用SciPy求解一组常微分方程
我正在尝试解决一组常微分方程,以模拟淀粉在淀粉酶(酶的一种)作用下的水解过程。当我尝试解这组方程时,我遇到了一个
lsoda-- at current t (=r1), mxstep (=i1) steps
taken on this call before reaching tout
In above message, I1 = 500
In above message, R1 = 0.6333483931400E+00
Excess work done on this call (perhaps wrong Dfun type).
Run with full_output = 1 to get quantitative information.
错误。代码如下:
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import numpy as np
import math
from scipy.integrate import odeint
def TemperatureProfile (timeandtemp,t) :
if t==0.0 :
T = 273.15+55
elif t>0.0 and t<9.0 :
T = 273.15+55+t
elif t>= 9.0 and t<=9.0+timeandtemp[0][0] :
T = 273.15+ timeandtemp[0][1]
elif t>9.0+timeandtemp[0][0] and t<9.0+timeandtemp[0][0]+8.0 :
T = 273.15+ timeandtemp[0][1]+t
elif t>=9.0+timeandtemp[0][0]+8.0 and t<=9.0+timeandtemp[0][0]+8.0+timeandtemp[1][0] :
T = 273.15+timeandtemp[1][1]
elif t>9.0+timeandtemp[0][0]+8.0+timeandtemp[1][0] and t<9.0+timeandtemp[0][0]+8.0+timeandtemp[1][0]+6:
T = 273.15+ timeandtemp[1][1]+t
elif t>=9.0+timeandtemp[0][0]+8.0+timeandtemp[1][0]+6 and t<9.0+timeandtemp[0][0]+8.0+timeandtemp[1][0]+6 :
T = 273.15+timeandtemp[2][1]
return T
def Flux (ci,temperature) :
'''
Constants
'''
K_gel1 = 5.7*10**31
K_gel2 = 3.1*10**14
k_glc = 0.023
k_glc2 = 2.9*10**-8
k_alphamal = 0.389
k_betamal = 0.137
k_alphamal2 = 1.2*10**-7
k_betamal2 = 8.4*10**-8
k_mlt = 0.117
k_mlt2 = 1.5*10**-8
k_dex = 0.317
k_degalpha = 6.9*10**30
E_degalpha = 224.2
k_degbeta = 7.6*10**60
E_degbeta = 410.7
starchUG = ci[0]
starchG = ci[1]
glc = ci[2]
mal = ci[3]
mlt = ci[4]
dex = ci[5]
enzA = ci[6]
enzB = ci[7]
'''
Relative activities
'''
if temperature < 336.15 :
a_alpha = -0.0011*temperature**3 + 1.091*temperature**2 - 352.89*temperature + 38008.3
a_beta = 0.049*temperature - 13.9
else :
a_alpha = 0.0055*temperature**3 - 5.663*temperature**2+ 1941.9*temperature- 221864
a_beta = 0.374*temperature + 128.3
'''
Equations
'''
if temperature < 333.15 :
r_gel = K_gel1*starchUG
else :
r_gel = K_gel2*starchUG
r_glc = k_glc*a_alpha*glc
r_mal = (k_alphamal*a_alpha+k_betamal*a_beta)*starchG
r_mlt = k_mlt*a_alpha*starchG
r_dex = k_dex*a_alpha*starchG
r_glc2 = k_glc2*a_alpha*dex
r_mal2 = k_alphamal2*a_alpha*dex+k_betamal2*a_beta*dex
r_mlt2 = k_mlt2*a_alpha*dex
r_degA = k_degalpha*math.exp(-E_degalpha/(8.3145*temperature))*enzA
r_degB = k_degbeta*math.exp(-E_degbeta/(8.3145*temperature))*enzB
r_acA = k_degalpha*math.exp(-E_degalpha/(8.3145*temperature))*a_alpha*enzA
r_acB = k_degbeta*math.exp(-E_degbeta/(8.3145*temperature))*a_beta*enzB
return np.array([r_glc,r_mal,r_mlt, r_dex, r_glc2, r_mal2, r_mlt2, r_degA, r_degB, r_acA, r_acB])
def Secmembre (ci,t,tempProf) :
temperature = TemperatureProfile(tempProf,t)
mS = np.mat([
[-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 1, -1, -1, -1, 0, 0, 0]])
mS = np.transpose(mS)
mS = np.array(mS)
metabos = Flux(ci,temperature)
varMetabos = np.dot(metabos,mS)
return varMetabos
# Initial conditions
initCond = []
temperatureProfile = ((30,64),(30,72),(5,78)) # Temperature steps in minutes
t0 = 0
tf = temperatureProfile[0][0]+temperatureProfile[1][0]+temperatureProfile[2][0]+9+8+6
nbPoints = 100
timeProfile = np.linspace(t0,tf,nbPoints)
print timeProfile
initCond.append(113.5)
initCond.append(0)
initCond.append(4)
initCond.append(5)
initCond.append(0)
initCond.append(0)
initCond.append(80000)
initCond.append(80000)
initCond=np.array(initCond)
result = odeint(Secmembre,initCond,timeProfile,args=(temperatureProfile,))
在使用Matlab时,我通常用ode23tb来解决这个问题。我觉得odeint可能不是合适的解法,但我不知道该用哪个。也许有人对这种类型的方程比较熟悉?
1 个回答
-1
总体来说,Python,特别是scipy,在解决常微分方程(ODE)方面没有MATLAB那么优化。
关于你的错误:
lsoda-- at current t (=r1), mxstep (=i1) steps
taken on this call before reaching tout
In above message, I1 = 500
In above message, R1 = 0.6333483931400E+00
Excess work done on this call (perhaps wrong Dfun type).
Run with full_output = 1 to get quantitative information.
这个求解器的工作方式是它内部定义了一些步骤。比如说,你在0(初始值)、1、2、3等点调用求解器。求解器会在这些点返回值,但实际上,它可能会先在0处解决ODE,然后是0.00000001、0.00000002、0.00000003、0.0001、0.01、0.1、0.5、1等等(这些数字只是举例)。不过,求解器会限制它能进行的步骤数量(这很合理,以避免可能的无限循环或非常长的运行时间)。
你的错误提示是说,求解器已经达到了最大步骤数 [mxstep (=i1(=500))],但仍然没有在其内部定义的可接受误差范围内得到结果,因此它停止并抛出了错误。
建议:调整你的求解器,让它可以进行更多步骤(在odeint中设置nstep),如果不行,可以调整内部的容忍度,也许你不需要那么精确,最后考虑使用ode而不是odeint(这样有更多选项)。
进一步阅读: