﻿ 时标上一类二维动力系统的非振荡解的存在性 Existence of Nonoscillatory Solutions of Two-Dimensional Time Scale Systems

Vol. 08  No. 12 ( 2019 ), Article ID: 33381 , 8 pages
10.12677/AAM.2019.812226

Existence of Nonoscillatory Solutions of Two-Dimensional Time Scale Systems

Ting Li, Feng Chen

School of Mathematics and Information Science, Guangxi University, Nanning Guangxi

Received: Nov. 15th, 2019; accepted: Dec. 3rd, 2019; published: Dec. 10th, 2019

ABSTRACT

The study of time-scale theory combines the theory of differential equations with the theory of difference equations effectively, and the results obtained are more extensively used than those of differential equations and difference equations. In this paper, we employ Schauder’s fixed point theorem, Knaster’s fixed point theorem and double improper integrals to establish the existence of nonoscillatory solutions to the two-dimensional time scale system composed of first-order dynamic equations, and verify the main obtained results through some examples.

Keywords:Two-Dimensional System, Nonoscillatory Solutions, Time Scales, Existence

1. 引言

$\left\{\begin{array}{l}{x}^{\Delta }\left(t\right)=f\left(t,y\left(t\right)\right)\\ {y}^{\Delta }\left(t\right)=g\left(t,x\left(t\right)\right)\end{array}$ (1.1)

$\left\{\begin{array}{l}{x}^{\prime }\left(t\right)=a\left(t\right)f\left(y\left(t\right)\right)\\ {y}^{\prime }\left(t\right)=b\left(t\right)g\left(x\left(t\right)\right)\end{array}$ (1.2)

$\left\{\begin{array}{l}{x}^{\Delta }\left(t\right)=a\left(t\right)f\left(y\left(t\right)\right)\\ {y}^{\Delta }\left(t\right)=b\left(t\right)g\left(z\left(t\right)\right)\\ {z}^{\Delta }\left(t\right)=c\left(t\right)h\left(x\left(t\right)\right)\end{array}$ (1.3)

2. 几个引理

$\begin{array}{l}{S}^{+}:=\left\{\left(x,y\right)\in S:\underset{t\to \infty }{\mathrm{lim}}x\left(t\right)y\left(t\right)>0\right\}\\ {S}^{-}:=\left\{\left(x,y\right)\in S:\underset{t\to \infty }{\mathrm{lim}}x\left(t\right)y\left(t\right)<0\right\}\end{array}$

$\begin{array}{l}{S}_{B,B}^{+}:=\left\{\left(x,y\right)\in {S}^{+}:\underset{t\to \infty }{\mathrm{lim}}x\left(t\right)=c,\underset{t\to \infty }{\mathrm{lim}}y\left(t\right)=d\right\}\\ {S}_{B,\infty }^{+}:=\left\{\left(x,y\right)\in {S}^{+}:\underset{t\to \infty }{\mathrm{lim}}x\left(t\right)=c,\underset{t\to \infty }{\mathrm{lim}}y\left(t\right)=\infty \right\}\end{array}$

$\begin{array}{l}{S}_{\infty ,B}^{+}:=\left\{\left(x,y\right)\in {S}^{+}:\underset{t\to \infty }{\mathrm{lim}}x\left(t\right)=\infty ,\underset{t\to \infty }{\mathrm{lim}}y\left(t\right)=d\right\}\\ {S}_{\infty ,\infty }^{+}:=\left\{\left(x,y\right)\in {S}^{+}:\underset{t\to \infty }{\mathrm{lim}}x\left(t\right)=\infty ,\underset{t\to \infty }{\mathrm{lim}}y\left(t\right)=\infty \right\}\end{array}$

3. 主要结论

${\int }_{{t}_{0}}^{\infty }f\left(s,L+{\int }_{{t}_{0}}^{s}g\left(u,{K}_{1}\right)\Delta u\right)\Delta s<\infty$ (3.1)

${\int }_{{t}_{0}}^{\infty }g\left(s,{K}_{2}\right)\Delta s<\infty$ (3.2)

$g\left(t,u\right)$ 关于u的单调性和对(1.1)的第二个方程由 ${t}_{1}$ 到t进行积分，得

$y\left(t\right)=y\left({t}_{1}\right)+{\int }_{{t}_{1}}^{t}g\left(s,x\left(s\right)\right)\Delta s\ge y\left({t}_{1}\right)+{\int }_{{t}_{1}}^{t}g\left(s,\frac{c}{2}\right)\Delta s$ (3.3)

$t\to \infty$ 时，可得

${\int }_{{t}_{1}}^{\infty }g\left(s,\frac{c}{2}\right)\Delta s\le d-y\left({t}_{1}\right)<\infty$

$c\ge x\left(t\right)\ge x\left({t}_{1}\right)+{\int }_{{t}_{1}}^{t}f\left(s,y\left({t}_{1}\right)+{\int }_{{t}_{1}}^{s}g\left(u,\frac{c}{2}\right)\Delta u\right)\Delta s$

$t\to \infty$ 时，有

${\int }_{{t}_{1}}^{t}f\left(s,y\left({t}_{1}\right)+{\int }_{{t}_{1}}^{s}g\left(u,\frac{c}{2}\right)\Delta u\right)\Delta s\le c-x\left({t}_{1}\right)<\infty$

$Y=\left\{x\left(t\right)\in X:\frac{{K}_{1}}{2}\le x\left(t\right)\le {K}_{1},t\ge {t}_{1}\right\}$

$\left(Fx\right)\left(t\right)={K}_{1}-{\int }_{t}^{\infty }f\left(s,L+{\int }_{{t}_{1}}^{s}g\left(u,x\left(u\right)\right)\Delta u\right)\Delta s$

1) 先证F为自映射算子，

${K}_{1}\ge \left(Fx\right)\left(t\right)\ge {K}_{1}-{\int }_{t}^{\infty }f\left(s,L+{\int }_{{t}_{1}}^{s}g\left(u,{K}_{1}\right)\Delta u\right)\Delta s\ge \frac{{K}_{1}}{2}$

2) 再证F在Y上连续，取Y上的一组函数列 ${x}_{n}$，使得 ${x}_{n}\to x$$x\in Y$

$|\left(F{x}_{n}\right)\left(t\right)-\left(Fx\right)\left(t\right)|\le {\int }_{t}^{\infty }|f\left(s,L+{\int }_{{t}_{1}}^{s}g\left(u,{x}_{n}\left(u\right)\right)\Delta u\right)-f\left(s,L+{\int }_{{t}_{1}}^{s}g\left(u,x\left(u\right)\right)\Delta u\right)|\Delta s$

3) 最后证明FY相对紧致，由于

$0<{\left(Fx\right)}^{\Delta }\left(t\right)\le f\left(t,L+{\int }_{{t}_{1}}^{t}g\left(u,{K}_{1}\right)\Delta u\right)<\infty$

$\stackrel{¯}{x}\left(t\right)={K}_{1}-{\int }_{t}^{\infty }f\left(s,L+{\int }_{{t}_{1}}^{s}g\left(u,\stackrel{¯}{x}\left(u\right)\right)\Delta u\right)\Delta s$

${\stackrel{¯}{x}}^{\Delta }\left(t\right)=f\left(t,L+{\int }_{{t}_{1}}^{t}g\left(u,\stackrel{¯}{x}\left(u\right)\right)\Delta u\right)$

$\stackrel{¯}{y}\left(t\right)=L+{\int }_{{t}_{1}}^{t}g\left(u,\stackrel{¯}{x}\left(u\right)\right)\Delta u$

$t\to \infty$ 时，有 $\stackrel{¯}{x}\left(t\right)\to {K}_{1}$$\stackrel{¯}{y}\left(t\right)\to \beta$，其中 $0<{K}_{1},\beta <\infty$，故 $\left(\stackrel{¯}{x},\stackrel{¯}{y}\right)$ 是(1.1)的非振荡，因此 ${S}_{B,B}^{+}\ne \varnothing$，证毕。

${\int }_{{t}_{0}}^{\infty }f\left(s,{\int }_{{t}_{0}}^{s}g\left(u,{K}_{1}\right)\Delta u\right)\Delta s<\infty$ (3.4)

${\int }_{{t}_{0}}^{\infty }g\left(s,{K}_{2}\right)\Delta s=\infty$ (3.5)

$\frac{c}{2}\le x\left(t\right)=x\left({t}_{1}\right)+{\int }_{{t}_{1}}^{t}f\left(s,y\left(s\right)\right)\Delta s\le c$

$t\to \infty$ 时，有

${\int }_{{t}_{1}}^{\infty }f\left(s,y\left(s\right)\right)\Delta s\le c-x\left({t}_{1}\right)<\infty$ (3.6)

$g\left(t,u\right)$ 关于u的单调性和对(1.1)的第二个方程由 ${t}_{1}$ 到t进行积分，得

$y\left({t}_{1}\right)+{\int }_{{t}_{1}}^{t}g\left(s,c\right)\Delta s\ge y\left(t\right)=y\left({t}_{1}\right)+{\int }_{{t}_{1}}^{t}g\left(s,x\left(s\right)\right)\Delta s$

$t\to \infty$ 时，有

${\int }_{{t}_{1}}^{\infty }g\left(s,c\right)\Delta s=\infty$ (3.7)

${\int }_{{t}_{1}}^{\infty }f\left(s,{\int }_{{t}_{1}}^{s}g\left(u,\frac{c}{2}\right)\Delta u\right)\Delta s\le {\int }_{{t}_{1}}^{\infty }f\left(s,y\left(s\right)\right)\Delta s<\infty$

$\Omega =\left\{x\left(t\right)\in X:K\le x\left(t\right)\le 2K,t\ge {t}_{1}\right\}$

$\left(Fx\right)\left(t\right)=K+{\int }_{{t}_{1}}^{t}f\left(s,{\int }_{{t}_{1}}^{s}g\left(u,x\left(u\right)\right)\Delta u\right)\Delta s$

$K\le \left(Fx\right)\left(t\right)\le K+{\int }_{{t}_{1}}^{t}f\left(s,{\int }_{{t}_{1}}^{s}g\left(u,2K\right)\Delta u\right)\Delta s\le 2K$

$\stackrel{¯}{x}\left(t\right)=K+{\int }_{{t}_{1}}^{t}f\left(s,{\int }_{{t}_{1}}^{s}g\left(u,\stackrel{¯}{x}\left(u\right)\right)\Delta u\right)\Delta s$

${\stackrel{¯}{x}}^{\Delta }\left(t\right)=f\left(t,{\int }_{{t}_{1}}^{t}g\left(u,\stackrel{¯}{x}\left(u\right)\right)\Delta u\right)$

$\stackrel{¯}{y}\left(t\right)={\int }_{{t}_{1}}^{t}g\left(u,\stackrel{¯}{x}\left(u\right)\right)\Delta u$

$t\to \infty$ 时，有 $\stackrel{¯}{x}\left(t\right)\to \alpha$$\stackrel{¯}{y}\left(t\right)\to \infty$，其中 $0<\alpha <\infty$，故 $\left(\stackrel{¯}{x},\stackrel{¯}{y}\right)$ 是(1.1)的非振荡解，因此 ${S}_{B,\infty }^{+}\ne \varnothing$，证毕。

${\int }_{{t}_{0}}^{\infty }g\left(s,{\int }_{{t}_{0}}^{s}f\left(u,{K}_{1}\right)\Delta u\right)\Delta s<\infty$

${\int }_{{t}_{0}}^{\infty }f\left(s,{K}_{2}\right)\Delta s=\infty$

4. 例子

$\left\{\begin{array}{l}{x}^{\Delta }\left(t\right)=\frac{q+1}{{q}^{2}{t}^{3}}{e}_{\frac{ty\left(t\right)-t+1}{\left(q-1\right){t}^{2}+\left(2-q\right)t}}\left(t,1\right),\\ {y}^{\Delta }\left(t\right)=\frac{1}{q\left(2{t}^{2}-1\right)}x\left(t\right).\end{array}$ (4.1)

$\begin{array}{c}{\int }_{1}^{\infty }g\left(t,K\right)\Delta t={\int }_{1}^{\infty }\frac{K}{q\left(2{t}^{2}-1\right)}\Delta t\\ =\frac{K}{q}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{2{t}^{2}-1}\left(q-1\right)t\le \text{}\frac{\left(q-1\right)K}{q}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{t}\text{}\\ =\frac{\left(q-1\right)K}{q}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{{q}^{n}}=K<\infty \end{array}$

$\begin{array}{l}{\int }_{1}^{\infty }\frac{q+1}{{q}^{2}{t}^{3}}f\left(t,\frac{1}{2}+{\int }_{1}^{t}\frac{1}{q\left(2{t}^{2}-1\right)}g\left(u,\frac{1}{2}\right)\Delta u\right)\Delta t\le {\int }_{1}^{\infty }\frac{q+1}{{q}^{2}{t}^{3}}f\left(t,1\right)\Delta t\\ ={\int }_{1}^{\infty }\frac{q+1}{{q}^{2}{t}^{3}}{e}_{\frac{1}{t+\left(q-1\right)\left(t-1\right)t}}\left(t,1\right)\Delta t\\ =\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{q+1}{{q}^{2}{t}^{3}}{e}_{\frac{1}{t+\left(q-1\right)\left(t-1\right)t}}\left(t,1\right)\left(q-1\right)t\\ =\frac{{q}^{2}-1}{{q}^{2}}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{{t}^{2}}\mathrm{exp}\left\{\frac{1}{\left(q-1\right)s}{\int }_{1}^{t}\mathrm{ln}\left(1+\left(q-1\right)s\right)\frac{\frac{1}{s}}{1+\left(q-1\right)\left(s-1\right)s}\Delta s\right\}\end{array}$

$\begin{array}{l}<\frac{{q}^{2}-1}{{q}^{2}}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{{t}^{2}}\mathrm{exp}\left\{\mathrm{ln}q{\int }_{1}^{t}\frac{1}{\left(q-1\right)s}\Delta s\right\}\\ =\frac{{q}^{2}-1}{{q}^{2}}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{{t}^{2}}\mathrm{exp}\left\{\mathrm{ln}q\underset{m=0}{\overset{n}{\sum }}1\right\}\\ =\frac{{q}^{2}-1}{{q}^{2}}\underset{t\in {\left[1,\infty \right)}_{{q}^{{ℕ}_{0}}}}{\sum }\frac{1}{t}\\ =\frac{{q}^{2}-1}{{q}^{2}}\underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{q}^{n}}=\frac{q+1}{q}<\infty \end{array}$

$\left\{\begin{array}{l}{x}^{\Delta }\left(t\right)=\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{t\left({t}^{2}+1\right)\left[{\left(\sqrt{t}+1\right)}^{4}+1\right]}y\left(t\right),\\ {y}^{\Delta }\left(t\right)={e}_{\frac{\left(x\left(t\right)-1\right)\left({t}^{2}+1\right)+1}{{t}^{2}+1+{t}^{2}\left(1+2\sqrt{t}\right)}}\left(t,1\right).\end{array}$ (4.2)

$\begin{array}{c}{\int }_{1}^{\infty }g\left(t,1\right)\Delta t={\int }_{1}^{\infty }\frac{{e}_{1}\left(t,1\right)}{{e}_{1-\frac{1}{{t}^{2}+1}}\left(t,1\right)}\Delta t\\ ={\int }_{1}^{\infty }{e}_{\frac{\frac{1}{{t}^{2}+1}}{1+\left(1+2\sqrt{t}\right)\left(1-\frac{1}{{t}^{2}+1}\right)}}\left(t,1\right)\Delta t\\ \ge \text{}{\int }_{1}^{\infty }\mathrm{exp}\left\{{\int }_{1}^{t}\frac{1}{1+2\sqrt{s}}\mathrm{ln}\left(1\right)\Delta s\right\}\Delta t=\infty \end{array}$

$\begin{array}{c}{\int }_{1}^{\infty }f\left(t,{\int }_{1}^{t}g\left(u,1\right)\Delta u\right)\Delta t\\ ={\int }_{1}^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{t\left({t}^{2}+1\right)\left[{\left(\sqrt{t}+1\right)}^{4}+1\right]}\left({\int }_{1}^{t}\frac{{e}_{1}\left(u,1\right)}{{e}_{1-\frac{1}{{u}^{2}+1}}\left(u,1\right)}\Delta u\right)\Delta t\\ ={\int }_{1}^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{t\left({t}^{2}+1\right)\left[{\left(\sqrt{t}+1\right)}^{4}+1\right]}\left({\int }_{1}^{t}{e}_{\frac{1}{{u}^{2}+1+{u}^{2}\left(1+2\sqrt{u}\right)}}\left(u,1\right)\Delta u\right)\Delta t\\ ={\int }_{1}^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{t\left({t}^{2}+1\right)\left[{\left(\sqrt{t}+1\right)}^{4}+1\right]}\left({\int }_{1}^{t}\mathrm{exp}\left\{{\int }_{1}^{u}\frac{1}{1+2\sqrt{s}}\mathrm{ln}\left(2+2\sqrt{s}\right)\frac{1}{{s}^{2}+1+{s}^{2}\left(1+2s\right)}\Delta s\right\}\Delta u\right)\Delta t\end{array}$

$\begin{array}{c}\le {\int }_{1}^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{t\left({t}^{2}+1\right)\left[{\left(\sqrt{t}+1\right)}^{4}+1\right]}\left({\int }_{1}^{t}u\Delta u\right)\Delta t\\ \le {\int }_{1}^{\infty }\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{t\left({t}^{2}+1\right)\left[{\left(\sqrt{t}+1\right)}^{4}+1\right]}{t}^{2}\Delta t\\ \le {\int }_{1}^{\infty }\frac{1}{t{\left(\sqrt{t}+1\right)}^{2}}\frac{t+{\left(\sqrt{t}+1\right)}^{2}}{{\left(\sqrt{t}+1\right)}^{2}}\Delta t\\ \le {\int }_{1}^{\infty }\frac{2}{t\sigma \left(t\right)}\Delta t=2<\infty \end{array}$

Existence of Nonoscillatory Solutions of Two-Dimensional Time Scale Systems[J]. 应用数学进展, 2019, 08(12): 1971-1978. https://doi.org/10.12677/AAM.2019.812226

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