我正在尝试从this paper实现公式16:
上面的方程式16是粘性波动方程式,它应该从大开始并随着时间的流逝而消亡,最终逼近0。但是,当我运行模拟时,它似乎爆炸了。如果您看下面的图片(迭代0-4),它似乎工作正常(即,正弦波看上去越来越小,这很好)。但是,在5.6和更高的位置开始爆炸(您可以看到y轴表示压力以数量级增加)。
这里是输出:
这里是使用Finite Difference Method的代码:
import numpy as np
import math
import matplotlib
import matplotlib.pyplot as plt
import time
x_mesh_param = 1000
# function that increments time steps (solves for the next state of p)
# returns 'p2'
# assuming no boundary conditions because we only care about what a hydrophone
# receives. Don't care about what happens to wave beyond hydrophone.
def iterate(p0,p1,x_mesh,dt,v,c0 ):
# step 3 on pg. 9
dx=x_mesh[1]-x_mesh[0]
p2 = np.zeros(p1.shape) # this is where we store next state of p
for i in range(1,len(x_mesh)-1):
p_txx = (p1[i+1]-p0[i+1]-2*(p1[i]-p0[i])+p1[i-1]-p0[i-1])/ (dt* dx**2)
p_xx = (p1[i+1]-2*p1[i]+p1[i-1]) / (dx**2)
p2[i] = (dt)**2 * (4.0/3 * v * p_txx + c0**2*p_xx)+2*p1[i]-p0[i]
# what happens at p2[0] (begin) and p2[-1] (end)
# forward difference (NO BOUNDARY conditions)
p_txx = (p1[2]-p0[2]-2*(p1[1]-p0[1])+p1[0]-p0[0])/(dt*dx**2)
p_xx = (p1[2]-2*p1[1]+p1[0]) / (dx**2)
#p2[0] = (dt)**2 * (4.0/3 * v * p_txx + c0**2*p_xx)+2*p1[0]-p0[0]
# taylor series
#p2[0]=1/(1-1/dt+1/(2*(dt)**2))* ((1-1/dt+1/((dt)**2))*p2[1]-p2[2]/(2*(dt**2)))
#p2[0] = p1[1] # NEUMANN
# if you comment out lines 34,37,39 you get Dirichlet Conditions
# backward difference
p_txx = (p1[-1]-p0[-1]-2*(p1[-2]-p0[-2])+p1[-3]-p0[-3])/(dt*dx**2)
p_xx = (p1[-1]-2*p1[-2]+p1[-3]) / (dx**2)
#p2[-1] = (dt)**2 * (4.0/3 * v * p_txx + c0**2*p_xx)+2*p1[-1]-p0[-1]
# Dirichlet if line 46 commented out
return p2
def firstIteration(p1,x_mesh,dt,v,c0 ):
# step 3 on pg. 9
dx=x_mesh[1]-x_mesh[0]
p2 = np.zeros(p1.shape) # this is where we store next state of p
for i in range(1,len(x_mesh)-1):
p_txx = 0
p_xx = (p1[i+1]-2*p1[i]+p1[i-1]) / (dx**2)
p2[i] = (dt)**2 * (4.0/3 * v * p_txx + c0**2*p_xx)+p1[i] # assuming p1[i]-p0[i]=0 (coming from assumption that d/dt (p(x,0)) =0 (initial cond. of motion of fluid)
# what happens at p2[0] (begin) and p2[-1] (end)
# forward difference (NO BOUNDARY conditions)
p_txx = 0
p_xx = (p1[2]-2*p1[1]+p1[0]) / (dx**2)
#p2[0] = (dt)**2 * (4.0/3 * v * p_txx + c0**2*p_xx)+p1[0]
# taylor series
#p2[0]=1/(1-1/dt+1/(2*(dt)**2))* ((1-1/dt+1/((dt)**2))*p2[1]-p2[2]/(2*(dt**2)))
#p2[0] = p1[1] # NEUMANN
# if you comment out lines 34,37,39 you get Dirichlet Conditions
# backward difference
p_txx = 0
p_xx = (p1[-1]-2*p1[-2]+p1[-3]) / (dx**2)
#p2[-1] = (dt)**2 * (4.0/3 * v * p_txx + c0**2*p_xx)+p1[-1]
return p2
def simulate():
states = []
L=100
#dt=.02
dt=0.001
v = .001/998
c0 = 1480
x_mesh = np.linspace(0,L,x_mesh_param)
state_initial = np.array([math.sin(x/20*math.pi) for x in x_mesh])
states.append(state_initial)
states.append(firstIteration(states[0], x_mesh,dt,v,c0 ))
for i in range(50):
new_state = iterate(states[-2], states[-1], x_mesh,dt,v,c0)
states.append(new_state)
return states
if __name__=="__main__":
states = simulate()
L=100
x_mesh = np.linspace(0,L,x_mesh_param)
counter=0
for s in states:
fig, ax = plt.subplots()
ax.plot(x_mesh, s)
ax.set(xlabel='distance (m)', ylabel='pressure (atm)',
title='Pressure vs. distance, iteration = '+str(counter))
ax.grid()
plt.show()
print counter
counter = counter + 1
您将看到有一个主程序调用simulate()。然后Simulation()调用firstIteration()尝试处理边界条件。然后其余的由iterate()处理。我基本上是使用中央,向前和向后的差异来计算波动方程的导数。
主程序只是在不同的时间步长上打印出整个图形。
从数学上讲,您的离散方程组具有随着时间呈指数增长的解。浮点数的最低有效数字的数字噪声将呈指数增长,直到它接管了您要寻找的解决方案。
如果P
是由当前和前两个时间步长的N
压力值组成的向量(3N值的数组),则代码中所有值的加法和减法都可以用矩阵形式重写(表示为一步):
P := D @ P
其中D
是形状(3N,3N)的矩阵。如果P0
是系统的初始状态,则在n
时间步长之后的系统状态为[]
Pn = np.linalg.matrix_power(D, n) @ P0
这里,matrix_power
是矩阵求幂,即matrix_power(D, 3) == D @ D @ D
。如果D
的特征值的实部> 1,则解将呈指数增长。您可以使用np.linalg.eigvals
检查特征值,最好使用较小的矩阵,例如N=100
。如果幸运的话,减小步长dx
和dt
可能足以使特征值进入非正值范围。如果不是,则必须从
D
中消除较大的特征值:
import numpy as np evals, evecs = np.linalg.eig(D) evals_1 = np.where(evals.real > 1, 0, evals) D_1 = evecs @ np.diag(evals_1) @ np.linalg.inv(evecs)
例如,使用小的矩阵(特征值1.01和0.99):
D = np.array([[ 1. , -0.01], [-0.01, 1. ]]) # result from eigenvalue elimination as above D_1 = np.array([[0.495, 0.495], [0.495, 0.495]])
尝试使用与特征值0.99的特征向量相差很小的初始向量:
>>> matpow = np.linalg.matrix_power >>> matpow(D, 200) @ [1, 1.001] array([0.13038866, 0.13770467]) >>> matpow(D_1, 200) @ [1, 1.001] array([0.13404666, 0.13404666]) >>> matpow(D, 2000) @ [1, 1.001] array([-219643.102525, 219643.102525]) >>> matpow(D_1, 2000) @ [1, 1.001] array([1.86468848e-09, 1.86468848e-09])
在经过200个时间步长之后,数字是相似的,但是在经过2000个时间步长之后,原始矩阵的输出就发散了。[如果遇到查找特征值和特征向量的麻烦(对于大型矩阵来说可能很慢,则可以通过用
matpow(D_1, n)
替换为]来大大加快后续计算的速度)>
Pn = evecs @ np.diag(evals_1**n) @ np.linalg.inv(evecs) @ P0