An asymmetric model two-dimensional oscillator

Abstract

We present a novel 2D oscillator with an asymmetric design and investigate its stable vibration utilizing the Ms-DTM method. Initially, we obtain the equations of motion for the proposed system. Subsequently, by employing Taylor expansion of $sinhz$ and $coshz$ , the derived nonlinear equations are transformed into linear ones, which we solve analytically using the eigenvalues-eigenfunctions technique. Additionally, we solve the obtained nonlinear system using the Ms-DTM method. Lastly, we examine the stability of the nonlinear system by visualizing the closed nature of the phase portrait.

Introduction

Differential equations for nonlinear oscillators are used in many areas of science such as: Physics, Structures, and Engineering. The study of these equations is an important topic for researchers. Significant efforts have been made to study them in the literature, and a large number of techniques (analytical and numerical) have been introduced and used to solve such systems.^1–3

Two key characteristics of nonlinear oscillators, which produce complex motion, are that motion becomes more significant as amplitude grows due to the nonlinearity, and in certain situations, the frequency will also change as amplitude changes. These kinds of nonlinear oscillators are common in the real world, and it’s important to remember that linked nonlinear oscillators are studied in a wide range of scientific fields, including biology and economics. The systems literature has been studied extensively.^4–8 One notable example is the van der Pol oscillator, a nonlinearly damping oscillator that was first presented in the 1920s by Balthasar van der Pol (1889–1959). Since it is difficult to find an analytical solution for such nonlinear oscillators, researchers have turned to a variety of numerical techniques.^9–12

In Reference 13, the researchers investigate the dynamic analysis of beam vibrations in the presence of a preload discontinuity. Additionally, they introduce a new and accurate equivalent function to effectively address this widely recognized nonlinearity. In another study Alborz ¹⁴ deals with the analytical solutions of nonlinear oscillators under initial velocity, and by applying the law of conservation of energy transforms the initial state of the system into a state where the velocity of the oscillator is zero.

If the velocity of the oscillator is zero and is outside the origin of motion, we can determine the relationship between frequency and amplitude. The studies of oscillators remain the prime and classic example to understand many physical, chemical as well as biological systems. In fact, numerous systems associated with interacting/coupled oscillators have been used to describe the cooperative activities of several systems.^15–21 The equation of motion of symmetric oscillator in two dimensions can be written as

$\ddot{x}+x=0$

$\ddot{y}+y=0$

The above equations can be rewritten with the force term as

$\ddot{x}=f_1\left(x\right)$

$\ddot{y}=f_2\left(y\right)$

Here, the forces can be expressed as $f_1\left(x\right)=\frac{dV}{dx}$ and $f_2\left(y\right)=\frac{dV}{dy}$ . However, if the forces are different from each other then the corresponding equation of motion can be written as

$\ddot{x}=f_1\left(x,y\right)$

$\ddot{y}=f_2\left(x,y\right)$

In the above equation, $f_1\left(x,y\right)\ne f_2\left(x,y\right)$ .

The equation of motion as states above would represent an asymmetric oscillator. In order to understand the asymmetric oscillator, one needs to study the variation of $x$ and $y$ jointly. It is worth mentioning here that the one may not find the exact analytical solution for some of these types of systems. However, one needs to use some approximation to find the partial analytical result. In other word, a graphical representation of $x$ versus $y$ or vice versa need not be circular in any case. It may be either prolate or oblate. Such systems may further be studied by combining the approximated analytical results with their numerical counterpart.

Description of the physical system

Consider asymmetric two-dimensional oscillator with the following Hamiltonian

$H=\frac{1}{2}\left[p_x^2+p_y^2+x^2+y^2\right]+\gamma xsinh\beta y.$

(1)

Here for simplicity, we assumed that $m=\omega =1$ . In the above relation, $x,y,p_x,and\hspace{0.17em}{p}_y$ are the usual classical positions and momentum variables.

The Lagrangain and Hamiltonian are related to each other as

$H={\displaystyle \sum }_{i=1}{p}_i{\dot{q}}_i-L.$

(2)

with $i=1,2;\hspace{0.17em}x,y$ .

So, we can write equation (2) as

$H=p_x\dot{x}+p_y\dot{y}-L.$

(3)

From equations (1) and (3) we have

$L=p_x\dot{x}+p_y\dot{y}-H.$

Now, using $\frac{d{q}_i}{dt}=\frac{\partial H}{\partial p_i}$ . So, for $q_i=x,and\hspace{0.17em}y$ , respectively, we have

$\frac{dx}{dt}=\frac{\partial H}{\partial p_x}\to \dot{x}=p_x.$

(4a)

$\frac{dy}{dt}=\frac{\partial H}{\partial p_y}\to \dot{y}=p_y.$

(4b)

On substituting equations (4a), (4b) into equation (3) and simplifying one gets

$L=\frac{1}{2}\left[{\dot{x}}^2+{\dot{y}}^2-x^2-y^2\right]-\gamma xsinh\beta y.$

(5)

Finally, the equations of motion (EOM’s) can be derived from equation (5) using the following formula $\frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_i}=0$ . So, for $i=x,and\hspace{0.17em}y$ , respectively, we obtained the following two EOM’s

$\ddot{x}+x+\gamma \hspace{0.17em}\sinh \hspace{0.17em}\beta y=0,$

(6)

$\ddot{y}+y+\gamma \beta x\hspace{0.17em}\cosh \hspace{0.17em}\beta y=0.$

(7)

Analytical solution of the model

Consider the following two equations

$\ddot{x}+x+10\hspace{0.17em}\sinh \left(0.1y\right)=0,$

(8)

$\ddot{y}+y+x\hspace{0.17em}\cosh \left(0.1y\right)=0.$

(9)

These two equations are non-linear due to the presence of ( $\sinh$ and $\cosh \hspace{0.17em})$ .

But the expansions of them are

$\sinh \hspace{0.17em}z=z+\frac{z^3}{3!}+\frac{z^5}{5!}+\dots$

(10)

$\cosh \hspace{0.17em}z=1+\frac{z^2}{2!}+\frac{z^4}{4!}+\dots$

(11)

Substituting equations (10) and (11) into equations (8) and (9) and keeping only terms up to power $o^{2}z=0.1y$ , we get

$\ddot{x}+x+y=0$

(12)

$\ddot{y}+y+x=0$

(13)

where equations (12) and (13) are linear and can be solved analytically as follows:

Let

$\frac{dx}{dt}=u_1,$

(14)

$\frac{dy}{dt}=u_2.$

(15)

So, equations (12) and (13) become

$\frac{d{u}_1}{dt}=-x-y,$

(16)

$\frac{d{u}_2}{dt}=-y-x.$

(17)

The system equations (16) and (17) can be written in matrix form as

$\frac{d}{dt}\left[\begin{array}{l}x\\ u_1\\ y\\ u_2\end{array}\right]=\left[\begin{array}{llll}0& 1& 0& 0\\ -1& 0& -1& 0\\ 0& 0& 0& 1\\ -1& 0& -1& 0\end{array}\right]\left[\begin{array}{l}x\\ u_1\\ y\\ u_2\end{array}\right].$

(18)

To solve equation (18), we need to find the eigenvalues and eigenvectors of matrix $A$ where

$A=\left[\begin{array}{llll}0& 1& 0& 0\\ -1& 0& -1& 0\\ 0& 0& 0& 1\\ -1& 0& -1& 0\end{array}\right]$

$\det \left(A-\lambda I\right)=0\stackrel{yields}{\to }{\lambda }^4+2{\lambda }^2=0$

Let

$\epsilon ={\lambda }^2\stackrel{yields}{\to }{\epsilon }_1=0\hspace{0.17em}\text{and}\hspace{0.17em}{\epsilon }_2=-2.$

As a result, we have four eigenvalues ${\lambda }_1=0,{\lambda }_2=0,{\hspace{0.17em}\lambda }_3=i\sqrt{2},and\hspace{0.17em}{\lambda }_4=-i\sqrt{2}$ and the corresponding normalized eigenvectors, respectively, are

$\left[\begin{array}{l}\frac{-1}{\sqrt{2}}\\ 0\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right],\left[\begin{array}{l}\frac{1}{2}\\ \frac{-1}{2}\\ \frac{-1}{2}\\ \frac{1}{2}\end{array}\right],\left[\begin{array}{l}-\frac{i\sqrt{2}}{2}\\ 1\\ \frac{-i\sqrt{2}}{2}\\ 1\end{array}\right],\text{and}\hspace{0.17em}\left[\begin{array}{l}\frac{i\sqrt{2}}{2}\\ 1\\ \frac{i\sqrt{2}}{2}\\ 1\end{array}\right]$

Finally, the general solution is

$\left[\begin{array}{l}x\\ u_1\\ y\\ u_2\end{array}\right]=C_1\left[\begin{array}{l}\frac{-1}{\sqrt{2}}\\ 0\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right]\exp \left(0t\right)+C_2\left\{\left[\begin{array}{l}\frac{-1}{\sqrt{2}}\\ 0\\ \frac{1}{\sqrt{2}}\\ 0\end{array}\right]\text{texp}\left(0t\right)+\left[\begin{array}{l}\frac{1}{2}\\ \frac{-1}{2}\\ -\frac{1}{2}\\ \frac{1}{2}\end{array}\right]\exp \left(0t\right)\right\}+C_3\left[\begin{array}{l}-\frac{i\sqrt{2}}{2}\\ 1\\ \frac{-i\sqrt{2}}{2}\\ 1\end{array}\right]\exp \left(i\sqrt{2}t\right)+C_4\left[\begin{array}{l}\frac{i\sqrt{2}}{2}\\ 1\\ \frac{i\sqrt{2}}{2}\\ 1\end{array}\right]\exp \left(-i\sqrt{2}t\right).$

(19)

From equation (19), we can find $x\left(t\right),u_1\left(t\right),y\left(t\right),{\hspace{0.17em}\text{and}\hspace{0.17em}u}_2\left(t\right)$

$x\left(t\right)=-\frac{C_3}{\sqrt{2}}+\frac{C_4}{2}+\frac{i{C}_2{e}^{-i\sqrt{2}t}}{\sqrt{2}}-\frac{i{C}_1{e}^{i\sqrt{2}t}}{\sqrt{2}}-\frac{C_{4}t}{\sqrt{2}}$

(20a)

$u_1\left(t\right)-\frac{C_4}{2}+C_2{e}^{-i\sqrt{2}t}+C_1{e}^{i\sqrt{2}t},$

(20b)

$y\left(t\right)=\frac{C_3}{\sqrt{2}}-\frac{C_4}{2}+\frac{i{C}_2{e}^{-i\sqrt{2}t}}{\sqrt{2}}-\frac{i{C}_1{e}^{i\sqrt{2}t}}{\sqrt{2}}+\frac{C_{4}t}{\sqrt{2}},$

(20c)

$u_2\left(t\right)=\frac{C_3}{2}+C_2{e}^{-i\sqrt{2}t}+C_1{e}^{i\sqrt{2}t}$

(20d)

Also using the initial conditions $x\left(0\right)=0,u_1\left(0\right)=1,y\left(0\right)=0,{\hspace{0.17em}and\hspace{0.17em}u}_2\left(0\right)=1$ , we get $C_1=C_2=0$ , ${C_3=C}_4=\frac{1}{2}$

$x\left(t\right)=\frac{i{e}^{-i\sqrt{2}t}}{2\sqrt{2}}-\frac{i{e}^{i\sqrt{2}t}}{2\sqrt{2}}=\frac{\sin \hspace{0.17em}\left(\sqrt{2}t\right)}{\sqrt{2}},$

(21)

$u_1\left(t\right)=\frac{1}{2}{e}^{-i\sqrt{2}t}+\frac{1}{2}{e}^{i\sqrt{2}t}=\cos \left(\sqrt{2}t\right),$

(22)

$y\left(t\right)=\frac{i{e}^{-i\sqrt{2}t}}{2\sqrt{2}}-\frac{i{e}^{i\sqrt{2}t}}{2\sqrt{2}}=\frac{\sin \hspace{0.17em}\left(\sqrt{2}t\right)}{\sqrt{2}},$

(23)

$u_2\left(t\right)=\frac{1}{2}{e}^{-i\sqrt{2}t}+\frac{1}{2}{e}^{i\sqrt{2}t}=\cos \left(\sqrt{2}t\right).$

(24)

In Figure 1, we simulate both the position $x\left(t\right)$ and the velocity $u_1\left(t\right)$ against time, respectively, while the phase plots are shown in Figure 2 for both $x\hspace{0.17em}\text{and}\hspace{0.17em}y\hspace{0.17em}\text{directions}$ .

Figure 1.

Simulation of analytical solution of the system equations (21) and (22) versus time.

Figure 2.

Simulation of analytical phase portrait of the system equations (21) and (24).

Solution of the system via Ms-DTM

In this section, we aimed to solve the EOM’s given in equations (6) and (7). These equations are nonlinear second order equations and they can’t be solved analytically. One of the semi-analytical methods that can be used to solve them is the Multi- Step Differential Transformation Method (Ms- DTM).

The Ms- DTM

This section is reserved for the Ms-DTM for obtaining the numerical solution of DEs. For this purpose, we consider the following nonlinear initial value problem

$f\left(t,y,y^{\prime },\dots ,y^{\left(p\right)}\right)=0,y^{\left(p\right)}\left(0\right)=c_k,\text{for}\hspace{0.17em}k=0,1,\dots ,p-1..$

(25)

Let’s assume that we want to find the solution of the problem equation (25) on a given $\left[0,T\right]$ interval.

In the real applications, the approximate solution of the problem equation (25) via the DTM can be expressed by the following finite series

$y\left(t\right)={\displaystyle \sum }_{n=0}^N{a}_n{t}^n.$

(26)

The multi-step approach considers the establishment of the approximate solution by a different point of view. Suppose that the interval $\left[0,T\right]$ is divided into $M$ subintervals ${[t}_0,t_1],{[t}_1,t_2],\dots ,{[t}_{M-1},t_M]$ of equal step size $h=\frac{T}{M},$ where ${\hspace{0.17em}t}_m=m.h$ . The basic steps of the DTM are as follows.

First, over the first interval $\left[0,t_1\right],$ using the initial conditions ${y_1}^{\left(k\right)}\left(0\right)=c_k,$ the DTM is applied to equation (25), thus the following approximate solution is obtained

$y_1\left(t\right)={\displaystyle \sum }_{n=0}^K{a}_{1n}{t}^n,t\in \left[0,t_1\right],$

(27)

For $m\ge 2$ and at each subinterval ${[t}_{m-1},t_m]$ , the initial conditions ${y_m}^{\left(k\right)}\left(t_{m-1}\right)={y_{m-1}}^{\left(k\right)}\left(t_{m-1}\right)$ are used and then the DTM is applied to equation (25) over the interval ${[t}_{m-1},t_m]$ , where $t_0$ is altered with $t_{m-1}$ .

The operation is repeated again and again and for the solution of $y\left(t\right),$ a sequence of approximate solutions $y_m\left(t\right),m=1,2,\dots ,M,$ is generated as follows

$y_m\left(t\right)={\displaystyle \sum }_{n=0}^N{a}_{mn}{\left(t-t_{m-1}\right)}^n,t\in \left[t_m,t_{m-1}\right],$

(28)

where $N=K.M$ . In actual fact, the Ms- DTM suggests the following solution

$y\left(t\right)=\{\begin{array}{l}{y}_1\left(t\right),t\in \left[0,t_1\right]\hspace{0.17em}\\ {\hspace{0.17em}y}_2\left(t\right),t\in \left[t_1,t_2\right]\\ ⋮\\ {\hspace{0.17em}y}_m\left(t\right),t\in \left[t_{M-1},t_M\right]\end{array}$

(29)

The Ms- DTM has a computationally simple performance for every desired value of $h$ . If the step size $h$ equals to $T$ , then it is easily observed this approach reduces to the classical DTM.

In the next section, the principal advantage of the above algorithm is that the obtained series solution converges for wide range of time is to be seen.

Simulation results and discussion

In this section, the equations of motion of the two-dimensional oscillator given in equations (6) and (7) are solved by an efficient, fast, and high accurate method, namely the multi-step differential transform method. The simulation results for the system given in equations (6) and (7) are presented for $\gamma =10$ and $\beta =0.1$ .

Now let us consider the conditions $x\left(0\right)=0,u_1\left(0\right)=1,y\left(0\right)=0,{\hspace{0.17em}\text{and}\hspace{0.17em}u}_2\left(0\right)=1$ . In view of the multi-step differential transform method, given in the previous section, and by using differential transform formulas and applying the differential transform to equations (6) and (7), we obtain

$\{\begin{array}{l}\left(k+1\right)\left(k+2\right)X_i\left(k\right)+X_i\left(k\right)+\gamma F\left(k\right)=0\\ \left(k+1\right)\left(k+2\right)Y_i\left(k\right)+Y_i\left(k\right)+\gamma \beta {\displaystyle \sum }_{l=0}^k{X}_i\left(l\right)G\left(k-l\right)=0\end{array}.$

(30)

or

$\{\begin{array}{l}{X}_i\left(k\right)=-\frac{X_i\left(k\right)+\gamma F\left(k\right)}{\left(k+1\right)\left(k+2\right)}\\ Y_i\left(k\right)=-\frac{Y_i\left(k\right)+\gamma \beta {\displaystyle \sum }_{l=0}^k{X}_i\left(l\right)G\left(k-l\right)}{\left(k+1\right)\left(k+2\right)}\end{array},$

(31)

where $X\left(k\right),Y\left(k\right),F\left(k\right)$ and $G\left(k\right)$ are the differential transformations of $x\left(t\right)$ , $y\left(t\right),\sinh \left(\beta y\left(t\right)\right)$ , and $\cosh \left(\beta y\left(t\right)\right)$ , respectively. It is known that $F\left(k\right)$ and $G\left(k\right)$ are defined as

$F\left(k\right)=\{\begin{array}{ll}\sinh \left(\beta Y_i\left(0\right)\right)& k=0\\ \beta {\displaystyle \sum }_{m=0}^{k-1}\frac{k-m}{k}G\left(m\right)Y_i\left(k-m\right)& k\ge 1\end{array}$

and

$G\left(k\right)=\{\begin{array}{ll}\cosh \left(\beta Y_i\left(0\right)\right)& k=0\\ \beta {\displaystyle \sum }_{m=0}^{k-1}\frac{k-m}{k}F\left(m\right)Y_i\left(k-m\right)& k\ge 1\end{array}$

from. ²²

The transformed initial conditions at $t_0=0$ are

$\{\begin{array}{l}{X}_i\left(1\right)=X_i^{\prime }\left(t_i\right),i=1,2,\dots ,K\le M\\ Y_i\left(1\right)=Y_i^{\prime }\left(t_i\right),i=1,2,\dots ,K\le M\\ X_i\left(0\right)=X_{i-1}\left(t_i\right),{\mathrm{X}}_0\left(0\right)=0,{\mathrm{X}}_0\left(1\right)=1\hspace{0.17em}\\ Y_i\left(0\right)=Y_{i-1}\left(t_i\right),Y_0\left(0\right)=0,{\mathrm{Y}}_0\left(1\right)=1\hspace{0.17em}\end{array}.$

(32)

After solving equation (31) and using the equation (32) the solution for the system given in equations (6) and (7) can be obtained for any each time step. For this case, the following initial conditions were considered: $x\left(0\right)=y\left(0\right)=0$ and $x^{\prime }\left(0\right)=y^{\prime }\left(0\right)=1$ . Thus, we obtain the following solutions for the time step 0.1. It should be note that the solutions for $x\left(t\right)$ and $x^{\prime }\left(t\right)$ in other intervals are not given here because they would take up too much space.

$x\left(t\right)=\{\begin{array}{ll}0.+1.\hspace{0.17em}\left(0.+t\right)\hspace{0.17em}-\hspace{0.17em}0.333333\hspace{0.17em}(0.+t)\wedge 3& 0.\le t\hspace{0.17em}<\hspace{0.17em}0.1\\ 0.0996667+0.99\hspace{0.17em}\left(-0.1+t\right)\hspace{0.17em}-\hspace{0.17em}0.0996675\hspace{0.17em}(-0.1+t)\wedge 2-\hspace{0.17em}0.330008\hspace{0.17em}(-0.1+t)\wedge 3& 0.1\le t\hspace{0.17em}<\hspace{0.17em}0.2\\ \begin{array}{l}0.19734+0.960166\hspace{0.17em}\left(-0.2+t\right)\hspace{0.17em}-\hspace{0.17em}0.197346\hspace{0.17em}(-0.2+t)\wedge 2-\hspace{0.17em}0.320086\hspace{0.17em}(-0.2+t)\wedge 3\\ 0.291063+0.911094\hspace{0.17em}\left(-0.3+t\right)\hspace{0.17em}-\hspace{0.17em}0.291083\hspace{0.17em}(-0.3+t)\wedge 2-\hspace{0.17em}0.303762\hspace{0.17em}(-0.3+t)\wedge 3\\ \begin{array}{l}0.378958+0.843765\hspace{0.17em}\left(-0.4+t\right)\hspace{0.17em}-\hspace{0.17em}0.379003\hspace{0.17em}(-0.4+t)\wedge 2-\hspace{0.17em}0.281353\hspace{0.17em}(-0.4+t)\wedge 3\\ ⋮\\ \begin{array}{l}0.42961\hspace{0.17em}-\hspace{0.17em}0.802654\hspace{0.17em}\left(-19.5+t\right)\hspace{0.17em}-\hspace{0.17em}0.424418\hspace{0.17em}(-19.5+t)\wedge 2+0.267849\hspace{0.17em}(-19.5+t)\wedge 3\\ 0.345368\hspace{0.17em}-\hspace{0.17em}0.879502\hspace{0.17em}\left(-19.6+t\right)\hspace{0.17em}-\hspace{0.17em}0.340092\hspace{0.17em}(-19.6+t)\wedge 2+0.293433\hspace{0.17em}(-19.6+t)\wedge 3\\ \begin{array}{l}0.25431\hspace{0.17em}-\hspace{0.17em}0.938717\hspace{0.17em}\left(-19.7+t\right)\hspace{0.17em}-\hspace{0.17em}0.24896\hspace{0.17em}(-19.7+t)\wedge 2+0.313137\hspace{0.17em}(-19.7+t)\wedge 3\\ \begin{array}{l}0.158262\hspace{0.17em}-\hspace{0.17em}0.979115\hspace{0.17em}\left(-19.8+t\right)-\hspace{0.17em}0.152847\hspace{0.17em}{\left(-19.8+t\right)}^2+0.326574\hspace{0.17em}{\left(-19.8+t\right)}^3\\ 0.0591486\hspace{0.17em}-\hspace{0.17em}0.999887\hspace{0.17em}\left(-19.9+t\right)\hspace{0.17em}-\hspace{0.17em}0.0536752\hspace{0.17em}(-19.9+t)\wedge 2+0.333482\hspace{0.17em}(-19.9+t)\wedge 3\end{array}\end{array}\end{array}\end{array}\end{array}& \begin{array}{l}0.2\le t<0.3\\ 0.3\le t\hspace{0.17em}<\hspace{0.17em}0.4\\ \begin{array}{l}0.\hspace{0.17em}4\le t\hspace{0.17em}<\hspace{0.17em}0.5\\ ⋮\\ \begin{array}{l}19.5\le t<19.6\\ 19.6\le t<19.7\\ \begin{array}{l}19.7\le t<19.8\\ \begin{array}{l}19.8\le t<19.9\\ 19.9\le t<20\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}$

$x^{\prime }\left(t\right)=\{\begin{array}{ll}1\hspace{0.17em}-\hspace{0.17em}1.\hspace{0.17em}t\wedge 2& 0.\le t\hspace{0.17em}<\hspace{0.17em}0.1\\ -0.330008\hspace{0.17em}\left(-3.03033+0.0040304\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)& 0.1\le t\hspace{0.17em}<\hspace{0.17em}0.2\\ \begin{array}{l}-0.320086\hspace{0.17em}\left(-3.12633+0.0330817\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ -0.303762\hspace{0.17em}\left(-3.30433+0.116526\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ \begin{array}{l}-0.281353\hspace{0.17em}\left(-3.59661+0.294142\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ ⋮\\ \begin{array}{l}0.267849\hspace{0.17em}\left(1199.55\hspace{0.17em}-\hspace{0.17em}120.169\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ \begin{array}{l}0.293433\hspace{0.17em}\left(1194.92\hspace{0.17em}-\hspace{0.17em}119.918\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ 0.313137\hspace{0.17em}\left(1192.6\hspace{0.17em}-\hspace{0.17em}119.79\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ \begin{array}{l}0.326574\hspace{0.17em}\left(1191.66\hspace{0.17em}-\hspace{0.17em}119.736\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\\ 0.333482\hspace{0.17em}\left(1191.44\hspace{0.17em}-\hspace{0.17em}119.722\hspace{0.17em}t+3\hspace{0.17em}t\wedge 2\right)\end{array}\end{array}\end{array}\end{array}\end{array}& \begin{array}{l}0.2\le t<0.3\\ 0.3\le t\hspace{0.17em}<\hspace{0.17em}0.4\\ \begin{array}{l}0.\hspace{0.17em}4\le t\hspace{0.17em}<\hspace{0.17em}0.5\\ ⋮\\ \begin{array}{l}19.5\le t<19.6\\ \begin{array}{l}19.6\le t<19.7\\ 19.7\le t<19.8\\ \begin{array}{l}19.8\le t<19.9\\ 19.9\le t<20\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}$

The behavior of $x\left(t\right)$ and x'(t) against time are presented with $\gamma =10$ and $\beta =0.1$ in Figure 3, while the phase plots are shown in Figure 4.

Figure 3.

Simulation of Ms- DTM results for both position $x\left(t\right)$ and velocity $u_1=\frac{dx}{dt}$ versus time.

Figure 4.

Simulation of Ms- DTM results for phase portrait for $u_1=\frac{dx}{dt}$ versus $x\left(t\right)$ and $y\left(t\right)$ versus $u_2=\frac{dy}{dt}$ .

Our results presented in Figures 1–4 show that Ms- DTM method used gives an excellent agreement with the analytical technique used.

Furthermore, the proposed model Hamiltonian can be considered as Hermitian as well as PT invariant satisfying the condition

$\left[H,PT\right]=0.$

(33)

Here $P$ stands for the Parity operator and $T$ stands for the time reversal operator satisfying the relations

$Px{P}^{-1}=-x\hspace{0.17em};\hspace{0.17em}P\hspace{0.17em}y\hspace{0.17em}P\wedge -1=-y.$

(34)

$P\hspace{0.17em}p{P}^{-1}=-p.$

(35)

$T\hspace{0.17em}i{T}^{-1}=-I.$

(36)

In this analogy intection term can be written as

$f\left(y,x\right)=ix\hspace{0.17em}(-isinh\left(\hspace{0.17em}\beta y\right).$

(37)

Further, it is easy to see that probability function reflects asymmetric nature. In Figures 5(a) and (b), we present probability function with and without interaction term. ²³

Figure 5.

Probability function (a) without and (b) with interaction term.

Conclusion

In this study, a two-dimensional asymmetric system has been considered. The equations of motion of the system were derived and solved analytically and using Ms- DTM method. We have noticed that the nonlinear equations of motion (i.e., equations (6) and (7)) can be transferred to a linear system as equations (12) and (13) by using Taylor expansion presented in equations (10) and (11). The linear system given in equations (12) and (13) can be solved analytically using the eigenvalues-eigenfunctions technique. In addition, the system given in equations (6) and (7) has been solved using the matrix method known as Ms-DTM. Finally, simulation results have been given in Figures 1–4.