Python 用Fipy估计偏微分方程系统的参数

我正在使用Fipy解决一个涉及两个参数或常数的偏微分方程系统，因此我想知道是否也可以在Fipy中估计这些参数，或者其他库中有哪些更适合于此

注意：我知道scipy有一些函数（针对MLE的optimize.minimize），但我不确定将它们应用于Fipy的代码是否足够

更新：对于下面的PDE系统，我想估计两个未知参数：“Beta”和“m”

在Fipy中求解此PDE的函数如下所示：

import scipy as sci
import fipy as fipy
import numpy as np
from fipy import *

# Grid
nx = 100
ny = 100

dx = 1.
dy = dx

mesh = Grid2D(nx=nx, ny=ny, dx=dx, dy=dy)

x = mesh.cellCenters[0]
y = mesh.cellCenters[1]

# Setting variable of results and adding inicial conditions

u = CellVariable(name="Individual 1", mesh=mesh, value=0.)
u.setValue(1., where=(50. < x) & (70. > x) & (50. < y) & (70. > y))

v = CellVariable(name="Individual 2", mesh=mesh, value=0.)
v.setValue(1., where=(40. < x) & (60. > x) & (40. < y) & (60. > y))

p = CellVariable(name= "Marks Individual 1", mesh=mesh, value=0.)
p.setValue(1., where=(50. < x) & (70. > x) & (50. < y) & (70. > y))

q = CellVariable(name= "Marks Individual 2", mesh=mesh, value=0.)
q.setValue(1., where=(40. < x) & (60. > x) & (40. < y) & (60. > y))

# Plotting inicial conditions
if __name__ == '__main__':
    viewer = Viewer(u, v, datamin=0., datamax=1.)
    viewer.plot()

# Setting PDE 
def HRMLE(params):
    m = params[0]
    beta = params[1]
    D = 1.

    CU = CellVariable(mesh=mesh, rank=1)
    CU[:]= 1.
    CU.setValue(-1., where = (x > 60.) * [[[1], [0]]])
    CU.setValue(-1., where = (y > 60.) * [[[0], [1]]])

    CV = CellVariable(mesh=mesh, rank=1)
    CV[:]=1.
    CV.setValue(-1., where = (x > 50.) * [[[1], [0]]])
    CV.setValue(-1., where = (y > 50.) * [[[0], [1]]])

    # Transient formulation
    eqU = TransientTerm() == DiffusionTerm(coeff=D) -\
                     ConvectionTerm(coeff=CU*q.value*beta) 

    eqV = TransientTerm() == DiffusionTerm(coeff=D) -\
                     ConvectionTerm(coeff=CV*p.value*beta) 

    eqP = TransientTerm() == u*(1 + m*q) - p

    eqQ = TransientTerm() == v*(1 + m*p) - q

    # Solving Transient term
    timeStepDuration = 1.
    steps = 50
    t = timeStepDuration * steps

    for step in range(steps):
        eqU.solve(var=u, dt=timeStepDuration)
        eqV.solve(var=v, dt=timeStepDuration)
        eqP.solve(var=p, dt=timeStepDuration)
        eqQ.solve(var=q, dt=timeStepDuration)

    # Plotting results
    #if __name__ == '__main__':
    #    vieweru = Viewer(u, datamin=0., datamax=1.)
    #    viewerv = Viewer(v, datamin=0., datamax=1.)
    #    vieweru.plot()
    #    viewerv.plot()

    loglink = np.sum(np.log(u.value)) + np.sum(np.log(v.value))
    return(loglink)

---

结果显示的值总是接近参数的初始值，这就是我不确定我的过程是否正确的原因。任何建议都将不胜感激。

好吧，我已经意识到，为了使函数最大化，必须将函数的输出乘以-1，然后将其最小化。因此，我的函数的输出应该是：

loglink = np.sum(np.log(u.value)) + np.sum(np.log(v.value))*-1
return(loglink)

---

另一方面，实际上最大化仅适用于网格中的点列表，而不是u1和u2的所有值

这是一个非常开放的问题，通过一个具体的例子可以改进。我添加了一个例子。非常感谢您的关注。在开始使用优化器检查解决方案是否符合您的预期以及环境是否合理之前，最好先查看“Beta”和“m”环境。根据模拟的挂钟持续时间，在“Beta”、“m”域上尝试一个非常粗糙的网格。

loglink = np.sum(np.log(u.value)) + np.sum(np.log(v.value))*-1
return(loglink)