Regular behavior of subharmonic in space functions of the zero kind

IF 1 Q1 MATHEMATICS
M. Zabolotskyi, T.M. Zabolotskyi, S. Tarasyuk, Yu.M. Hal
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引用次数: 0

Abstract

Let $u$ be a subharmonic in $\mathbb{R}^m$, $m\geq 3$, function of the zero kind with Riesz measure $\mu$ on negative axis $Ox_1$, $n(r,u)=\mu\left(\{x\in\mathbb{R}^m \colon |x|\leq r\}\right)$, \[N(r,u)=(m-2)\int_1^r n(t,u)/t^{m-1}dt,\] $\rho(r)$ is a proximate order, $\rho(r)\to\rho$ as $r\to+\infty$, $0<\rho<1$. We found the asymptotic of $u(x)$ as $|x|\to+\infty$ by the condition $N(r,u)=\left(1+o(1)\right)r^{\rho(r)}$, $r\to+\infty$. We also investigated the inverse relationship between a regular growth of $u$ and a behavior of $N(r,u)$ as $r\to+\infty$.
空间零类次谐函数的规律行为
让 $u$ 是 $mathbb{R}^m$ 中的次谐波,$m\geq 3$,在负轴 $Ox_1$ 上具有 Riesz 量 $\mu$ 的零类函数,$n(r、u)=\mu\left(\{x\in\mathbb{R}^m \colon |x|\leq r\}/right)$, \[N(r,u)=(m-2)\int_1^r n(t,u)/t^{m-1}dt,\] $\rho(r)$ 是一个近似阶,$\rho(r)\to\rho$为 $r\to+\infty$,$0<\rho<1$。我们通过条件 $N(r,u)=\left(1+o(1)\right)r^{rho(r)}$, $r\to+\infty$ 发现了 $u(x)$ 在 $|x|\to+\infty$ 时的渐近线。我们还研究了 $u$ 的规律增长与 $N(r,u)$ 在 $r\to+\infty$ 时的行为之间的反比关系。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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来源期刊
CiteScore
1.90
自引率
12.50%
发文量
31
审稿时长
25 weeks
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