where γ gives the width of the nonlinearities. With this, we can use Eqs. (11) and (12) to find ρ ( x, t ) = exp( − ξ $^{2}$/ 6 γ $^{2}$) / √ χ and ζ ( ξ ) = ∫ exp( − ξ $^{2}$/ 3 γ $^{2}$) dξ , respectively. The potentials are now given by with f ( t ) e h ( t ) given by Eqs. (16) and (17). Here we use a = ∫ χ − $^{2}$dt , which leads us to h ( t ) = 0 . In this case, we can obtain solutions of the form with the choice µ$_{1}$ = µ$_{2}$ = G$_{11}$ = − 1 , G$_{12}$ = G$_{21}$ = − 2 , and G$_{22}$ = 0 , in the Eqs. (3-4). In this case, we note that ψ$_{1}$ describes a group of atoms similar to the group considered in [21],(but without quintic interaction) which interacts with the other group of atoms described by ψ$_{2}$ . This last group, however, does not self-interact. Here, the general solution has the form As before, we see that the temporal evolution of the solutions are directly related to χ − 1 / $^{2}$, which can be obtained from Eq. (16). For f = 1 and for f = 1 + ϵ cos( ω$_{0}$t ) , we can get periodic and quasiperiodic solutions, respectively, as show before in the former example. In Fig. 3 we show the potential for the case (a) periodic and (b) quasiperiodic, respectively. The solution (26) is plotted in Fig. 4, for the,(a) periodic, and (b) quasiperiodic, cases, respectively. As before, we do not depict the case ψ$_{2}$ , since it is similar to ψ$_{1}$ . We have studied numerically the propagation of the initial states ψ$_{1}$ ( x, 0) e ψ$_{2}$ ( x, 0) , given by Eqs. (26) and (27), at t = 0 , after introducing aleatory perturbation of 3% in the profile of the initial states. With this, we could verify that the solutions (26) and (27) are stable against such aleatory perturbations. Example # 3 – Next, we consider the case with µ$_{1}$ = µ$_{2}$ = G$_{11}$ = 0 , G$_{12}$ = − 2 , G$_{21}$ = 2 , and G$_{22}$ = − 1 . We use this in Eqs. (3-4). This leads to two interesting analytical solutions: one field leading to a bright-like solution, and the other giving rise to a dark-like solution. They are given explicitly by 6 A$_{1}$ ( x, t ) = 1 √$_{2}$ tanh( ζ ( x, t )) , (28) ψ$_{2}$ ( x, t ) = e $^{iη}$(2 χ ) − 1 / $^{2}$exp( − ξ $^{2}$/ 6 γ $^{2}$)sech( ζ ( x, t )) . (27) ψ$_{1}$ ( x, t ) = e $^{iη}$χ − 1 / $^{2}$exp( − ξ $^{2}$/ 6 γ $^{2}$)sech( ζ ( x, t )) , (26) A$_{1}$ ( x, t ) = sech( ζ ( x, t )) , (24) A$_{2}$ ( x, t ) = 1 √$_{2}$ sech( ζ ( x, t )) , (25) v$_{j}$ = f ( t ) x 2 + h ( t ) − µ$_{j}$ exp( − 2 ξ $^{2}$/ 3 γ $^{2}$) , (23)