillating coefficients on a class of open rectifiable Jordan curve. | Turk J Math 24 (2000) , 251 – 257. ¨ ITAK ˙ c TUB On Non-Homogeneous Riemann Boundary Value Problem Kadir Kutlu Abstract In this paper we consider non-homogeneous Riemann boundary value problem with unbounded oscillating coefficients on a class of open rectifiable Jordan curve. Key Words: Curve, Cauchy type integral, singular integral, Riemann problem. In [1], homogeneous Riemann boundary value problem was studied when curve γ satisfies the condition θ(δ) ≈ δ and G is an oscillating function at the end points of the curve. In this work we investigate the non-homogeneous Riemann problem in the same case and we will use terminology and notations introduced in [1]. We need the following class of functions for our future references: H a1 (µ1 , ν1) + H a2 (µ2 , ν2 ) = {g ∈ Cγ : g = g1 + g2 , gk ∈ Cγ , Ωagkk (ξ) = O(ξ −νk ), ωgakk (δ, ξ) = O(δ µk ξ −µk −νk )} where k = 1; 2, µk ∈ (0, 1], νk ∈ [0, 1), δ, ξ ∈ (0, d], δ ≤ ξ, d = diam γ. Lemma 1. [3] Suppose that γ satisfies θ(δ) ≈ δ, G(t)=exp(2πif(t)), Ωaf k (ξ) = O (ln 1ξ ), ωfak (δ, ξ) = O ( δξ ), δ ≤ ξ, k = 1, 2, g ∈ Ha1 (µ1 , ν1) + Ha2 (µ2 , ν2 ) and suppose 1991 AMS subject classification: Primary 30E20, 30E25; secondary 45E05 251 KUTLU h is holomorphic in C\γ, continuously extendable to γˆ from both sides, h(z)6=0 for all z∈ C\γ, h± (t)6=0 for all t∈ γˆ , h+ (t) = G(t)h− (t), g/h+ ∈ L (γ). Then the function Φ0 (z) = h(t) 2πi Z γ g(τ ) dτ, z 6∈ γ h+ (τ )(τ − z) (1) is holomorphic in C\γ, continuously extendable to γˆ from both sides and satisfies the homogeneous boundary conditions We introduce the quantity k 1 Re z→ak ln | z − ak | Z ∆G = lim γ f(τ ) dτ, z 6∈ γ τ −z (2) k and if ∆G is finite introduce k ∆k , if ∆G ∈ Z h Gk i æ0k = ∆G + 1, if ∆kG 6∈ Z (3) k=1,2. Lemma 2. Suppose that γ satisfies θ(δ) ≈ δ, G and g are as in lemma1 and 0 0 χ0 (z) = (z − a1 )−æ1 (z − a2 )−æ2 exp Z γ f(τ ) dτ, z 6∈ γ. τ −z Then g/χ+ 0 is integrable on γ. Proof. It is obvious that g/χ+ 0 is bounded on a