方程

对一维Schrödinger-Boussinesq方程

{i∂tψ−ψxx−uψ=0uxx+uxxxx+3(u2)xx−3utt+(∣ψ∣2)xx=0\left\{ \begin{aligned} i \partial_t \psi - \psi_{xx} -u\psi &= 0 \\ u_{xx}+u_{xxxx}+3\left(u^2\right)_{xx}-3u_{tt} + \left( \left| \psi \right|^2 \right)_{xx} &= 0 \end{aligned} \right.

取周期边界. 令ψ=p+iq\psi = p + iq, vxx=∂tuv_{xx}=\partial_t u. 则方程变成

{∂tp=−qxx−uq∂tq=pxx+up∂tu=vxx∂tv=13(u+uxx+3u2+∣ψ∣2)\left\{ \begin{aligned} \partial_t p &= -q_{xx} - uq \\ \partial_t q &= p_{xx} + up \\ \partial_t u &= v_{xx} \\ \partial_t v &= \frac{1}{3}\left(u+u_{xx}+3u^2+\left|\psi\right|^2\right) \end{aligned} \right.

记质量为

∫Ω∣u(x,t)∣2dx\int_{\Omega} | u(x,t)|^2 \text{d}x

有质量守恒 记能量为

E(p,q,u,v)=∫Ω13u2+∣∇v∣2+23∣∇ψ∣2+13∣∇u∣2−23u3+23u∣ψ∣2dxE\left(p,q,u,v\right)=\int_{\Omega} \frac{1}{3}u^2+|\nabla v|^2+\frac{2}{3}|\nabla \psi |^2+\frac{1}{3}|\nabla u|^2-\frac{2}{3}u^3+\frac{2}{3}u|\psi |^2 \text{d}x

有能量守恒.

VP-AVF

空间采取有限差分, 对(u,v)→(p,q)(u,v)\rightarrow (p,q)的顺序进行基于变量的AVF方法后, 格式为

{δt+pn=−Dq12−un+1q12δt+qn=Dp12+un+1p12δt+un=Dv12δt+vn=13(u12+Du12+∣ψn∣2+(un)2+(un+1)2+unun+1)\left\{ \begin{aligned} \delta_t^+ p^n &= -Dq^{\frac{1}{2}}-u^{n+1} q^{\frac{1}{2}} \\ \delta_t^+ q^n &= Dp^{\frac{1}{2}}+u^{n+1} p^{\frac{1}{2}} \\ \delta_t^+ u^n &= Dv^{\frac{1}{2}} \\ \delta_t^+ v^n &= \frac{1}{3} \left(u^{\frac{1}{2}}+D u^{\frac{1}{2}} +\left| \psi^{n} \right|^2+(u^n)^2+(u^{n+1})^2+u^nu^{n+1}\right) \end{aligned} \right.

其中DD表示2阶中心差分离散.

GRD-AVF

在基于格点的AVF方法中, 不妨采取(i−1)→(i)→(i+1)(i-1) \rightarrow (i) \rightarrow (i+1)的顺序, 则格式有

{δt+Pin=−1h2(Qi−1n+1+Qi+1n−2Qi12)−Uin+1(Qin+Qin+1)δt+Qin=1h2(Pi−1n+1+Pi+1n−2Pi12)+Uin+1(Pin+Pin+1)δt+Uin=1h2(Vi+1n+Vi−1n+1−Vin−Vin+1)δt+Vin=23Ui12+13h2(Ui−1n+1+Ui+1n−Uin+1−Uin)+13∣Ψin∣2+13((Uin)2+(Uin+1)2+UinUin+1)\left\{ \begin{aligned} \delta_t^+ P_i^n &= -\frac{1}{h^2} \left(Q_{i-1}^{n+1}+Q_{i+1}^n-2Q_i^{\frac{1}{2}}\right) - U_i^{n+1} (Q_i^n + Q_i^{n+1}) \\ \delta_t^+ Q_i^n &= \frac{1}{h^2} \left(P_{i-1}^{n+1}+P_{i+1}^n-2P_i^{\frac{1}{2}}\right) + U_i^{n+1} (P_i^n + P_i^{n+1}) \\ \delta_t^+ U_i^n &= \frac{1}{h^2}(V_{i+1}^n+V_{i-1}^{n+1}-V_i^n-V_i^{n+1}) \\ \delta_t^+ V_i^n &= \frac{2}{3}U_i^{\frac{1}{2}} + \frac{1}{3h^2}\left(U_{i-1}^{n+1}+U_{i+1}^{n}-U_i^{n+1}-U_i^n\right)+\frac{1}{3}\left|\Psi_{i}^{n}\right|^2 +\frac{1}{3}\left( (U_{i}^n)^2+(U_{i}^{n+1})^2+U_{i}^nU_{i}^{n+1} \right) \end{aligned} \right.

则将前两式合并后可得数值格式

{(i+τUin+1−τh2)Ψin+1=iΨin+τh2(Ψin−Ψi+1n−Ψi−1n+1)−τUin+1ΨinUin+1+τh2Vin+1=τh2(Vi+1n+Vi−1n+1−Vin+1)+UinVin+1+τ3h2(1−h2−h2Uin)Uin+1−τ3(Uin+1)2=Vin+τ3Uin+τ3h2(Ui−1n+1+Ui+1n−Uin)+τ3∣Ψin∣2+τ3(Uin)2\left\{ \begin{aligned} \left(\text{i}+ \tau U_i^{n+1}-\frac{\tau}{h^2}\right) \Psi_i^{n+1}&= \text{i}\Psi_i^n +\frac{\tau}{h^2}\left(\Psi_i^n-\Psi_{i+1}^n-\Psi_{i-1}^{n+1}\right) -\tau U_i^{n+1}\Psi_i^n\\ U_i^{n+1}+\frac{\tau}{h^2}V_i^{n+1} &= \frac{\tau}{h^2}(V_{i+1}^n+V_{i-1}^{n+1}-V_i^{n+1})+U_i^n \\ V_i^{n+1} + \frac{\tau}{3h^2}\left(1-h^2-h^2U_i^n \right)U_i^{n+1} - \frac{\tau}{3}\left(U_i^{n+1}\right)^2 &= V_i^n+\frac{\tau}{3}U_i^n+\frac{\tau}{3h^2}\left( U_{i-1}^{n+1}+U_{i+1}^n-U_i^n\right)+\frac{\tau}{3}\left|\Psi_{i}^{n}\right|^2 +\frac{\tau}{3}\left(U_i^n\right)^2 \end{aligned} \right.

数值验证

精确解算例

  1. 平面波解

u=2ψ=ei(x−t)u=2\\ \psi = e^{\text{i}(x-t)}

  1. 孤子波解

ψ=A⋅sech(kx−ωt)tanh⁡(kx−ωt)exp⁡(i(σx+δt))u=B⋅sech2(kx−ωt)\psi = A \cdot \text{sech} (kx-\omega t) \tanh (kx-\omega t) \exp \left(\text{i} (\sigma x + \delta t) \right)\\ u = B \cdot \text{sech}^2 (kx-\omega t)

问题

需要启动层