数学物理学报, 2019, 39(3): 560-569 doi:

论文

一类变延迟中立型微分方程梯形方法的渐近估计

张根根,1, 王晚生3, 肖爱国2

Asymptotic Estimation of the Trapezoidal Method for a Class of Neutral Differential Equation with Variable Delay

Zhang Gengen,1, Wang Wansheng3, Xiao Aiguo2

通讯作者: 张根根, E-mail: zhanggen036@163.com

收稿日期: 2018-02-1  

基金资助: 国家自然科学基金.  11701110

Received: 2018-02-1  

Fund supported: the NSFC.  11701110

摘要

该文研究了一类变延迟中立型微分方程梯形方法的稳定性,并借助于一个泛函不等式得到了数值解的渐近估计.此渐近估计对数值解的性态不仅比数值渐近稳定性描述得更加精确,而且能给出非稳定情形数值解的上界估计式.

关键词: 中立型延迟微分方程 ; 梯形方法 ; 渐近估计 ; 渐近稳定性

Abstract

In this paper, we investigate the stability of the trapezoidal method for a class of neutral differential equation with variable delay and obtain the asymptotic estimation of numerical solution with the aid of a functional inequality. The asymptotic estimation is more accurate than asymptotic stability in describing the behaviours of the numerical solution, and gives the upper bound estimates of the numerical solution for the nonstable case.

Keywords: Neutral delay differential equation ; Trapezoidal method ; Asymptotic estimation ; Asymptotic stability

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本文引用格式

张根根, 王晚生, 肖爱国. 一类变延迟中立型微分方程梯形方法的渐近估计. 数学物理学报[J], 2019, 39(3): 560-569 doi:

Zhang Gengen, Wang Wansheng, Xiao Aiguo. Asymptotic Estimation of the Trapezoidal Method for a Class of Neutral Differential Equation with Variable Delay. Acta Mathematica Scientia[J], 2019, 39(3): 560-569 doi:

1 引言

中立型延迟微分系统常出现在生态学、物理学、自动控制等科学与工程领域中[1, 7],其特征是系统右端函数不仅依赖于过去时间的状态,而且还依赖于过去时间状态的变化率.只有极少数中立型延迟微分方程能够获得解析解表达式,其数值分析文献主要研究了线性问题[1, 3, 6]和一些非线性问题[1, 10-12]的稳定性以及基于Lipschitz条件的非线性问题数值解的收敛性.

目前,仅有少量文献研究方程解的渐近估计.鉴于渐近估计不仅比渐近稳定性描述解的性态更加精确,而且能给出非稳定情形解的上界估计式. Iserles在文献[6]中构造了自治中立型比例延迟微分系统Dirichlet级数形式的解,获得了解析解的上界估计式及渐近稳定的充分条件. Liu等[8-9]考虑了非自治中立型比例延迟微分系统,获得了解析解的渐近估计. Čermák等[2]考虑了用梯形方法求解非自治非中立型变延迟微分方程,借助于一个泛函不等式获得了数值解的上界估计式及其渐近稳定的一个充分条件. Zhang等[13]考虑了用$ \theta $ -方法求解非自治中立型比例延迟微分方程,获得了数值解的上界估计式及其渐近稳定的一个充分条件.

本文在文献[2, 13]的基础上,研究非自治变延迟中立型微分方程梯形方法的渐近估计,获得了数值解的上界估计式及其渐近稳定的一个充分条件,并用数值算例验证了所获结果.因为变延迟中立型微分方程更一般化且更难分析处理,故这样的推广是很有必要的.

2 离散格式

考虑一类变延迟中立型微分方程

$ \begin{equation} y'(t ) = a(t)y(t)+b(t)y(\theta(t))+ c(t)y^\prime(\phi(t)), \ t\geq t_0, \end{equation} $

其中$ a(t), b(t), c(t) $是区间$ [t_0, \infty) $上的非零复值连续函数, $ \theta(t), \phi(t) $是两个可微函数且严格单调递增,进一步, $ \theta(t_0) = t_0, \phi(t_0) = t_0 $,对任意的$ t > t_0 $满足$ \theta(t) < t, \phi(t) < t $.$ \psi(t) = \theta^{-1}(t) $, $ \xi(t) = \phi^{-1}(t) $.对方程(2.1)两边同时积分得

$ \begin{equation} y(t_{n+1}) -y(t_{n}) = \int\nolimits_{t_{n}}^{t_{n+1}} a(s)y(s){\rm d}s + \int\nolimits_{t_{n}}^{t_{n+1}} b(s)y(\theta(s)){\rm d}s + \int\nolimits_{t_{n}}^{t_{n+1}} c(s)y^\prime(\phi(s)){\rm d}s, \end{equation} $

其中$ t_n = t_0 + nh $$ h > 0 $为给定步长.记$ u_n : = u(t_n) $$ (u = y, a, b, c, \theta, \phi) $, $ \overline{ v}_n : = (v_n - t_0)/h $$ (v = \theta, \phi) $, $ w(x) = c'(\xi(x)) (\xi'(x))^2 + c(\xi(x)) \xi''(x) $,以及$ w_{\phi_{n }} : = w(\phi_{n }) $.用梯形方法对(2.2)式右端积分进行离散得

$ \begin{eqnarray} \int\nolimits_{t_{n}}^{t_{n+1}} a(s)y(s){\rm d}s &\approx &\frac{1}{2} h (a_ny_n +a_{n+1}y_{n+1}), \\ \int\nolimits_{t_{n}}^{t_{n+1}} b(s)y(\theta(s)){\rm d}s & = &\int\nolimits_{\theta_n}^{\theta_{n+1}} b(\psi(x)) y(x) \psi'(x){\rm d}x \\ & = &\frac{1}{2} (\theta_{n+1}-\theta_n ) ( b_n y^h(\theta_n) \psi'(\theta_n) + b_{n+1} y^h(\theta_{n+1}) \psi'(\theta_{n+1})) \\ &\approx &h b_n ( \beta_n y_{[\overline{ \theta }_n]} +\alpha_n y_{[\overline{ \theta }_n]+1} ) , \\ \int\nolimits_{t_{n}}^{t_{n+1}} c(s)y'(\phi(s)){\rm d}s & = & c(\xi(x))y(x)\xi'(x)|_{\phi_n}^{\phi_{n+1}} -\int\nolimits_{\phi_n}^{\phi_{n+1}} w(x) y(x) {\rm d}x \\ & = &c_{n+1}y^h(\phi_{n+1})\xi'(\phi_{n+1}) - c_n y^h(\phi_n)\xi'(\phi_n)\\ && -\frac{1}{2} (\phi_{n+1} - \phi_n) ( w_{\phi_{n+1 }} y^h(\phi_{n+1}) + w_{\phi_{n }} y^h(\phi_n) ) \\ &\approx & c_n ( \hat{\alpha_n} y_{[\overline{ \phi }_n]+1} + \hat{\beta_n} y_{[\overline{ \phi }_n]}), \end{eqnarray} $

其中, $ [\cdot] $为取整函数, $ y^h(v_{n+1}) $, $ y^h(v_n) $$ y_{[\overline{v}_n]} $$ y_{[\overline{v}_n]+1} $$ (v = \theta, \phi) $的线性组合,即

综上所述,由方程(2.2)和(2.3)得到如下递推关系

$ \begin{equation} y_{n+1} = R_n y_n + h S_n(\beta _n y_{[\overline{ \theta }_n]} + \alpha_n y_{[\overline{ \theta }_n]+1}) + T_n (\hat{\beta_n} y_{[\overline{ \phi }_n]} + \hat{\alpha_n} y_{[\overline{ \phi }_n]+1}) , \end{equation} $

其中

$ \begin{equation} R_n = \frac{2+h a_n}{2-h a_{n+1}}, \quad S_n = \frac{2b_n}{2-h a_{n+1}}, \quad T_n = \frac{2c_n }{2-h a_{n+1}} . \end{equation} $

3 渐近估计

本节将构造方程(2.4)的解$ y_n $的渐近估计.考虑不等式

$ \begin{equation} |S_n |h (|\beta_n| \varrho_{[\overline{ \theta }_n]} + |\alpha_n | \varrho_{[\overline{ \theta }_n]+1}) + | T_n | (|\hat{\beta_n}| \varrho_{[\overline{ \phi }_n]} + |\hat{\alpha_n}| \varrho_{[\overline{ \phi }_n]+1}) \leq (1-|R_n|)\varrho_n, \end{equation} $

并假设

$ \begin{equation} \begin{array}{lll} \tilde{S}: = \sup\limits_{n\in Z^+}(\left| S_n\right|) < \infty, \; \; \tilde{R}: = \sup\limits_{n\in Z^+}{\left| R_n\right|}< 1, \; \; \tilde{T}: = \sup\limits_{n\in Z^+}(\left| T_n\right|) < \infty, \\ \eta: = \sup\limits_{n\in Z^+}(| \alpha_n|+| \beta_n|) <\infty, \quad \tilde{\eta}: = \sup\limits_{n\in Z^+}(| \hat{\alpha_n}|+|\hat{\beta_n} |)<\infty , \end{array} \end{equation} $

$ \begin{equation} \tilde{\gamma}: = \frac{h\tilde{S}\eta +\tilde{T} \tilde{\eta}}{1-\tilde{R}}. \end{equation} $

不等式(3.1)在本文中起着极其重要的作用,只须找到(3.1)式的一个显式解,即可构造出方程(2.4)解$ y_n $的渐近估计.为此考虑辅助方程(参见文献[2, 4])

$ \begin{equation} \varphi(\tau(t)) = \nu \varphi(t), \quad \nu = \tau'(t_0), \ t\geq t_0. \end{equation} $

在区间$ [t_0, \infty) $上,当$ \tau(t) \in C^2([t_0, \infty)) $, $ \tau(t_0) = t_0 $, $ \tau(t) < t $时, $ \tau'(t) > 0 $$ \tau'(t_0) < 1 $,则方程(3.4)的解$ \varphi $存在唯一、连续可微、严格递增并满足$ \varphi'(t_0) = 1 $,且

$ \begin{equation} \varphi(t) = \lim\limits_{n\rightarrow \infty } \nu^{-n} ( \tau^n(t)- t_0 ), \quad \nu = \tau'(t_0), \ t\geq t_0, \end{equation} $

其中$ \tau^n(t) $$ \tau(t) $的第$ n $步迭代.利用这些性质可以得到下述引理.

引理3.1[2]  设$ \tau(t) \in C^2([t_0, \infty)) $,对任意$ t > t_0 $, $ \tau(t_0) = t_0 $, $ \tau(t) < t $, $ \tau'(t) > 0 $以及$ \tau'(t) $$ [t_0, \infty) $上递减且$ \tau'(t_0) < 1 $,则由(3.5)式定义的函数$ \varphi(t) $是方程(3.4)的解, $ \varphi'(t) $非负连续且在区间$ [t_0, \infty) $上单调递减,并满足$ \frac{\varphi'(t)}{\varphi(t)} \leq \frac{1}{t-t_0}. $

定义

$ \begin{equation} \tau(t) = \left\{ \begin{array}{ll} \max(\theta(t), \phi(t)), \quad &{\hbox{当}}\; \; \tilde{\gamma} \geq 1 , \\ \min (\theta(t), \phi(t)), &{\hbox{当}}\; \; 0< \tilde{\gamma}< 1, \end{array}\right. \end{equation} $

显然, $ \tau(t) $满足引理3.1中的所有假设条件.

定理3.1 假设(3.2)式成立, $ \tilde{\gamma} $$ \varphi(t) $分别由(3.3)与(3.5)式给定, $ t^\star \geq t_0 $是方程$ t- \tau(t) = h $的唯一实根且$ k^\star = [(t^\star -t_0)/h] +1, $则序列

$ \begin{equation} \varrho_n = \left\{ \begin{array}{ll} (\varphi(t_0 +(n-k^\star )h) )^{-\log_{\nu} \tilde{\gamma}}, \quad &{\hbox{当}} \; \; \tilde{\gamma} \geq1, \\ (\varphi(t_0 +(n+k^\star )h) )^{-\log_{\nu} \tilde{\gamma}}, &{\hbox{当}} \; \; 0< \tilde{\gamma}< 1, \end{array} \right. \end{equation} $

定义了方程(3.1)的一个正解,即$ \varrho_n > 0 $且满足方程(3.1).

 令$ \overline{ \tau }_n = (\tau_n- t_0)/h $.$ \tilde{\gamma}\geq 1 $时, $ \varrho_n $是一个递增序列,则利用(3.4), (3.6)和(3.7)式得

即对所有的$ n \geq k^\star $,当$ \tilde{\gamma}\geq 1 $时,不等式(3.1)成立.类似可证得当$ 0 < \tilde{\gamma} < 1 $时,对所有的$ n\in N^+ $,不等式(3.1)成立.

至此,我们显式给出了不等式(3.1)的一个解序列,进而能得到方程(2.4)解$ y_n $的渐近估计.

定理3.2 设$ y_n $是方程(2.4)的一个解,如果(3.2)式成立, $ \tilde{\gamma} $$ \varphi(t) $分别由(3.3)与(3.5)式给定且$ \nu = \tau'(t_0) $,则

$ \begin{equation} y_n = O\left((\varphi(n))^{-\log_{\nu}\tilde{\gamma}}\right), \quad {\hbox{当}}\ n\rightarrow \infty . \end{equation} $

特别地,当$ 0 < \tilde{\gamma} < 1 $时,其解渐近稳定.

 首先,对方程(2.4)作变换$ z_n = y_n/\varrho_n $, $ \varrho_n $由(3.7)式给定,则

$ \begin{eqnarray} z_{n+1}\varrho_{n+1} & = & R_n z_n \varrho_n + h S_n(\beta _n z_{[\overline{ \theta }_n]}\varrho_{[\overline{ \theta }_n]} + \alpha_n z_{[\overline{ \theta }_n]+1}\varrho_{[\overline{ \theta }_n]+1}) \\ &&+ T_n (\hat{\beta_n} z_{[\overline{ \phi }_n]}\varrho_{[\overline{ \phi }_n]} + \hat{\alpha_n} z_{[\overline{ \phi }_n]+1}\varrho_{[\overline{ \phi }_n]+1}). \end{eqnarray} $

$ n\rightarrow \infty $时,我们需要证明方程(3.9)的任意解$ z_n $均有界.取$ \sigma_0 > \max \left\{\frac{1+\nu }{1-\nu }, \frac{ \tau^{-1}(t_0+k^\star h)- t_0 }{h}\right\} $, $ \sigma_0 \in Z^+ $且定义区间$ I_0: = [[\overline{ \tau }_{\sigma_0}] $, $ \sigma_0]\cap Z^+ $,及$ \sigma_{m+1} : = [\frac{ \tau^{-1}(t_0+(\sigma_m -1)h)-t_0 }{h}] $,这里$ m = 0, 1, \cdots $, $ I_{m+1}: = [\sigma_m, \sigma_{m+1}]\cap Z^+ $,记

考虑任意的$ n^\star \in I_{m+1} $, $ n^\star > \sigma _m $,我们希望通过$ z_k $来估计$ z_{n^\star} $,其中$ k \in \bigcup\limits_{j = 0}^m I_j $,分三种情况进行讨论:

(ⅰ)假设$ R_{n^{\star}-1} = 0 $,则

对上式两边取绝对值且由不等式(3.1)估计$ |z_{n^\star} | $

$ \tilde{\gamma} \geq 1 $时, $ \varrho_n $是一个递增序列,故$ \left|z_{n^\star}\right| \leq B_m \frac{\varrho_{n^\star-1}}{\varrho_{n^\star}} \leq B_m $.$ 0 < \tilde{\gamma} < 1 $时, $ \varrho_n $是一个递减序列,利用中值定理,二项式定理及定理3.2,得

其中$ K_1 > 0 $为正常数.

(ⅱ)假设对任意的$ n \in [\sigma_m, n^\star-1] \cap Z^+ $满足$ R_n \neq 0 $,则(3.9)式两边同乘$ \prod\limits_{l = \sigma_m}^{n}\frac{1}{R_l} $

其中$ \triangle $是向前差分.两边从$ \sigma_m $$ n^\star-1 $求和得

对上式两边取绝对值得

$ \begin{equation} \left|z_{n^{\star}} \right|\leq B_m \left( \frac{\varrho_{\sigma_m}}{\varrho_{n^{\star}}} \prod\limits_{l = \sigma_m}^{n^{\star}-1}{\left| R_l \right|} + \frac{1}{\varrho_{n^{\star}}} \sum\limits_{i = \sigma_m}^{n^{\star}-1}(1-| R_i|)\varrho_i \prod\limits_{l = i+1}^{n^{\star}-1}{| R_l |} \right). \end{equation} $

又因为

所以把上式代入(3.10)式且部分求和得

$ \tilde{\gamma} \geq 1 $时, $ \varrho_n $是一个递增序列且$ \triangle \varrho_i\geq 0 $,故$ \left|z_{n^{\star}} \right|\leq B_m $.$ 0 < \tilde{\gamma} < 1 $时, $ \varrho_n $是一个递减序列且$ \triangle \varrho_i\leq 0 $, $ \triangle \varrho_i\geq \triangle \varrho_{i-1} $,利用中值定理及$ \varphi'(t) $单调递减性得

其中$ \overline{K}_2 $是一个适当的正实数(由(3.4)式可知),再利用不等式$ \frac{-\triangle \varrho_i}{1-\left|R_i \right|} \leq \frac{-\triangle \varrho_{\sigma_m}}{1- \tilde{R} } $

其中$ K_2 $的一个正常数.

(ⅲ)假设$ R_{n^{\star}-1} \neq 0 $且存在$ k \in [\sigma_m, n^\star-2] \cap Z^+ $使得$ R_k = 0 $.定义$ \sigma^\star : = \sup(k, k \in [\sigma_m, n^{\star}-2] \cap Z^+ $$ R_n = 0 $),则

对方程(3.9)两边同乘$ \prod\limits_{l = \sigma^{\star}+1}^{n}\frac{1}{R_l} $,且从$ \sigma^{\star}+1 $$ n^{\star}-1 $求和得

$ \sigma^\star $的定义知$ R_{\sigma^\star} = 0 $,则由情况(i)结论得到估计式

又由(3.1)式得

上式即为对不等式(3.10)右端作相应修改(用$ \sigma^\star+1 $代替$ \sigma_m $).采用情况(ⅱ)相同技巧,得

其中$ K_3 $是一个正常数.

综上三种情况(ⅰ), (ⅱ), (ⅲ),对任意的$ n^{\star} \in I_{m+1} $, $ n^{\star} > \sigma_m $

因此$ B_{m+1}\leq B_m (1+ O(1/\sigma_m)) $.$ m \rightarrow \infty $时,它保证了$ \prod\limits_{j = 1}^m(1+ 1/\sigma_j) $是收敛的.再由引理3.1可得

$ \sigma_m \geq \nu^{-m}(\sigma_0- \frac{1+\nu}{1-\nu}) $及相应的无穷积收敛.所以当$ m \rightarrow \infty $时,序列$ {B_m} $一致有界,即(3.8)式成立.

注3.1 当$ \theta(t) = \phi(t) $时, (2.4)式可以改写为

$ \begin{equation} y_{n+1} = R_n y_n + (h S_n \beta_n + T_n \hat{\beta_n}) y_{\lfloor\overline{ \theta }_n\rfloor} + (h S_n \alpha_n + T_n \hat{\alpha_n}) y_{\lfloor\overline{ \theta }_n\rfloor+1}), \end{equation} $

考虑不等式

并假设

$ \begin{equation} \begin{array}{lll} \tilde{R}: = \sup\limits_{n\in Z^+}{\left| R_n\right|}< 1, \; \; S_{\alpha}: = \sup\limits_{n\in Z^+}(| h S_n \beta_n + T_n \hat{\beta_n} |) < \infty, \\ S_{\beta}: = \sup\limits_{n\in Z^+}(| h S_n \alpha_n + T_n \hat{\alpha_n}|) < \infty, \end{array} \end{equation} $

$ \begin{equation} \gamma: = \frac{ S_{\alpha} +S_{\beta}}{1-\tilde{R}}, \end{equation} $

则可以得到方程(3.11)解的渐近估计

$ \begin{equation} y_n = O\left((\varphi(n))^{-\log_{\nu}\gamma}\right), \quad {\hbox{当}}\ n\rightarrow \infty . \end{equation} $

特别地,当$ 0 < \gamma < 1 $时,其解渐近稳定.

注3.2 当$ \theta(t) = pt, \phi(t) = \lambda t $均取比例延迟时,辅助方程(3.4)变为

引理3.1与定理3.1的所有假设条件显然成立.此时,定理3.2与注3.1均是文献[13]中相关结论的直接推广,所以变延迟项$ \theta(t), \phi(t) $更具一般性.

例3.1 考虑中立型延迟微分方程

其中$ 0 < \varepsilon < 1 $, $ a(t), b(t), c(t) $均是区间$ [1, \infty) $上的非零复值连续函数.令$ t_{n } = 1+ nh $, $ \overline{\varepsilon}_n = \frac{ t_n^\varepsilon -1}{h} $, $ w_{\varepsilon_{n }} = c'_n (t_{n }^{1-\varepsilon})^2 + c_n (1-\varepsilon) t_{n }^{1-2\varepsilon} $,则(2.4)式可以改写为

$ \begin{equation} y_{n+1} = R_n y_n + (h S_n \beta_n + T_n \hat{\beta_n}) y_{[ \overline{\varepsilon}_n ]} + (h S_n \alpha_n + T_n \hat{\alpha_n}) y_{[ \overline{\varepsilon}_n ]+1}, \end{equation} $

其中$ R_n, S_n, T_n $由(2.5)式给定,且

不难验证,当$ \theta(t) = \phi(t) = t^\varepsilon $时,引理3.1与定理3.2的所有假设条件均成立,其辅助泛函方程(3.4)为

且解$ \varphi(t) = \log t $.借助于上述泛函方程,则可得类似于(3.8)与(3.14)式的渐近估计.

推论3.1 设$ y_n $是方程(3.15)的一个解,如果(3.2)式成立且$ \tilde{\gamma} $是由(3.3)式给定,则

推论3.2  设$ y_n $是方程(3.15)的一个解,如果(3.12)式成立且$ \gamma $是由(3.13)式给定,则

4 数值算例

本节将用一个数值例子来验证已获理论结果的正确性及方法的有效性.

例4.1 考虑初值问题

$ \begin{equation} y'(t) = -10y(t)- 15y(\frac{t}{2})+ 0.6 y'(\frac{t}{2}), \; \; y(0) = 1, \; \; t \geq 0. \end{equation} $

取步长$ h = 0.05 $,容易计算得$ |R| = 0.6 $, $ |S| = 12 $, $ |T| = 0.48 $, $ \eta = 1 $, $ \tilde{\eta} = 2 $,且$ \tilde{\gamma} = 3.9 $$ \gamma = 3.15 $.所以本文所得渐近估计(3.8)和(3.14)分别为

$ \begin{equation} y_n = O(n^{1.96347412397489}), \qquad {\hbox{当}} \, n \rightarrow \infty, \end{equation} $

$ \begin{equation} y_n = O(n^{1.65535182861255}), \qquad {\hbox{当}} \, n \rightarrow \infty, \end{equation} $

即分别存在适当的实常数$ L_1 > 0, L_2 > 0 $使得

因为$ \gamma > 1 $,所以

由定理3.2的证明可知,取$ L_2 = B_0 $,且

$ \sigma_0 = 20 $,则$ L_2 = B_0 = 0.10490479511068 $,用同样的方法计算得到$ L_1 = 0.05527556369595 $,渐近估计分别为

至此,我们给出了方程(3.16)的数值解$ y_{n} $的上界估计式

图 1分别给出了$ (t, \log_{10} f_1(t)) $, $ (t, \log_{10} f_2(t)) $$ (t, \log_{10} y^h(t)) $的图形,其中$ t\in [1, 500] $, $ y^h(t) $$ y_{n} $$ y_{n+1} $的线性组合.

图 1

图 1   $ y^h(t) $为数值解, $ f_1(t), f_2(t) $分别为渐近估计(3.8)和(3.14)


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