根据论文,他们告诉我可以用RK四阶数值求解。在
如你所见,最后两个方程是耦合的,我构造了一个关联xn和yn的矩阵(概率),其中n=1..(n-对的计数,这里n等于4):向量([x1,x2,…,xn,y1,y2,…,yn])=概率.dot(向量([x1,x2,…,xn,y1,y2,…,yn]),其中质数是时间微分。但是每一步我都有额外的求和项(un*xn,yn也一样),这是我面临的第一个问题,不知道如何处理。在
我写了一段代码,出现了很多我无法处理的错误。在
当我试图独自应付时,我会非常感谢你的帮助。在
上面显示我的代码:
导入库
import numpy as np
import math
import scipy.constants as sc
from scipy.sparse import diags
from scipy.integrate import ode
import matplotlib.pyplot as plt
from matplotlib import mlab
初始数据和常数
^{pr2}$*构造与dif/eq/概率关联的矩阵*
k = np.array([hn_nInc*np.ones(n-1),hn*np.ones(n),hn_nDec*np.ones(n-1)])
offset = [-1,0,1]
Probability = diags(k,offset).toarray() # bn(tk)=xn(tk)+iyn(tk)
xt0_list = [0] * n
yt0_list = [0] * n
*dif右侧。等式.*
# dimen_param = [un,vn,zn,vzn] [tn]
# x_list = [x1,...,xn] [tn]
# y_list = [y1,...,yn] [tn]
def fun(dimen_param, x_list, y_list):
return dimen_param[1]
def fvn(dimen_param, x_list, y_list):
return -(x_list[len(x_list)-1]**2 + y_list[len(y_list)-1]**2)- wn_prime*dimen_param[1] + Omega_n * (1-np.e ** (-ksi_n * dimen_param[0]))*np.e ** (-ksi_n * dimen_param[0])
def fzn(dimen_param, x_list, y_list):
return dimen_param[3]
def fvzn(dimen_param, x_list, y_list):
return -wn_prime * dimen_param[3]-(wn_sqr ** 2) * dimen_param[2] - 1
def fxn(dimen_param, x_list, y_list):
return Probability.dot(y_list)
def fyn(dimen_param, x_list, y_list):
return -Probability.dot(x_list)
#xv = [dimen_param, x_list, y_list]
def f(xv):
k_d = xv[0:4]
k_x = xv[4:4+len(xt0_list)]
k_y = xv[4+len(xt0_list):4+len(xt0_list)+len(yt0_list)]
return ([fun(k_d, k_x, k_y),fvn(k_d, k_x, k_y),fzn(k_d, k_x, k_y),fvzn(k_d, k_x, k_y),fxn(k_d, k_x, k_y),fyn(k_d, k_x, k_y)])
*四阶龙格-库塔方法的实现*
def RK4(f, dimen_paramT0, xt0_list, yt0_list):
T = np.linspace(0, 1. / step, 1. / step +1)
xvinit = np.concatenate([dimen_paramT0, xt0_list, yt0_list])
xv = np.zeros( (len(T), len(xvinit)) )
xv[0] = xvinit
for i in range(int(1. / step)):
k1 = f(xv[i])
k2 = f(xv[i] + step/2.0*k1)
k3 = f(xv[i] + step/2.0*k2)
k4 = f(xv[i] + step*k3)
xv[i+1] = xv[i] + step/6.0 *( k1 + 2*k2 + 2*k3 + k4)
return T, xv
*正在运行*
print RK4(f, dimen_paramT0, xt0_list, yt0_list)
目前的问题是:
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-104-be8ed2e50d37> in <module>()
----> 1 RK4(f, dimen_paramT0, xt0_list, yt0_list)
<ipython-input-103-8c48cf5efe73> in RK4(f, dimen_paramT0, xt0_list, yt0_list)
7 for i in range(int(1. / step)):
8 k1 = f(xv[i])
----> 9 k2 = f(xv[i] + step/2.0*k1)
10 k3 = f(xv[i] + step/2.0*k2)
11 k4 = f(xv[i] + step*k3)
TypeError: can't multiply sequence by non-int of type 'float'
k1
是一个pythonlist
,其中multiply表示“repeat”,因此显然这对浮动没有意义
^{pr2}$您需要
numpy.ndarray
的向量行为,其中所以确保
f
的返回语句是:根据记录,
scipy
已经有了一个RK4 solver,不需要实现自己的。在相关问题 更多 >
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