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title: 工作日志
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# BDF1和BDF2的BLUSGS实现

对于微分方程
$$
\frac{\partial Q}{\partial t} = R(Q)
$$


BDF1和BDF2的理论公式如下
$$
\begin{align}
Q^{n+1} - Q^{n} &= \Delta t R^{n+1} \\
Q^{n+1} - \frac{4}{3} Q^{n} + \frac{1}{3} Q^{n-1} &= \frac{2}{3} \Delta t R^{n+1} \\
\end{align}
$$

## BDF1的BLUSGS实现

BLUSGS算法描述见：Sun, Y.; Wang, Z. & Liu, Y. Efficient Implicit Non-linear LU-SGS Approach for Compressible Flow Computation Using High-order Spectral Difference Method, *Comm. Comput. Phys.,* **2009***, 5*, 760-778 

- BDF1的原始形式
  $$
  \begin{align}
  \frac{ Q^{n+1} - Q^{n} }{\Delta t} &= R^{n+1} \\
  \frac{ Q^{n+1} - Q^{n} }{\Delta t} - \left ( R^{n+1} - R^{n} \right ) &= R^{n} \\
  \end{align}
  $$
  
- 线化$R^{n+1}-R^{n}$
  $$
  \begin{align}
  R^{n+1} - R^{n} = \frac{\partial R_c}{\partial Q_c} \Delta Q^{n+1}_c + \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{n+1}_{nb}
  \end{align}
  $$
  
- 将线化$R$代入原始形式
  $$
  \begin{align}
  \Delta Q^{n+1} &= Q^{n+1} - Q^{n} \\
  \frac{ \Delta Q^{n+1} }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \Delta Q^{n+1}_c - \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{n+1}_{nb} &= R^n_c \\
  \left ( \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{n+1}_c - \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{n+1}_{nb} &= R^n_c \\
  \end{align}
  $$
  
- Gauss-Seidel加速
  $$
  \begin{align}
  \left ( \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{k+1}_c - \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{k}_{nb} &= R^n_c \\
  \Delta Q^{k+1}_c &= Q^{k+1} - Q^{n} \\
  \end{align}
  $$
  
- 再次线化$R^{k}-R^{n}$
  $$
  R^{k} - R^{n} = \frac{\partial R_c}{\partial Q_c} \Delta Q^{k}_c + \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{k}_{nb}
  $$
  
- 将线化$R$代入，整理
  $$
  \begin{align}
  \left ( \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{k+1}_c &= R^k_c - \frac{\partial R_{c}}{\partial Q_{c}} \Delta Q^{k}_{c} \\
   &= R^k_c + \left ( \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{k}_{c} - \frac{ I }{\Delta t} \Delta Q^{k}_{c} \\
  \left ( \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \left ( \Delta Q^{k+1}_c - \Delta Q^{k}_{c} \right ) &= R^k_c - \frac{ I }{\Delta t} \Delta Q^{k}_{c} \\
  \left ( \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \tilde{\Delta} Q^{k+1}_c &= R^k_c - \frac{ I }{\Delta t} \Delta Q^{k}_{c} \\
  \tilde{\Delta} Q^{k+1}_c &= Q^{k+1}_c - Q^{k}_c \\
  \end{align}
  $$
  

最后两个方程是最终的BDF1的BLUSGS实现。

## BDF2的BLUSGS实现

BDF2的实现与BDF1类似。

- BDF2的原始形式
  $$
  \begin{align}
  \frac{ Q^{n+1} - \frac{4}{3} Q^{n} + \frac{1}{3} Q^{n-1} }{ \frac{2}{3} \Delta t } &= R^{n+1} \\
  \frac{3}{2} \left ( \frac{ \Delta Q^{n+1} }{ \Delta t } - \frac{1}{3} \frac{ \Delta Q^{n} }{ \Delta t } \right ) - \left ( R^{n+1} - R^{n} \right ) &= R^{n} \\
  \Delta Q^{n+1} &= Q^{n+1} - Q^{n} \\
  \end{align}
  $$
  
- 线化$R^{n+1}-R^{n}$
  $$
  \begin{align}
  R^{n+1} - R^{n} = \frac{\partial R_c}{\partial Q_c} \Delta Q^{n+1}_c + \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{n+1}_{nb}
  \end{align}
  $$
  
- 将线化$R$代入原始形式
  $$
  \begin{align}
  \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{n+1}_c - \frac{1}{2} \frac{ I }{\Delta t} \Delta Q^{n}_c - \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{n+1}_{nb} &= R^n_c \\
  \end{align}
  $$
  
- Gauss-Seidel加速
  $$
  \begin{align}
  \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{k+1}_c - \frac{1}{2} \frac{ I }{\Delta t} \Delta Q^{n}_c - \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{k}_{nb} &= R^n_c \\
  \Delta Q^{k+1}_c &= Q^{k+1} - Q^{n} \\
  \end{align}
  $$
  
- 再次线化$R^{k}-R^{n}$
  $$
  R^{k} - R^{n} = \frac{\partial R_c}{\partial Q_c} \Delta Q^{k}_c + \sum \frac{\partial R_{c}}{\partial Q_{nb}} \Delta Q^{k}_{nb}
  $$
  
- 将线化$R$代入，整理
  $$
  \begin{align}
  \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{k+1}_c - \frac{1}{2} \frac{ I }{\Delta t} \Delta Q^{n}_c &= R^k_c - \frac{\partial R_{c}}{\partial Q_{c}} \Delta Q^{k}_{c} \\
   &= R^k_c + \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \Delta Q^{k}_{c} - \frac{3}{2} \frac{ I }{\Delta t} \Delta Q^{k}_{c} \\
  \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \left ( \Delta Q^{k+1}_c - \Delta Q^{k}_{c} \right ) - \frac{1}{2} \frac{ I }{\Delta t} \Delta Q^{n}_c &= R^k_c - \frac{3}{2} \frac{ I }{\Delta t} \Delta Q^{k}_{c} \\
  \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \tilde{\Delta} Q^{k+1}_c &= R^k_c - \frac{3}{2} \frac{ I }{\Delta t} \Delta Q^{k}_{c} + \frac{1}{2} \frac{ I }{\Delta t} \Delta Q^{n}_c \\
  \left ( \frac{3}{2} \frac{ I }{\Delta t} - \frac{\partial R_c}{\partial Q_c} \right ) \tilde{\Delta} Q^{k+1}_c &= R^k_c - \frac{1}{2 \Delta t} \left ( 3 I Q^{k}_{c} - 4 I Q^{n}_c + I Q^{n-1}_c \right ) \\
  \tilde{\Delta} Q^{k+1}_c &= Q^{k+1}_c - Q^{k}_c \\
  \end{align}
  $$
  

最后两个方程是最终的BDF2的BLUSGS实现。