小编Mll*_*che的帖子

我正在数值求解一阶微分方程组的x(t).该系统是:

dy/dt=(C)\*[(-K\*x)+M*A]

我已经实现了Forward Euler方法来解决这个问题,如下所示:这是我的代码:

import matplotlib
import numpy as np
from numpy import *
from numpy import linspace
from matplotlib import pyplot as plt


C=3
K=5
M=2
A=5
#------------------------------------------------------------------------------
def euler (f,x0,t):
    n=len (t)
    x=np.array ([x0*n])
    for i in xrange (n-1):
        x[i+1] = x[i] + ( t[i+1] - t[i] ) * f( x[i], t[i] )
    return x



#---------------------------------------------------------------------------------          
if __name__=="__main__":
    from pylab import *

    def f(x,t): 
        return (C)*[(-K*x)+M*A]

    a,b=(0.0,10.0)
    n=200
    x0=-1.0
    t=linspace (a,b,n)

    #numerical solutions
    x_euler=euler(f,x0,t)

    #compute true solution values …

我刚开始用Python编程,对Numpy软件包很新......我仍然试图了解它.我试图用euler方法解决一个函数.

这是我的代码:

Z=4
B=8
U=1
C=4

a,b=(0.0,10.0)
n=2000
x0=-1.0
t=linspace (a,b,n)
#-----------------------------------------------------------------------------
def euler (f,x0,t):
    n=len (t)
    x=np.array(n*[x0,])
    for i in xrange (n-1):
        float (x[i] + ( t[i+1] - t[i] ) * f( x[i], t[i] ))
    return x



#---------------------------------------------------------------------------------          
if __name__=="__main__":


    def f(x,t): 
        return float((Z)*[-(1/6)*B*C*x^3+0.5*U*t^2])


    #numerical solutions
    x_euler=euler(f,x0,t)


    #figure
    plt.plot (t,x_euler, "b")
    xlabel (t)
    ylabel (x)
    legend ("Euler")

    show()

对于这些问题,我不同意类似的解决方案.这是我的追溯:

Traceback (most recent call last):
  File "C:\Python27\testeuler.py", line 45, in <module>
    x_euler=euler(f,x0,t)
  File "C:\Python27\testeuler.py", line 31, in euler
    float (x[i] …