在Python中使用odeint实现积分数学方程 [英] implement an integration math equation using odeint in Python

问题描述

我正在尝试使用scipy.odeint函数在python中求解以下方程式.

I am trying to solve the following equation in python using the scipy.odeint function.

目前，我能够实现这种形式的方程式

Currently I am able to implement this form of the equation

在python中使用以下脚本:

in python using the following script:

def dY(y1, x):
    a = 0.001
    yin = 1
    C = 0.01
    N = 1
    dC = C/N
    b1 = 0
    return (a/dC)*(yin-y1)+b1*dC

x = np.linspace(0,20,1000)
y0 = 0
res = odeint(dY, y0, x)
plt.plot(t,res, '-')
plt.show()

我的第一个方程式的问题是"i".我不知道如何对方程进行积分，但仍然能够提供当前和以前的"y"(yi-1和yi)值. "i"只是一个在0..100范围内的序列号.

My problem with the first equation is 'i'. I don't know how to integrate the equation and still be able to provide the current and previous 'y'(yi-1 and yi) values. 'i' is simply a sequence number that is within a range of 0..100.

原始等式为:

我用y，x，a，b和C重写了

Which I rewrote using y,x,a,b and C

Edit2: 我编辑了Pierre de Buyl的代码，并更改了N值.幸运的是，我有一个验证表来验证结果.不幸的是，结果并不相等.

I edited Pierre de Buyl' code and changed the N value. Luckily I have a validation table to validate the outcome against. Unfortunately, the results are not equal.

这是我的验证表:

这是numpy的输出:

and here is the numpy output:

使用的代码:

def dY(y, x):
    a = 0.001
    yin = 1
    C = 0.01
    N = 3
    dC = C/N
    b1 = 0.01
    y_diff = -np.copy(y)
    y_diff[0] += yin
    y_diff[1:] += y[:-1]
    return (a/dC)*(y_diff)+b1*dC

x = np.linspace(0,20,11)
y0 = np.zeros(3)
res = odeint(dY, y0, x)
plt.plot(x,res, '-')

如您所见，值相差0.02 ..

as you can see the values are different by an offset of 0.02..

我错过了导致偏移量的东西吗?

Am I missing something that results in this offset?

推荐答案

该方程是一个耦合"常微分方程(请参见

The equation is a "coupled" ordinary differential equation (see "System of ODEs" on Wikipedia.

变量是一个包含y[0]，y[1]等的向量.要求解ODE，必须将向量作为初始条件，函数dY也必须返回向量.

The variable is a vector containing y[0], y[1], etc. To solve the ODE you must feed a vector as the initial condition and the function dY must return a vector as well.

我已修改您的代码以实现此结果:

I have modified your code to achieve this result:

def dY(y, x):
    a = 0.001
    yin = 1
    C = 0.01
    N = 1
    dC = C/N
    b1 = 0
    y_diff = -np.copy(y)
    y_diff[0] += yin
    y_diff[1:] += y[:-1]
    return (a/dC)*y_diff+b1*dC

我已经将部分y[i-1] - y[i]编写为NumPy向量运算，并且对坐标y[0]作了特殊情况(在您的表示法中为y1，但在Python中数组从0开始).

I have written the part y[i-1] - y[i] as a NumPy vector operation and special cased the coordinate y[0] (that is the y1 in your notation but arrays start at 0 in Python).

对于所有yi使用初始值为0的解决方案是

The solution, using an initial value of 0 for all yi is

x = np.linspace(0,20,1000)
y0 = np.zeros(4)
res = odeint(dY, y0, x)
plt.plot(x,res, '-')

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