我试图通过在python中使用gekko来估计以下一组微分方程的参数，但出现了一些错误。这些是我正在使用的方程式：
使用的代码是：
from gekko import GEKKO
#experimental data
t_data=[0,2,4,6,8,10,12,14,16,18,20]
x_data=[0.0844,0.2068,0.8046,1.8019,2.4655,2.7467,2.7765,2.6966,2.6529,2.6605,2.5464]
L_data=[0,18.194,36.389,540.069,958.987,1326.418,1069.154,1195.935,1256.966,1422.267,1267.442]
c_data=[9.845,9.4193,9.0340,7.6427,5.9962,5.2468,4.1849,4.4343,4.2462,3.8870,3.6511]
s_data=[5.0619,4.3798,2.6438,0.6220,0.557,0.492,0.4268,0.415,0.4017,0.395,0.3906]
m = GEKKO(remote=False)
m.time = t_data
x = m.CV(value=x_data); x.FSTATUS = 1 # fit to measurements
L = m.CV(value=L_data); L.FSTATUS = 1
c = m.CV(value=c_data); c.FSTATUS = 1
s = m.CV(value=s_data); s.FSTATUS = 1
umax = m.FV(); umax.STATUS = 1 # adjustable parameters
kc = m.FV(); kc.STATUS = 1
smin = m.FV(); smin.STATUS = 1
ks = m.FV(); ks.STATUS = 1
kd = m.FV(); kd.STATUS = 1
a = m.FV(); a.STATUS = 1
b = m.FV(); b.STATUS = 1
yxc = m.FV(); yxc.STATUS = 1
q = m.FV(); q.STATUS = 1
yxs = m.FV(); yxs.STATUS = 1
# differential equations
m.Equation(x.dt() == umax*(c/(kc+c))*((s-smin)/(ks+s-smin))*x-kd*x )
m.Equation(L.dt() == a*x.dt()+b*x)
m.Equation(c.dt() == (-1/yxc)*x.dt()-q*x*(c/(kc+c)))
m.Equation(s.dt() == (-1/yxs)*umax*(c/(kc+c))*((s-smin)/(ks+s-smin))*x)
#Global Options
m.options.IMODE = 5
m.options.EV_TYPE = 2
m.options.NODES = 3
m.options.SOLVER = 3
m.solve()
执行代码后，出现以下错误：
**Error**: Exception: Access Violation
At line 2683 of file MUMPS/src/dmumps_part2.F
Traceback: not available, compile with -ftrace=frame or -ftrace=full
**Error**: 'results.json' not found. Check above for additional error details
Traceback (most recent call last):
File "C:\Users\vcham\Desktop\2020\belgium\python\parameter estimation bg\test2.py", line 56, in <module>
m.solve()
File "C:\Users\vcham\anaconda3\lib\site-packages\gekko\gekko.py", line 2145, in solve
self.load_JSON()
File "C:\Users\vcham\anaconda3\lib\site-packages\gekko\gk_post_solve.py", line 13, in load_JSON
f = open(os.path.join(self._path,'options.json'))
FileNotFoundError: [Errno 2] No such file or directory: 'C:\\Users\\vcham\\AppData\\Local\\Temp\\tmplna43pbhgk_model1\\options.json'
请为我提出克服这些错误的解决方案或估算参数的方法。提前致谢！！！
python大神给出的解决方案
尝试解决问题时，带有线性求解器MUMPS的IPOPT求解器崩溃。获得成功解决方案的一种方法是使用m.options.SOLVER=1切换到APOPT求解器，并在FV参数上设置上限和下限。
from gekko import GEKKO
#experimental data
t_data=[0,2,4,6,8,10,12,14,16,18,20]
x_data=[0.0844,0.2068,0.8046,1.8019,2.4655,2.7467,2.7765,2.6966,2.6529,2.6605,2.5464]
L_data=[0,18.194,36.389,540.069,958.987,1326.418,1069.154,1195.935,1256.966,1422.267,1267.442]
c_data=[9.845,9.4193,9.0340,7.6427,5.9962,5.2468,4.1849,4.4343,4.2462,3.8870,3.6511]
s_data=[5.0619,4.3798,2.6438,0.6220,0.557,0.492,0.4268,0.415,0.4017,0.395,0.3906]
m = GEKKO(remote=False)
m.time = t_data
x = m.CV(value=x_data); x.FSTATUS = 1 # fit to measurements
L = m.CV(value=L_data); L.FSTATUS = 1
c = m.CV(value=c_data); c.FSTATUS = 1
s = m.CV(value=s_data); s.FSTATUS = 1
umax = m.FV(1,lb=0.1,ub=10); umax.STATUS = 1 # adjustable parameters
kc = m.FV(1,lb=0.1,ub=10); kc.STATUS = 1
smin = m.FV(1,lb=0.1,ub=10); smin.STATUS = 1
ks = m.FV(1,lb=0.1,ub=10); ks.STATUS = 1
kd = m.FV(1,lb=0.1,ub=10); kd.STATUS = 1
a = m.FV(1,lb=0.1,ub=10); a.STATUS = 1
b = m.FV(1,lb=0.1,ub=10); b.STATUS = 1
yxc = m.FV(1,lb=0.1,ub=10); yxc.STATUS = 1
q = m.FV(1,lb=0.1,ub=10); q.STATUS = 1
yxs = m.FV(1,lb=0.1,ub=10); yxs.STATUS = 1
# differential equations
m.Equation(x.dt() == umax*(c/(kc+c))*((s-smin)/(ks+s-smin))*x-kd*x )
m.Equation(L.dt() == a*x.dt()+b*x)
m.Equation(c.dt() == (-1/yxc)*x.dt()-q*x*(c/(kc+c)))
m.Equation(s.dt() == (-1/yxs)*umax*(c/(kc+c))*((s-smin)/(ks+s-smin))*x)
#Global Options
m.options.IMODE = 5
m.options.EV_TYPE = 2
m.options.NODES = 3
m.options.SOLVER = 1
m.solve()
print('umax: ' + str(umax.value[0]))
print('kc: ' + str(kc.value[0]))
print('smin: ' + str(smin.value[0]))
print('ks: ' + str(ks.value[0]))
print('kd: ' + str(kd.value[0]))
print('a: ' + str(a.value[0]))
print('b: ' + str(b.value[0]))
print('yxc: ' + str(yxc.value[0]))
print('q: ' + str(q.value[0]))
print('yxs: ' + str(yxs.value[0]))
import matplotlib.pyplot as plt
plt.subplot(4,1,1)
plt.plot(t_data,x.value)
plt.plot(t_data,x_data)
plt.subplot(4,1,2)
plt.plot(t_data,L.value)
plt.plot(t_data,L_data)
plt.subplot(4,1,3)
plt.plot(t_data,c.value)
plt.plot(t_data,c_data)
plt.subplot(4,1,4)
plt.plot(t_data,s.value)
plt.plot(t_data,s_data)
plt.show()
我对所有参数应用了0.1的下限和10.0的上限，但是由于某些解决方案在边界处完成，因此可能需要修改这些参数。我建议您扩大（*）的界限，直到最佳解决方案没有在界限处完成。报告的解决方案是：
umax: 10.0*
kc: 0.1*
smin: 0.11236933654
ks: 6.2350087285
kd: 0.20089406528
a: 10.0*
b: 10.0*
yxc: 1.6785099655
q: 0.1*
yxs: 6.7395055839
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