## Oscillation Analysis of a Kind of Systems with Piecewise Continuous Arguments

Liu Ying, Gao Jianfang,

School of Mathematical Sciences, Harbin Normal University, Harbin 150025

 基金资助: 国家自然科学基金.  12001143哈尔滨师范大学学术创新项目.  HSDSSCX2020-33

 Fund supported: the NSFC.  12001143the Academic Innovation Project of Harbin Normal University.  HSDSSCX2020-33

Abstract

In this paper, we mainly use $\theta$-method to analyze the oscillation of differential equations with piecewise continuous arguments of retarded type, and discuss the oscillation and non-oscillation of analytic solution and numerical solution. The sufficient conditions for the numerical methods to preserve the oscillation of the equation under the condition of the analytic solution oscillation are obtained. Meanwhile, some numerical experiments are given.

Keywords： Piecewise continuous arguments ; Delay differential system ; Numerical solutions ; Oscillation ; θ-methods

Liu Ying, Gao Jianfang. Oscillation Analysis of a Kind of Systems with Piecewise Continuous Arguments. Acta Mathematica Scientia[J], 2022, 42(3): 826-838 doi:

## 1 引言

$$$Y(n+1)+(QP^{-1}-e^P-QP^{-1}e^P)Y(n)=0.$$$

$$$\lambda^{2}-(e^{a}+e^{c})\lambda+e^{a+c}-\frac {bd}{ac}(e^{a}-1)(e^{c}-1)=0.$$$

$\rm (1)$ 方程(2.1)的每个解振动;

$\rm (2)$ 方程(2.2)的每个解振动;

$\rm (3)$ 特征方程(2.3)无正根.

由定理$2.2$, 方程(2.1)存在一个非振动解的充分必要条件是其特征方程(2.3)

$a>0$时, 由于$h<h_3=\frac{1}{\theta a}$, 有$1+\frac{ha}{1-h\theta a}>1$.又由于$b\geq 0$, 所以有

$a<0$时, 有$1+\frac{ha}{1-h\theta a}<1$.又由于$b\geq 0$, 所以有

$Y_{km+l+1}>0$.同理可得, 如果$Y_{km}<0$, 那么$Y_{km+l+1}<0$.定理3.1得证.

$\lambda_1>0$, $\lambda_2\leq0$时, 根据根与系数的关系可得

$\rm (2)$ 方程(3.3)存在一个非振动解当且仅当$bd\geq C(m)$.

## 4 方程振动性的保持性

$p-1 \leq {e^{a}-1}, q-1 \leq {e^{c}-1},$

$(2)$ 同理可证.当$-1<z<0$时, $f(z)$单调递减, 即

$(3) $$a>0, c<0, e^{a}+e^{c}<2, p\geq e^{a}, q\geq e^{c}, q<1, p>1, p+q<2 时, 有 p-1 \geq {e^{a}-1}, q-1 \geq {e^{c}-1}, 结论得证. (4) 同理可证.当 0<z<1 时, f(z) 单调递减, 即 则有 ac>0, p>1, q>1, 或者 ac>0, p<1, q<1, 要证 只需证 z(a, c)\leq z(p, q) , 即 (5)$$ c>a>0, p\geq e^{a}, q\leq e^{c}, q>p>1$时,

$p-1 \geq {e^{a}-1}, q-1 \leq {e^{c}-1},$

$(6)$ 同理可证.当$z>1$时, $f(z)$单调递增, 即

$(7) $$a>c>0, p\leq e^{a}, q\geq e^{c}, p>q>1 时, 有 p-1 \leq {e^{a}-1}, q-1 \geq {e^{c}-1}, 结论得证. (8) 同理可证. 引理4.2[26] 对任意的 m>|a| , 有 \rm (1)$$ a>0$,

$\rm (2)$$a<0$,

## 5 数值算例

$$$\left\{\begin{array}{l} Y'(t)={\left(\begin{array}{cc} 2&0\\ 0&-1 \end{array}\right)}Y(t)+{\left(\begin{array}{cc} 0&1\\ 1&0 \end{array}\right)}Y([t]), \qquad t\geq0, \\ Y(0)={\left(\begin{array}{cc} 1\\ 1 \end{array}\right)}, \end{array} \right.$$$

### 图 1

$$$\left\{\begin{array}{l} Y'(t)={\left(\begin{array}{cc} 3&0\\ 0&3 \end{array}\right)}Y(t)+{\left(\begin{array}{cc} 0&-2\\ -4&0 \end{array}\right)}Y([t]), \qquad t\geq0, \\ Y(0)={\left(\begin{array}{cc} 1\\ 1 \end{array}\right)}, \end{array} \right.$$$

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