﻿ 潘勒韦IV型差分方程亚纯解唯一性 The Unicity of the Meromorphic Solutions of Painlevé IV Difference Equations

Pure Mathematics
Vol. 08  No. 06 ( 2018 ), Article ID: 27466 , 8 pages
10.12677/PM.2018.86080

The Unicity of the Meromorphic Solutions of Painlevé IV Difference Equations

Meijuan Zhang1, Shanhua Lin2

1College of Mathematics and Informatics, Fujian Normal University, Fuzhou Fujian

2College of Mathematics and Computer Science, Quanzhou Normal University, Quanzhou Fujian

Received: Oct. 15th, 2018; accepted: Oct. 27th, 2018; published: Nov. 8th, 2018

ABSTRACT

In this paper, we use Nevanlinna theory to discuss the unicity problems of the finite-order transcendental meromorphic solution of Painlevé IV difference equations with another meromorphic function that share three values.

Keywords:Complex Difference Equation, Meromorphic Solution, Unicity

1福建师范大学数学与信息学院，福建 福州

2泉州师范学院数学与计算机科学学院，福建 泉州

1. 引言

${\rho }_{2}\left(f\right)=\underset{r\to \infty }{\mathrm{lim}\mathrm{sup}}\frac{{\mathrm{log}}^{+}{\mathrm{log}}^{+}T\left(r,f\right)}{\mathrm{log}r}$

 均是f的小函数，记 $R\left(z,f\right)=\frac{P\left(z,f\right)}{Q\left(z,f\right)}$ 是关于z亚纯，关于f有理的函数，其中 ，并且 $P\left(z,f\right),Q\left(z,f\right)$ 的次数分别为 $\mathrm{deg}P\left(z,f\right)=m,\mathrm{deg}Q\left(z,f\right)=n$

$f\left(z+1\right)f\left(z-1\right)=R\left(z,f\right)$

1991年，Ramani [9] 考察了潘勒韦IV型差分方程 $\left(f\left(z+1\right)+f\left(z\right)\right)\left(f\left(z-1\right)+f\left(z\right)\right)=R\left(z\right)$，其中该方程的所有系数均为常数且 $R\left(z\right)$ 是关于z有理的函数。

 (1.1)

$\varphi \left(z,f\right)=\left(f\left(z+1\right)+f\left(z\right)\right)\left(f\left(z-1\right)+f\left(z\right)\right)Q\left(z,f\right)-P\left(z,f\right)$ (1.2)

2. 引理

$H\left(z,f\right)=\underset{\lambda \in J}{\sum }{a}_{\lambda }\left(z\right)\underset{j=1}{\overset{{\tau }_{\lambda }}{\prod }}f{\left(z+{\delta }_{\lambda ,j}\right)}^{{\mu }_{\lambda ,j}},$ (2.1)

${\mathrm{deg}}_{f}H\left(z,f\right)=\underset{\lambda \in J}{\mathrm{max}}\left\{{d}_{\lambda }\right\}.$

$m\left(r,\frac{f\left(z+c\right)}{f\left(z\right)}\right)=ο\left(\frac{T\left(r,f\right)}{{r}^{1-{\rho }_{2}-\epsilon }}\right)=S\left(r,f\right)$

$L\left(z,g\right)+D\left(z\right)=0,$

$|\frac{L\left(z,g\right)}{g\left(z\right)}|=\frac{1}{|g\left(z\right)|}\underset{\gamma }{\sum }{b}_{\gamma }\left(z\right)g{\left(z\right)}^{{l}_{0}}g{\left(z+{c}_{1}\right)}^{{l}_{1}}\cdots g{\left(z+{c}_{\nu }\right)}^{{l}_{\nu }}\le {\underset{\gamma }{\sum }|{b}_{\gamma }\left(z\right)||\frac{g\left(z+{c}_{1}\right)}{g\left(z\right)}|}^{{l}_{1}}\cdots {|\frac{g\left(z+{c}_{\nu }\right)}{g\left(z\right)}|}^{{l}_{\nu }},$

$\begin{array}{c}m\left(r,\frac{L}{g}\right)\le m\left(r,{\underset{\gamma }{\sum }|{b}_{\gamma }\left(z\right)||\frac{g\left(z+{c}_{1}\right)}{g\left(z\right)}|}^{{l}_{1}}\cdots {|\frac{g\left(z+{c}_{\nu }\right)}{g\left(z\right)}|}^{{l}_{\nu }}\right)\le \underset{\gamma }{\sum }m\left(r,{b}_{\gamma }\left(z\right){|\frac{g\left(z+{c}_{1}\right)}{g\left(z\right)}|}^{{l}_{1}}\cdots {|\frac{g\left(z+{c}_{\nu }\right)}{g\left(z\right)}|}^{{l}_{\nu }}\right)\\ =\underset{\gamma }{\sum }\frac{1}{\text{2π}}{\int }_{0}^{\text{2π}}{\mathrm{log}}^{+}|{b}_{\gamma }\left(r{e}^{i\theta }\right){|\frac{g\left(r{e}^{i\theta }+{c}_{1}\right)}{g\left(r{e}^{i\theta }\right)}|}^{{l}_{1}}\cdots {|\frac{g\left(r{e}^{i\theta }+{c}_{\nu }\right)}{g\left(r{e}^{i\theta }\right)}|}^{{l}_{\nu }}|\text{d}\theta \\ \le \underset{\gamma }{\sum }\left(m\left(r,{b}_{\gamma }\left(z\right)\right)+{l}_{1}m\left(r,\frac{g\left(z+{c}_{1}\right)}{g\left(z\right)}\right)+\cdots +{l}_{\nu }m\left(r,\frac{g\left(z+{c}_{\nu }\right)}{g\left(z\right)}\right)\right)=S\left(r,f\right),\end{array}$

${I}_{1}=\frac{1}{\text{2π}}{\int }_{0}^{\text{2π}}{\mathrm{log}}^{+}|\frac{1}{g\left(r{\text{e}}^{i\theta }\right)}|\text{d}\theta \le m\left(r,\frac{D}{g}\right)+m\left(r,\frac{1}{D}\right)=m\left(r,\frac{L}{g}\right)+m\left(r,\frac{1}{D}\right)=S\left(r,f\right).$ (2.2)

${I}_{\text{2}}=\frac{1}{2\text{π}}{\int }_{0}^{\text{2π}}{\mathrm{log}}^{+}|\frac{1}{g\left(r{e}^{i\theta }\right)}|\text{d}\theta =0.$ (2.3)

$m\left(r,\frac{1}{f-a}\right)={I}_{1}+{I}_{2}=S\left(r,f\right).$

${\mathrm{deg}}_{f}H\left(z,f\right)<\mathrm{max}\left\{\mathrm{deg}P\left(z,f\right),\mathrm{deg}Q\left(z,f\right)\right\},$

$m\left(r,f\right)=S\left(r,f\right)$

i) 若h为p次多项式，则 

ii) 若h为超越亚纯函数，则 $\rho =\mu =\infty$

$\underset{j=1}{\overset{n}{\sum }}{f}_{j}\equiv 1,$

$\underset{j=1}{\overset{n}{\sum }}N\left(r,\frac{1}{{f}_{j}}\right)+\left(n-1\right)\underset{j=1}{\overset{n}{\sum }}\stackrel{¯}{N}\left(r,{f}_{j}\right)<\left(\lambda +ο\left(1\right)\right)T\left(r,{f}_{k}\right)\left(r\in I,k=1,2,\cdots ,n-1\right),$

3. 定理1.1的证明

$m\left(r,\frac{1}{f-{e}_{1}}\right)=S\left(r,f\right),m\left(r,\frac{1}{f-{e}_{2}}\right)=S\left(r,f\right).$ (3.1)

$\begin{array}{c}T\left(r,f\right)\le N\left(r,f\right)+N\left(r,\frac{1}{f-{e}_{1}}\right)+N\left(r,\frac{1}{f-{e}_{2}}\right)+S\left(r,f\right)\\ \le N\left(r,g\right)+N\left(r,\frac{1}{g-{e}_{1}}\right)+N\left(r,\frac{1}{g-{e}_{2}}\right)+S\left(r,f\right)\\ \le 3T\left(r,g\right)+S\left(r,f\right),\end{array}$ (3.2)

 (3.3)

$\rho \left(f\right)=\rho \left(g\right).$ (3.4)

$\frac{f-{e}_{1}}{g-{e}_{1}}={e}^{A},\frac{f-{e}_{2}}{g-{e}_{2}}={e}^{B},$ (3.5)

$T\left(r,{e}^{A}\right)\le 4T\left(r,f\right)+S\left(r,f\right),$

$T\left(r,{e}^{B}\right)\le 4T\left(r,f\right)+S\left(r,f\right).$ (3.6)

$C=B-A$，则由(3.5)式可得

$f={e}_{1}+\left({e}_{2}-{e}_{1}\right)\frac{{e}^{B}-1}{{e}^{C}-1},$ (3.7)

$f={e}_{\text{2}}+\left({e}_{2}-{e}_{1}\right)\frac{{e}^{A}-1}{{e}^{C}-1}{e}^{C}.$ (3.8)

$T\left(r,f\right)\le T\left(r,{e}^{C}\right)+T\left(r,{e}^{B}\right)+S\left(r,f\right),$ (3.9)

${e}^{B}-1=D{B}_{1},{e}^{C}-1=D{C}_{1},$

$T\left(r,f\right)=m\left(r,\frac{1}{f-{e}_{1}}\right)+N\left(r,\frac{1}{f-{e}_{1}}\right)+S\left(r,f\right)=N\left(r,\frac{1}{{B}_{1}}\right)+S\left(r,f\right).$ (3.10)

$T\left(r,f\right)=m\left(r,f\right)+N\left(r,f\right)=N\left(r,\frac{1}{{C}_{1}}\right)+S\left(r,f\right).$ (3.11)

$T\left(r,{e}^{B}\right)\le N\left(r,{e}^{B}\right)+N\left(r,\frac{1}{{e}^{B}}\right)+N\left(r,\frac{1}{{e}^{B}-1}\right)+S\left(r,{e}^{B}\right)=N\left(r,\frac{1}{{e}^{B}-1}\right)+S\left(r,f\right)\le T\left(r,{e}^{B}\right)+S\left(r,f\right),$

$N\left(r,\frac{1}{{e}^{B}-1}\right)=N\left(r,\frac{1}{{B}_{1}}\right)+N\left(r,\frac{1}{D}\right)+S\left(r,f\right),$

$T\left(r,{e}^{B}\right)=N\left(r,\frac{1}{{B}_{\text{1}}}\right)+N\left(r,\frac{1}{D}\right)+S\left(r,f\right).$ (3.12)

$T\left(r,{e}^{C}\right)=N\left(r,\frac{1}{{C}_{\text{1}}}\right)+N\left(r,\frac{1}{D}\right)+S\left(r,f\right).$ (3.13)

$T\left(r,{e}^{C}\right)=T\left(r,{e}^{B}\right)+S\left(r,f\right).$ (3.14)

$T\left(r,{e}^{C}\right)=T\left(r,{e}^{A}\right)+S\left(r,f\right).$ (3.15)

$\rho \left({e}^{A}\right)=\rho \left({e}^{B}\right)=\rho \left({e}^{C}\right)=\rho \left(f\right),$ (3.16)

$\mathrm{deg}\left(A\right)=\mathrm{deg}\left(B\right)=\mathrm{deg}\left(C\right)=d.$ (3.17)

$\begin{array}{l}\left[2{e}_{1}+\left({e}_{2}-{e}_{1}\right)\left(\frac{{e}^{\stackrel{¯}{B}}-1}{{e}^{\stackrel{¯}{C}}-1}+\frac{{e}^{B}-1}{{e}^{C}-1}\right)\right]\left[2{e}_{1}+\left({e}_{2}-{e}_{1}\right)\left(\frac{{e}^{\underset{_}{B}}-1}{{e}^{\underset{_}{C}}-1}+\frac{{e}^{B}-1}{{e}^{C}-1}\right)\right]\underset{j=0}{\overset{n}{\sum }}{b}_{j}{\left[{e}_{1}+\left({e}_{2}-{e}_{1}\right)\frac{{e}^{B}-1}{{e}^{C}-1}\right]}^{j}\\ =\underset{i=0}{\overset{m}{\sum }}{a}_{i}{\left[{e}_{1}+\left({e}_{2}-{e}_{1}\right)\frac{{e}^{B}-1}{{e}^{C}-1}\right]}^{i}\end{array}$

$\begin{array}{l}\left\{2{e}_{1}\left({e}^{\stackrel{¯}{C}}-1\right)\left({e}^{C}-1\right)+\left({e}_{2}-{e}_{1}\right)\left[\left({e}^{\stackrel{¯}{B}}-1\right)\left({e}^{C}-1\right)+\left({e}^{B}-1\right)\left({e}^{\stackrel{¯}{C}}-1\right)\right]\right\}\\ \text{ }\cdot \left\{2{e}_{1}\left({e}^{\underset{_}{C}}-1\right)\left({e}^{C}-1\right)+\left({e}_{2}-{e}_{1}\right)\left[\left({e}^{\underset{_}{B}}-1\right)\left({e}^{C}-1\right)+\left({e}^{B}-1\right)\left({e}^{\underset{_}{C}}-1\right)\right]\right\}\\ \text{ }\cdot \underset{j=0}{\overset{n}{\sum }}{b}_{j}{\left[{e}_{1}\left({e}^{C}-1\right)+\left({e}_{2}-{e}_{1}\right)\left({e}^{B}-1\right)\right]}^{j}{\left({e}^{C}-1\right)}^{m-j}\\ =\left({e}^{\stackrel{¯}{C}}-1\right)\left({e}^{\underset{_}{C}}-1\right)\underset{i=0}{\overset{m}{\sum }}{a}_{i}{\left[{e}_{1}\left({e}^{C}-1\right)+\left({e}_{2}-{e}_{1}\right)\left({e}^{B}-1\right)\right]}^{i}{\left({e}^{C}-1\right)}^{m-i+2}\end{array}$ (3.18)

$\stackrel{¯}{B}=B\left(z+1\right)=B\left(z\right)+{s}_{1}\left(z\right)$$\underset{_}{B}=B\left(z-1\right)=B\left(z\right)+{s}_{2}\left(z\right)$$\stackrel{¯}{C}=C\left(z+1\right)=C\left(z\right)+{t}_{1}\left(z\right)$$\underset{_}{C}=C\left(z-1\right)=C\left(z\right)+{t}_{2}\left(z\right)$ 。其中 ${s}_{1}\left(z\right),{s}_{2}\left(z\right),{t}_{1}\left(z\right),{t}_{2}\left(z\right)$ 均为次数至多是 $d-1$ 的多项式，于是(3.18)整理可得

$\underset{\mu =0}{\overset{m}{\sum }}\underset{\lambda =0}{\overset{m+2}{\sum }}{M}_{\mu ,\lambda }{e}^{\mu B+\lambda C}+4{e}_{2}^{2}\underset{j=0}{\overset{n}{\sum }}{b}_{j}{e}_{2}{}^{j}-\underset{i=0}{\overset{m}{\sum }}{a}_{i}{e}_{2}^{i}=0,$ (3.19)

 (3.20)

$-\underset{\mu =0}{\overset{m}{\sum }}\underset{\lambda =0}{\overset{m+2}{\sum }}\frac{{M}_{\mu ,\lambda }{e}^{\mu B+\lambda C}}{\varphi \left(z,{e}_{2}\right)}\equiv 1,$ (3.21)

$\mathrm{deg}\left(\mu B+\lambda C\right)=d.$

$T\left(r,{e}^{\mu B+\lambda C}{e}^{-\mu A}\right)=T\left(r,{e}^{-\mu A}\right)+S\left(r,f\right)=\mu T\left(r,{e}^{A}\right)+S\left(r,f\right),$

$T\left(r,{e}^{\mu B+\lambda C}{e}^{-\mu A}\right)=T\left(r,{e}^{\left(\mu +\lambda \right)C}\right)+S\left(r,f\right)=\left(\mu +\lambda \right)T\left(r,{e}^{A}\right)+S\left(r,f\right),$

$N\left(r,\frac{1}{{f}_{k}^{\ast }}\right)=S\left(r,{f}_{l}^{\ast }\right),\stackrel{¯}{N}\left(r,{f}_{k}^{\ast }\right)=S\left(r,{f}_{l}^{\ast }\right),$

$\underset{k=1}{\overset{q}{\sum }}N\left(r,\frac{1}{{f}_{k}^{\ast }}\right)+\left(q-1\right)\underset{k=1}{\overset{q}{\sum }}\stackrel{¯}{N}\left(r,{f}_{k}^{\ast }\right)<\left(\tau +ο\left(1\right)\right)T\left(r,{f}_{l}^{\ast }\right),$

$-\frac{{M}_{\mu ,\lambda }}{\varphi \left(z,{e}_{2}\right)}{e}^{\mu B+\lambda C}\equiv 1.$

$\rho \left(-\frac{{M}_{\mu ,\lambda }}{\varphi \left(z,{e}_{2}\right)}\right)=\rho \left({e}^{\mu B+\lambda C}\right)$，矛盾，故假设不成立。

The Unicity of the Meromorphic Solutions of Painlevé IV Difference Equations[J]. 理论数学, 2018, 08(06): 596-603. https://doi.org/10.12677/PM.2018.86080

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