在Python中求解微分方程的麻烦

Question

我正在尝试完成this，其中我必须解决五个普通问题 使用odeint的微分方程，并复制该任务中给出的数字。

这是我的代码：

import scipy as sp
import scipy.interpolate as ip
import numpy as np
import matplotlib.pyplot as pl

d = 8.64
Mu1 = 4.95*10**2
Mu2 = 4.95*10**(-2)
vs = 0.12
vd = 1.23
w = 10**(-3)
k1 = 2.19*10**(-4)
k2 = 6.12*10**(-5)
k3 = 0.997148
k4 = 6.79*10**(-2)

p0 = 1.00
sigmas0 = 2.01
sigmad0 = 2.23
alphas0 = 2.20
alphad0 = 2.26

hs = (sigmas0-(sigmas0**(2)-k3*alphas0*(2*sigmas0-alphas0))**(1/2))/k3
cs = (alphas0-hs)/2
ps = k4*(hs**2)/cs

t_points = [ 1000, 1850, 1950, 1980, 2000, 2050, 2080, 2100, 2120, 2150, 2225, 2300, 2500, 5000 ]
y_points = [ 0.0, 0.0, 1.0, 4.0, 5.0, 8.0, 10.0, 10.5, 10.0, 8.0, 3.5, 2.0, 0.0, 0.0 ]

t1 = np.array(t_points)
y1 = np.array(y_points)

new_length = 1000
new_t = np.linspace(t1.min(), t1.max(), new_length)
new_y2 = ip.pchip_interpolate(t1, y1, new_t)

pl.plot(t_points,y_points,'o', new_t,new_y2)
pl.show()

ft = sp.interpolate.interp1d(new_t, new_y2)

def equations(x, t1):

    p = x[0]
    alphad = x[1]
    alphas = x[2]
    sigmad = x[3] 
    sigmas = x[4]

    dpdt = (ps-p)/d + ft/Mu1
    dalphaddt = (1/vd)*(k2-w*(alphad-alphas))
    dalphasdt = (1/vs)*(w*(alphad-alphas)-k2)
    dsigmaddt = (1/vd)*(k1-w*(sigmad-sigmas))
    dsigmasdt = (1/vs)*(w*(sigmad-sigmas)-k1-(ps-p)/d*Mu2)

    return [dpdt, dalphaddt, dalphasdt, dsigmaddt, dsigmasdt]

solve =  sp.integrate.odeint(equations, [p0, alphad0, alphas0, sigmad0, sigmas0], t1)

似乎这部分：

dpdt = (ps-p)/d + ft/Mi1

是错误的，我不知道如何解决。

错误提示：

  

TypeError：/：“ interp1d”和“ float”不支持的操作数类型。

任何想法和建议都值得赞赏。

编辑：当我应用meowgoesthedog建议的更改时，出现错误：

Traceback (most recent call last):

  File "<ipython-input-5-324757833872>", line 1, in <module>
    runfile('E:/Data/Project 2/project2.py', wdir='E:/Data/Project 2')

  File "D:\Programs\Anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 668, in runfile
    execfile(filename, namespace)

  File "D:\Programs\Anaconda3\lib\site-packages\spyder_kernels\customize\spydercustomize.py", line 108, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)

  File "E:/Data/Project 2/project2.py", line 59, in <module>
    solve =  odeint(equations, [p0, alphad0, alphas0, sigmad0, sigmas0], t1)

  File "D:\Programs\Anaconda3\lib\site-packages\scipy\integrate\odepack.py", line 233, in odeint
    int(bool(tfirst)))

  File "E:/Data/Project 2/project2.py", line 51, in equations
    dpdt = (ps-p)/d + ft(t1)/Mu1

  File "D:\Programs\Anaconda3\lib\site-packages\scipy\interpolate\polyint.py", line 79, in __call__
    y = self._evaluate(x)

  File "D:\Programs\Anaconda3\lib\site-packages\scipy\interpolate\interpolate.py", line 664, in _evaluate
    below_bounds, above_bounds = self._check_bounds(x_new)

  File "D:\Programs\Anaconda3\lib\site-packages\scipy\interpolate\interpolate.py", line 696, in _check_bounds
    raise ValueError("A value in x_new is above the interpolation "

ValueError: A value in x_new is above the interpolation range.

`

Answer 1

根据interp1d's documentation：

  

ynew = f(xnew) # use interpolation function returned by interp1d

它返回一个函数/可调用对象，该对象采用值x并返回内插值f(x)。对于您的情况"x" = t：

dpdt = (ps-p)/d + ft(t1)/Mu1   # pass t1 to ft to obtain interpolated value

---

更新

1. 此新错误是由于odeint在f(t)的值t <{> 。这对于纠错是必需的，并且没有防止t_points这样做的选项。但是，我们可以使用odeint，在提供的样本之外外推 f(t)：

   InterpolatedUnivariateSpline

   与from scipy.interpolate import InterpolatedUnivariateSpline ... ft = InterpolatedUnivariateSpline(t1, y1, k=1) 一样，它返回具有相同签名的函数。但是，应用此修复程序后，结果将变为：

   

   那当然是不正确的。

2. 您已在函数外部将interp1d声明为常量。实际上，它们是hs, cs, ps和alpha*变量的函数，因此必须在每次调用sigma*时进行评估：

   equation

   现在，结果与练习中的图形差不多了。

   

3. 您将def equations(x, t): p = x[0] alphad = x[1] alphas = x[2] sigmad = x[3] sigmas = x[4] hs = (sigmas-(sigmas**(2)-k3*alphas*(2*sigmas-alphas))**(1/2))/k3 cs = (alphas-hs)/2 ps = k4*(hs**2)/cs dpdt = (ps-p)/d + ft(t)/Mu1 dalphaddt = (1/vd)*(k2-w*(alphad-alphas)) dalphasdt = (1/vs)*(w*(alphad-alphas)-k2) dsigmaddt = (1/vd)*(k1-w*(sigmad-sigmas)) dsigmasdt = (1/vs)*(w*(sigmad-sigmas)-k1-(ps-p)/d*Mu2) return [dpdt, dalphaddt, dalphasdt, dsigmaddt, dsigmasdt] 作为水平轴变量传递给了t1。它只有14个元素，对于平滑输出来说太少了。改为通过odeint：

   new_t

   结果现在完全符合预期！