Numerical solution of the general coupled nonlinear Schrödinger equations on unbounded domains

The numerical solution of the general coupled nonlinear Schrödinger equations on unbounded domains is considered by applying the artificial boundary method in this paper. In order to design the local absorbing boundary conditions for the coupled nonlinear Schrödinger equations, we generalize the unified approach previously proposed [J. Zhang et al., Phys. Rev. E 78, 026709 (2008)]. Based on the methodology underlying the unified approach, the original problem is split into two parts, linear and nonlinear terms, and we then achieve a one-way operator to approximate the linear term to make the wave out-going, and finally we combine the one-way operator with the nonlinear term to derive the local absorbing boundary conditions. Then we reduce the original problem into an initial boundary value problem on the bounded domain, which can be solved by the finite difference method. The stability of the reduced problem is also analyzed by introducing some auxiliary variables. Ample numerical examples are presented to verify the accuracy and effectiveness of our proposed method.

Figure 1

Comparison of the numerical solution with the exact solution: (a) $\gamma =1/2$, $\beta =1$; (b) $\beta =2/3$ at time $T=3.0$; (c) $\gamma =1$, $\beta =1$; and (d) $\beta =2$ at time $T=1.5$.

Figure 2

Reflection ratios for different times with parameters $\gamma =1$ and $\beta =2/3$.

Figure 3

The evolution of the $\left|\psi \right|$ and $\left|\phi \right|$ for $\gamma =1/2$, $\beta =1/2$: (a), (b) ${\alpha }_1={\alpha }_2=0.1$, $v_1=-4$, $v_2=4$, $x_1=-10$ and $x_2=10$; (c), (d) ${\alpha }_1=0.1$, ${\alpha }_2=0.3$, $v_1=-4$, $v_2=5$, $x_1=-5$ and $x_2=5$.

Figure 4

Mass of the CNLSEs for $\gamma =1/2$, $\beta =1/2$: (a) ${\alpha }_1={\alpha }_2=0.1$, $v_1=-4$, $v_2=4$, $x_1=-10$, and $x_2=10$; (b) ${\alpha }_1=0.1$, ${\alpha }_2=0.3$, $v_1=-4$, $v_2=5$, $x_1=-5$, and $x_2=5$.

Figure 5

Evolution of the $\left|\psi \right|$ and $\left|\phi \right|$ for $\gamma =1/2$, $\beta =1$, ${\alpha }_1={\alpha }_2=1$, $v_1=2$, $v_2=-2$, $x_1=-9$, and $x_2=-15$.

Figure 6

Evolution of the $\left|\psi \right|$ and $\left|\phi \right|$ for $\gamma =1$, $\beta =2$: (a, b) ${\alpha }_1={\alpha }_2=0.1$, $v_1=-4$, $v_2=4$, $x_1=-10$, and $x_2=10$; (c, d) ${\alpha }_1={\alpha }_2=0.1$, $v_1=-4$, $v_2=4$, $x_1=-15$, and $x_2=5$.