Some Results on Difference Riccati Equations and Delay Differential Equations

Wang Qiong,1, Long Fang,2, Wang Jun,3

 基金资助: 国家自然科学基金.  11771090国家自然科学基金.  11571049国家自然科学基金.  11271227国家自然科学基金.  11461070上海自然科学基金.  17ZR1402900教育部长江学者和创新团队发展计划项目.  IRT1264山东大学基本科研业务费专项资金.  2017JC019

 Fund supported: the NSFC.  11771090the NSFC.  11571049the NSFC.  11271227the NSFC.  11461070the Natural Science Foundation of Shanghai.  17ZR1402900the PCSIRT.  IRT1264the Fundamental Research Funds of Shandong University.  2017JC019

Abstract

We investigate difference Riccati equations with rational coefficients and delay differential equations with constant coefficients. For difference Riccati equations with some relation among coefficients, we prove that every transcendental meromorphic solution is of order no less than one. We also consider the rational solutions for delay differential equations.

Keywords： Difference Riccati equation ; Meromorphic solution ; Delay differential equation ; Rational solution

Wang Qiong, Long Fang, Wang Jun. Some Results on Difference Riccati Equations and Delay Differential Equations. Acta Mathematica Scientia[J], 2019, 39(4): 832-838 doi:

1 引言与主要结果

$$$\label{eq-1.4}\omega(z)[\omega(z+1)-\omega(z-1)]+a\omega'(z)=b\, \omega(z),$$$

$$$\label{eq-1.5} \omega(z+1)-\omega(z-1)+a(z)\frac{\omega' (z)}{\omega (z)}=\frac{P(z, \omega)}{Q(z, \omega)},$$$

$$$\label{eq-1.6} \begin{array}{l}P(\omega)=b_{p}\omega^{p}(z)+b_{p-1}\omega^{p-1}(z)+\cdots+b_{0}, \\ Q(\omega)=c_{q}\omega^{q}(z)+c_{q-1}\omega^{q-1}(z)+\cdots+c_{0}, \end{array}$$$

2 定理1.2的证明

$f(z+2)=\frac{\tilde{b}(z_0+1)+O(z-z_0)}{\tilde{b}(z_0)/\tilde{d}(z_0)+O(z-z_0)+\tilde{d}(z_0+1)+O(z-z_0)}\\~~~~~~~~~~~~~~=\frac{\tilde{b}(z_0+1)\tilde{d}(z_0)+O(z-z_0)}{\tilde{b}(z_0)+\tilde{d}(z_0)\tilde{d}(z_0+1)+O(z-z_0)}.$

$$$\begin{array}{ll} \omega'(z)&=o(1), \, \, \omega(z+1)=d_0+o(1), \, \, \omega(z-1)=d_0+o(1), \\ P(\omega)&=P(d_0)+o(1), \, \, Q(\omega)=Q(d_0)+o(1).\\ \end{array}$$$

$$$\label{eq-5.3} \omega(z)=\frac{F(z)}{G(z)}.$$$

$$$\label{eq-5.4} \begin{array}{ll} &Q(\omega)[(\overline{F}\underline{G}-\underline{F}\overline{G})GF+a(F'G-FG')\overline{G}\underline{G}]=P(\omega)\overline{G}\underline{G}GF, \end{array}$$$

$$$\label{eq-5.5} P(\omega)=\big(\sum \limits^{p}_{j=0} b_jF^{j}G^{p-j}\big)/G^{p}, \quad Q(\omega)=\big(\sum \limits^{q}_{j=0}c_jF^{j}G^{q-j}\big)/G^{q}.$$$

$$$\label{eq-5.7} C(z)(A(z)+B(z))=D(z),$$$

$a=0$,我们得到$\deg(C(A+B))\leq 3n+m-2+np+mk+n(q-k)$$\deg(D)=n(q+3)+m+ml+n(p-l)$.再次由(3.7)式,可知

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