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使用Runge Kutta 4th Order解决Python中的ODE系统,意外错误

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我在解决Python中的物理问题时遇到了问题 . 这是一个ODE系统 . ODE System在某些时间点,必须注入一些剂量以提高值A.

我必须用四阶龙格库塔法解决这个问题 . 我以为我得到了它,但每次都会出现计算错误 .

RuntimeWarning: overflow encountered

在短暂的计算时间之后,这些值会上升到无穷大......另一个问题是,在第3个循环中,Tr得到一个物理上不可能的负值 .

这是我的代码:

a = 10.
v = 8.
x = 0.8
p = 1.02
u1 = 0.2
u2 = 0.1
ur = 0.25
c = 0.0001

N_0 = a
T1_0 = 0.001
T2_0 = 0.0001
Tr_0 = 0.001
A_0 = 0.

t = 0.
h = 1.

while (t <= int(np.amax(time))):

  for i in range(len(time)): 
     if t == time[i]:        
        A_0 = A_0 + dosis[i]

  print('INITAL VALUE: ',' t =', t,' A =', A_0,' N =', N_0,' T1 =', T1_0,' T2 =', T2_0,' Tr =', Tr_0) 

  a1 = -(A_0) * (T1_0 + T2_0 + Tr_0)
  b1 = -(N_0) + a - (N_0 * A_0 * ((T1_0 / (1. + u2 * T2_0)) + c))-(p * N_0 * A_0 * (T2_0 + c)) - (x * N_0 * A_0 *(Tr_0 + c))
  c1 = -(T1_0) + ((v * N_0 * A_0) / (1. + ur * Tr_0)) * ((T1_0 / (1. + u2 * T2_0)) + c)
  d1 = -(T2_0) + (p * ((v * N_0 * A_0) / (1. + ur * Tr_0)) * ((T2_0 + c) / (1. + u1 * (T1_0 / (1. + u2 * T2_0)))))
  e1 = -(Tr_0) + (x * v * N_0 * A_0 * (Tr_0 + c))

  A_1 = A_0 + (h / 2.) * a1
  N_1 = N_0 + (h / 2.) * b1
  T1_1 = T1_0 + (h / 2.) * c1
  T2_1 = T2_0 + (h / 2.) * d1
  Tr_1 = Tr_0 + (h / 2.) * e1

  a2 = -(A_1) * (T1_1 + T2_1 + Tr_1)
  b2 = -(N_1) + a - (N_1 * A_1 * ((T1_1 / (1. + u2 * T2_1)) + c))-(p * N_1 * A_1 * (T2_1 + c)) - (x * N_1 * A_1 *(Tr_1 + c))
  c2 = -(T1_1) + ((v * N_1 * A_1) / (1. + ur * Tr_1)) * ((T1_1 / (1. + u2 * T2_1)) + c)
  d2 = -(T2_1) + (p * ((v * N_1 * A_1) / (1. + ur * Tr_1)) * ((T2_1 + c) / (1. + u1 * (T1_1 / (1. + u2 * T2_1)))))
  e2 = -(Tr_1) + (x * v * N_1 * A_1 * (Tr_1 + c))   

  A_2 = A_0 + (h / 2.) * a2
  N_2 = N_0 + (h / 2.) * b2
  T1_2 = T1_0 + (h / 2.) * c2
  T2_2 = T2_0 + (h / 2.) * d2
  Tr_2 = Tr_0 + (h / 2.) * e2

  a3 = -(A_2) * (T1_2 + T2_2 + Tr_2)
  b3 = -(N_2) + a - (N_2 * A_2 * ((T1_2 / (1. + u2 * T2_2)) + c))-(p * N_2 * A_2 * (T2_2 + c)) - (x * N_2 * A_2 *(Tr_2 + c))
  c3 = -(T1_2) + ((v * N_2 * A_2) / (1. + ur * Tr_2)) * ((T1_2 / (1. + u2 * T2_2)) + c)
  d3 = -(T2_2) + (p * ((v * N_2 * A_2) / (1. + ur * Tr_2)) * ((T2_2 + c) / (1. + u1 * (T1_2 / (1. + u2 * T2_2)))))
  e3 = -(Tr_2) + (x * v * N_2 * A_2 * (Tr_2 + c)) 

  A_3 = A_0 + h * a3
  N_3 = N_0 + h * b3
  T1_3 = T1_0 + h * c3
  T2_3 = T2_0 + h * d3
  Tr_3 = Tr_0 + h * e3

  a4 = -(A_3) * (T1_3 + T2_3 + Tr_3)
  b4 = -(N_3) + a - (N_3 * A_3 * ((T1_3 / (1. + u2 * T2_3)) + c))-(p * N_3 * A_3 * (T2_3 + c)) - (x * N_3 * A_3 *(Tr_3 + c))
  c4 = -(T1_3) + ((v * N_3 * A_3) / (1. + ur * Tr_3)) * ((T1_3 / (1. + u2 * T2_3)) + c)
  d4 = -(T2_3) + (p * ((v * N_3 * A_3) / (1. + ur * Tr_3)) * ((T2_3 + c) / (1. + u1 * (T1_3 / (1. + u2 * T2_3)))))
  e4 = -(Tr_3) + (x * v * N_3 * A_3 * (Tr_3 + c))

  t = t + h
  A_0 = A_0 + (h / 6.) * (a1 + (2. * a2) + (2. * a3) + a4)
  N_0 = N_0 + (h / 6.) * (b1 + (2. * b2) + (2. * b3) + b4)
  T1_0 = T1_0 + (h / 6.) * (c1 + (2. * c2) + (2. * c3) + c4)
  T2_0 = T2_0 + (h / 6.) * (d1 + (2. * d2) + (2. * d3) + d4)
  Tr_0 = Tr_0 + (h / 6.) * (e1 + (2. * e2) + (2. * e3) + e4)

这是当前的输出:

INITAL VALUE:   t = 0.0  A = 0.0  N = 10.0  T1 = 0.001  T2 = 0.0001  Tr = 0.001

INITAL VALUE:   t = 1.0  A = 0.1  N = 10.0  T1 = 0.0003750000000000001  T2 = 3.7500000000000003e-05  Tr = 0.0003750000000000001

INITAL VALUE:   t = 2.0  A = 0.2975779058326524  N = 9.9799116753568  T1 = 0.09021366809218588  T2 = 0.029730294675718548  Tr = 0.04000932816912045

INITAL VALUE:   t = 3.0  A = 0.20021726558830752  N = 9.428758582263455  T1 = 4.6119100465532386  T2 = 3.8633993244370335  Tr = -7.920033484370073

INITAL VALUE:   t = 4.0  A = 40682939.2164535  N = 737615666179.137  T1 = -32.736202523709856  T2 = 265493.88168389443  Tr = -5887693774039.941

INITAL VALUE:   t = 5.0  A = inf  N = nan  T1 = -inf  T2 = inf  Tr = -inf

INITAL VALUE:   t = 6.0  A = nan  N = nan  T1 = nan  T2 = nan  Tr = nan

INITAL VALUE:   t = 7.0  A = nan  N = nan  T1 = nan  T2 = nan  Tr = nan

INITAL VALUE:   t = 8.0  A = nan  N = nan  T1 = nan  T2 = nan  Tr = nan

INITAL VALUE:   t = 9.0  A = nan  N = nan  T1 = nan  T2 = nan  Tr = nan

INITAL VALUE:   t = 10.0  A = nan  N = nan  T1 = nan  T2 = nan  Tr = nan

我希望有人知道什么可以帮助我谢谢你的时间!

1 回答

  • 0

    我终于可以解决问题 . 但我必须将步长调整到h = 0.001,这样才不会出错 . 这需要程序3小时的计算...

    我可以pimp while循环使它更快一点吗?

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