Question

我编写了一个使用4阶Runge-Kutta积分计算钟摆速度和力的代码，遗憾的是，由于我收到此错误，我无法运行它：

 30: RuntimeWarning: overflow encountered in double_scalars
   th2 = th[j+1] + (ts/2)*om[j+1]
 33: RuntimeWarning: overflow encountered in double_scalars
   om2 = om[j+1] + (ts/2)*f2
 39: RuntimeWarning: invalid value encountered in double_scalars
   th3 = th2 + (ts/2)*om2
 40: RuntimeWarning: invalid value encountered in sin
   f3 = (-q*om2 - sin(th2) + b*cos(od)*ts)

我不确定我错过了什么或者我的错误是什么。任何想法都会有所帮助。谢谢！

这是我的代码：

 from scipy import *
 from matplotlib.pyplot import *
 ##A pendulum simulation using fourth order 
 ##Runge-Kutta differentiation

 ts=.05 #time step size, dt_a
 td=20 #trial duration, dt_b
 te=int(td/ts) #no of timesteps

 q=0.5 #scaled damping coefficient or friction factor
 m=1 #mass
 g=9.81 #grav. acceleration
 l=9.81 #length of pendulum

 th=[((rand()*2)-1)*pi] #random initial angle, [-pi,pi]
 om=[0] #initial angular velocity
 b=0.9 #scaled initial (force)/ml or driving coef.
 od=0.66 #driving force freq


 for j in range(te):
     #Euler approximation
     th.append(th[j] + ts*om[j]) #storing theta values
     f1 = (-q*om[j] - sin(th[j]) + b*cos(od)*ts)
     #ts += ts 
     om.append(om[j] + ts*f1)    

     #ts=.05
     #1 at mid-interval
     th2 = th[j+1] + (ts/2)*om[j+1]

     f2 = (-q*om[j+1] - sin(th[j+1]) + b*cos(od)*ts)
     om2 = om[j+1] + (ts/2)*f2
     #ts += ts

     #ts=.05
     #2 at mid-interval

     th3 = th2 + (ts/2)*om2
     f3 = (-q*om2 - sin(th2) + b*cos(od)*ts)
     om3 = om2 + (ts/2)*f3
     #ts += ts


     #next time step

     th4 = th3 + (ts)*om3
     f4 = (-q*om3 - sin(th3) + b*cos(od)*ts)
     om4 = om3 + (ts)*f4

     ts += ts


     dth=(om[j] + 2*om[j+1] + 2*om2 + om3)/6
     dom=(f1 + 2*f2 + 2*f3 + f4)/6
     th[j+1] = th[j] + ts*dth #integral of angle 
     om[j+1] = om[j] + ts*dom #integral of velocity

 plot(th,om),xlabel('Angle'),ylabel('')
 show()

Answer 1

RK4 的实现存在误区。阶段中间点的状态值不会相互递增，而是全部基于步骤的基本状态。因此，对于 ODE 系统 th' = om，om'=f(t,th,om)

def f(t,th,om): return -q*om - sin(th) + b*cos(od*t)
for j in range(te):
     k_th1 = om[j]
     k_om1 = f(t[j],th[j],om[j])   

     #1 at mid-interval
     k_th2 = om[j] + (ts/2)*k_om1
     k_om2 = f(t[j]+(dt/2),th[j]+(dt/2)*k_th1,om[j]+(dt/2)*k_om1)

     #2 at mid-interval
     k_th3 = om[j] + (ts/2)*k_om2
     k_om3 = f(t[j]+(dt/2),th[j]+(dt/2)*k_th2,om[j]+(dt/2)*k_om2)

     #next time step
     k_th4 = om[j] + ts*k_om3
     k_om4 = f(t[j]+dt,th[j]+dt*k_th3,om[j]+dt*k_om3)

     dth=(k_om1 + 2*k_om2 + 2*k_om3 + k_om4)/6
     dom=(k_th1 + 2*k_th2 + 2*k_th3 + k_th4)/6
     th.append(th[j] + ts*dth) #integral of angle 
     om.append(om[j] + ts*dom) #integral of velocity

摆锤整合。溢出。蟒蛇

1 个答案: