Python中具有无限初始条件的ODE [英] ODEs with infinite initlal condition in python

问题描述

我有一个二阶微分方程，我想用python解决.问题在于，对于其中一个变量，我没有0的初始条件，而只有无穷大的值.能否告诉我应该为scipy.integrate.odeint提供哪些参数?能解决吗?

I have a second order differential equation that I want to solve it in python. The problem is that for one of the variables I don't have the initial condition in 0 but only the value at infinity. Can one tell me what parameters I should provide for scipy.integrate.odeint ? Can it be solved?

等式:

Theta需要在时间上找到.它的一阶导数在t=0处等于零. Theta在t=0处未知，但在足够大的时间变为零.所有其余的都是已知的.由于近似值I可以设置为零，因此消除了二阶导数，这应该使问题更容易解决.

Theta needs to be found in terms of time. Its first derivative is equal to zero at t=0. theta is not known at t=0 but it goes to zero at sufficiently large time. all the rest is known. As an approximate I can be set to zero, thus removing the second order derivative which should make the problem easier.

推荐答案

这远不是一个完整的答案，而是应OP的要求张贴在这里.

This is far from being a full answer, but is posted here on the OP's request.

我在评论中描述的方法就是所谓的拍摄方法，它允许将边界值问题转换为初始值问题.为了方便起见，我将您的函数theta重命名为y.要用数字方式求解方程，您首先需要使用两个辅助函数z1 = y和z2 = y'将其转换为一阶系统，因此您的当前方程式

The method I described in the comment is what is known as a shooting method, that allows converting a boundary value problem into an initial value problem. For convenience, I am going to rename your function theta as y. To solve your equation numerically, you would first turn it into a first order system, using two auxiliary function, z1 = y and z2 = y', and so your current equation

I y'' + g y' + k y = f(y, t)

将被系统迷惑

z1' = z2
z2' = f(z1, t) - g z2 - k z1

您的边界条件是

z1(inf) = 0
z2(0) = 0

因此，我们首先设置函数以计算新矢量函数的导数:

So first we set up the function to compute the derivative of your new vectorial function:

def deriv(z, t) :
    return np.array([z[1],
                     f(z[0], t) - g * z[1] - k * z[0]])

如果我们有条件z1[0] = a，我们可以在t = 0和t = 1000之间进行数值求解，并在最后一次获得y的值，例如

If we had a condition z1[0] = a we could solve this numerically between t = 0 and t = 1000, and get the value of y at the last time as something like

def y_at_inf(a) :
    return scipy.integrate.odeint(deriv, np.array([a, 0]),
                                  np.linspace(0, 1000, 10000))[0][-1, 0]

所以现在我们所需要知道的是a的值使t = 1000在我们穷人的无穷大t = 1000处具有

So now all we need to know is what value of a makes y = 0 at t = 1000, our poor man's infinity, with

a = scipy.optimize.root(y_at_inf, [1])

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