Numpy 对于某些函数,欧拉能比龙格库塔更好吗?
我试图解决史蒂文·斯特罗加茨的非线性动力学和混沌的练习。在练习2.8.3、2.8.4和2.8.5中,对于初始值问题dx/dt=-x,预计将分别实施Euler方法、改进Euler方法和Runge-Kutta(四阶)方法;x(0)=1以查找x(1) 从分析上看,答案是1/e。我在寻找每种方法的误差。令我大吃一惊的是,与改进的Euler和Runge Kutta相比,我在Euler中得到的误差更小 我的代码看起来像这样。抱歉这么寒酸Numpy 对于某些函数,欧拉能比龙格库塔更好吗?,numpy,numerical-methods,differential-equations,runge-kutta,Numpy,Numerical Methods,Differential Equations,Runge Kutta,我试图解决史蒂文·斯特罗加茨的非线性动力学和混沌的练习。在练习2.8.3、2.8.4和2.8.5中,对于初始值问题dx/dt=-x,预计将分别实施Euler方法、改进Euler方法和Runge-Kutta(四阶)方法;x(0)=1以查找x(1) 从分析上看,答案是1/e。我在寻找每种方法的误差。令我大吃一惊的是,与改进的Euler和Runge Kutta相比,我在Euler中得到的误差更小 我的代码看起来像这样。抱歉这么寒酸 from scipy.integrate import odeint
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt
to = 0
xo = 1
tf = 1
deltaT = np.zeros([5])
errorE = np.zeros([5])
errorIE = np.zeros([5])
errorRK = np.zeros([5])
for j in range(0,5):
n = pow(10,j)
deltat = (tf - to)/(n)
print ("delta t is",deltat)
deltaT[j] = deltat
t = np.linspace(to,tf,n)
xE = np.zeros([n])
xIE = np.zeros([n])
xRK = np.zeros([n])
xE[0] = xo
xIE[0] = xo
xRK[0] = xo
for i in range (1,n):
#Regular Euler
xE[i] = deltat*(-xE[i-1]) + xE[i-1]
#Improved Euler
IEintermediate = deltat*(-xIE[i-1]) + xIE[i-1]
xIE[i] = xIE[i-1] - deltat*(xIE[i-1] + IEintermediate)/2
#Runge-Kutta fourth order
k1 = -deltat*xRK[i-1]
k2 = -deltat*(xRK[i-1] + k1/2)
k3 = -deltat*(xRK[i-1] + k2/2)
k4 = -deltat*(xRK[i-1] + k3)
xRK[i] = xRK[i-1] + (k1 + 2*k2 + 2*k3 + k4)/6
print (deltat,xE[i],xIE[i],xRK[i])
errorE[j] = np.exp(-1) - xE[n-1]
errorIE[j] = np.exp(-1) - xIE[n-1]
errorRK[j] = np.exp(-1) - xRK[n-1]
错误:
对于delT=1.0
- 欧拉误差为-0.63212055888285577
- I.欧拉误差为-0.63212055888285577
- RK误差为-0.63212055888285577
- 欧拉误差-0.019541047828557645
- I.欧拉误差-0.039348166379443716
- RK错误-0.03869055002863331
- 欧拉-0.0018501964782845493
- 一、欧拉-0.003703427083890265
- RK-0.0036972498815148747
- 欧拉-0.0001840470877806366
- I.欧拉-0.000368124801483849635
- RK-0.00036806344222467535
- 欧拉-1.839504510836587e-05
- 一、欧拉-3.67903967520844e-05
- RK-3.678978357835039e-05
这合法吗?如果没有,为什么会发生这种情况?您只在执行步长为
h=1/n
的积分步骤,因此您可以计算
exp(-(n-1)/n)=1/e*exp(1/n)
它有近似值
1/e + 1/e*1/n
报告的误差值正好是一阶的,因此会被一阶欧拉方法明显扭曲。更准确地说,Euler值是
(1-1/n)^(n-1) = exp((n-1)*(-1/n-1/(2n^2)+O(1/n^3))
= 1/e*exp(1/(2n)+..)
= 1/e + h/(2e) + ...
如果您调整代码以执行额外步骤以达到时间1
,您将获得正确的错误图片
delta t is 1.0
Euler 0.0 0.367879441171
imp. Euler 0.5 -0.132120558829
Runge-Kutta 4 0.375 -0.00712055882856
delta t is 0.1
Euler 0.3486784401 0.0192010010714
imp. Euler 0.368540984834 -0.00066154366211
Runge-Kutta 4 0.367879774412 -3.33241056083e-07
delta t is 0.01
Euler 0.366032341273 0.00184709989821
imp. Euler 0.367885618716 -6.17754474969e-06
Runge-Kutta 4 0.367879441202 -3.09130498977e-11
delta t is 0.001
Euler 0.367695424771 0.000184016400479
imp. Euler 0.367879502531 -6.13592486265e-08
Runge-Kutta 4 0.367879441171 -4.05231403988e-15
delta t is 0.0001
Euler 0.367861046433 1.83947385133e-05
imp. Euler 0.367879441785 -6.13176398545e-10
Runge-Kutta 4 0.367879441171 -2.6645352591e-15