10

我真的不知道从哪里开始解决这个问题，因为我对此没有太多经验，但需要使用计算机解决项目的这一部分。

我有一个二阶 ODE，即：

m = 1220

k = 35600

g = 17.5

a = 450000

b 介于 1000 和 10000 之间，增量为 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 使得解决方案永远不会是积极的。我知道图表应该是什么样子，但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：

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()

这并没有真正产生任何东西。

有什么想法吗？谢谢

4

2 回答 2

10

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

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

becomes:

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

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]'
                   ))

Now you can feed it to the odeint as such:

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

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

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)
于 2013-11-05T00:48:38.710 回答
6

我不认为你可以解决你所说的问题：你的初始条件，x = 0并x' > 0暗示解决方案对于非常接近起点的某些值是积极的。所以没有任何b解决方案永远不会是积极的......

撇开这一点不谈，要求解二阶微分方程，首先需要将其重写为两个一阶微分方程的系统。定义y = x'我们可以将您的单个方程重写为：

x' = y
y' = -b/m*y - k/m*x - a/m*x**3 - g

x[0] = 0, y[0] = 5

所以你的函数应该是这样的：

def fun(z, t, m, k, g, a, b):
    x, y = z
    return np.array([y, -(b*y + (k + a*x*x)*x) / m - g])

你可以解决和绘制你的方程：

m, k, g, a = 1220, 35600, 17.5, 450000
tmax, dt = 10, 0.01
t = np.linspace(0, tmax, num=np.round(tmax/dt)+1)
for b in xrange(1000, 10500, 500):
    print 'Solving for b = {}'.format(b)
    sol = odeint(fun, [0, 5], t, args=(m, k, g, a, b))[..., 0]
    plt.plot(t, sol, label='b = {}'.format(b))
plt.legend()

在此处输入图像描述

于 2013-11-05T00:42:44.490 回答