电磁粒子模拟中电荷守恒的电流分配方案满足的统一公式
付梅艳,卢朓,朱湘疆

Unified Formulation of Charge-conserving Current Assignment in Electromagnetic Particle-in-Cell Simulation
Meiyan Fu,Tiao Lu,Xiangjiang Zhu
表 4 在七个元胞情形下, V方法, E方法和U方法给出的电流密度公式
$J$ V method $(\times \frac{q}{2\Delta t})$ E method $(\times \frac{q}{2\Delta t})$ U method $(\times \frac{q}{2\Delta t})$
$ \xi_1$ $ \Delta x_1 (1-2y^{n}-\Delta y \frac{\Delta x_1}{\Delta x}) $ $ (0.5-x^n)(0.5-y^n) $ $ (0.5-x^n)(0.5-y^n) $
$ \xi_2$ $ \Delta x_1 \Delta y {\Delta x_1}/ (\Delta x)$ $ (0.5-x^n)(0.5-y^n) $ $ (0.5-x^n)(0.5-y^n) $
$ \xi_3$ $ [\Delta y (2-\Delta x_1)\frac{\Delta x_1}{\Delta x} + (2-\Delta x \frac{\Delta y_2}{\Delta y})(0.5-y^{n}-\Delta y \frac{\Delta x_1}{\Delta x})] $ $ (2+x^n-x^{n+1})(0.5-y^n)$ $ (1.5+x^n)(0.5-y^n)$
$ \xi_4$ $ \Delta x_1 (1+2y^{n}+\Delta y \frac{\Delta x_1}{\Delta x}) $ $ (0.5-x^{n})(2+y^{n}-y^{n+1}) $ $ (0.5-x^n)(1.5+y^n) $
$ \xi_5$ $ \Delta x_2 (0.5-y^{n}-\Delta y \frac{\Delta x_1}{\Delta x}) $ $ (x^{n+1}-0.5)(0.5-y^{n}) $ $ 0 $
$ \xi_6$ $ \Delta x \frac{\Delta y_2}{\Delta y}(0.5-y^{n}-\Delta y \frac{\Delta x_1}{\Delta x}) $ $ (x^{n+1}-0.5)(0.5-y^{n}) $ $ 0 $
$ \xi_7$ $ [\Delta x_2 (1.5+y^{n}+\Delta y \frac{\Delta x_1}{\Delta x}) + (x^{n+1}-0.5-\Delta x \frac{\Delta y_2}{\Delta y})(2.5-y^{n+1})] $ $ (x^{n+1}-0.5)(2+y^{n}-y^{n+1}) $ $ (x^{n+1}-0.5)(2.5-y^{n+1}) $
$ \xi_8$ $ (y^{n+1}-0.5) (2.5-\Delta x \frac{\Delta y_2}{\Delta y}-x^{n+1}) $ $ (2+x^{n}-x^{n+1})(y^{n+1}-0.5) $ $ (2.5-x^{n+1})(y^{n+1}-0.5)$
$ \xi_9$ $ (y^{n+1}-0.5) (-0.5+\Delta x \frac{\Delta y_2}{\Delta y}+x^{n+1}) $ $ (x^{n+1}-0.5)(y^{n+1}-0.5) $ $ (x^{n+1}-0.5)(y^{n+1}-0.5) $
$ \xi_{10}$ $ (y^{n+1}-0.5) (-0.5-\Delta x \frac{\Delta y_2}{\Delta y}+x^{n+1}) $ $ (x^{n+1}-0.5)(y^{n+1}-0.5) $ $ (x^{n+1}-0.5)(y^{n+1}-0.5) $
$ \xi_{11}$ $ 0 $ $ (0.5-x^{n})(y^{n+1}-0.5) $ $ 0 $
$ \xi_{12}$ $ 0 $ $ (0.5-x^{n})(y^{n+1}-0.5) $ $ 0 $
其中: $\Delta x = x^{n+1}-x^{n}$, $\Delta y = y^{n+1}-y^{n}$, $\Delta x_1 = 0.5-x^{n}$, $\Delta y_1 = \Delta y {\Delta x_1}/{\Delta x}, $
$x_{M_1} = 0.5$, $y_{M_1} = y^{n}+\Delta y_1$, $\Delta y_2 = 0.5-y_{M_1}$, $\Delta x_2 = \Delta x {\Delta y_2}/{\Delta y}.$