西南师范大学学报(自然科学版)   2019, Vol. 44 Issue (7): 1-7.  DOI: 10.13718/j.cnki.xsxb.2019.07.001
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  • 三阶非线性模糊差分方程动力学行为分析    [PDF全文]
    张千宏 , 王贵英     
    贵州财经大学 数学与统计学院, 贵阳 550025
    摘要:研究一类三阶非线性模糊差分方程正解的存在性及渐近行为 $ {x_{n + 1}} = \frac{{{x_{n - 2}}}}{{A + {x_{n - 2}}{x_{n - 1}}{x_n}}},n = 0,1, \cdots $ 其中(xn)是正模糊数数列,A及初始值x-2x-1x0是正模糊数.最后给出数值例子以验证理论结论的正确性.
    关键词模糊差分方程    平衡点    有解性    持久性    渐近稳定    

    差分方程(系统)作为微分方程及时滞微分方程的离散形式,在经济学、生态学、计算机、系统工程学等学科上有很多应用,取得了不少成果[1-9].模糊差分方程是差分方程的一种推广形式,系统中参数及初始值为模糊数,解为模糊数数列.最近有关模糊差分方程解的动力学行为研究已引起部分学者的关注,取得了一些有意义的成果[10-22],基于此,本文进一步讨论如下三阶非线性模糊差分方程

    $ {x_{n + 1}} = \frac{{{x_{n - 2}}}}{{A + {x_{n - 2}}{x_{n - 1}}{x_n}}},n = 0,1, \cdots $ (1)

    其中:A是正模糊数;初始值x-2x-1x0是正模糊数.

    为方便起见,首先给出下列定义:

    定义1 A为模糊数,如果A$\mathbb{R}$→[0, 1]满足(ⅰ)-(ⅳ)

    (ⅰ)A是正规的,即存在$x \in \mathbb{R}$使得A(x)=1;

    (ⅱ) A是模糊凸的,即对所有t∈[0, 1],$ {x_1}, {x_2} \in \mathbb{R}$使得

    $ A\left( {t{x_1} + \left( {1 - t} \right){x_2}} \right) \ge \min \left\{ {A\left( {{x_1}} \right),A\left( {{x_2}} \right)} \right\} $

    (ⅲ) A是上半连续的;

    (ⅳ) A的支撑,${\rm{supp}}A = \overline {\bigcup\nolimits_{\alpha \in \left( {0, 1} \right]} {{{\left[A \right]}_\alpha }} } = \overline {\left\{ {x:A\left( x \right) > 0} \right\}} $是紧的.

    Aα-截集表示为${\left[A \right]_\alpha } = \left\{ {x \in \mathbb{R}:A\left( x \right) \ge \alpha } \right\}$α∈[0, 1],显然[A]α是闭区间.如果${\rm{supp}}A \subset \left( {0, \infty } \right)$,则模糊数是正的.显然如果A是正实数,那么A是模糊数且[A]α=[AA],α∈(0,1].即A是平凡的模糊数.

    定义2 设AB是模糊数,[A]α=[AlαArα],[B]α=[BlαBrα],α∈(0,1],模糊数空间的范数为:

    $ \left\| A \right\| = \mathop {\sup }\limits_{\alpha \in \left( {0,1} \right]} \max \left\{ {\left| {{A_{l,\alpha }}} \right|,\left| {{A_{r,\alpha }}} \right|} \right\} $

    距离为

    $ D\left( {A,B} \right) = \mathop {\sup }\limits_{\alpha \in (0,1]} \max \left\{ {\left| {{A_{l,\alpha }} - {B_{l,\alpha }}} \right|,\left| {{A_{r,\alpha }} - {B_{r,\alpha }}} \right|} \right\} $

    根据文献[11, 13]给出模糊数数列有界和持久性定义如下:

    定义3  如果存在一个正实数M(或N)使得

    $ {\rm{supp}}{x_n} \subset \left[ {M,\infty } \right),n = 1,2, \cdots $

    $ \operatorname{supp} x_{n} \subset(0, N], n=1,2, \cdots $

    则称模糊数数列(xn)是持久的(或有界的).

    如果存在正实数MN>0使得

    $ \operatorname{supp} x_{n} \subset[M, N], n=1,2, \cdots $

    则称(xn)是有界和持久的.如果范数‖xn‖,n=1,2,…,是无界数列,则称(xn),n=1,2,…,是无界数列.

    定义4 如果正模糊数数列(xn)满足方程(1),则称xn是方程(1)的正解.如果正模糊数x满足

    $ x=\frac{x}{A+x^{3}} $

    则称正模糊数x是(1)式的正平衡点.

    定义5 设正模糊数数列(xn),正模糊数x使得

    $ \left[x_{n}\right]_{\alpha}=\left[L_{n, \alpha}, \mathbb{R}_{n,\alpha} \right], n=0,1,2, \cdots, \alpha \in(0,1] $ (2)

    $ {\left[ x \right]_\alpha } = \left[ {{L_\alpha },{\mathbb{R}_\alpha }} \right],\alpha \in \left( {0,1} \right] $ (3)

    如果limn→∞D(xnx)=0,则n→∞时,模糊数数列(xn)关于D收敛于模糊数x.

    定义6 (ⅰ)设方程(1)有正平衡点x,如果对任意ε>0,存在δ=δ(ε)>0,使得方程(1)的每一个正解xn,满足D(x-ix)≤δi=0,1,2,有D(xnx)≤εn>0,则称方程(1)的正平衡点x是稳定的.

    (ⅱ)如果它是稳定的且当n→∞时,方程(1)的每一个正解关于D收敛于正平衡点x,则称方程(1)的正平衡点x是渐近稳定的.

    1 主要结论

    首先研究方程(1)正解的存在性,需要下面的引理.

    引理1[23]  设$f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \times {\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + }$是连续的,ABCD是模糊数.那么

    $ {\left[ {f\left( {A,B,C,D} \right)} \right]_\alpha } = f\left( {{{\left[ A \right]}_\alpha },{{\left[ B \right]}_\alpha },{{\left[ C \right]}_\alpha },{{\left[ D \right]}_\alpha }} \right),\alpha \in \left( {0,1} \right] $ (4)

    引理2[23]  设uE~(模糊数空间),[u]α=[u-(α),u+(α)],α∈(0,1].那么u-(α)与u+(α)可以看成(0,1]上的函数,满足

    (ⅰ) u-(α)非减和左连续;

    (ⅱ) u+(α)非增和左连续;

    (ⅲ) u-(1)≤u+(1).

    反之,对任意定义在(0,1]满足上面(ⅰ)-(ⅲ)的函数a(α)和b(α),存在uE~,使得对任意α∈(0,1],[u]α=[a(α),b(α)].

    定理1 考虑方程(1),其中A是正模糊数.那么对任意正模糊数x-2x-1x0,(1)式存在唯一正解xn.

     证明类似于命题2.1[14],略.

    下面给出方程(1)模糊正解的性质,需要下面的引理.

    引理3 对于差分系统

    $ {y_{n + 1}} = \frac{{{y_{n - 2}}}}{{q + {z_{n - 2}}{z_{n - 1}}{z_n}}},{z_{n + 1}} = \frac{{{z_{n - 2}}}}{{p + {y_{n - 2}}{y_{n - 1}}{y_n}}},n = 0,1, \cdots $ (5)

    其中pq及初始值y-iz-ii=0,1,2是正实数.以下结论成立:

    (ⅰ)系统(5)的每一个正解(ynzn)满足当n=3k+ii∈{1,2,3},k=1,2,…,

    $ 0 < {y_n} < \frac{1}{{{q^{k + 1}}}}\max \left\{ {{y_{ - 2}},{y_{ - 1}},{y_0}} \right\},0 < {z_n} < \frac{1}{{{p^{k + 1}}}}\max \left\{ {{z_{ - 2}},{z_{ - 1}},{z_0}} \right\} $ (6)

    (ⅱ)如果

    $ p > 1,q > 1 $ (7)

    n→∞,系统(5)的每一个正解(ynzn)收敛于系统(5)的平衡点(0,0).

    (ⅲ)如果

    $ p < 1,q < 1 $ (8)

    那么(0,0)与$\left( {\sqrt[3]{{1 - p}}, \sqrt[3]{{1 -q}}} \right)$是不稳定的.

     (ⅰ)设{(ynzn)}是系统(5)的正解.因为yn>0与zn>0,当n≥-2,由系统(5)推出

    $ y_{n} \leqslant \frac{1}{q} y_{n-3}, z_{n} \leqslant \frac{1}{p} z_{n-3}, n=1,2,3, \cdots $ (9)

    vnwn是下面系统的解,

    $ v_{n}=\frac{1}{q} v_{n-3}, w_{n}=\frac{1}{p} w_{n-3}, n \geqslant 1 $ (10)

    使得

    $ v_{-2}=y_{-2}, v_{-1}=y_{-1}, v_{0}=y_{0}, w_{-2}=z_{-2}, w_{-1}=z_{-1}, w_{0}=z_{0} $ (11)

    我们用归纳法证明

    $ y_{n} \leqslant v_{n}, z_{n} \leqslant w_{n}, n=3 k+i, i=1,2,3, k \in \mathbb{N}_{+} $ (12)

    假设当k=m≥1,(12)式成立,那么由(9)式得

    $ y_{3(m+1)+i} \leqslant \frac{1}{q} y_{3 m+i} \leqslant \frac{1}{q} v_{3 m+i}=v_{3(m+1)+i}, \\ z_{3(m+1)+i} \leqslant \frac{1}{p} z_{3 m+i} \leqslant \frac{1}{p} w_{3 m+i}=w_{3(m+1)+i} $ (13)

    故(12)式是正确的.由(10)式与(11)式有,当n=3k+ii=1,2,3,k≥1时

    $ \left\{\begin{array}{l}{v_{3 k+i}=\frac{1}{q^{k+1}} v_{i-3}} \\ {w_{3 k+i}=\frac{1}{p^{k+1}} w_{i-3}}\end{array}\right. $ (14)

    那么由(9),(12),(14)式可知(6)式成立.

    (ⅱ)由(6)和(7)式,显然limn→∞yn=0,limn→∞zn=0.

    (ⅲ)由(6)和(8)式,显然(0,0)是局部不稳定的.接下来证明平衡点$\left( {\sqrt[3]{{1 - p}}, \sqrt[3]{{1 -q}}} \right)$是局部不稳定的.容易得到系统(5)关于平衡点$\left( {\sqrt[3]{{1 - p}}, \sqrt[3]{{1 -q}}} \right)$的线性化系统

    $ {\mathit{\Phi }_{n + 1}} = G{\mathit{\Phi }_n} $ (15)

    其中

    $ {\mathit{\Phi }_n} = \left( \begin{array}{l} {y_n}\\ {y_{n - 1}}\\ {y_{n - 2}}\\ {z_n}\\ {z_{n - 1}}\\ {z_{n - 2}} \end{array} \right),G = \left( {\begin{array}{*{20}{c}} 0&0&{\frac{1}{q}}&\eta &\eta &\eta \\ 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ \beta &\beta &\beta &0&0&{\frac{1}{p}}\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0 \end{array}} \right) $

    这里$\eta =- \sqrt[3]{{\left( {1 -p} \right){{\left( {1 -q} \right)}^2}}}$$\beta =- \sqrt[3]{{{{\left( {1 -p} \right)}^2}\left( {1 -q} \right)}}$.(15)式的特征方程为

    $ P\left( \lambda \right) = {\lambda ^6} - \eta \beta {\lambda ^4} - \left( {2\eta \beta + \frac{1}{p} - \frac{1}{q}} \right){\lambda ^3} - 3\eta \beta {\lambda ^2} - 2\eta \beta \eta - \eta \beta + \frac{1}{{pq}} $ (16)

    由(16)式有6×6矩阵

    $ \sum\nolimits_{6 \times 6} =\\ {\left[ {\begin{array}{*{20}{c}} 0&{ - \left( {2\eta \beta + \frac{1}{p} - \frac{1}{q}} \right)}&{ - 2\eta \beta }&0&0&0\\ 1&{ - \eta \beta }&{ - 3\eta \beta }&{ - \eta \beta + \frac{1}{{pq}}}&0&0\\ 0&0&{ - \left( {2\eta \beta + \frac{1}{p} - \frac{1}{q}} \right)}&{ - 2\eta \beta }&0&0\\ 0&0&{ - \eta \beta }&{ - 3\eta \beta }&{ - \eta \beta + \frac{1}{{pq}}}&0\\ 0&0&0&{ - \left( {2\eta \beta + \frac{1}{p} - \frac{1}{q}} \right)}&{ - 2\eta \beta }&0\\ 0&0&0&0&{ - 3\eta \beta }&{ - \eta \beta + \frac{1}{{pq}}} \end{array}} \right]} $

    显然不是所有Δk>0(其中Δkk=1,2,…,6,是矩阵∑6×6k阶顺序主子式).因此由定理1.3.2[1]知正平衡解$\left( {\sqrt[3]{{1 - p}}, \sqrt[3]{{1 -q}}} \right)$是局部不稳定的.

    定理2 考虑模糊差分方程(1),其中Ax-ii=0,1,2是正模糊数.那么下面的命题成立.

    (ⅰ)如果对任意α∈(0,1]使得

    $ A_{l, \alpha}>p>1 $ (17)

    那么当n=3k+ii=1,2,3,$k \in {\mathbb{N}_ + }$时,方程(1)的每一个正解xn满足,

    $ {\left[ {{x_n}} \right]_\alpha } = \left[ {{L_{n,\alpha }},{\mathbb{R}_{n,\alpha }}} \right],0 < {L_{n,\alpha }} \leqslant \frac{1}{{{p^{k + 1}}}}\mathop {\max }\limits_{1 \leqslant i \leqslant 3} \left\{ {{\mathbb{R}_{ - i + 1,\alpha }}} \right\},\\ 0 < {\mathbb{R}_{n,\alpha }} \leqslant \frac{1}{{{p^{k + 1}}}}\mathop {\max }\limits_{1 \leqslant i \leqslant 3} \left\{ {{\mathbb{R}_{ - i + 1,\alpha }}} \right\} $

    (ⅱ)如果(17)式成立,那么当n→∞,方程(1)的每一个正解xn关于D收敛于平衡点x.

     (ⅰ)设xn是方程(1)具有初始值x-2x-1x0的正解,使得

    $ {\left[ {{x_n}} \right]_\alpha } = \left[ {{L_{n,\alpha }},{\mathbb{R}_{n,\alpha }}} \right],\alpha \in (0,1],n = - 2, - 1,0, \cdots $ (18)

    那么按照文献[6]命题2.1的方法,有$\left( {{L_{n, \alpha }}, {\mathbb{R}_{n, \alpha }}} \right)$n=0,1,…,满足下面带参数的常差分系统

    $ {L_{n + 1,\alpha }} = \frac{{{L_{n - 2,\alpha }}}}{{{A_{r,\alpha }} + {\mathbb{R}_{n - 2,\alpha }}{\mathbb{R}_{n - 1,\alpha }}{\mathbb{R}_{n,\alpha }}}},{\mathbb{R}_{n + 1,\alpha }} = \frac{{{\mathbb{R}_{n - 2,\alpha }}}}{{{A_{l,\alpha }}+{L_{n - 2,\alpha }}{L_{n - 1,\alpha }}{L_{n,\alpha }}}} $ (19)

    因(17)式满足,由引理3(ⅰ)知Lnα${\mathbb{R}_{n, \alpha }}$满足

    $ 0 < {L_{n,\alpha }} \leqslant \frac{1}{{{p^{k + 1}}}}\mathop {\max }\limits_{1 \leqslant i \leqslant 3} \left\{ {{\mathbb{R}_{ - i + 1,\alpha }}} \right\},0 < {\mathbb{R}_{n,\alpha }} \leqslant \frac{1}{{{p^{k + 1}}}}\mathop {\max }\limits_{1 \leqslant i \leqslant 3} \left\{ {{\mathbb{R}_{ - i + 1,\alpha }}} \right\},\alpha \in \left( {0,1} \right] $

    (ⅱ)因(18)式成立,那么按照文献[6]的命题2.3的方法,有唯一的平衡点x,其中

    $ {[x]_\alpha } = \left[ {{L_\alpha },{\mathbb{R}_\alpha }} \right] = \left[ {0,0} \right],\alpha \in \left( {0,1} \right] $ (20)

    xn是方程(1)的正解,使得(18)式成立.因(17)式成立,应用引理3(ⅱ)于系统(19),有

    $ \mathop {\lim }\limits_{n \to \infty } {L_{n,\alpha }} = {L_\alpha } = 0,\mathop {\lim }\limits_{n \to \infty } {\mathbb{R}_{n,\alpha }} = {\mathbb{R}_\alpha } = 0 $ (21)

    由(21)式得

    $ \mathop {\lim }\limits_{n \to \infty } D\left( {{x_n},x} \right) = \mathop {\lim }\limits_{n \to \infty } \mathop {\sup }\limits_{\alpha \in \left( {0,1} \right]} \left\{ {\max \left\{ {\left| {{L_{n,\alpha }} - {L_\alpha }} \right|,\left| {{\mathbb{R}_{n,\alpha }} - {\mathbb{R}_\alpha }} \right|} \right\}} \right\} = 0 $

    定理2(ⅱ)得证.

    定理3 考虑模糊差分方程(1),若A是正实数(平凡模糊数)使得0 < A < 1,x-ii=0,1,2是正模糊数.那么存在不稳定的平衡点0与$x = \sqrt[3]{{1 -A}}$.

     显然0是方程(1)的平衡点.对方程(1)的唯一正平衡点x,有下面的关系

    $ L_{\alpha}=\frac{L_{\alpha}}{A+\mathbb{R}_{\alpha}^{3}}, \mathbb{R}_{\alpha}=\frac{\mathbb{R}_{\alpha}}{A+L_{\alpha}^{3}} $ (22)

    由此有

    $ {[x]_\alpha } = \left[ {{L_\alpha },{\mathbb{R}_\alpha }} \right],{L_\alpha } = {\mathbb{R}_\alpha } = \sqrt[3]{{1 - A}} $

    应用引理3(ⅲ)于系统(19),平衡点0与$x = \sqrt[3]{{1 -A}}$是不稳定的.

    2 数值例子

    为验证结论的有效性,给出以下例子.

     考虑三阶非线性模糊差分方程

    $ x_{n+1}=\frac{x_{n-2}}{A+x_{n-2} x_{n-1} x_{n}}, n=0,1, \cdots $ (23)

    其中A是正模糊数,且

    $ A(x)=\left\{\begin{array}{ll}{x-1.2} & {1.2 \leqslant x \leqslant 2.2} \\ {-0.5 x+2.1} & {2.2 \leqslant x \leqslant 4.2}\end{array}\right. $ (24)

    初始值x-2x-1x0满足

    $ {x_{ - 2}}\left( x \right) = \left\{ \begin{array}{l} 10x - 2\;\;\;\;\;\;\;\;0.2 \le x \le 0.3\\ - \frac{{10}}{3}x + 2\;\;\;\;\;0.3 \le x \le 0.6 \end{array} \right.\\ {x_{ - 1}}\left( x \right) = \left\{ \begin{array}{l} 5x - 5\;\;\;\;\;\;\;\;\;\;0.1 \le x \le 0.3\\ - 5x + 2.5\;\;\;\;\;0.3 \le x \le 0.5 \end{array} \right. $ (25)
    $ {x_0}\left( x \right) = \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} 10x - 1\\ - 5x + 2 \end{array}&\begin{array}{l} 0.1 \le x \le 0.2\\ 0.2 \le x \le 0.4 \end{array} \end{array}} \right. $ (26)

    由(25)式有

    $ {\left[ A \right]_\alpha } = \left[ {1.2 + \alpha ,4.2 - 2\alpha } \right],\alpha \in \left( {0,1} \right] $ (27)

    所以$\overline {\bigcup\nolimits_{\alpha \in \left( {0, 1} \right]} {{{\left[A \right]}_\alpha }} } = \left[{1.2, 4.2} \right]$.

    由(25)和(26)式,对α∈(0,1],

    $ \begin{array}{l} {\left[ {{x_{ - 2}}} \right]_\alpha } = \left[ {0.2 + \frac{1}{{10}}\alpha ,0.6 - \frac{3}{{10}}\alpha } \right]\\ {\left[ {{x_{ - 1}}} \right]_\alpha } = \left[ {0.1 + \frac{1}{5}\alpha ,0.5 - \frac{1}{5}\alpha } \right]\\ {\left[ {{x_0}} \right]_\alpha } = \left[ {0.1 + \frac{1}{{10}}\alpha ,0.4 - \frac{1}{5}\alpha } \right] \end{array} $ (28)

    $\overline {\bigcup\nolimits_{\alpha \in \left( {0, 1} \right]} {{{\left[{{x_{-2}}} \right]}_\alpha }} } = \left[{0.2, 0.6} \right]$$\overline {\bigcup\nolimits_{\alpha \in \left( {0, 1} \right]} {{{\left[{{x_{-1}}} \right]}_\alpha }} } = \left[{0.1, 0.5} \right]$$\overline {\bigcup\nolimits_{\alpha \in \left( {0, 1} \right]} {{{\left[{{x_0}} \right]}_\alpha }} } = \left[{0.1, 0.4} \right]$.

    由(23)和(27)式得含参数α∈(0,1]的差分系统

    $ \begin{array}{*{20}{c}} {{L_{n + 1,\alpha }} = \frac{{{L_{n - 2,\alpha }}}}{{4.2 - 2\alpha + {\mathbb{R}_{n - 2,\alpha }}{\mathbb{R}_{n - 1,\alpha }}{\mathbb{R}_{n,\alpha }}}}} \\ {{\mathbb{R}_{n + 1,\alpha }} = \frac{{{\mathbb{R}_{n - 2,\alpha }}}}{{1.2 + \alpha + {L_{n - 2,\alpha }}{L_{n - 1,\alpha }}{L_{n,\alpha }}}}} \end{array} $ (29)

    因此Alα>1,对任意α∈(0,1],即条件(17)满足,所以由定理2,方程(23)的每一个正解xn是有界和持久的.另外,方程(23)有唯一平衡解$\bar x = 0$,且当n→∞时方程(23)的每一个正解xn关于D收敛到唯一平衡点0(图 1-图 4).

    图 1 系统(29)的解
    图 2 α=0,系统(29)的解
    图 3 α=0.5,系统(29)的解
    图 4 α=1,系统(29)的解
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    Dynamical Behavior of a Third-Order Nonlinear Fuzzy Difference Equation
    ZHANG Qian-hong , WANG Gui-ying     
    School of Mathematics and Statistics, Guizhou University of Finance and Economics, Guiyang 550025, China
    Abstract: This paper is concerned with the existence, asymptotic behavior of the positive solutions of a third order fuzzy nonlinear difference equation $ {x_{n + 1}} = \frac{{{x_{n-2}}}}{{A + {x_{n-2}}{x_{n-1}}{x_n}}}, n = 0, 1, \cdots $ where (xn) is a sequence of positive fuzzy numbers, A and the initial values x-2, x-1, x0 are positive fuzzy numbers. Finally an illustrative example is given to demonstrate the effectiveness of the results obtained.
    Key words: fuzzy difference equation    equilibrium point    bounded    persistence    asymptotic stability    
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