🎯要点
🎯任意维度求解器，绘制三维投影结果 | 🎯解二维静电场、静磁场 | 🎯狄利克雷、诺依曼条件几何矩阵算子 | 🎯算法模拟低溫半导体材料 | 🎯计算曲面达西流 | 🎯电子结构计算和原子模拟 | 🎯算法模拟量子波函数和电子束
📜泊松方程 | 本文 - 用例
📜Python火焰锋动力学和浅水表面波浪偏微分方程
📜Python数值和符号算法计算及3D视图物理数学波形方程
📜Python射频电磁肿瘤热疗数学模型和电磁爆炸性变化统计推理模型
📜达西流用例：Python和R水力电导率和达西流神经算子
🍇Python最低阶有限差分泊松方程
我们想要解泊松方程：
∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 ∂x2∂2u+∂y2∂2u=0
在 [ 0 , 1 ] × [ 0 , 1 ] [0,1] \times[0,1] [0,1]×[0,1] 方域中，具有边界条件
u ( x , 0 ) = x , u ( x , 1 ) = x − 1 , u ( 0 , y ) = − y , u ( 1 , y ) = 1 − y . u(x, 0)=x, \quad u(x, 1)=x-1, \quad u(0, y)=-y, \quad u(1, y)=1-y . u(x,0)=x,u(x,1)=x−1,u(0,y)=−y,u(1,y)=1−y.
我们将使用最低阶有限差分表示：
∂ 2 u ∂ x 2 ( x i , y j ) ≃ 1 Δ x ( ∂ u ∂ x ( x i + 1 , y j ) − ∂ u ∂ x ( x i − 1 , y j ) ) ≃ 1 Δ x ( 1 Δ x ( u ( x i + 1 , y j ) − u ( x i , y j ) ) − 1 Δ x ( u ( x i , y j ) − u ( x i − 1 , y j ) ) ) \begin{gathered} \frac{\partial^2 u}{\partial x^2}\left(x_i, y_j\right) \simeq \frac{1}{\Delta x}\left(\frac{\partial u}{\partial x}\left(x_{i+1}, y_j\right)-\frac{\partial u}{\partial x}\left(x_{i-1}, y_j\right)\right) \\ \simeq \frac{1}{\Delta x}\left(\frac{1}{\Delta x}\left(u\left(x_{i+1}, y_j\right)-u\left(x_i, y_j\right)\right)-\frac{1}{\Delta x}\left(u\left(x_i, y_j\right)-u\left(x_{i-1}, y_j\right)\right)\right) \end{gathered} ∂x2∂2u(xi,yj)≃Δx1(∂x∂u(xi+1,yj)−∂x∂u(xi−1,yj))≃Δx1(Δx1(u(xi+1,yj)−u(xi,yj))−Δx1(u(xi,yj)−u(xi−1,yj)))
📜有限差分用例：Python微磁学磁倾斜和西塔规则算法
最终简化为，
∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ≃ ( 1 Δ 2 ( u i + 1 , j + u i − 1 , j + u i , j + 1 + u i , j − 1 − 4 u i , j ) ) \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} \simeq\left(\frac{1}{\Delta^2}\left(u_{i+1, j}+u_{i-1, j}+u_{i, j+1}+u_{i, j-1}-4 u_{i, j}\right)\right) ∂x2∂2u+∂y2∂2u≃(Δ21(ui+1,j+ui−1,j+ui,j+1+ui,j−1−4ui,j))
最直接的方法是直接求解线性系统：
import numpy as np
import matplotlib as ml
import matplotlib.pyplot as pp
def boundary(grid):x = np.linspace(0,1,len(grid))grid[0,:] = np.interp(x,[0,1],[0,1])grid[:,-1] = np.interp(x,[0,1],[1,0])grid[-1,:] = np.interp(x,[0,1],[-1,0])grid[:,0] = np.interp(x,[0,1],[0,-1])
def poisson_direct(gridsize,set_boundary):A = np.zeros(shape=(gridsize,gridsize,gridsize,gridsize),dtype='d')b = np.zeros(shape=(gridsize,gridsize),dtype='d')dx = 1.0 / (gridsize - 1)for i in range(1,gridsize-1):for j in range(1,gridsize-1):A[i,j,i-1,j] = A[i,j,i+1,j] = A[i,j,i,j-1] = A[i,j,i,j+1] = 1/dx**2A[i,j,i,j] = -4/dx**2for i in range(0,gridsize):A[0,i,0,i] = A[-1,i,-1,i] = A[i,0,i,0] = A[i,-1,i,-1] = 1set_boundary(b)return np.linalg.tensorsolve(A,b)
sol = poisson_direct(25,boundary)
为了展示解，我们需要将矩阵的通常解释（行、列、从上到左）与在笛卡尔平面上绘图的想法联系起来。
pp.imshow(sol.T,cmap=ml.cm.Blues,interpolation='none',origin='lower')
将其变成我们将再次使用的函数。
def showsol(sol):pp.imshow(sol.T,cmap=ml.cm.Blues,interpolation='none',origin='lower')
showsol(poisson_direct(51,boundary))
让我们尝试一种迭代方法：我们将泊松方程转化为扩散方程来求解
∂ u ∂ t = ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 \frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} ∂t∂u=∂x2∂2u+∂y2∂2u
并通过正向时间中心空间差分求解收敛
u i , j n + 1 = u i , j n + Δ t Δ 2 ( u i + 1 , j n + u i − 1 , j n + u i , j + 1 n + u i , j − 1 n − 4 u i , j n ) u_{i, j}^{n+1}=u_{i, j}^n+\frac{\Delta t}{\Delta^2}\left(u_{i+1, j}^n+u_{i-1, j}^n+u_{i, j+1}^n+u_{i, j-1}^n-4 u_{i, j}^n\right) ui,jn+1=ui,jn+Δ2Δt(ui+1,jn+ui−1,jn+ui,j+1n+ui,j−1n−4ui,jn)
对于最大稳定时间步 Δ t = Δ 2 / 4 \Delta t=\Delta^2 / 4 Δt=Δ2/4 ，导出雅可比方法
u i , j n + 1 = 1 4 ( u i + 1 , j n + u i − 1 , j n + u i , j + 1 n + u i , j − 1 n ) u_{i, j}^{n+1}=\frac{1}{4}\left(u_{i+1, j}^n+u_{i-1, j}^n+u_{i, j+1}^n+u_{i, j-1}^n\right) ui,jn+1=41(ui+1,jn+ui−1,jn+ui,j+1n+ui,j−1n)
这相当于用最近邻的平均值替换每个网格值。这与调和函数理论是一致的。
def jacobi(grid):newgrid = np.zeros(shape=grid.shape,dtype=grid.dtype)newgrid[1:-1,1:-1] = 0.25 * (grid[1:-1,:-2] + grid[1:-1,2:] +grid[:-2,1:-1] + grid[2:,1:-1])newgrid[0,:] = grid[0,:]newgrid[-1,:] = grid[-1,:]newgrid[:,0] = grid[:,0]newgrid[:,-1] = grid[:,-1]return newgrid
我们从随机配置开始，应用边界条件，然后迭代。我们间歇性地绘制解，结果显示它收敛了。
