用Fipy估计偏微分方程系统的参数

Question

我正在用Fipy解一个偏微分方程系统，它包含两个参数或常数，所以我想知道在Fipy中是否也可以估计这些参数，或者其他库中的哪些更适合于这一点。你知道吗

注意：我知道scipy有一些功能(优化.最小化对于MLE），但我不确定将它们应用于Fipy代码是否足够。你知道吗

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中求解这个偏微分方程的函数如下：

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)

最后，对于最大可能性标准，我想最大化：

设置初始值，并使用scipy

mb = [0., .5]
mb

results = sci.optimize.minimize(HRMLE, mb, method='Nelder-Mead')
results

结果显示，参数的值总是接近我的初始值，这就是我不确定我的过程是否正确的原因。如有任何建议，我们将不胜感激。你知道吗

Answer 1

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。所以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，实际上最大化只针对网格中的点列表，而不是所有的u1和u2值。你知道吗