激波管问题python

激波管问题是一个经典的一维非定常流动问题。Python可以用来解决激波管问题。下面是一个使用Python解决激波管问题的示例代码：

import numpy as np
import matplotlib.pyplot as plt

# 定义初始条件
rhoL = 1.0
uL = 0.0
pL = 1.0
rhoR = 0.125
uR = 0.0
pR = 0.1
nx = 81
dx = 0.25
dt = 0.0002
gamma = 1.4

# 初始化数组
x = np.linspace(-10, 10, nx)
rho = np.ones(nx) * rhoL
rho[int((nx-1)/2):] = rhoR
u = np.ones(nx) * uL
u[int((nx-1)/2):] = uR
p = np.ones(nx) * pL
p[int((nx-1)/2):] = pR
E = p/(gamma-1) + rho*u**2/2
t = 0

# 定义计算函数
def get_flux(rho, u, p):
    F = np.zeros((3, nx))
    E = p/(gamma-1) + rho*u**2/2
    F[0] = rho*u
    F[1] = rho*u**2 + p
    F[2] = rho*u*E + u*p
    return F

# 进行时间循环
for n in range(100):
    rho_n = rho.copy()
    u_n = u.copy()
    p_n = p.copy()
    E_n = E.copy()
    F_n = get_flux(rho_n, u_n, p_n)
    for i in range(1, nx-1):
        rho[i] = 0.5 * (rho_n[i+1] + rho_n[i-1]) - dt/(2*dx)*(F_n[0][i+1] - F_n[0][i-1])
        u[i] = 0.5 * (u_n[i+1] + u_n[i-1]) - dt/(2*dx)*(F_n[1][i+1] - F_n[1][i-1] + (p_n[i+1] - p_n[i-1])/rho[i])
        p[i] = 0.5 * (p_n[i+1] + p_n[i-1]) - dt/(2*dx)*(F_n[2][i+1] - F_n[2][i-1] + u_n[i+1]*p_n[i+1]/rho[i+1] - u_n[i-1]*p_n[i-1]/rho[i-1])
    E = p/(gamma-1) + rho*u**2/2
    F = get_flux(rho, u, p)
    for i in range(1, nx-1):
        rho[i] = rho_n[i] - dt/dx*(F[0][i] - F[0][i-1])
        u[i] = u_n[i] - dt/dx*(F[1][i] - F[1][i-1] + (p[i] - p[i-1])/rho[i])
        p[i] = p_n[i] - dt/dx*(F[2][i] - F[2][i-1] + u[i]*(p[i] - p[i-1])/rho[i])
    E = p/(gamma-1) + rho*u**2/2
    t += dt

# 画图
plt.plot(x, rho, 'r', label='Density')
plt.plot(x, u, 'g', label='Velocity')
plt.plot(x, p, 'b', label='Pressure')
plt.legend()
plt.show()

这个示例代码使用了Lax-Wendroff数值方法来求解激波管问题。它可以计算激波管问题的密度、速度和压力随时间的变化。你可以根据需要修改初始条件、网格数和时间步长等参数来进行计算。