在本节中,我们将在模型中使用几个著名的分形分数导数(4).
6.1幂律分形分数导数
在本小节中,我们替换了(4)通过幂律分形分数导数:
$$\开始{聚集}{}_{0}^{\mathsf{FF-P}}\mathcal{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{t}}}\马塔尔{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{H}}{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{E}}$$
(60)
其中函数幂律分形分数阶导数\(\ phi(t)\)定义为[11]
$$开始{对齐}{}^{mathsf{FF-P}}_{0}{\mathcal{I}}{{t}^{\boldsymbol{\alpha}},\psi}\phi(t)=\frac{1}{\Gamma(n-{\bolsymbol{\alfa}}}}{(t-{eta})^{1-n+{boldsymbol{\alpha}}},\quad n-1<{boldsymbol{\ alpha},\spi\le-n\in\mathbb{n},\ end{aligned}$$
(61)
和
$$\begin{aligned}\frac{d}{dt^{\psi}}\phi(u)=\lim_{n\rightarrow\infty}\frac{\phi。\结束{对齐}$$
(62)
通过在公式(60)然后放置\(t=t{n+1}\)在结果中,我们得到了以下递归形式:
$$\开始{aligned}&\开始{arigned}\mathcal{T}(T)_{n+1}(t)&=\mathcal{T}(T)_{1} (0)\\&&quad{}+\frac{\tau}{\Gamma({\boldsymbol{\alpha})}\sum_{j=0}^{n}/int_{j}^{t_{j+1}}{eta}^{\tau-1}\bigl(\mathcal{t}({\eta})\bigl(1-\mathcal{t}({\eta})\bigr)-\beta _{12}\mathcal{t}({\eta})\mathcal{H}({\eta})-\beta{13}\mathcal{t}({\eta})\mathcal{E}({\eta})\bigr)\,d{\eta},\end{aligned}\\&&\mathcal{高}_{n+1}(t)=\mathcal{高}_{2} (0)+\frac{\tau}{\Gamma({\boldsymbol{\alpha}})}\sum_{j=0}^{n}\int_{t_{j}}^{t_{j+1}}{\eta}^{\tau-1}\bigl bigr)-\beta_{21}\mathcal{t}({\eta})\mathcal{H}(}\eta{)\bigr]\bigr{电子}_{n+1}(t)=\mathcal{电子}_{3} (0)+\frac{\tau}{\Gamma({\eta})+s_{3}}-\beta_{31}\mathcal{t}。\结束{对齐}$$
现在让我们定义函数
$$\开始{聚集}\mathcal{克}_{1} ({\eta})={\eta}^{\tau-1}\bigl(\mathcal{T}({\eta})\bigr),\\\mathcal{克}_{2} ({\eta})={\eta}^{\tau-1}\bigl{克}_{3} ({\eta})={\eta}^{\tau-1}\biggl(\frac{k_{3}\mathcal{T}({\eta}al{E}({\eta})\biggr)。\结束{聚集}$$
(63)
这些函数可以在\([t{\zeta},t{\zeta+1}]\)作为
$$\马塔尔{克}_{i} ({\eta})=\frac{{\eta}-t{\zeta-1}}{t{\zeta}-t{\ zeta-1}}\mathcal{克}_{i} (t_{\zeta})-\frac{\eta}-t_{j}}{t_{\zeta}-t_{\zeta-1}}\mathcal{克}_{i} (t{\zeta-1})$$
(64)
因此,我们获得
$$\开始{aligned}&\mathcal{T}(T)_{n+1}=\mathcal{T}(T)_{0}+\frac{\tau(\Delta t)^{{\boldsymbol{\alpha}}}{\Gamma({\bolsymbol{\alfa}}+2)}\sum_{\zeta=0}^{n}\bigl[\mathcal{克}_{1} (t{\zeta})\bigl[(-\zeta+1+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+{\bolssymbol}}+2)\\&\hphantom{\mathcal{T}(T)_{n+1}=}{}-(-\zeta+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+2{\bolssymbol}}+2)\bigr]\end{aligned}$$
(65)
$$\开始{aligned}&\hphantom{\mathcal{T}(T)_{n+1}=}{}-\mathcal{克}_{1} (t{\zeta-1})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}+1}-{高}_{n+1}=\mathcal{高}_{0}+\frac{\tau(\Delta t)^{{\boldsymbol{\alpha}}}{\Gamma({\bolsymbol{\alfa}}+2)}\sum_{\zeta=0}^{n}\bigl[\mathcal{克}_{2} (t{\zeta})\bigl[(-\zeta+1+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+{\bolssymbol}}+2)\\&\hphantom{\mathcal{高}_{n+1}=}{}-(-\zeta+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+2{\bolssymbol}}+2)\bigr]\end{aligned}$$
(66)
$$\开始{aligned}&\hphantom{\mathcal{高}_{n+1}=}{}-\mathcal{克}_{2} (t{\zeta-1})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}+1}-(-\zeta+n$$
(67)
$$\开始{aligned}&\mathcal{电子}_{n+1}=\mathcal{电子}_{0}+\frac{\tau(\Delta t)^{{\boldsymbol{\alpha}}}{\Gamma({\bolsymbol{\alfa}}+2)}\sum_{\zeta=0}^{n}\bigl[\mathcal{克}_{3} (t{\zeta})\bigl[(-\zeta+1+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+{\bolssymbol}}+2)\\&\hphantom{\mathcal{电子}_{n+1}=}{}-(-\zeta+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+2{\bolssymbol}\alpha}+2)\bigr]\\&\hphantom{\mathcal{电子}_{n+1}=}{}-\mathcal{克}_{3} (t{\zeta-1})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}+1}-(-\zeta+n$$
(68)
哪里\(\mathcal{克}_{j} \),\(泽塔=1,2,3\),在中给出(63).
6.2数值模拟
数字10–12演示模型中状态变量的变化(60)应用方案时(65)对于不同的值\(tau\ in(0,1]\)、和\(α=0.95\)在本次模拟中,我们考虑了模型中的以下值:\(β{12}=1\),\(β{13}=2.5),\(k{2}=0.6),\(β{21}=1.5\),\(k{3}=4.5\),\(s{3}=1\),\(a{31}=0.2)、和\(d_{3}=0.5\)。在我们的数值模拟中,\(t{\mathrm{final}}=500\)和\(hbar=0.001)。在图中10,我们把\((\mathcal{T}(T),\mathcal{H}(T),\mathcal{E}(c))|_{T=0}=(0.1,0.1,0.1)\)在这种情况下,模型表现出混沌吸引子行为。此外,通过施加初始条件\((mathcal{T}(T),mathcal}H}(T),mathcal{E}(c))|_{T=0}=(0.3,0.3,0.3))和\(β{12}=0.745)该模型显示了图中的极限环行为11。此外,对于\(\mathcal{T}(0)=0.3517\),\(数学{H}(0)=0.1115),\(数学{E}(0)=0.4951)、和\(β{12}=0.920),图12确定了解中的周期轨道。
6.3指数衰减律分形分数导数
在本部分中,我们替换了(4)通过具有指数衰减定律的分形分数导数:
$$\开始{聚集}{}_{0}^{\mathsf{FF-E}}\mathcal{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{t}}}\马塔尔{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{H}}{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{E}}$$
(69)
其中函数具有指数衰减律的分形分数导数\({\phi}(t)\)定义为[11]
$${}^{mathsf{FF-E}}_{0}{\mathcal{I}}_}^{\boldsymbol{\alpha}},\psi}{\phi}\eta})\exp\biggl[-\frac{{\boldsymbol{\alpha}}{n-{\bolssymbol}\alpha{}}(t-{\eta{)\biggr]d{\eta},\quad n-1<{\bodsymbol{\alfa}},\psi\le n\in\mathbb{n}$$
(70)
和\(\frac{d}{dt^{psi}}\phi(u)\)在中引入(62).
将Caputo–Fabrizio积分应用于等式(69),我们获得
$$\begin{collected}{\mathcal{T}}(T)={\matchcal{T}{(0)+\frac{\gamma_{1}}{\mathsf{M}({\boldsymbol{\alpha}})}\mathcal{F}(F)_{1} \bigl(t,\mathcal{t}(t{F}(F)_{1} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}{F}(F)_{2} \bigl(t,\mathcal{t}(t{F}(F)_{2} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}{F}(F)_{3} \bigl(t,\mathcal{t}(t{F}(F)_{3} \bigl({\eta},\mathcal{T}({\ta}。\结束{聚集}$$
(71)
设置\(t=t{n+1}\)在(71),基于中提出的方案[11],我们得到
$$\开始{聚集}{\mathcal{T}(T)_{n+1}}={\mathcal{T}}(0)+\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha})}\mathcal{F}(F)_{1} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )+\frac{\tau{\boldsymbol{\alpha}}}{\mathsf{M}({\bolssymbol}\alpha{})}\int_{0}^{t_{n+1}}\mathcal{F}(F)_{1} \bigl({\eta},\mathcal{T}({\ta}{高}_{n+1}}={\mathcal{H}}(0)+\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha})}\mathcal{F}(F)_{2} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )+\frac{\tau{\boldsymbol{\alpha}}}{\mathsf{M}({\bolssymbol}\alpha{})}\int_{0}^{t_{n+1}}\mathcal{F}(F)_{2} \bigl({\eta},\mathcal{T}({\ta}{电子}_{n+1}}={\mathcal{E}}(0)+\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha})}\mathcal{F}(F)_{3} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )+\frac{\tau{\boldsymbol{\alpha}}}{\mathsf{M}({\bolssymbol}\alpha{})}\int_{0}^{t_{n+1}}\mathcal{F}(F)_{3} \bigl({\eta},\mathcal{T}({\ta}。\结束{聚集}$$
(72)
计算两者之间的差异\({\mathcal{T}(T)_{n+1}}\)和\({\mathcal{T}(T)_{n}}\)产量
$$\开始{聚集}{\mathcal{T}(T)_{n+1}}-{\mathcal{T}(T)_{n} }=\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha}})}\bigl[\mathcal{F}(F)_{1} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\mathcal{F}(F)_{1} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\bigr]\\hphantom{{\mathcal{T}(T)_{n+1}}-{\mathcal{T}(T)_{n} }=}{}+\frac{\tau{\boldsymbol{\alpha}}{\mathsf{M}{F}(F)_{1} \bigl({\eta},\mathcal{T}({\ta}{高}_{n+1}}-{\马塔尔{高}_{n} }=\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha}})}\bigl[\mathcal{F}(F)_{2} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\mathcal{F}(F)_{2} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\bigr]\\hphantom{{\mathcal{高}_{n+1}}-{\马塔尔{高}_{n} }=}{}+\frac{\tau{\boldsymbol{\alpha}}{\mathsf{M}{F}(F)_{2} \bigl({\eta},\mathcal{T}({\ta}{电子}_{n+1}}-{\马塔尔{电子}_{n} }=\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha}})}\bigl[\mathcal{F}(F)_{3} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\mathcal{F}(F)_{3} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\bigr]\\hphantom{{\mathcal{电子}_{n+1}}-{\马塔尔{电子}_{n} }=}{}+\frac{\tau{\boldsymbol{\alpha}}{\mathsf{M}{F}(F)_{3} \bigl({\eta},\mathcal{T}({\ta}。\结束{聚集}$$
(73)
因此,可以使用以下迭代程序确定问题的近似解:
$$\开始{聚集}{\mathcal{T}(T)_{n+1}}={\mathcal{T}(T)_{n} }+\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha}})}\bigl[\mathcal{F}(F)_{1} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\mathcal{F}(F)_{1} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\bigr]\\hphantom{{\mathcal{T}(T)_{n+1}}=}{}+\frac{\tau{\boldsymbol{\alpha}}}{\mathsf{M}({\bolsymbol{\alfa})}\biggl[\frac}3\Delta}{2}\mathcal{F}(F)_{1} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\压裂{\Delta}{2}\mathcal{F}(F)_{1} (t{n-1},x{n-1{,mathcal{高}_{n-1},\mathcal{电子}_{n-1})\biggr],\\{mathcal{高}_{n+1}}={\mathcal{高}_{n} }+\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha}})}\bigl[\mathcal{F}(F)_{2} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\mathcal{F}(F)_{2} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\bigr]\\hphantom{{\mathcal{高}_{n+1}}=}{}+\frac{\tau{\boldsymbol{\alpha}}}{\mathsf{M}({\bolsymbol{\alfa})}\biggl[\frac}3\Delta}{2}\mathcal{F}(F)_{2} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\压裂{\Delta}{2}\mathcal{F}(F)_{2} (t{n-1},\mathcal{T}(T)_{n-1},y{n-1{,\mathcal{电子}_{n-1})\biggr],\\{mathcal{电子}_{n+1}}={\mathcal{电子}_{n} }+\frac{\tau}{\mathsf{M}({\boldsymbol{\alpha}})}\bigl[\mathcal{F}(F)_{3} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\mathcal{F}(F)_{3} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\bigr]\\hphantom{{\mathcal{电子}_{n+1}}=}{}+\frac{\tau{\boldsymbol{\alpha}}}{\mathsf{M}({\bolsymbol{\alfa})}\biggl[\frac}3\Delta}{2}\mathcal{F}(F)_{3} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )-\压裂{\Delta}{2}\mathcal{F}(F)_{3} (t{n-1},\mathcal{T}(T)_{n-1},\mathcal{高}_{n-1},\mathcal{电子}_{n-1})\biggr],\end{聚集}$$
(74)
哪里\(\mathcal{F}(F)_{j} \),\(泽塔=1,2,3\),在中给出(63).
6.4数值模拟
数字13–15演示模型中状态变量的变化(69)when方案(74)应用于不同的值\(tau\ in(0,1]\)和\(α=0.95\)在本次模拟中,我们考虑了模型中的以下值:\(β{12}=1\),\(β{13}=2.5),\(k{2}=0.6),\(β{21}=1.5\),\(k{3}=4.5\),\(s{3}=1\),\(a{31}=0.2)、和\(d_{3}=0.5\)。在我们的数值模拟中,\(t{\mathrm{final}}=500\)和\(hbar=0.001)。在图中13,我们采取\((\mathcal{T}(T),\mathcal{H}(T),\mathcal{E}(c))|_{T=0}=(0.1,0.1,0.1)\)在这种情况下,模型表现出混沌吸引子行为。此外,对于\((mathcal{T}(T),mathcal}H}(T),mathcal{E}(c))|_{T=0}=(0.3,0.3,0.3))和\(β{12}=0.745),该模型显示了图中的极限环行为14此外\(\mathcal{T}(0)=0.3517\),\(数学{H}(0)=0.1115),\(数学{E}(0)=0.4951)、和\(β{12}=0.920),图15确定了解中的周期轨道。
6.5Mittag-Lefler定律分形分数导数
在本小节中,我们在(4). 所以,我们实现了
$$\begin{collected}{}_{0}^{\mathsf{FF-ABC}}\mathcal{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{t}}}\马塔尔{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{H}}{D}(D)_{t} ^{{\boldsymbol{\alpha}},\tau}{\mathcal{E}}$$
(75)
其中函数的Mittag-Leffer定律的分形分数导数\({\phi}(t)\)定义为[11]
$$\开始{aligned}&{0}^{\mathsf{FF-ABC}}\mathcal{D}(D)_{t} ^{{黑体符号{\alpha}},\tau}{\phi}(t)\\&\quad=\frac{\mathsf{B}({\boldsymbol{\alfa}})}{n-{{\bolsymbol{\alba}}}}\frac}{d^{\psi}}\int_{0}^{t}{\ph}\alpha}}{n-{\boldsymbol{\alpha{}}(t-{\eta})^{\bolsymbol{\alfa}}\biggr]d{\eta},四元n-1<{\boldsymbol{\alpha}},\psi\len\in\mathbb{n},\ end{aligned}$$
(76)
和\(\frac{d}{dt^{psi}}\phi(u)\)在中引入(62).
将Atangana–Baleanu积分应用于(75),我们获得
$$开始{聚集}{\mathcal{T}}(T)={\matchcal{T}{(0)+\frac{\taut^{\tau-1}(1-{\boldsymbol{\alpha}})}{\Gamma{F}(F)_{1} \bigl(t,\mathcal{t}(t ^{t}\delta^{\tau-1}\mathcal{F}(F)_{1} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}}(1-{\boldsymbol{\alpha}})}{\Gamma({\bolssymbol}\alpha{}){\mathcal{F}(F)_{2} \bigl(t,\mathcal{t}(t),\mathcal{H}(t),\mathcal{E}(t),x_{4}(t)\bigr)\\hphantom{\mathcal{H}}(t)=}{F}(F)_{2} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}}(1-{\boldsymbol{\alpha}})}{\Gamma({\bolssymbol}\alpha{}){\mathcal{F}(F)_{3} \bigl(t,\mathcal{t}(t ^{t}\delta^{\tau-1}\mathcal{F}(F)_{3} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}(}\ta}),\mathcal{E}。\结束{聚集}$$
(77)
根据中提出的方案[11],我们得到
$$\开始{对齐}&\开始{对齐}{\mathcal{T}(T)_{n+1}}&={\mathcal{T}(T)_{0}}+\frac{\tau{t{n}^{alpha-1}}(1-{\boldsymbol{\alpha}})}{\Gamma{F}(F)_{1} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )\\\\quad{}+\frac{\tau{\boldsymbol{\alpha}}}{\Gamma({\boldsymbol{\alpha})\mathsf{B}({\boldsymbol{\alpha})}\sum_{\zeta=0}^{n}\ int _{t_{\zeta}}^{t_{j+1}}\ delta ^{\tau-1}\mathcal{F}(F)_{1} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}{高}_{n+1}}&={\mathcal{高}_{0}}+\frac{\tau{t{n}^{alpha-1}}(1-{\boldsymbol{\alpha}})}{\Gamma{F}(F)_{2} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )\\\\quad{}+\frac{\tau{\boldsymbol{\alpha}}}{\Gamma({\boldsymbol{\alpha})\mathsf{B}({\boldsymbol{\alpha})}\sum_{\zeta=0}^{n}\ int _{t_{\zeta}}^{t_{j+1}}\ delta ^{\tau-1}\mathcal{F}(F)_{2} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}{电子}_{n+1}}&={\mathcal{电子}_{0}}+\frac{\tau{t{n}^{alpha-1}}(1-{\boldsymbol{\alpha}})}{\Gamma{F}(F)_{3} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )\\\\quad{}+\frac{\tau{\boldsymbol{\alpha}}}{\Gamma({\boldsymbol{\alpha})\mathsf{B}({\boldsymbol{\alpha})}\sum_{\zeta=0}^{n}\ int _{t_{\zeta}}^{t_{j+1}}\ delta ^{\tau-1}\mathcal{F}(F)_{3} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}(}\ta}),\mathcal{E}。\end{aligned}\end{alinged}$$
(78)
现在使用拉格朗日多项式分段插值公式(64),我们获得
$$\开始{aligned}&\mathcal{T}(T)_{n+1}=\mathcal{T}(T)_{0}+\frac{\tau{t{n}^{\alpha-1}}{F}(F)_{1} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )\\&\hphantom{\mathcal{T}(T)_{n+1}=}{}+\frac{\tau\hbar^{{\boldsymbol{\alpha}}}{\Gamma{F}(F)_{1} (t_{\zeta},\mathcal){T}(T)_{\zeta},\mathcal{高}_{\zeta},\mathcal{E}_{\zeta})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}(-\zeta+n+{\bold symbol}\alpha}+2)\\&\hphantom{\mathcal{T}(T)_{n+1}=}{}-(-\zeta+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+2{\bolssymbol}\alpha}+2)\bigr]\\&\hphantom{\mathcal{T}(T)_{n+1}=}{}-{t{\zeta-1}^{\alpha-1}}\mathcal{F}(F)_{1} (t{\zeta-1},\mathcal{T}(T)_{\zeta-1},\mathcal{高}_{\zeta-1},\mathcal{E}_{\zeta-1})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}+1}-(-\zeta+n)^{\boldsymbol{\alpha}}}}(-\zeta+n+{\boldsymbol{\alpha}}+1)\bigr]\bigr],\\\\begin{collected}\mathcal{高}_{n+1}=\mathcal{高}_{0}+\frac{tau{t{n}^{alpha-1}}(1-{boldsymbol{alpha}})}{\Gamma({boldsymbol{\alpha}){\mathcal{F}(F)_{2} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )\\\hphantom{\mathcal{高}_{n+1}=}{}+\frac{\tau\hbar^{{\boldsymbol{\alpha}}}{\Gamma{F}(F)_{2} (t_{\泽塔},\mathcal{T}(T)_{\zeta},\mathcal{高}_{\zeta},\mathcal{E}_{\zeta})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}}(-\zeta+n+{\boldsymbol{\alpha}}+2)\\\\hphantom{\mathcal{高}_{n+1}=}{}-(-\zeta+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+2{\bolssymbol}\alpha}+2)\bigr]\\hphantom{\mathcal{高}_{n+1}=}{}-{t{\zeta-1}^{\alpha-1}}\mathcal{F}(F)_{2} (t{\zeta-1},\mathcal{T}(T)_{\zeta-1},\mathcal{H}_{\zeta-1}、\mathcal{E}_{\zeta-1-})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}+1}-(-\zeta+n)铝{电子}_{n+1}=\mathcal{电子}_{0}+\frac{tau{t{n}^{alpha-1}}(1-{boldsymbol{alpha}})}{\Gamma({boldsymbol{\alpha}){\mathcal{F}(F)_{3} (t{n},\mathcal{T}(T)_{n} ,\mathcal{高}_{n} ,\mathcal{电子}_{n} )\\\hphantom{\mathcal{电子}_{n+1}=}{}+\frac{\tau\hbar^{{\boldsymbol{\alpha}}}{\Gamma{F}(F)_{3} (t_{\泽塔},\mathcal{T}(T)_{\zeta},\mathcal{高}_{\zeta},\mathcal{E}_{\zeta})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}(-\zeta+n+{\bold symbol}\alpha}+2)\\hphantom{\mathcal{电子}_{n+1}=}{}-(-\zeta+n)^{{\boldsymbol{\alpha}}}(-\zeta+n+2{\bolssymbol}\alpha}+2)\bigr]\\hphantom{\mathcal{电子}_{n+1}=}{}-{t{\zeta-1}^{\alpha-1}}\mathcal{F}(F)_{3} (t{\zeta-1},\mathcal{T}(T)_{\zeta-1},\mathcal{高}_{\zeta-1},\mathcal{E}_{\zeta-1})\bigl[(-\zeta+1+n)^{\boldsymbol{\alpha}}+1}-(-\zeta+n)#^{\bloldsymbol{\alfa}}}(-\ zeta+n+{\bold symbol}\alpha{}+1)\bigr]\bigr],\end{聚集}\end{aligned}$$
(79)
哪里
$$\开始{聚集}\mathcal{F}(F)_{1} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}\beta_{13}\mathcal{T}({\eta})\mathcal{E}(}\eta{),\\mathcal{F}(F)_{2} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}})\mathcal{H}({\eta})\ bigr],\\mathcal{F}(F)_{3} \bigl({\eta},\mathcal{T}({\ta}),\mathcal{H}T}({\eta})\mathcal{E}。\结束{聚集}$$
(80)
6.6数值模拟
数字16–18演示模型中状态变量的变化(75)when方案(79)用于不同的值\(tau\ in(0,1]\)和\(α=0.95\)在本次模拟中,我们考虑了模型中的以下值:\(β{12}=1\),\(β{13}=2.5),\(k{2}=0.6),\(β{21}=1.5\),\(k{3}=4.5\),\(s{3}=1\),\(a{31}=0.2)、和\(d_{3}=0.5\)。在我们的数值模拟中,\(t{\mathrm{final}}=500\)和\(hbar=0.001)。在图中16,我们采取\((\mathcal{T}(T),\mathcal{H}(T),\mathcal{E}(c))|_{T=0}=(0.1,0.1,0.1)\)在这种情况下,模型表现出混沌吸引子行为。此外,通过考虑\((mathcal{T}(T),mathcal}H}(T),mathcal{E}(c))|_{T=0}=(0.3,0.3,0.3))和\(β{12}=0.745)该模型显示了图中的极限环行为17,用于\(\mathcal{T}(0)=0.3517\),\(数学{H}(0)=0.1115),\(数学{E}(0)=0.4951)、和\(β{12}=0.920),我们在图中所示的解中得到了周期轨道轨迹18.
在图中19和20,我们还研究了当两个参数\(β{12})和\(c{3}\)分别进行更改。在这两幅图中,我们可以看到,每个参数的变化都会导致变量行为发生一些有意义和有建设性的变化。作为一个生物学结论,我们可以指出这样一个事实,即只有当效应细胞的募集率大于癌细胞的失活率时,才能实现模型的稳定性。换句话说,如果免疫系统无法检测和攻击癌细胞,那么必须使用有效的治疗来控制肿瘤的生长。