Matlab ode45和Scipy odeint之间的差异:同一模型的结果不同

Question

编辑

因此，我能够生成正确的图，但这是在将采样间隔调整为以下之后：

t_list = np.linspace(0, 30, 100)

将X打印为：

[1.         1.         1.         1.         1.         1.
1.         1.         1.         1.         1.         1.
1.         1.         1.         1.         1.         1.
0.91265299 0.8107059  0.7366542  0.68370578 0.64633005 0.62021062
0.6020953  0.58960093 0.58101775 ...]

但这就引出了一个问题：为什么这个系统如此依赖于采样间隔？

结束编辑

我正在尝试使用scipy将一个简单的matlab微分方程系统重新创建为python。我不知道为什么我在scipy执行中收到RuntimeWarning: invalid value encountered in double_scalars。我在odeint调用中是否缺少一个可选参数？

python:

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

def model(y0, t):
x = y0[0]
y = y0[1]
z = y0[2]
if t <= 10:
sys_input = 1.0
else:
sys_input = 0.75
a = 1.0
b = 1.0
c = 1.0
E = 1.0
dxdt = sys_input - a * E * (x ** 0.5)
dydt = a * E * (x ** 0.5) - b * (y ** 0.5)
dzdt = b * (y ** 0.5) - c * (z ** 0.5)
return [dxdt, dydt, dzdt]

t_list = np.linspace(0, 30, 31)
# Initial conditions vector
yi = [1.0, 1.0, 1.0]
ret = odeint(model, y0=yi, t=t_list)
X = ret[:, 0]
print(X)

并打印以下内容：

<input>:18: RuntimeWarning: invalid value encountered in double_scalars
<input>:19: RuntimeWarning: invalid value encountered in double_scalars
[ 1.  1.  1.  1. nan nan nan nan nan nan nan nan nan nan nan nan nan nan
nan nan nan nan nan nan nan nan nan nan nan nan nan]

其中，如下matlab代码生成连续结果：

tspan = [0,30];
x0 = 1.0; % Initial value of x
y0 = 1.0; % Initial value of y
z0 = 1.0; % Initial value of z
initial_values = [x0; y0; z0]; % Initial value of the vector w
[T,R] = ode45(@(t,w) diff_eq(t,w),tspan,initial_values);
X = R(:,1);
Y = R(:,2);
Z = R(:,3);
for i = 1: length(X)
if(mod(i, 10)==0 &&  i > 1)
disp(' ');
end
fprintf('X[%i] = %.2f, ', i, X(i));
end
disp(' ');

function dw_vectordt = diff_eq(t,w_vector) 
x = w_vector(1);
y = w_vector(2);
z = w_vector(3);
if (t<=10)
sys_input= 1.0;    
else
sys_input=0.75;
end
a = 1.0;
b = 1.0;
c = 1.0;
E = 1.0;
dxdt = sys_input-a*E*x^(0.5);
dydt = a*E*x^(0.5)-b*y^(0.5);
dzdt = b*y^(0.5)-c*z^(0.5);
dw_vectordt = [dxdt; dydt; dzdt];
end

打印语句返回：

X[1] = 1.00, X[2] = 1.00, X[3] = 1.00, X[4] = 1.00, X[5] = 1.00, X[6] = 1.00, X[7] = 1.00, X[8] = 1.00, X[9] = 1.00,  
X[10] = 1.00, X[11] = 1.00, X[12] = 1.00, X[13] = 1.00, X[14] = 1.01, X[15] = 0.99, X[16] = 0.92, X[17] = 0.82, X[18] = 0.78, X[19] = 0.74,  
X[20] = 0.71, X[21] = 0.68, X[22] = 0.66, X[23] = 0.64, X[24] = 0.63, X[25] = 0.61, X[26] = 0.60, X[27] = 0.59, X[28] = 0.59, X[29] = 0.58,  
X[30] = 0.58, X[31] = 0.57, X[32] = 0.57, X[33] = 0.57, X[34] = 0.57, X[35] = 0.57, X[36] = 0.56, X[37] = 0.56, X[38] = 0.56, X[39] = 0.56,  
X[40] = 0.56, X[41] = 0.56, X[42] = 0.56, X[43] = 0.56, X[44] = 0.56, X[45] = 0.56, X[46] = 0.56, X[47] = 0.56, X[48] = 0.56, X[49] = 0.56,  
X[50] = 0.56, X[51] = 0.56, X[52] = 0.56, X[53] = 0.56,

Answer 1

您正在处理的ODE系统可能很僵硬。当y0的元素在积分过程中变为负数时，您遇到的RuntimeWarning由平方根运算引发。出现这种情况是因为积分器的时间步长太大，并且您使用的解算器不太适合潜在的刚性系统。通过t_list增加提供给t的元素的数量可能会减少时间步长，因此允许变通方法。为了更好地理解正在发生的事情，我鼓励您使用下面的片段，该片段使用SciPy推荐的较新的solve_ivpAPI。特别感兴趣的是关键字自变量method和max_step。使用RK23、RK45、DOP853和LSODA中的任何一个作为一种方法都会导致开箱即用的较差的解估计，因为所有这些解算器都是从非刚性方法开始的。LSODA应该检测刚度并切换到刚度积分器，但由于时间步长快速增加，它无法做到这一点。对于所有这些方法，设置max_step=0.5允许以潜在的计算时间为代价处理ODE系统。或者，使用Radau或BDF将开箱即用，因为这些解算器可以处理刚性ODE系统。但是，建议手动提供系统的雅可比矩阵，否则用有限差分近似。

import numpy as np
from scipy.integrate import solve_ivp

def model(t, y0):
x, y, z = y0
sys_input = 1 if t <= 10 else 0.75
a, b, c, E = 1, 1, 1, 1
return (sys_input - a * E * np.sqrt(x),
a * E * np.sqrt(x) - b * np.sqrt(y),
b * np.sqrt(y) - c * np.sqrt(z))

sol = solve_ivp(model, t_span=(0, 30), y0=(1, 1, 1))
print(sol)