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[英]Solving system of coupled differential equations using Runge-Kutta in python
[英]Coupled system of 4 differential equations - Python
我在圖中得到了 4 個微分方程的耦合系統。 我有 4 個函數(xG;yG;gamma;beta)及其衍生物。 它們都是同一個自變量t的function。
我正在嘗試用 odeint 解決它。 問題是,為了做到這一點,我認為我需要以每個二階導數不依賴於其他二階導數的方式來表達系統。 這涉及到一定數量的數學,這肯定會讓我在某個地方出錯(我試過了。)。
你知道我怎么能:
我附上我的測試代碼
謝謝
import numpy
import math
from numpy import loadtxt
from pylab import figure, savefig
import matplotlib.pyplot as plt
# Use ODEINT to solve the differential equations defined by the vector field
from scipy.integrate import odeint
def vectorfield(w, t, p):
"""
Defines the differential equations for the coupled system.
Arguments:
w : vector of the state variables:
w = [Xg, Xg1 Yg, Yg1, Gamma, Gamma1, Beta, Beta1]
t : time
p : vector of the parameters:
p = [m, rAG, Ig,lcavo]
"""
#Xg is position ; Xg1 is the first derivative ; Xg2 is the second derivative (the same for the other functions)
Xg, Xg1, Yg, Yg1, Gamma, Gamma1, Beta, Beta1 = w
Xg2=-(Ig*Gamma2*math.cos(Beta))/(rAG*m*(-math.cos(Gamma)*math.sin(Beta)+math.sin(Gamma)*math.cos(Beta)))
Yg2=-(Ig*Gamma2*math.sin(Beta))/(rAG*m*(-math.cos(Gamma)*math.sin(Beta)+math.sin(Gamma)*math.cos(Beta)))-9.81
Gamma2=((Beta2*lcavo*math.sin(Beta))+(Beta1**2*lcavo*math.cos(Beta))+(Xg2)-(Gamma1**2*rAG*math.cos(Gamma)))/(rAG*math.sin(Gamma))
Beta2=((Yg2)+(Gamma2*rAG*math.cos(Gamma))-(Gamma1**2*rAG*math.sin(Gamma))+(Beta1**2*lcavo*math.sin(Beta)))/(lcavo*math.cos(Beta))
m, rAG, Ig,lcavo, Xg2, Yg2, Gamma2, Beta2 = p
# Create f = (Xg', Xg1' Yg', Yg1', Gamma', Gamma1', Beta', Beta1'):
f = [Xg1,
Xg2,
Yg1,
Yg2,
Gamma1,
Gamma2,
Beta1,
Beta2]
return f
# Parameter values
m=2.722*10**4
rAG=2.622
Ig=3.582*10**5
lcavo=4
# Initial conditions
Xg = 0.0
Xg1 = 0
Yg = 0.0
Yg1 = 0.0
Gamma=-2.52
Gamma1=0
Beta=4.7
Beta1=0
# ODE solver parameters
abserr = 1.0e-8
relerr = 1.0e-6
stoptime = 5.0
numpoints = 250
#create the time values
t = [stoptime * float(i) / (numpoints - 1) for i in range(numpoints)]
Deltat=t[1]
# Pack up the parameters and initial conditions:
p = [m, rAG, Ig,lcavo, Xg2, Yg2, Gamma2, Beta2]
w0 = [Xg, Xg1, Yg, Yg1, Gamma, Gamma1, Beta, Beta1]
# Call the ODE solver.
wsol = odeint(vectorfield, w0, t, args=(p,),
atol=abserr, rtol=relerr)
您需要將所有二階導數重寫為一階導數並一起求解 8 ODE:
然后你需要所有導數的初始條件,但似乎你已經有了。 僅供參考,您的代碼沒有運行( line 71: NameError: name 'Xg2' is not defined
),請檢查一下。
此外,有關更多信息,請參閱以數字方式求解二階 ODE 。
編輯#1:第一步,您需要解耦方程組。 雖然您可以手動解決它,但我不推薦,所以讓我們使用sympy
模塊:
import sympy as sm
from sympy import symbols
# define symbols. I assume all the variables are real-valued, this helps the solver. If not, I believe the result will be the same, but just calculated slower
Ig, gamma, gamma1, gamma2, r, m, beta, beta1, beta2, xg2, yg2, g, l = symbols('I_g, gamma, gamma1, gamma2, r, m, beta, beta1, beta2, xg2, yg2, g, l', real = True)
# define left hand sides as expressions
# 2nd deriv of gamma
g2 = (beta2 * l * sm.sin(beta) + beta1**2 *l *sm.cos(beta) + xg2 - gamma1**2 *r * sm.cos(gamma))/(r*sm.sin(gamma))
# 2nd deriv of beta
b2 = (yg2 + gamma2 * r * sm.cos(gamma) - gamma1**2 *r * sm.sin(gamma) + beta1**2 *l *sm.sin(beta))/(l*sm.cos(beta))
# 2nd deriv of xg
x2 = -Ig*gamma2*sm.cos(beta)/(r*m*(-sm.sin(beta)*sm.cos(gamma) + sm.sin(gamma)*sm.cos(beta)))
# 2nd deriv of yg
y2 = -Ig*gamma2*sm.sin(beta)/(r*m*(-sm.sin(beta)*sm.cos(gamma) + sm.sin(gamma)*sm.cos(beta))) - g
# now let's solve the system of four equations to decouple second order derivs
# gamma2 - g2 means "gamma2 - g2 = 0" to the solver. The g2 contains gamma2 by definition
# one could define these equations the other way, but I prefer this form
result = sm.solve([gamma2-g2,beta2-b2,xg2-x2,yg2-y2],
# this line tells the solver what variables we want to solve to
[gamma2,beta2,xg2,yg2] )
# print the result
# note that it is long and ugly, but you can copy-paste it as python code
for res in result:
print(res, result[res])
現在我們已經解耦了所有的二階導數。 例如, beta2
的表達式是
所以它(以及所有其他二階導數也)具有形式
請注意,不依賴於xg
或yg
。
並且要解決的整個 ODE 系統是
現在所有 ODE 都依賴於四個變量,這些變量不是任何事物的導數。 此外,由於xg
和yg
是退化的,因此也只有 6 個方程而不是 8 個。但是,可以將這兩個方程以與gamma
和beta
相同的方式重寫,以獲得 8 個方程的完整系統,並將其整合在一起。
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