﻿ 一类费马型微分–差分方程解的问题 Solution on Differential-Difference Equation of Fermat-Type

Pure Mathematics
Vol. 10  No. 03 ( 2020 ), Article ID: 34508 , 5 pages
10.12677/PM.2020.103023

Solution on Differential-Difference Equation of Fermat-Type

Bingmao Deng

School of Financial Mathematics and Statics, Guangdong University of Finance, Guangzhou Guangdong

Received: Feb. 17th, 2020; accepted: Mar. 4th, 2020; published: Mar. 13th, 2020

ABSTRACT

In this paper, we mainly discussed entire solutions with finite order of the following Fermat type differential-difference equation ${\left[{f}^{\prime }\left(z\right)\right]}^{2}+{\left[{\Delta }_{c}^{k}f\left(z\right)-f\left(z\right)\right]}^{2}=1$ and obtained some interesting results.

Keywords:Fermat-Type Equation, Entire Function, Differential-Difference Equation

1. 引言及主要结果

,

$\underset{j=0}{\overset{k}{\sum }}{b}_{j}{a}^{j}=\frac{1}{i}$,$\underset{j=0}{\overset{k}{\sum }}{b}_{j}{\left(-a\right)}^{j}=-\frac{1}{i}$.

2012年，Liu等人 [11] 研究了以下费马型差分–微分方程，并获得了以下结论。

${\left[{f}^{\prime }\left(z\right)\right]}^{2}+{\left[f\left(z+c\right)-f\left(z\right)\right]}^{2}=1$. (1.1)

$f\left(z\right)=\mathrm{sin}\left(2z+Bi\right)/2$,

${\left[{f}^{\prime }\left(z\right)\right]}^{2}+{\left[{\Delta }_{c}^{k}f\left(z\right)-f\left(z\right)\right]}^{2}=1$. (1.2)

$f\left(z\right)\equiv ±1$，或者

，其中 $a,b,d$ 是常数，并且满足 ${\text{e}}^{kac}={\left(-1\right)}^{k}$

${\left({\text{e}}^{ac}-1\right)}^{k}\ne 1$,$a=i{\left({\text{e}}^{ac}-1\right)}^{k}-i$.

$f\left(z\right)\equiv ±1$，或者

$f\left(z\right)=\frac{1}{3}\mathrm{sin}\left(3z+bi\right)+d$，其中 $b,d$ 是常数， $c=\left(\pi +n\pi \right)/3$

2. 一些引理

$\underset{i=1}{\overset{n}{\sum }}{f}_{i}\left(z\right){\text{e}}^{{g}_{i}\left(z\right)}\equiv 0$ ;

$T\left(r,{f}_{j}\right)=S\left(r,{\text{e}}^{{g}_{h}-{g}_{l}}\right)$,$r\to \infty$,$r\notin E$,

${f}_{i}\left(z\right)\equiv 0\left(1\le i\le n\right)$

$f\left(z\right)={z}^{k}P\left(z\right){\text{e}}^{Q\left(z\right)}$,

3. 定理3的证明

$f\left(z\right)$ 是方程(1.2)的有穷级整函数解，将(1.3)改写成下式

。 (3.1)

(3.2)

${f}^{\prime }\left(z\right)=\frac{{\text{e}}^{p\left(z\right)}+{\text{e}}^{-p\left(z\right)}}{2}$, (3.3)

${\Delta }_{c}^{k}f\left(z\right)-f\left(z\right)=\frac{{\text{e}}^{p\left(z\right)}-{\text{e}}^{-p\left(z\right)}}{2i}$. (3.4)

${\left[{\Delta }_{c}^{k}f\left(z\right)-f\left(z\right)\right]}^{\prime }=\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{f}^{\prime }\left(z+jc\right)-{f}^{\prime }\left(z\right)=\frac{{p}^{\prime }\left(z\right)}{2i}\left[{\text{e}}^{p\left(z\right)}+{\text{e}}^{-p\left(z\right)}\right]$. (3.5)

${f}^{\prime }\left(z+jc\right)=\frac{1}{2}\left[{\text{e}}^{p\left(z+jc\right)}+{\text{e}}^{-p\left(z+jc\right)}\right]$. (3.6)

$\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}\left[{\text{e}}^{p\left(z+jc\right)}+{\text{e}}^{-p\left(z+jc\right)}\right]-\left[{\text{e}}^{p\left(z\right)}+{\text{e}}^{-p\left(z\right)}\right]=-i{p}^{\prime }\left(z\right)\left[{\text{e}}^{p\left(z\right)}+{\text{e}}^{-p\left(z\right)}\right]$. (3.7)

$\left[i{p}^{\prime }\left(z\right)-1+{\left(-1\right)}^{k}\right]\left[{\text{e}}^{p\left(z\right)}+{\text{e}}^{-p\left(z\right)}\right]+\underset{j=1}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}\left[{\text{e}}^{p\left(z+jc\right)}+{\text{e}}^{-p\left(z+jc\right)}\right]=0$. (3.8)

，则对任意的 $0\le j，均有

$\rho \left({\text{e}}^{p\left(z+jc\right)-p\left(z+lc\right)}\right)=m-1\ge 1$, 且 $\rho \left({\text{e}}^{p\left(z+jc\right)+p\left(z+lc\right)}\right)=m\ge 2$.

(3.8)式两边同时乘以 ${\text{e}}^{p\left(z\right)}$，并化简，可得

$\left[ia-1+{\left(-1\right)}^{k}\right]\left[{\text{e}}^{2p\left(z\right)}+1\right]+\underset{j=1}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}\left[{\text{e}}^{p\left(z\right)+p\left(z+jc\right)}+{\text{e}}^{p\left(z\right)-p\left(z+jc\right)}\right]=0$.

$\left[ia-1+{\left(-1\right)}^{k}\right]\left[{\text{e}}^{2p\left(z\right)}+1\right]+\underset{j=1}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}\left[{\text{e}}^{2p\left(z\right)+ajc}+{\text{e}}^{-ajc}\right]=0$. (3.9)

${\text{e}}^{2p\left(z\right)}\left[\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{\text{e}}^{ajc}+ia-1\right]+\left[\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{\text{e}}^{-ajc}+ia-1\right]=0$. (3.10)

$\left\{\begin{array}{l}\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{\text{e}}^{ajc}+ia-1=0;\\ \underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{\text{e}}^{-ajc}+ia-1=0.\end{array}$

${\text{e}}^{2p\left(z\right)}=-\frac{\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{\text{e}}^{-ajc}+ia-1}{\underset{j=0}{\overset{k}{\sum }}{\left(-1\right)}^{k-j}{C}_{k}^{j}{\text{e}}^{ajc}+ia-1}$,

$\left\{\begin{array}{l}{\left({\text{e}}^{ac}-1\right)}^{k}+ia-1=0;\\ {\left({\text{e}}^{-ac}-1\right)}^{k}+ia-1=0.\end{array}$ (3.11)

$f\left(z\right)=\frac{{\text{e}}^{az+b}-{\text{e}}^{-az-b}}{2a}+d=\frac{1}{ai}\mathrm{sin}\left[aiz+bi\right]+d$.

Solution on Differential-Difference Equation of Fermat-Type[J]. 理论数学, 2020, 10(03): 162-166. https://doi.org/10.12677/PM.2020.103023

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