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Singular integral operators
Open AccessArticle

New definition of a singular integral operator

Annals of Communications in Mathematics 2023

, 6 (4)

, 220-224

DOI: https://doi.org/10.62072/acm.2023.060402

AbstractLet D be a connected bounded domain in R^2, S be its boundary which is closed, connected, and smooth, or S=(-∞,∞). Let Φ(z) be the function defined as Φ(z)=1/(2πi) ∫S(f(s)ds)/(s-z), where f∈L^1(S) and z=x+iy. The singular integral operator Af is defined as Af: =1/(iπ) ∫S(f(s)ds)/(s-t), where t∈S. This new definition simplifies the proof of the existence of Φ(t). Necessary and sufficient conditions are given for f∈L^1(S) to be the boundary value of an analytic function in D. The Sokhotsky-Plemelj formulas are derived for f∈L^1(S). Our new definition allows one to treat singular boundary values of analytic functions.
Open AccessArticle

On the Riemann Problem

Annals of Communications in Mathematics 2024

, 7 (4)

, 451-454

DOI: https://doi.org/10.62072/acm.2024.070411

AbstractThe Riemann problem is stated as follows: find an analytic in a domain D+ ∪ D− function Φ(z) such that (∗) Φ+(t) = G(t)Φ−(t) + g(t), t ∈ S. Here S is the boundary of D+, D− complements the complex plane to D+ ∪ S, the functions G = G(t) and g = g(t) belong to Hμ(S), the space of H¨older-continuous functions. The theory of problem (*) is developed also for continuous G. If G = 1, S ∈ C∞ and g is a tempered distribution, then problem (∗) has a solution in tempered distributions. It is proved that problem (∗) for G ∈ Lp(S) and g a tempered distribution does not make sense. It is proved that if G ∈ C∞(S), G̸ = 0 on S, and g is a distribution of the class D′, then the Riemann problem makes sense and a method for solving this problem is given. It is proved that if S = R = (−∞, ∞) and G ∈ Lp(R), where p ≥ 1 is a fixed number, then | ln G| does not belong to Lq (R) for any q ≥ 1.