Meromorphic solutions of Bi-Fermat type partial differential and difference equations

IF 1.4 3区 数学 Q1 MATHEMATICS
Yingchun Gao, Kai Liu
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引用次数: 0

Abstract

Fermat type functional equation with four terms

$$\begin{aligned} f(z)^{n}+g(z)^{n}+h(z)^{n}+k(z)^{n}=1 \end{aligned}$$

is difficult to solve completely even if \(n=2,3\), in which the certain type of the above equation is also interesting and significant. In this paper, we first to consider the Bi-Fermat type quadratic partial differential equation

$$\begin{aligned} f(z_{1},z_{2})^{2}+\left( \frac{\partial f(z_{1},z_{2})}{\partial z_{1}}\right) ^{2}+g(z_{1},z_{2})^{2}+\left( \frac{\partial g(z_{1},z_{2})}{\partial z_{1}}\right) ^{2}=1 \end{aligned}$$

in \(\mathbb {C}^{2}\). In addition, we consider the Bi-Fermat type cubic difference equation

$$\begin{aligned} f(z)^{3}+g(z)^{3}+f(z+c)^{3}+g(z+c)^{3}=1 \end{aligned}$$

in \(\mathbb {C}\) and obtain partial meromorphic solutions on the above equation.

比-费马型偏微分方程和差分方程的同态解
具有四个项的费马型函数方程 $$\begin{aligned} f(z)^{n}+g(z)^{n}+h(z)^{n}+k(z)^{n}=1 \end{aligned}$$即使在 \(n=2,3\)的情况下也很难完全求解,其中上述方程的某种类型也很有趣且意义重大。在本文中,我们首先考虑 Bi-Fermat 型二次偏微分方程 $$begin{aligned} f(z_{1},z_{2})^{2}+\left( \frac{\partial f(z_{1}、z_{2})}{partial z_{1}}\right) ^{2}+g(z_{1},z_{2})^{2}+left( ( frac{partial g(z_{1},z_{2})}{partial z_{1}}\right) ^{2}=1 \end{aligned}$$ in \(\mathbb {C}^{2}\).此外,我们还考虑了 Bi-Fermat 型立方差分方程 $$\begin{aligned} f(z)^{3}+g(z)^{3}+f(z+c)^{3}+g(z+c)^{3}=1 \end{aligned}$$ in \(\mathbb {C}/),并得到了上述方程的部分分形解。
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
Analysis and Mathematical Physics
Analysis and Mathematical Physics MATHEMATICS, APPLIED-MATHEMATICS
CiteScore
2.70
自引率
0.00%
发文量
122
期刊介绍: Analysis and Mathematical Physics (AMP) publishes current research results as well as selected high-quality survey articles in real, complex, harmonic; and geometric analysis originating and or having applications in mathematical physics. The journal promotes dialog among specialists in these areas.
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