Question

我有三个耦合的ODE，我在python中使用RK4方法来解决它们给定的初始条件。当我运行它们时，我收到以下错误：

RuntimeWarning：在exp
中遇到溢出

对我可能做错的任何帮助都将非常感激。

以下是代码（最初粘贴here）：

#simple exponential potential
# u = K*phi'/H0; v = omega_matter**(1/3)*(1+z); w = l*K*phi' - ln((K**2)*V0/H0**2)
# f,g,h are functions of derivation of u,v,w respectively derieved w.r.t t*H0 = T

import matplotlib.pyplot as plt
import numpy as np
import math

def f(T,u,v,w):
    return -3*u*((v**3 + (u**2)/6 + np.exp(-w)/3)**0.5) + l*np.exp(-w)

def g(T,u,v,w):
    return -v*(v**3 + (u**2)/6 + np.exp(-w)/3)**0.5

def h(T,u):
    return l*u

p = 0.1
q = 1.0
dh = 0.1
n = (q-p)/dh
u = [0.0]
v = [1100]
T = [0.00001]
w = [-44.927]
l = 1.3

for i in range(0,int(n)):
    k1 = f(T[i],u[i],v[i],w[i])
    r1 = g(T[i],u[i],v[i],w[i])
    print k1, r1
    s1 = h(T[i],u[i])
    print s1
    k2 = f(T[i] + 0.5*dh,u[i] + k1*dh,v[i] + k1*dh,w[i] + k1*dh)
    r2 = g(T[i] + 0.5*dh,u[i] + r1*dh,v[i] + r1*dh,w[i] + r1*dh)
    s2 = h(T[i] + 0.5*dh,u[i] + s1*dh)
    print k2,r2,s2
    k3 = f(T[i] + 0.5*dh,u[i] + k2*dh,v[i] + k2*dh,w[i] + k2*dh)
    r3 = g(T[i] + 0.5*dh,u[i] + r2*dh,v[i] + r2*dh,w[i] + r2*dh)
    s3 = h(T[i] + 0.5*dh,u[i] + s2*dh)
    k4 = f(T[i] + dh,u[i] + dh*k3,v[i] + dh*k3,w[i] + k3*dh)
    r4 = g(T[i] + dh,u[i] + r3*dh,v[i] + dh*r3,w[i] + r3*dh)
    s4 = h(T[i] + dh,u[i] + dh*s3)
    T == T.append(T[i] + dh)
    u == u.append(u[i] + (dh/6)*(k1 + 2.0*k2 + 2.0*k3 + k4))
    v == v.append(v[i] + (dh/6)*(r1 + 2.0*r2 + 2.0*r3 + r4))
    w == w.append(w[i] + (dh/6)*(s1 + 2.0*s2 + 2.0*s3 + s4))

plt.plot(T,u, '-b')
plt.plot(T,v, '-r')
plt.plot(T,w, '-g')   
plt.title('quintessence cosmological model')
plt.show()

Answer 1

您的问题似乎来自于f或g函数调用。将-w的值传递给该代码中的numpy.exp函数时，请注意它。如果它评估为足够大的正数，则调用exponential function就好像很容易导致溢出。谷歌告诉我，e ^ 44.927将评估为3.2474927e + 19。

您可能需要为w选择较小的绝对值才能运行此代码而不会出现溢出错误。

（请参阅Python-How to determine largest/smallest int/long/float/complex numbers my system can handle了解如何确定Python中允许的最大整数值是多少。）

RuntimeWarning：在exp中遇到溢出

1 个答案: