In this section, we will discuss the dynamical behavior through bifurcation and sensitivity.
Bifurcation analysis
From (18), we can write as
$$begin{aligned} F^{”}=frac{alpha _1kappa ^2+alpha _4kappa -kappa omega }{alpha _1K^2}F+frac{alpha _3kappa +alpha _5kappa }{alpha _1K^2}F^3-frac{alpha _2}{alpha _1K^2}F^5. end{aligned}$$
(59)
Let (frac{alpha _1kappa ^2+alpha _4kappa -kappa omega }{alpha _1K^2}=A), (frac{alpha _3kappa +alpha _5kappa }{alpha _1K^2}=B) and (frac{alpha _2}{alpha _1K^2}=C), then we get
$$begin{aligned} F^{”}=AF+BF^3-CF^5. end{aligned}$$
(60)
Using the Galilean transformation on (60), then we get dynamical system
$$begin{aligned} {left{ begin{array}{ll} F^{‘}=H, \ H^{‘}=AF+BF^3-CF^5. end{array}right. } end{aligned}$$
(61)
Case 1: (A>0), (B>0), and (C>0).
We take different values of parameters (alpha _1 = 0.3), (alpha _2 = 0.5), (alpha _3 = 0.4), (alpha _4 = 0.2), (alpha _5 = 0.1), (omega = 0.1), (K = 0.5), and (kappa = 0.4), then get five equilibrium points (0, 0), (0.81, 0), ((-0.81,0)), (0.51i, 0), and ((-0.51 i,0)). As we see in Fig. 7 (0, 0) is the saddle and (0.81, 0), ((-0.81,0)) are center points.
Phase portrait diagram when (A>0), (B>0), and (C>0).
Case 2: (A<0), (B<0), and (C<0).
We take different values of parameters (alpha _1 = 0.3), (alpha _2 = -5), (alpha _3 = 0.2), (alpha _4 = 1), (alpha _5 = 0.3), (omega = -5), (K = 2), and (kappa = -1.5), then we get five equilibrium points (0, 0), (0.17, 0), ((-0.17,0)), (0.10i, 0), and ((-0.10 i,0)). As we analyze in Fig. 8, (0, 0) is the center point.
Phase portrait diagram when (A<0), (B<0), and (C<0).
Case 3: (A<0), (B>0), and (C>0).
We take different values of parameters (alpha _1 = 0.5), (alpha _2 =2), (alpha _3 = 0.6), (alpha _4 = -3), (alpha _5 = 0.4), (omega = 5), (K = 1), and (kappa = 0.8), then we get five equilibrium points (0, 0), ((1-0.89 i,0)), ((-1+0.89 i,0)), ((1+0.88 i,0)), and ((-1-0.88 i,0)). As we see in Fig. 9 (0, 0) is the center point.
Phase portrait diagram when (A<0), (B>0), and (C>0).
Case 4: (A>0), (B<0), and (C>0).
We take various values of parameters (alpha _1 = 0.5), (alpha _2 =2), (alpha _3 = -1), (alpha _4 = 3.0), (alpha _5 = -0.5), (omega = 1.0), (K = 1), and (kappa = 1.2), then we get five equilibrium points (0, 0), (1.43i, 0), ((-1.43 i,0)), (1.06, 0), and ((-1.06 i,0)). As we see in Fig. 10 (0, 0) is the saddle point and (1.06, 0), ((-1.06,0)) are the center points.
Phase portrait diagram when (A>0), (B<0), and (C>0).
Case 5: (A>0), (B>0), and (C<0).
We take diverse values of parameters (alpha _1 = 0.5), (alpha _2 =-2), (alpha _3 = 1.0), (alpha _4 = 3.0), (alpha _5 = 0.5), (omega = 1.0), (K = 1.0), and (kappa = 1.2), then get five equilibrium points (0, 0), ((0.63-0.92 i,0)), ((-0.63+0.92 i,0)), ((0.63+0.92 i,0)), and ((-0.63-0.92 i,0)). As we analyze in Fig. 11, (0, 0) is the saddle point.
Phase portrait diagram when (A>0), (B>0), and (C<0).
Case 6: (A<0), (B<0), and (C>0).
We take various values of parameters (alpha _1 = 1), (alpha _2 =3), (alpha _3 = -2), (alpha _4 = -4), (alpha _5 = -1), (omega = 8.0), (K = 0.8), and (kappa = 1.0), then we get five equilibrium points (0, 0), ((0.81-1.03 i,0)), ((-0.81+1.03 i,0)), ((0.81+1.03 i,0)), and ((-0.81-1.03 i,0)). As we see in Fig. 12, (0, 0) is the center point.
Phase portrait diagram when (A<0), (B<0), and (C>0).
Case 7: (A<0), (B>0), and (C<0).
We take various values of parameters (alpha _1 = 1.0), (alpha _2 =-5.21), (alpha _3 = 1.5), (alpha _4 = 22.69), (alpha _5 = 3.57), (omega = 10.0), (K = 1.0), and (kappa = 1.2), then we get five equilibrium points (0, 0), (1.57i, 0), ((-1.57 i,0)), (1.14, 0), and ((-1.14,0)). As we see in Fig. 13, (0, 0) is the center point and (1.14, 0), ((-1.14,0)) are the saddle points.
Phase portrait diagram when (A<0), (B>0), and (C<0).
Case 8: (A>0), (B<0), and (C<0).
We take various values of parameters (alpha _1 = 0.5), (alpha _2 =-3.0), (alpha _3 = -1.0), (alpha _4 = 4.0), (alpha _5 = -0.5), (omega = 1.0), (K = 1.0), and (kappa = 1.2), then we get five equilibrium points (0, 0), ((0.86-0.67 i,0)), ((-0.86+0.67 i,0)), ((0.86+0.67 i,0)), and ((-0.86-0.67 i,0)). As we see in Fig. 14, (0, 0) is the saddle point.
Phase portrait diagram when (A>0), (B<0), and (C<0).
Sensitivity analysis
For sensitivity analysis we will decompose the Eq. (60) into dynamical system as,
$$begin{aligned} {left{ begin{array}{ll} F^{‘} = H, \ H^{‘} = AF + BF^3 – CF^5. end{array}right. } end{aligned}$$
(62)
The system (62) explores how variations in initial conditions and parameters (alpha _1 = 0.8), (kappa = 2), (alpha _2 = 0.4), (K = 3), (alpha _3 = 0.3), (alpha _4 = 0.5), (alpha _5 = 0.1), and (alpha _6 = 0.5) influence the system’s behavior. The system is governed by nonlinear terms that can yield to complex dynamics, making it highly sensitive to small variation in initial conditions. In Fig. 15a, where the system starts with (0.40, 0) and a slight perturbation is introduced by changing the initial conditions to (0.45, 0.01), there is a significant divergence between the two trajectories. This shows the system’s high sensitivity to initial conditions, characteristic of unstable or chaotic systems where small changes lead to large variations in the results.
In contrast, Fig. 15b present the system with different initial conditions (1.60, 1.02) and a slight changes in the initial values. The two curves in this plot show much less divergence, showing that the system is less sensitive to changes in initial conditions. This present that the system is operating in a more stable region, where minor variations do not significantly affect its trajectory. Thus, the comparison between the two figures indicates the system shows sensitive.
Sensitivity behavior of perturbed system (42) assuming the initial condition (a) (0.40, 0) for blue solid line and (0.45,0.01) for red dotted curve, (b)(1.60, 1.02) for blue solid line and (1.75, 1.4) red dotted curve.