Question

我有两个像这样的微分方程：

al=arctan(ri')
ri''=(vz-C_alz(ri,al))*(1+ri' **2)/C_aln(ri,al)

初始值为：ri(l=0)=2.5, ri'(l=0)=0

vz是关于l和C_alz的函数，C_aln是关于ri和al的两个函数。这三个复杂的数学表达式通过代码1计算，结果是正确的，ri的轨迹可以用作对代码2结果的引用。

# -*- coding: utf-8 -*-
"""
Created on Fri Dec 18 14:41:07 2015
@author: marswang
code 1
"""

from numpy.linalg import inv
from sympy import sqrt, sin, cos, tan, atan, diff
from numpy import array
import sympy as sp
import matplotlib.pyplot as pl
import numpy as np
def r_trajektor (L,R):  
    B= np.array([[1, L[0], L[0]**2,L[0]**3 ], [0, 1, 2*L[0], 3*L[0]**2], [1, L[1], L[1]**2, L[1]**3], [0, 1, 2*L[1], 3*L[1]**2]])
    X= np.array([[R[0]], [0], [R[1]], [0]])
    A=np.dot(inv(B), X)
    return A    

a=7.6
rh_s=2420                          
rh_m=2285                          
the_0=11                           
Rc=49                              
vc= 0                             
vp=10

# two points in the r-trajektory and 1. derivative:
# point1 l0=0,r(l0)=2.5 and r'(l0)=0 
# point2 l1=36,r(l1)=22.5 and r'(l1)=0
L= array([[0], [36]])
R= array([[2.5], [22.5]])
# A symbolic derivation expression of vg
ri=sp.Symbol('ri') 
al=sp.Symbol('al')
l=sp.Symbol('l')  
A=r_trajektor(L,R)
# convert 'list' to 'float'
rir=float(A[0])+float(A[1])*l+float(A[2])*l**2+float(A[3])*l**3
dri=diff(rir,l)
alr=atan(diff(rir,l))
dal=diff(alr,l)
hi=a*sqrt((1-sin(the_0+al))/(1+a/(sqrt(2)*ri)))
hir=hi.subs([(al,alr),(ri,rir)])
C_aln=(1-ri**2/(Rc**2))*diff(hi,al)+(a**2)*ri/(Rc**2)*sin(the_0+al)
print C_aln
C_alz=1-(rh_s*ri**2)/(rh_m*Rc**2)+((1-(ri**2)/(Rc**2))*diff(hi,ri)-2*ri*hi/(Rc**2)-(a**2)/(Rc**2)*cos(the_0+al))*tan(al)
print C_alz
vg=((vp-vc)/(diff(al,l)*C_aln+C_alz)).subs([(al,alr),(ri,rir)])
vz=10/vg
print vz


# the function has to be worked with the arrays from numpy too.
# Thus have to lambdify the expression

l_val = np.linspace(0.0,36.0,100.0)

fr=sp.lambdify(l,rir,"numpy")
r_val=fr(l_val)
## plot
pl.figure(figsize=(10,10),dpi=98)
p1 =pl.subplot(2, 1, 1)
p1.plot(l_val,r_val,"g-",label="ri",)
p1.axis([0.0, 40.0,0.0,25.0])
p1.set_ylabel("ri in mm",fontsize=14)
p1.set_xlabel("l in mm",fontsize=14)
p1.legend(loc=4)

现在我想解决这两个微分方程，并使用代码2从计算出的ri代码1反向获取l与vz相关的图形（轨迹）

# -*- coding: utf-8 -*-
"""
Created on Sat Jan 02 13:15:19 2016
@author: marswang
code 2
"""
from sympy import sqrt, sin, cos, tan, atan
import matplotlib.pyplot as pl
import numpy as np
from scipy.integrate import odeint
a=7.6                              
rh_s=2420                          
rh_m=2285                          
the_0=11                           
Rc=49                              
vc= 0                             
vp=10

def func(ri,l):
    vz=((-0.00257201646090535*l**2 + 0.0925925925925926*l + 9.25185853854297e-17)*(-0.00633069554352353*sqrt((-sin(atan(-0.00257201646090535*l**2 + 0.0925925925925926*l + 9.25185853854297e-17) + 11) + 1)/(1 + 3.8*sqrt(2)/(-0.000857338820301783*l**3 + 0.0462962962962963*l**2 + 9.25185853854297e-17*l + 2.5)))*(-0.000857338820301783*l**3 + 0.0462962962962963*l**2 + 9.25185853854297e-17*l + 2.5) + 14.44*sqrt(2)*sqrt((-sin(atan(-0.00257201646090535*l**2 + 0.0925925925925926*l + 9.25185853854297e-17) + 11) + 1)/(1 + 3.8*sqrt(2)/(-0.000857338820301783*l**3 + 0.0462962962962963*l**2 + 9.25185853854297e-17*l + 2.5)))*(-(-0.000857338820301783*l**3 + 0.0462962962962963*l**2 + 9.25185853854297e-17*l + 2.5)**2/2401 + 1)/((1 + 3.8*sqrt(2)/(-0.000857338820301783*l**3 +
    0.0462962962962963*l**2 + 9.25185853854297e-17*l + 2.5))*(-0.000857338820301783*l**3 + 0.0462962962962963*l**2 + 9.25185853854297e-17*l + 2.5)**2) - 0.0240566430653894*cos(atan(-0.00257201646090535*l**2 + 0.0925925925925926*l + 9.25185853854297e-17) + 11)) - 484*(-0.000857338820301783*l**3 + 0.0462962962962963*l**2 +
    9.25185853854297e-17*l + 2.5)**2/1097257 + 1)    
    ri0=ri[0]
    ri1=ri[1]       
    C_aln=0.0240566430653894*ri0*sin(atan(ri1) + 11) - 3.8*sqrt((-sin(atan(ri1) + 11) + 1)/(1 + 3.8*sqrt(2)/ri0))*(-ri0**2/2401 + 1)*cos(atan(ri1) + 11)/(-sin(atan(ri1) + 11) + 1)
    C_alz=-484*ri0**2/1097257 + (-0.00633069554352353*ri0*sqrt((-sin(atan(ri1) + 11) + 1)/(1 +
    3.8*sqrt(2)/ri0)) - 0.0240566430653894*cos(atan(ri1) + 11) + 14.44*sqrt(2)*sqrt((-sin(atan(ri1) + 11) + 1)/(1 + 3.8*sqrt(2)/ri0))*(-ri0**2/2401 + 1)/(ri0**2*(1 + 3.8*sqrt(2)/ri0)))*tan(atan(ri1)) + 1    
    ri2=(vz-C_alz)*(1+ri1**2)/C_aln/
    return [ri1, ri2]

init = np.array([2.5, 0.1])
l = np.linspace(0.0,36.0,100.0)
sol=odeint(func, init, l)
pl.plot(l, sol[:,0], color='b')
pl.legend()
pl.xlabel(" l in mm ", fontsize=14)
pl.ylabel(" ri in mm ", fontsize=14)
pl.show()

我已多次检查vz，C_aln，C_alz的表达式。这些表达式直接来自代码1.但代码2的结果是不正确的，我得到的总是与引用不同。 ri的参考结果应该与第一张图片类似：

但代码2中'ri'的计算结果如下：

我不确定在这种情况下是否正确使用odeint，这两个图形之间出现此错误的原因是什么？

我是Python世界的新手，任何帮助都会非常感激。谢谢。

Answer 1

维度为2且阶数为1的系统的ODE函数应返回维数为2的向量。

对于ODE User，ODE函数应遵循方案

python3 manage.py sqlflush

Python：使用odeint求解二阶线性微分方程

1 个答案: