1. Introduction
2. Basic definitions and preliminaries
3. The physical system
3.1. 1D system
Figure 1. A bead on a Lemniscate. |
3.2. 2D system
4. Numerical procedure
• | for the 1D system: $\begin{eqnarray}\begin{array}{c}\left\{\begin{array}{l}{}_{a}^{{\rm{ABC}}}{{\rm{D}}}_{t}^{\alpha }{x}_{1}={x}_{2},\\ {}_{a}^{{\rm{ABC}}}{{\rm{D}}}_{t}^{\alpha }\tfrac{1}{1+{x}_{1}^{4}}{x}_{2}=\tfrac{2{x}_{1}^{3}}{{\left(1+{x}_{1}^{4}\right)}^{2}}{x}_{2}^{2}+{\epsilon }^{2}\tfrac{1-{x}_{1}^{4}}{{\left(1+{x}_{1}^{4}\right)}^{2}}{x}_{1},\end{array}\right.\end{array}\end{eqnarray}$ |
• | for the 2D system: $\begin{eqnarray}\begin{array}{c}\left\{\begin{array}{l}{}_{a}^{{\rm{ABC}}}{{\rm{D}}}_{t}^{\alpha }{r}_{1}={r}_{2},\\ {}_{t}^{{\rm{ABC}}}{{\rm{D}}}_{b}^{\alpha }\tfrac{1}{1+{r}_{1}^{4}}{r}_{2}=-\tfrac{{r}_{1}-{r}_{1}^{5}}{{\left(1+{r}_{1}^{4}\right)}^{2}}{\phi }_{2}^{2}+\tfrac{2{r}_{1}^{3}}{{\left(1+{r}_{1}^{4}\right)}^{2}}{r}_{2}^{2}+{\epsilon }^{2}\tfrac{{r}_{1}-{r}_{1}^{5}}{{\left(1+{r}_{1}^{4}\right)}^{2}},\\ {}_{a}^{{\rm{ABC}}}{{\rm{D}}}_{t}^{\alpha }{\phi }_{1}={\phi }_{2},\\ {}_{t}^{{\rm{ABC}}}{{\rm{D}}}_{b}^{\alpha }\tfrac{{r}_{1}^{2}}{1+{r}_{1}^{4}}{\phi }_{2}=0.\end{array}\right.\end{array}\,\end{eqnarray}$ Exerting the ABC integral operator [22] into equations ( $\begin{eqnarray}\left\{\begin{array}{l}{x}_{1}(t)={x}_{1}(a)+\tfrac{1-\alpha }{N(\alpha )}{x}_{2}(t)+\tfrac{\alpha }{{\rm{\Gamma }}(\alpha )N(\alpha )}{\displaystyle \int }_{a}^{t}{x}_{2}(\gamma )\\ \times \,{\left(t-\gamma \right)}^{\alpha -1}{\rm{d}}\gamma ,\\ \\ {x}_{2}(t)=\tfrac{1-\alpha }{N(\alpha )}\left(\tfrac{2{x}_{1}^{3}(t)}{1+{x}_{1}^{4}(t)}{x}_{2}^{2}(t)+{\epsilon }^{2}\tfrac{1-{x}_{1}^{4}(t)}{1+{x}_{1}^{4}(t)}{x}_{1}(t\right)\\ +\tfrac{\alpha (1+{x}_{1}^{4}(t))}{{\rm{\Gamma }}(\alpha )N(\alpha )}{\displaystyle \int }_{t}^{b}\left(\tfrac{2{x}_{1}^{3}(\gamma )}{{\left(1+{x}_{1}^{4}(\gamma \right)}^{2}}{x}_{2}^{2}(\gamma )+{\epsilon }^{2}\tfrac{1-{x}_{1}^{4}(\gamma )}{{\left(1+{x}_{1}^{4}(\gamma \right)}^{2}}{x}_{1}(\gamma \right)\\ \times \,{\left(\gamma -t\right)}^{\alpha -1}{\rm{d}}\gamma ,\end{array}\right.\end{eqnarray}$ $\begin{eqnarray}\left\{\begin{array}{l}{r}_{1}(t)={r}_{1}(a)+\tfrac{1-\alpha }{N(\alpha )}{r}_{2}(t)+\tfrac{\alpha }{{\rm{\Gamma }}(\alpha )N(\alpha )}{\displaystyle \int }_{a}^{t}{r}_{2}(\gamma )\\ {\left(t-\gamma \right)}^{\alpha -1}{\rm{d}}\gamma ,\\ \\ {r}_{2}(t)=\tfrac{1-\alpha }{N(\alpha )}\left(-\tfrac{{r}_{1}(t)-{r}_{1}^{5}(t)}{1+{r}_{1}^{4}(t)}{\phi }_{2}^{2}(t)+\tfrac{2{r}_{1}^{3}(t)}{1+{r}_{1}^{4}(t)}{r}_{2}^{2}(t)\right.\\ +\left.{\epsilon }^{2}\tfrac{{r}_{1}(t)-{r}_{1}^{5}(t)}{1+{r}_{1}^{4}(t)}\right)\\ +\tfrac{\alpha (1+{r}_{1}^{4}(t))}{{\rm{\Gamma }}(\alpha )N(\alpha )}{\displaystyle \int }_{t}^{b}\left(-\tfrac{{r}_{1}(\gamma )-{r}_{1}^{5}(\gamma )}{{\left(1+{r}_{1}^{4}(\gamma \right)}^{2}}{\phi }_{2}^{2}(\gamma )+\tfrac{2{r}_{1}^{3}(\gamma )}{{\left(1+{r}_{1}^{4}(\gamma \right)}^{2}}{r}_{2}^{2}(\gamma )\right.\\ +\left.{\epsilon }^{2}\tfrac{{r}_{1}(\gamma )-{r}_{1}^{5}(\gamma )}{{\left(1+{r}_{1}^{4}(\gamma \right)}^{2}}\right){\left(\gamma -t\right)}^{\alpha -1}{\rm{d}}\gamma ,\\ {\phi }_{1}(t)={\phi }_{1}(a)+\tfrac{1-\alpha }{N(\alpha )}{\phi }_{2}(t)+\tfrac{\alpha }{{\rm{\Gamma }}(\alpha )N(\alpha )}{\displaystyle \int }_{a}^{t}{\phi }_{2}(\gamma )\\ \times \,{\left(t-\gamma \right)}^{\alpha -1}{\rm{d}}\gamma ,\\ \\ {\phi }_{2}(t)=\tfrac{\alpha }{{\rm{\Gamma }}(\alpha )N(\alpha )}\left({r}_{1}^{2}(t)+\tfrac{1}{{r}_{1}^{2}(t)}\right)(b-t),\end{array}\right.\end{eqnarray}$ where x1(a), r1(a) and φ1(a) are the initial values and x2(b) = r2(b) = φ2(b) = 0. Now, we consider the length of time step ${{\ell }}_{M}=\tfrac{b-a}{M}$ for a uniform partition on [a, b] where M is a positive integer that can be selected optionally. Moreover, we represent the numerical approximations of xp(tq), rp(tq), and φp(tq) by ${x}_{p,q},{r}_{p,q}$, and ${\phi }_{p,q}$, respectively, where p = 1, 2, and ${t}_{q}\,=a+q{{\ell }}_{M},0\leqslant q\leqslant M$ is the time instant at the q-th node. Afterwards, the fractional Euler approach [31] is applied to discretize the convolution integrals in equations ( $\begin{eqnarray}\left\{\begin{array}{l}{X}_{1}-\tfrac{1-\alpha }{N(\alpha )}{X}_{2}-\tfrac{\alpha }{N(\alpha )}{A}_{M,\alpha }{X}_{2}={X}_{\mathrm{1,0}},\\ \\ {X}_{2}-\tfrac{1-\alpha }{N(\alpha )}{{\rm{\Lambda }}}_{1}({X}_{1},{X}_{2})-\tfrac{\alpha }{N(\alpha )}{{\rm{\Lambda }}}_{2}({X}_{1}){B}_{M,\alpha }{{\rm{\Lambda }}}_{3}({X}_{1},{X}_{2})=0,\end{array}\right.\end{eqnarray}$ $\begin{eqnarray}\left\{\begin{array}{l}{R}_{1}-\tfrac{1-\alpha }{N(\alpha )}{R}_{2}-\tfrac{\alpha }{N(\alpha )}{A}_{M,\alpha }{R}_{2}={R}_{\mathrm{1,0}},\\ \\ {R}_{2}-\tfrac{1-\alpha }{N(\alpha )}{{\rm{\Lambda }}}_{4}({R}_{1},{R}_{2},{{\rm{\Theta }}}_{2})-\tfrac{\alpha }{N(\alpha )}{{\rm{\Lambda }}}_{5}({R}_{1}){B}_{M,\alpha }{{\rm{\Lambda }}}_{6}\\ ({R}_{1},{R}_{2},{{\rm{\Theta }}}_{2})=0,\\ \\ {{\rm{\Theta }}}_{1}-\tfrac{1-\alpha }{N(\alpha )}{{\rm{\Theta }}}_{2}-\tfrac{\alpha }{N(\alpha )}{A}_{M,\alpha }{{\rm{\Theta }}}_{2}={{\rm{\Theta }}}_{\mathrm{1,0}},\\ \\ {{\rm{\Theta }}}_{2}-\tfrac{\alpha }{N(\alpha )}{{\rm{\Lambda }}}_{7}({R}_{1}){B}_{M,\alpha }=0,\end{array}\right.\end{eqnarray}$ where $\begin{eqnarray}\begin{array}{rcl}{A}_{M,\alpha } & = & {\left({B}_{M,\alpha }\right)}^{{\rm{T}}}={{\ell }}_{M}\left[\begin{array}{cccc}{\tau }_{0,\alpha } & 0 & \ldots & 0\\ {\tau }_{1,\alpha } & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0\\ {\tau }_{M,\alpha } & \ldots & {\tau }_{1,\alpha } & {\tau }_{0,\alpha }\end{array}\right],\\ {\tau }_{0,\alpha } & = & 1,\hspace{0.2cm}{\tau }_{q,\alpha }=\left(1+\displaystyle \frac{\alpha -1}{q}\right){\tau }_{q-1,\alpha },\hspace{0.2cm}q=1,2,\ldots ,\end{array}\end{eqnarray}$ $\begin{eqnarray}\begin{array}{rcl}{X}_{p} & = & \left[\begin{array}{c}{x}_{p,0}\\ \vdots \\ {x}_{p,M}\end{array}\right],\hspace{0.1cm}{R}_{p}=\left[\begin{array}{c}{R}_{p,0}\\ \vdots \\ {R}_{p,M}\end{array}\right],\hspace{0.1cm}{{\rm{\Theta }}}_{p}=\left[\begin{array}{c}{{\rm{\Theta }}}_{p,0}\\ \vdots \\ {{\rm{\Theta }}}_{p,M}\end{array}\right],\\ {X}_{p,0} & = & \left[\begin{array}{c}{x}_{p,0}\\ \vdots \\ {x}_{p,0}\end{array}\right],\hspace{0.1cm}{R}_{p,0}=\left[\begin{array}{c}{R}_{p,0}\\ \vdots \\ {R}_{p,0}\end{array}\right],\hspace{0.1cm}{{\rm{\Theta }}}_{p,0}=\left[\begin{array}{c}{{\rm{\Theta }}}_{p,0}\\ \vdots \\ {{\rm{\Theta }}}_{p,0}\end{array}\right],\hspace{0.1cm}p=1,2,\end{array}\end{eqnarray}$ $\begin{eqnarray}\begin{array}{c}\begin{array}{rcl}{{\rm{\Lambda }}}_{1}({X}_{1},{X}_{2}) & = & \left[\begin{array}{c}\tfrac{2{x}_{1,0}^{3}}{1+{x}_{1,0}^{4}}{x}_{2,0}^{2}+{\epsilon }^{2}\tfrac{1-{x}_{1,0}^{4}}{1+{x}_{1,0}^{4}}{x}_{1,0}\\ \vdots \\ \tfrac{2{x}_{1,M}^{3}}{1+{x}_{1,M}^{4}}{x}_{2,M}^{2}+{\epsilon }^{2}\tfrac{1-{x}_{1,M}^{4}}{1+{x}_{1,M}^{4}}{x}_{1,M}\end{array}\right],\\ {{\rm{\Lambda }}}_{3}({X}_{1},{X}_{2}) & = & \left[\begin{array}{c}\tfrac{2{x}_{1,0}^{3}}{{\left(1+{x}_{1,0}^{4}\right)}^{2}}{x}_{2,0}^{2}+{\epsilon }^{2}\tfrac{1-{x}_{1,0}^{4}}{{\left(1+{x}_{1,0}^{4}\right)}^{2}}{x}_{1,0}\\ \vdots \\ \tfrac{2{x}_{1,M}^{3}}{{\left(1+{x}_{1,M}^{4}\right)}^{2}}{x}_{2,M}^{2}+{\epsilon }^{2}\tfrac{1-{x}_{1,M}^{4}}{{\left(1+{x}_{1,M}^{4}\right)}^{2}}{x}_{1,M}\end{array}\right],\\ {{\rm{\Lambda }}}_{2}({X}_{1},{X}_{2}) & = & {\rm{diag}}\left[\begin{array}{ccc}1+{x}_{1,0}^{4} & \ldots & 1+{x}_{1,M}^{4}\end{array}\right],\end{array}\end{array}\end{eqnarray}$ as well as $\begin{eqnarray}\begin{array}{l}{{\rm{\Lambda }}}_{4}({R}_{1},{R}_{2},{{\rm{\Theta }}}_{2})\\ =\,\left[\begin{array}{c}-\tfrac{{r}_{\mathrm{1,0}}-{r}_{1,0}^{5}}{1+{r}_{1,0}^{4}}{\phi }_{2,0}^{2}+\tfrac{2{r}_{1,0}^{3}}{1+{r}_{1,0}^{4}}{r}_{2,0}^{2}+{\epsilon }^{2}\tfrac{{r}_{\mathrm{1,0}}-{r}_{1,0}^{5}}{1+{r}_{1,0}^{4}}\\ \vdots \\ -\tfrac{{r}_{1,M}-{r}_{1,M}^{5}}{1+{r}_{1,M}^{4}}{\phi }_{2,M}^{2}+\tfrac{2{r}_{1,M}^{3}}{1+{r}_{1,M}^{4}}{r}_{2,M}^{2}+{\epsilon }^{2}\tfrac{{r}_{1,M}-{r}_{1,M}^{5}}{1+{r}_{1,M}^{4}}\end{array}\right],\\ {{\rm{\Lambda }}}_{6}({R}_{1},{R}_{2},{{\rm{\Theta }}}_{2})\\ =\,\left[\begin{array}{c}-\tfrac{{r}_{\mathrm{1,0}}-{r}_{1,0}^{5}}{{\left(1+{r}_{\mathrm{1,0}}^{4}\right)}^{2}}{\phi }_{2,0}^{2}+\tfrac{2{r}_{1,0}^{3}}{{\left(1+{r}_{\mathrm{1,0}}^{4}\right)}^{2}}{r}_{2,0}^{2}+{\epsilon }^{2}\tfrac{{r}_{\mathrm{1,0}}-{r}_{1,0}^{5}}{{\left(1+{r}_{\mathrm{1,0}}^{4}\right)}^{2}}\\ \vdots \\ -\tfrac{{r}_{1,M}-{r}_{1,M}^{5}}{{\left(1+{r}_{1,M}^{4}\right)}^{2}}{\phi }_{2,M}^{2}+\tfrac{2{r}_{1,M}^{3}}{{\left(1+{r}_{1,M}^{4}\right)}^{2}}{r}_{2,M}^{2}+{\epsilon }^{2}\tfrac{{r}_{1,M}-{r}_{1,M}^{5}}{{\left(1+{r}_{1,M}^{4}\right)}^{2}}\end{array}\right],\end{array}\end{eqnarray}$ $\begin{eqnarray*}\begin{array}{c}\begin{array}{rcl}{{\rm{\Lambda }}}_{5}({R}_{1}) & = & {\rm{diag}}\left[\begin{array}{ccc}1+{r}_{1,0}^{4} & \ldots & 1+{r}_{1,M}^{4}\end{array}\right],\\ {{\rm{\Lambda }}}_{7}({R}_{1}) & = & {\rm{diag}}\left[\begin{array}{ccc}\left({r}_{1,0}^{2}+\tfrac{1}{{r}_{1,0}^{2}}\right)(b-0) & \ldots & \left({r}_{1,M}^{2}+\tfrac{1}{{r}_{1,M}^{2}}\right)(b-M)\end{array}\right].\end{array}\end{array}\end{eqnarray*}$ |
5. Simulation results and discussion
Figure 2. Simulation results of x(t) for the 1D case when x(0) = 0.1 and ϵ = 1. |
Figure 3. Simulation results of r(t) for the 2D case when r(0) = 0.1, φ(0) = 0 and ϵ = 1. |
Figure 4. Simulation results of φ(t) for the 2D case when r(0) = 0.1, φ(0) = 0 and ϵ = 1. |