import matplotlib.pyplot as plt import numpy as np a = 1.5 L = 1.0 T = 1.0 N = 32 dx = 2 / N alpha = 0.1 dt = alpha * dx art_visc_coeff = 0.6 u_prev = np.zeros(N) u = np.zeros(N) u_prev[0:int(N/2)] = 1.0 u_prev[int(N/2):N] = -1.0 t = 0.0 fig = plt.figure() plt.plot(u_prev) BC = 1 k = 0 while t < T: for i in range(1,N-1): visc_term = art_visc_coeff*dt*abs(a)/dx*(u_prev[i+1]-2*u_prev[i] +u_prev[i-1]) u[i] = u_prev[i] - dt*a * ( u_prev[i+1]- u_prev[i-1])/(2*dx) +visc_term if BC == 1: # wrong way to put numerical BC - sol at outlow remains 0. u[0] = 1.0 u[N-1] = -1. # WRONG elif BC == 2: # first order extrapolation u[N-1] = u[N-2] u[0] = 1 elif BC == 3: # 2nd order extrapolation u[0] = 1 u[N-1] = 2*u[N-2] - u[N-3] elif BC == 4: # upwind discretization of the eq u[N-1] = u_prev[N-1] - dt*a * ( u_prev[N-1]- u_prev[N-2])/dx u[0] = 1 plt.plot(u) for j in range(N): u_prev[j] = u[j] t = t + dt k = k + 1 plt.show()
时间: 2024-04-08 11:36:29 浏览: 131
import numpy as np import matplotlib.pyplot as plt import math
这段代码是用来模拟一维对流扩散方程的数值解。代码中使用了有限差分方法来离散化方程,然后使用迭代循环来求解。
首先,定义了一些参数,如长度L,时间T,空间网格数N,空间步长dx,时间步长dt等。然后初始化了初始条件u_prev,并设置边界条件BC。
在迭代循环中,首先计算了粘性项visc_term,并根据对流扩散方程进行更新计算。根据不同的边界条件BC,对边界点进行修正。然后将更新后的结果绘制出来,并更新u_prev为当前的解。重复以上步骤直到达到指定的时间T。
最后,使用plt.show()将结果绘制出来。
需要注意的是,这段代码中给出了几种不同的边界条件选择,通过修改BC的值可以切换不同的边界条件。具体的选择和实际问题相关,请根据需要进行适当的调整。
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