方程

对一维Schrödinger-Boussinesq方程

$$\left\{ \begin{aligned} i \partial_t \psi + \gamma \psi_{xx} -\xi u\psi &= 0 \\ u_{tt}-u_{xx}+\alpha u_{xxxx}-\left(f(u)\right)_{xx} -\omega \left( \left| \psi \right|^2 \right)_{xx} &= 0 \end{aligned} \right.$$

取周期边界. 令$\psi = p + iq$, $v_{xx}=\partial_t u$. 则方程变成

$$\left\{ \begin{aligned} \partial_t p &= -\gamma q_{xx} + \xi uq \\ \partial_t q &= \gamma p_{xx} - \xi up \\ \partial_t u &= v_{xx} \\ \partial_t v &= u-\alpha u_{xx}+f(u)+\omega \left|\psi\right|^2 \end{aligned} \right.$$

记质量为

$$\int_{\Omega} | \psi(x,t)|^2 \text{d}x$$

有质量守恒 记能量为

$$E\left(p,q,u,v\right)=\int_{\Omega} u^2+|\nabla v|^2+\frac{2\omega \gamma}{\xi}|\nabla \psi |^2+\alpha|\nabla u|^2+2F(u)+2\omega u|\psi |^2 \text{d}x$$

其中$F(u)$表示$f(u)$的积分, 一般取$f(u)=\theta u^2$. 这时有能量守恒.

能量守恒

VP-AVF

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

$$\left\{ \begin{aligned} \delta_t^+ p^n &= -D\gamma q^{\frac{1}{2}}+\xi u^{n+1} q^{\frac{1}{2}} \\ \delta_t^+ q^n &= D\gamma p^{\frac{1}{2}}-\xi u^{n+1} p^{\frac{1}{2}} \\ \delta_t^+ u^n &= Dv^{\frac{1}{2}} \\ \delta_t^+ v^n &= u^{\frac{1}{2}}-D\alpha u^{\frac{1}{2}} +\omega \left| \psi^{n} \right|^2+\frac{\theta}{3}\left((u^n)^2+(u^{n+1})^2+u^nu^{n+1}\right) \end{aligned} \right.$$

其中$D$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$(i-1) \rightarrow (i) \rightarrow (i+1)$的顺序, 则格式有

$$\left\{ \begin{aligned} \delta_t^+ P_i^n &= -\frac{\gamma}{h^2} \left(Q_{i-1}^{n+1}+Q_{i+1}^n-2Q_i^{\frac{1}{2}}\right) + \frac{\xi}{2} U_i^{n+1} (Q_i^n + Q_i^{n+1}) \\ \delta_t^+ Q_i^n &= \frac{\gamma}{h^2} \left(P_{i-1}^{n+1}+P_{i+1}^n-2P_i^{\frac{1}{2}}\right) - \frac{\xi}{2} U_i^{n+1} (P_i^n + P_i^{n+1}) \\ \delta_t^+ U_i^n &= \frac{1}{h^2}(V_{i+1}^n+V_{i-1}^{n+1}-V_i^n-V_i^{n+1}) \\ \delta_t^+ V_i^n &= U_i^{\frac{1}{2}} - \frac{\alpha}{h^2}\left(U_{i-1}^{n+1}+U_{i+1}^{n}-U_i^{n+1}-U_i^n\right)+\omega\left|\Psi_{i}^{n}\right|^2 +\frac{\theta}{3}\left( (U_{i}^n)^2+(U_{i}^{n+1})^2+U_{i}^nU_{i}^{n+1} \right) \end{aligned} \right.$$

数值格式

DP-AVF

将前两式合并后可得数值格式

$$\left\{ \begin{aligned} \left(\text{i}- \frac{\tau \xi U_i^{n+1}}{2}-\frac{\tau \gamma}{h^2}\right) \Psi_i^{n+1}&= \text{i}\Psi_i^n -\frac{\tau \gamma}{h^2}\left(\Psi_{i+1}^n+\Psi_{i-1}^{n+1}-\Psi_i^n\right) +\frac{\tau \xi U_i^{n+1}}{2}\Psi_i^n\\ U_i^{n+1}+\frac{\tau}{h^2}V_i^{n+1} &= \frac{\tau}{h^2}(V_{i+1}^n+V_{i-1}^{n+1}-V_i^{n+1})+U_i^n \\ V_i^{n+1} - \tau\left(\frac{\alpha}{h^2}+\frac{1}{2}+\frac{\theta}{3}U_i^n \right)U_i^{n+1} - \frac{\tau \theta}{3}\left(U_i^{n+1}\right)^2 &= V_i^n+\frac{\tau}{2}U_i^n-\frac{\tau \alpha}{h^2}\left( U_{i-1}^{n+1}+U_{i+1}^n-U_i^n\right)+\tau \omega\left|\Psi_{i}^{n}\right|^2 +\frac{\tau \theta}{3}\left(U_i^n\right)^2 \end{aligned} \right.$$

伴随格式

$$\left\{ \begin{aligned} \left(\text{i}- \frac{\tau \xi U_i^{n}}{2}-\frac{\tau \gamma}{h^2}\right) \Psi_i^{n+1}&= \text{i}\Psi_i^n -\frac{\tau \gamma}{h^2}\left(\Psi_{i-1}^n+\Psi_{i+1}^{n+1}-\Psi_i^n\right) +\frac{\tau \xi U_i^{n}}{2}\Psi_i^n\\ U_i^{n+1}+\frac{\tau}{h^2}V_i^{n+1} &= \frac{\tau}{h^2}(V_{i-1}^n+V_{i+1}^{n+1}-V_i^{n+1})+U_i^n \\ V_i^{n+1} - \tau\left(\frac{\alpha}{h^2}+\frac{1}{2}+\frac{\theta}{3}U_i^n \right)U_i^{n+1} - \frac{\tau \theta}{3}\left(U_i^{n+1}\right)^2 &= V_i^n+\frac{\tau}{2}U_i^n-\frac{\tau \alpha}{h^2}\left( U_{i+1}^{n+1}+U_{i-1}^n-U_i^n\right)+\tau \omega\left|\Psi_{i}^{n+1}\right|^2 +\frac{\tau \theta}{3}\left(U_i^n\right)^2 \end{aligned} \right.$$

启动层外推

$Dy_i^n=\frac{y_i^n+y_{i+2}^{n}-2y_{i+1}^{n}}{h^2}$

$$\left\{ \begin{aligned} \left(\text{i}- \frac{\tau \xi U_i^{n}}{2}+\frac{\tau \gamma}{2h^2}\right) \Psi_i^{n+1}&= \text{i}\Psi_i^n -\frac{\tau \gamma}{h^2}\left(\frac{1}{2}\Psi_{i}^n+\Psi_{i+2}^{n}-2\Psi_{i+1}^n\right) +\frac{\tau \xi U_i^{n}}{2}\Psi_i^n \\ U_i^{n+1}-\frac{\tau}{2h^2}V_i^{n+1}&= U_i^n+\frac{\tau}{h^2}\left(\frac{1}{2}V_i^n+V_{i+2}^n-2V_{i+1}^n\right)\\ V_i^{n+1} - \tau\left(-\frac{\alpha}{2h^2}+\frac{1}{2}+\frac{\theta}{3}U_i^n \right)U_i^{n+1} - \frac{\tau \theta}{3}\left(U_i^{n+1}\right)^2 &= V_i^n+\frac{\tau}{2}U_i^n-\frac{\tau \alpha}{h^2}\left( \frac{1}{2}U_{i}^{n}+U_{i+2}^n-2U_{i+1}^n\right)+\tau \omega\left|\Psi_{i}^{n}\right|^2 +\frac{\tau \theta}{3}\left(U_i^n\right)^2 \end{aligned} \right.$$

数值验证

精确解算例

1. 平面波解

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

2. 孤子波解 取$b_1=\delta +\frac{S^2}{4\gamma}$, $d_1=1-S^2$, $\mu=\sqrt{\frac{b_1}{\gamma}}$, $\zeta=x-St$, $S$和$\delta$是自由变量.
   • $\gamma \xi d_1=2b_1 (3\gamma\theta -\alpha \xi)$, $3\alpha\xi \neq \gamma\theta$, $4\alpha b_1 \neq \gamma d_1$.

   $$\left\{ \begin{aligned} \psi &= \pm\frac{6b_1}{\xi}\sqrt{\frac{\gamma\theta-\alpha\xi}{\gamma\omega}} \text{sech} (\mu\zeta) \tanh (\mu \zeta) \exp \left(\text{i} (\frac{S}{2\gamma} x + \delta t) \right)\\ u &= -\frac{6b_1}{\xi} \text{sech}^2 (\mu\zeta) \\ v &= \frac{12Sb_1}{\mu \xi}\left( \frac{x_r-x}{x_r-x_l}-\frac{1}{1+\text{e}^{2\mu\zeta}} \right)\qquad x\in [x_l,x_r], t \geq 0 \end{aligned} \right.$$

   • $3\alpha\xi \neq \gamma\theta$, $4\alpha b_1 \neq \gamma d_1$.

   $$\left\{ \begin{aligned} \psi &= \sqrt{\frac{6\alpha b_1}{\gamma^2\theta\omega}\left( \gamma d_1-4\alpha b_1 \right)} \text{sech} (\mu\zeta) \exp \left(\text{i} (\frac{S}{2\gamma} x + \delta t) \right)\\ u &= -\frac{6b_1}{\xi} \text{sech}^2 (\mu\zeta) \\ v &= \frac{4Sb_1}{\mu \xi}\left( \frac{x_r-x}{x_r-x_l}-\frac{1}{1+\text{e}^{2\mu\zeta}} \right)\qquad x\in [x_l,x_r], t \geq 0 \end{aligned} \right.$$

   • $\gamma = 1$, $\xi=1$, $\alpha=1$, $\theta=\frac{4}{3}$, $\omega=\frac{1}{18}$, $S=\sqrt{\frac{1}{5}}$, $\delta=\frac{1}{12}$, $\Omega = [-40,40]$, $T=10$. 此时$b_1=\frac{2}{15}$, $d_1=\frac{4}{5}$, $\mu=\sqrt{\frac{2}{15}}$, 应选择第1个精确解.
   • $\gamma = 1$, $\xi=1$, $\alpha=\frac{2}{3}$, $\theta=\frac{1}{9}$, $\omega=-\frac{1}{2}$, $S=\sqrt{\frac{22}{15}}$, $\delta=\frac{1}{3}$, $\Omega = [-64,64]$, $T=1$. 此时$b_1=\frac{7}{10}$, $d_1=-\frac{7}{15}$, $\mu=\sqrt{\frac{7}{10}}$, 应选择第1个精确解.

问题

1. 需要启动层
   • 暂时单线程不并行的情况下, 左边界用精确解代替.
   • 应该是要用别的数值格式来先算出启动层, 还在看.
   • Richardson Extrapolation/外推
2. 数值格式涉及二次函数, 两个根的问题
   • 先试了一下求根公式中正负号, 发现全正会发散, 全负时误差$10^{-1}$.
   • 双根取离上一步近的那个.
3. 目前能量误差$10^{-2}$.