我目前正在尝试使用一些通用的显式Runge-Kutta方法来编写一些python代码,以解决一阶ODE的任意系统,该方法由值alpha,gamma(均为维m的向量)和beta(值为L的下三角矩阵)定义用户传递的Butcher表的维度mxm)。我的代码似乎适用于单个ODE,并在几个不同的示例上对其进行了测试,但是我正努力将我的代码推广到矢量值ODE(即系统)。
特别是,我尝试使用我的代码中提供的Butcher Tableau值定义的Heun方法来求解Van der Pol振荡器ODE(简化为一阶系统),但收到错误
- “ RuntimeWarning:在double_scalars
f = lambda t,u: np.array(... etc)
中遇到溢出”和- “ RuntimeWarning:在添加
kvec[i] = f(t+alpha[i]*h,y+h*sum)
中遇到无效的值]
紧跟着我的解决方案向量明显爆炸了。请注意,下面注释掉的代码是我尝试并已正确解决的单个ODE的示例之一。谁能帮忙吗?这是我的代码:
import numpy as np
def rk(t,y,h,f,alpha,beta,gamma):
'''Runga Kutta iteration'''
return y + h*phi(t,y,h,f,alpha,beta,gamma)
def phi(t,y,h,f,alpha,beta,gamma):
'''Phi function for the Runga Kutta iteration'''
m = len(alpha)
count = np.zeros(len(f(t,y)))
kvec = k(t,y,h,f,alpha,beta,gamma)
for i in range(1,m+1):
count = count + gamma[i-1]*kvec[i-1]
return count
def k(t,y,h,f,alpha,beta,gamma):
'''returning a vector containing each step k_{i} in the m step Runga Kutta method'''
m = len(alpha)
kvec = np.zeros((m,len(f(t,y))))
kvec[0] = f(t,y)
for i in range(1,m):
sum = np.zeros(len(f(t,y)))
for l in range(1,i+1):
sum = sum + beta[i][l-1]*kvec[l-1]
kvec[i] = f(t+alpha[i]*h,y+h*sum)
return kvec
def timeLoop(y0,N,f,alpha,beta,gamma,h,rk):
'''function that loops through time using the RK method'''
t = np.zeros([N+1])
y = np.zeros([N+1,len(y0)])
y[0] = y0
t[0] = 0
for i in range(1,N+1):
y[i] = rk(t[i-1],y[i-1], h, f,alpha,beta,gamma)
t[i] = t[i-1]+h
return t,y
#################################################################
'''f = lambda t,y: (c-y)**2
Y = lambda t: np.array([(1+t*c*(c-1))/(1+t*(c-1))])
h0 = 1
c = 1.5
T = 10
alpha = np.array([0,1])
gamma = np.array([0.5,0.5])
beta = np.array([[0,0],[1,0]])
eff_rk = compute(h0,Y(0),T,f,alpha,beta,gamma,rk, Y,11)'''
#constants
mu = 100
T = 1000
h = 0.01
N = int(T/h)
#initial conditions
y0 = 0.02
d0 = 0
init = np.array([y0,d0])
#Butcher Tableau for Heun's method
alpha = np.array([0,1])
gamma = np.array([0.5,0.5])
beta = np.array([[0,0],[1,0]])
#rhs of the ode system
f = lambda t,u: np.array([u[1],mu*(1-u[0]**2)*u[1]-u[0]])
#solving the system
time, sol = timeLoop(init,N,f,alpha,beta,gamma,h,rk)
print(sol)
mu=100
的Van der Pol振荡器是一种快速慢速系统,在模式切换时具有非常陡峭的匝数,因此相当刚性。对于显式方法,这需要较小的步长,最小的合理步长为1e-5
至1e-6
。您已经获得了针对h=0.001
的极限循环的解决方案,其最大速度为150。