需要帮助解决python中的二阶非线性ODE [英] Need help solving a second order non-linear ODE in python

问题描述

我真的不知道从哪里开始这个问题，因为我对此没有很多经验，但是需要使用计算机来解决项目的这一部分.

I don't really know where to start with this problem, as I haven't had much experience with this but it is required to solve this part of the project using a computer.

我有一个二阶ODE，即:

I have a 2nd order ODE which is:

m = 1220

k = 35600

g = 17.5

a = 450000

和b在1000到10000之间，增量为500.

and b is between 1000 and 10000 with increments of 500.

x(0)= 0 

x'(0)= 5


m*x''(t) + b*x'(t) + k*x(t)+a*(x(t))^3 = -m*g

我需要找到最小的b，以便解永远不会是正数.我知道图形应该是什么样子，但是我只是不知道如何使用odeint来获得微分方程的解. 这是我到目前为止的代码:

I need to find the smallest b such that the solution is never positive. I know what the graph should look like, but I just don't know how to use odeint to get a solution to the differential equation. This is the code I have so far:

from    numpy    import    *    
from    matplotlib.pylab    import    *    
from    scipy.integrate    import    odeint

m = 1220.0
k = 35600.0
g  = 17.5
a = 450000.0
x0= [0.0,5.0]

b = 1000

tmax = 10
dt = 0.01

def fun(x, t):
    return (b*x[1]-k*x[0]-a*(x[0]**3)-m*g)*(1.0/m)
t_rk = arange(0,tmax,dt)   
sol = odeint(fun, x0, t_rk)
plot(t_rk,sol)
show()

实际上什么都不产生.

有什么想法吗?谢谢

推荐答案

要使用scipy.integrate.odeint求解二阶ODE，应将其编写为一阶ODE系统:

To solve a second-order ODE using scipy.integrate.odeint, you should write it as a system of first-order ODEs:

我先定义z = [x', x]，然后定义z' = [x'', x']，这就是您的系统！当然，您必须插入自己的真实关系:

I'll define z = [x', x], then z' = [x'', x'], and that's your system! Of course, you have to plug in your real relations:

x'' = -(b*x'(t) + k*x(t) + a*(x(t))^3 + m*g) / m

成为:

z[0]' = -1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g)
z[1]' = z[0]

z[0]' = -1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g)
z[1]' = z[0]

或者，只需将其命名为d(z):

Or, just call it d(z):

def d(z, t):
    return np.array((
                     -1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g),  # this is z[0]'
                     z[0]                                         # this is z[1]'
                   ))

现在您可以像这样将其喂入odeint:

Now you can feed it to the odeint as such:

_, x = odeint(d, x0, t).T

(_是我们制作的x'变量的空白占位符)

(The _ is a blank placeholder for the x' variable we made)

为了在x的最大值始终为负的约束条件下最小化b，可以使用scipy.optimize.minimize.我将通过实际上最大化x的最大值来实现它，但要遵守x保持为负的约束，因为我无法考虑如何在不求函数反转的情况下最小化参数.

In order to minimize b subject to the constraint that the maximum of x is always negative, you can use scipy.optimize.minimize. I'll implement it by actually maximizing the maximum of x, subject to the constraint that it remains negative, because I can't think of how to minimize a parameter without being able to invert the function.

from scipy.optimize import minimize
from scipy.integrate import odeint

m = 1220
k = 35600
g = 17.5
a = 450000
z0 = np.array([-.5, 0])

def d(z, t, m, k, g, a, b):
    return np.array([-1/m * (b*z[0] + k*z[1] + a*z[1]**3 + m*g), z[0]])

def func(b, z0, *args):
    _, x = odeint(d, z0, t, args=args+(b,)).T
    return -x.max()  # minimize negative max

cons = [{'type': 'ineq', 'fun': lambda b: b - 1000, 'jac': lambda b: 1},   # b > 1000
        {'type': 'ineq', 'fun': lambda b: 10000 - b, 'jac': lambda b: -1}, # b < 10000
        {'type': 'ineq', 'fun': lambda b: func(b, z0, m, k, g, a)}] # func(b) > 0 means x < 0

b0 = 10000
b_min = minimize(func, b0, args=(z0, m, k, g, a), constraints=cons)

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