Singular Solutions with Asymptotic Expansion of Linear Partial Differential Equations in the Complex Domain
Author(s) -
Sunao Ōuchi
Publication year - 1998
Publication title -
publications of the research institute for mathematical sciences
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.786
H-Index - 39
eISSN - 1663-4926
pISSN - 0034-5318
DOI - 10.2977/prims/1195144627
Subject(s) - mathematics , asymptotic expansion , domain (mathematical analysis) , mathematical analysis , partial differential equation , singular solution , linear differential equation , method of matched asymptotic expansions , differential equation
Consider the linear partial differential equation P (z, ∂z)u(z) = f(z) in C , where f(z) is not holomorphic on K = {z0 = 0}, but it has an asymptotic expansion with respect to z0 as z0 → 0 in some sectorial region. We show under some conditions on P (z, ∂z) that there exists a solution u(z) which has an asymptotic expansion of the same type as that of f(z). §0. Introduction Let P (z, ∂z) be a linear partial differential operator with holomorphic coefficients in a neighborhood Ω of z = 0 in C and K = {z0 = 0}. Consider the equation P (z, ∂z)u(z) = f(z), (0.1) where f(z) is holomorphic except on K, but f(z) has an asymptotic expansion f(z) ∼ ∑ n fn(z ′)zn 0 as z0 → 0 in some sectorial region with respect to z0. In the present paper we study the existence of solutions. Firstly we remark that if we require nothing about the behavior of u(z) near K, there exists a solution u(z) with singularities on K under some conditions on the principal symbol of P (z, ∂z). But the singularities of u(z) may be much stronger than those of f(z) (see [1], [2], [5] and [9]). Communicated by T. Kawai. Received April 18, 2001. Revised May 17, 2002 and March 3, 2003. 2000 Mathematics Subject Classification(s): Primary 35A20; Secondary 35B40, 35C20.
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