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Bifurcation Control of the Discrete FHN System
DOI: 10.12677/DSC.2022.113013, PDF, HTML, XML, 下载: 75  浏览: 112

Abstract: This paper investigates the control of Neimark-Sacker bifurcation of the discrete FHN system. A kind of state feedback controller is designed, which not only has a simple form, but also accurately and effectively controls the critical value of such bifurcation without changing the position of the fixed point, so as to realize the advance or delay of the bifurcation.

1. 引言

2. 主要内容

$f:\left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}x+\delta \left(x-\frac{{x}^{3}}{3}-y+I\right)\\ y+\frac{\delta }{\tau }\left(ax+b-y\right)\end{array}\right)$ (1)

2.1. 不动点及其Neimark-Sacker分支

$\left(\begin{array}{c}u\\ v\end{array}\right)\to \left(\begin{array}{cc}1.25& -0.25\\ 0.5625& 0.6875\end{array}\right)\left(\begin{array}{c}u\\ v\end{array}\right)+\left(\begin{array}{c}-\frac{0.25{u}^{3}}{3}+\alpha \left(u-\frac{{u}^{3}}{3}-v\right)\\ \alpha \left(2.25u-1.25v\right)\end{array}\right)$ (2)

$\left(\begin{array}{l}U\\ V\end{array}\right)\to \left(\begin{array}{cc}0.96875& 0.24804\\ -0.24804& 0.96875\end{array}\right)\left(\begin{array}{l}U\\ V\end{array}\right)+\left(\begin{array}{c}G\left(U,V\right)\\ K\left(U,V\right)\end{array}\right)$ (3)

$G\left(U,V\right)=\alpha \left(-0.1249999U+0.825525V\right)$,

$\begin{array}{c}K\left(U,V\right)=-0.2271283{\left(0.416025U+0.3669V\right)}^{3}+\alpha \left(-0.99215618U{}_{}{}^{}\\ \text{\hspace{0.17em}}\text{ }\text{ }-0.9085133{\left(0.416025U+0.3669V\right)}^{3}-0.125V\right)\end{array}$.

${\lambda }_{1,2}\left(\alpha \right)=0.125\alpha -0.96875±i\sqrt{0.984375{\alpha }^{2}+0.4921875\alpha +0.0615234375}$,

${|{\lambda }_{1，2}\left(\alpha \right)|}^{2}={\alpha }^{2}+0.25\alpha +1$.

$G\left(0,0\right)=0$, $K\left(0,0\right)=0$, $d={\frac{\text{d}|\lambda \left(\alpha \right)|}{\text{d}\alpha }|}_{\alpha =0}=0.125>0$,

$l=-\mathrm{Re}\left[\frac{\left(1-2\lambda \right){\stackrel{¯}{\lambda }}^{2}}{1-\lambda }{Q}_{11}{Q}_{20}\right]-\frac{1}{2}{|{Q}_{11}|}^{2}-{|{Q}_{02}|}^{2}\text{}+{\mathrm{Re}\left(\stackrel{¯}{\lambda }{Q}_{21}\right)|}_{\alpha =0}=-0.01763964<0$,

${Q}_{20}=\frac{1}{8}{\left({G}_{UU}-{G}_{VV}+2{K}_{UV}+i\left({K}_{UU}-{K}_{VV}-2{G}_{UV}\right)\right)|}_{\left(0,0\right)}=0$,

${Q}_{11}=\frac{1}{4}\left({G}_{UU}+{G}_{VV}+i\right){\left({K}_{UU}+{K}_{VV}\right)|}_{\left(0,0\right)}=0$,

${Q}_{02}=\frac{1}{8}{\left({G}_{UU}-{G}_{VV}-2{K}_{UV}+i\left({K}_{UU}-{K}_{VV}+2{G}_{UV}\right)\right)|}_{\left(0,0\right)}=0$,

$\begin{array}{c}{Q}_{21}=\frac{1}{16}{\left(\left({G}_{UUU}+{G}_{UVV}+{K}_{UUV}+{K}_{VVV}\right)+i\left({K}_{UUU}+{K}_{UVV}-{G}_{UUV}-{G}_{VVV}\right)\right)|}_{\left(0,0\right)}\\ =-\left(0.014112+0.102564\alpha \right)-i\left(0.016+0.116297\alpha \right)\end{array}$.

2.2. Neimark-Sacker分支的控制

$\left(\begin{array}{c}x\\ y\end{array}\right)\to \left(\begin{array}{c}x+\delta \left(x-\frac{{x}^{3}}{3}-y+0.3\right)+{A}_{11}x+{A}_{12}\left(y-0.3\right)\\ y+1.25\delta \left(1.8x+0.3-y\right)\end{array}\right)$ (4)

(i) 当 ${A}_{11}>0$$\frac{11{A}_{11}}{9}<{A}_{12}$$\frac{3-{A}_{11}}{9}\le {A}_{12}$ 时， $F\left(\delta \right)=0$ 在区间(0, 0.25)仅有一根 $\frac{0.25+1.25{A}_{11}+2.25{A}_{12}-\sqrt{\Delta }}{2}$

(ii) 当 ${A}_{11}<0$$\frac{11{A}_{11}}{9}>{A}_{12}$ 时， $F\left(\delta \right)=0$ 在区间(0, 0.25)仅有一根 $\frac{0.25+1.25{A}_{11}+2.25{A}_{12}+\sqrt{\Delta }}{2}$

(iii) 当 ${A}_{11}>0$$\frac{-1-5{A}_{11}}{9}<{A}_{12}<\mathrm{min}\left\{\frac{1-5{A}_{11}}{9},\frac{11{A}_{11}}{9}\right\}$$\Delta >0$ 时， $F\left(\delta \right)=0$ 在区间(0, 0.25)有两根 $\frac{0.25+1.25{A}_{11}+2.25{A}_{12}±\sqrt{\Delta }}{2}$

(iv) 当 ${A}_{11}=0$$-\frac{1}{9}<{A}_{12}<0$ 时， $F\left(\delta \right)=0$ 在区间(0, 0.25)仅有一根 $\frac{\sqrt{5.0625{A}_{12}^{2}+1.125{A}_{12}+0.0625}+0.25}{2}+\frac{2.25{A}_{12}}{2}$

(v) 当 $0<{A}_{11}<\frac{1}{6}$$\frac{11{A}_{11}}{9}={A}_{12}$ 时， $F\left(\delta \right)=0$ 在区间(0, 0.25)仅有一根 $\frac{-\sqrt{16{A}_{11}^{2}-2{A}_{11}+0.0625}+0.25}{2}+\frac{3.875{A}_{11}}{2}$

(I) 当 $0.25<{A}_{11}$$\frac{3-{A}_{11}}{9}<{A}_{12}<\frac{11{A}_{11}}{9}$ 时， $F\left(\delta \right)=0$ 在区间(0.25, 1)仅有一根 $\frac{0.25+1.25{A}_{11}+2.25{A}_{12}-\sqrt{\Delta }}{2}$

(II) 当 ${A}_{11}\le 0$$\frac{11{A}_{11}}{9}<{A}_{12}<\frac{3-{A}_{11}}{9}$ 时， $F\left(\delta \right)=0$ 在区间(0.25, 1)仅有一根 $\frac{0.25+1.25{A}_{11}\text{\hspace{0.17em}}+2.25{A}_{12}\text{\hspace{0.17em}}+\sqrt{\Delta }}{2}$

(III) 当 $\frac{1}{16}\le {A}_{11}$$\frac{1-5{A}_{11}}{9}<{A}_{12}<\mathrm{min}\left\{\frac{7-5{A}_{11}}{9},\frac{11{A}_{11}}{9},\frac{3-{A}_{11}}{9}\right\}$$\Delta >0$ 时， $F\left(\delta \right)=0$ 在区间(0.25, 1)有两根 $\frac{0.25+1.25{A}_{11}+2.25{A}_{12}±\sqrt{\Delta }}{2}$

(IV) 当 $0<{A}_{11}<\frac{1}{16}$$\frac{11{A}_{11}}{9}={A}_{12}$ 时， $F\left(\delta \right)=0$ 在区间(0.25, 1)仅有一根 $\frac{\sqrt{16{A}_{11}^{2}-2{A}_{11}+0.0625}+0.25}{2}+\frac{3.875{A}_{11}}{2}$

(V) 当 $0.25<{A}_{11}<1$${A}_{12}=\frac{3-{A}_{11}}{9}$ 时， $F\left(\delta \right)=0$ 在区间(0.25, 1)仅有一根 $\frac{1+0.875{A}_{11}-\sqrt{{A}_{11}^{2}-2{A}_{11}+1}}{2}$

(i) 显然 $F\left(0\right)>0$$F\left(0.25\right)<0$，由根的存在性定理得， $F\left(\delta \right)=0$ 在区间(0, 0.25)仅有一根 $\frac{0.25+1.25{A}_{11}}{2}+\frac{2.25{A}_{12}-\sqrt{\Delta }}{2}$

(ii) 易知 $F\left(0\right)<0$$F\left(0.25\right)>0$，从而 $F\left(\delta \right)=0$ 在区间(0, 0.25)有一根 $\frac{0.25+1.25{A}_{11}+2.25{A}_{12}+\sqrt{\Delta }}{2}$

(iii) 显然 $F\left(0\right)>0$$F\left(0.25\right)>0$，由 $\frac{-1-5{A}_{11}}{9}<{A}_{12}<\frac{1-5{A}_{11}}{9}$ 可得 $F\left(\delta \right)$ 的对称轴 $\frac{1+5{A}_{11}+9{A}_{11}}{8}\in \left(0,0.25\right)$

(iv) 由已知可得 $F\left(0\right)=0$$F\left(0.25\right)=-\frac{9{A}_{12}}{16}>0$$F\left(\delta \right)$ 的对称轴 $\frac{1+9{A}_{12}}{8}>0$，从而 $F\left(\delta \right)=0$ 在(0, 0.25)中存在一根 $\frac{0.25+2.25{A}_{12}+\sqrt{5.0625{A}_{12}^{2}+1.125{A}_{12}+0.0625}}{2}$

(v) 由已知可得 $F\left(\delta \right)$ 的对称轴 $\frac{1+5{A}_{11}+9{A}_{12}}{8}=\frac{1+16{A}_{11}}{8}<0.25$$F\left(0\right)>0$$F\left(0.25\right)=0$，从而可得 $F\left(\delta \right)=0$ 在(0,0.25)中存在一根 $\frac{0.25+3.875{A}_{11}-\sqrt{16{A}_{11}^{2}-2{A}_{11}+0.0625}}{2}$

$\left(-0.421875{A}_{11}+0.140625{A}_{12}-0.984375\right){\delta }^{*}+{A}_{11}^{2}+3.9375{A}_{11}<0$ (5)

${\delta }^{*}\ne 0.125+0.625{A}_{11}+1.125{A}_{12}$ (6)

${\delta }^{*}\ne {A}_{12}$ (7)

$1+{\delta }^{*}+{A}_{11}\ne 0$ (8)

$\delta ={\delta }^{*}$ 时，控制系统(4)的不动点 ${Z}_{0}=\left(0,0.3\right)$ 发生了Neimark-Sacker分支。

$\left(\begin{array}{l}{u}_{1}\\ {v}_{1}\end{array}\right)\to \left(\begin{array}{cc}1+{\delta }^{*}+{A}_{11}& -{\delta }^{*}+{A}_{12}\\ 2.25{\delta }^{*}& 1-1.25{\delta }^{*}\end{array}\right)\left(\begin{array}{l}{u}_{1}\\ {v}_{1}\end{array}\right)+\left(\begin{array}{c}-\frac{{\delta }^{*}}{3}{u}_{1}^{3}+\beta \left({u}_{1}-\frac{{u}_{1}^{3}}{3}-{v}_{1}\right)\\ \beta \left(2.25{u}_{1}-1.25{v}_{1}\right)\end{array}\right)$ (9)

$J=\left(\begin{array}{cc}1+{\delta }^{*}+{A}_{11}+\beta & -{\delta }^{*}+{A}_{12}-\beta \\ 2.25\left({\delta }^{*}+\beta \right)& 1-1.25\left({\delta }^{*}+\beta \right)\end{array}\right)$

$\begin{array}{c}{p}^{2}\left(0\right)-4q\left(0\right)=-3.9375{\left({\delta }^{*}\right)}^{2}+\left(4.5{A}_{11}+9{A}_{12}\right){\delta }^{*}+{A}_{11}^{2}\\ =-3.9375\left(\left(0.25+1.25{A}_{11}+2.25{A}_{12}\right){\delta }^{*}-{A}_{11}\right)+\left(4.5{A}_{11}+9{A}_{12}\right){\delta }^{*}+{A}_{11}^{2}\\ =\left(-0.421875{A}_{11}+0.140625{A}_{12}-0.984375\right){\delta }^{*}+{A}_{11}^{2}+3.9375{A}_{11},\end{array}$

$\left(\begin{array}{l}X\\ Y\end{array}\right)\to \left(\begin{array}{cc}-\frac{p\left(0\right)}{2}& \frac{\sqrt{4-{p}^{2}\left(0\right)}}{2}\\ -\frac{\sqrt{4-{p}^{2}\left(0\right)}}{2}& -\frac{p\left(0\right)}{2}\end{array}\right)\left(\begin{array}{l}X\\ Y\end{array}\right)+\left(\begin{array}{l}g\left(X,Y\right)\\ h\left(X,Y\right)\end{array}\right)$ (10)

$g\left(X,Y\right)=-\frac{4{\delta }^{*}{\left({\delta }^{*}-{A}_{12}\right)}^{2}}{3}{X}^{3}+\frac{\beta }{2\left({\delta }^{*}-{A}_{12}\right)}\left(\left(4.25{\delta }^{*}+{A}_{11}-2{A}_{12}\right)X-\frac{8{\left({\delta }^{*}-{A}_{12}\right)}^{3}}{3}{X}^{3}-\sqrt{4-{p}^{2}\left(0\right)}Y\right)$,

$\begin{array}{c}h\left(X,Y\right)=-\frac{4{\delta }^{*}\left(2.25{\delta }^{*}+{A}_{11}\right){\left({\delta }^{*}-{A}_{12}\right)}^{2}}{3\sqrt{4-{p}^{2}\left(0\right)}}{X}^{3}+\frac{\beta }{\sqrt{4-{p}^{2}\left(0\right)}}\left(\frac{1}{2\left({\delta }^{*}-{A}_{12}\right)}\left(25.1875{\left({\delta }^{*}\right)}^{2}\\ {}^{{}_{}{}^{}}+\left(4{A}_{11}-24.625{A}_{12}\right){\delta }^{*}+{A}_{11}^{2}+0.5{A}_{11}{A}_{12}+{A}_{12}^{2}\right)X\\ \text{\hspace{0.17em}}\text{ }\text{ }-\frac{4\left(2.25{\delta }^{*}+{A}_{11}\right){\left({\delta }^{*}-{A}_{12}\right)}^{2}}{3}{X}^{3}+\left(3.5{\delta }^{*}-0.25{A}_{12}\right)Y\right).\end{array}$

${P}_{20}=\frac{1}{8}{\left({g}_{XX}-{g}_{YY}+2{h}_{XY}+i\left({h}_{XX}-{h}_{YY}-2{g}_{XY}\right)\right)|}_{\left(0,0\right)}=0$,

${P}_{11}=\frac{1}{4}{\left({g}_{XX}+{g}_{YY}+i\left({h}_{XX}+{h}_{YY}\right)\right)|}_{\left(0,0\right)}=0$,

${P}_{02}=\frac{1}{8}{\left({g}_{XX}-{g}_{YY}-2{h}_{XY}+i\left({h}_{XX}-{h}_{YY}+2{g}_{XY}\right)\right)|}_{\left(0,0\right)}=0$,

$\begin{array}{c}{P}_{21}=\frac{1}{16}{\left(\left({g}_{XXX}+{g}_{XYY}+{h}_{XXY}+{h}_{YYY}\right)+i\left({h}_{XXX}+{h}_{XYY}-{g}_{XXY}-{g}_{YYY}\right)\right)|}_{\left(0,0\right)}\\ =-\frac{{\delta }^{*}{\left({\delta }^{*}-{A}_{12}\right)}^{2}}{2}\left(1+\frac{2.25{\delta }^{*}+{A}_{11}}{\sqrt{4-{p}^{2}\left(0\right)}}i\right).\end{array}$

$\beta =0$ 时，系统(10)的不动点(0, 0)发生了Neimark-Sacker分支。因此当 $\delta ={\delta }^{*}$ 时，控制系统(4)的不动点 ${Z}_{1}=\left(0,0.3\right)$ 发生了Neimark-Sacker分支，故定理成立。

3. 数值模拟

(a)(b)

Figure 1. (a) Bigurcation diagram of the fixed point ${Z}_{0}$ of the system (1) for $\delta \in \left(0.15,0.4\right)$ ; (b) Phase diagram of the periodic orbit for $\delta =0.251$

1) Neimark-Sacker分支的提前：取 ${A}_{11}=0$${A}_{11}=-\frac{1}{45}$${A}_{11}$${A}_{12}$ 满足引理(iv)条件： ${A}_{11}=0$$-\frac{1}{9}<-\frac{1}{45}={A}_{12}<0$。从而 ${\delta }^{*}=0.2$，因为 ${A}_{11}$${A}_{12}$ 还满足定理2的条件：

$\left(-0.421875{A}_{11}+0.140625{A}_{12}-0.984375\right){\delta }^{*}+{A}_{11}^{2}+3.9375{A}_{11}=-0.1975<0$$0.125+0.625{A}_{11}+1.125{A}_{12}=0.1\ne {\delta }^{*}$$0.2={\delta }^{*}\ne {A}_{12}$$1+{\delta }^{*}+{A}_{11}=1.2\ne 0$。所以当 $\delta =0.2$ 时，控制系统(4)的不动点 ${Z}_{0}$ 发生了Neimark-Sacker分支，即Neimark-Sacker分支提前发生。 $\delta \in \left(0.15,0.4\right)$$\left(\delta ,x\right)$ 平面上控制系统(4)不动点 ${Z}_{0}$ 的分支图如图2所示。

2) Neimark-Sacker分支的延迟：令 ${A}_{11}=-1$${A}_{11}=-\frac{2}{9}$${A}_{11}$${A}_{12}$ 满足引理(II)条件： ${A}_{11}<0$$\frac{11{A}_{11}}{9}=-\frac{11}{9}<{A}_{12}<\frac{4}{9}=\frac{3-{A}_{11}}{9}$。从而 ${\delta }^{*}=0.5$，又由于 ${A}_{11}$${A}_{12}$ 满足定理2条件：

$\left(-0.421875{A}_{11}+0.140625{A}_{12}-0.984375\right){\delta }^{*}+{A}_{11}^{2}+3.9375{A}_{11}=-3.234375<0$$0.125+0.625{A}_{11}+1.125{A}_{12}=-0.75\ne {\delta }^{*}$$0.5={\delta }^{*}\ne {A}_{12}$$1+{\delta }^{*}+{A}_{11}=0.5\ne 0$。故当 $\delta =0.5$ 时，控制系统(4)的不动点 ${Z}_{0}$ 发生了Neimark-Sacker分支，即Neimark-Sacker分支延迟发生，此时控制系统(4)不动点 ${Z}_{0}$ 分支图如图3所示。

Figure 2. Bigurcation diagram of the fixed point ${Z}_{0}$ of the system (4) for ${A}_{11}=0$, ${A}_{11}=-\frac{1}{45}$

Figure 3. Bigurcation diagram of the fixed point ${Z}_{0}$ of the system (4) for ${A}_{11}=-1$, ${A}_{11}=-\frac{2}{9}$

${A}_{11}$${A}_{12}$ 属于区域(i)~(v)时，不动点 ${Z}_{0}$ 提前发生Neimark-Sacker分支；当 ${A}_{11}$${A}_{12}$ 属于区域(I)~(V)时，不动点 ${Z}_{0}$ 延迟发生Neimark-Sacker分支。

Figure 4. Effective domain of the control parameters A11 and A12

4. 结论

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