求解非线性方程：为 Gibbs 自由能问题添加约束

Question

我正在尝试求解一个非线性系统，该系统将使用拉格朗日方法和指数公式来最小化吉布斯自由能。这些方程已经有指数形式的拉格朗日量 Y1...Y6，稍后将转换为化学物质的摩尔数 n1...n9。

问题在于 fsolve() 给出的答案差异很大，即使以相同的猜测重新运行问题，它也会给出不同的值。但主要问题是，我通过不同的猜测得出的所有解决方案都没有物理意义，因为在将 Ys 转换为 ns 之后，我得到质量的负值。

因此，根据所涉及的物理学，我可以确定所有 [n1...n9] >= 0。也可以确定所有最大值为[n1...n9].

如何将此添加到代码中？

import numpy as np
import scipy
from scipy.optimize import fsolve
import time
#
# "B" is the energy potentials of the species [C_gr , CO , CO2 , H2 , CH4 , H2O , N2* , SiO2* , H2S]
B = [-11.0, -309.3632404425132, -613.3667287153355, -135.61840658777166, -269.52018727412405, -434.67499662354476, -193.0773646004259, -980.0, -230.02942769438977]
# "a_atoms" is the number of atoms in the reactants [C, H, O, N*, S, SiO2*]  
# * Elements that doesn't react. '
a_atoms = [4.27311296e-02, 8.10688756e-02, 6.17738749e-02, 1.32864225e-01, 3.18931655e-05, 3.74477901e-04]
P_zero = 100.0 # Standard energy pressure
P_eq = 95.0 # Reaction pressure
# Standard temperature 298.15K, reaction temperature 940K.
#
start_time = time.time()
def GibbsEq(z):
# Lambda's exponentials:
    Y1 = z[0]
    Y2 = z[1] 
    Y3 = z[2]
    Y4 = z[3] 
    Y5 = z[4] 
    Y6 = z[5]
# Number of moles in each phase:
    N1 = z[6]
    N2 = z[7]
    N3 = z[8]
# Equations of energy conservation and mass conservation:
    F = np.zeros(9) 
    F[0] = (P_zero/P_eq) * N1 * ((B[1] * (Y1 * Y3) + B[2] * (Y1 * Y3**2) + B[4] * (Y1 * Y2**2)) + N2 * (B[0] * Y1)) - a_atoms[0]
    F[1] = (P_zero/P_eq) * N1 * (2 * B[3] * Y2**2 + 4 * B[4] * (Y1 * Y2**4) + 2 * B[5] * ((Y2**2) * Y3) + 2 * B[8] * ((Y2**2) * Y5)) - a_atoms[1]
    F[2] = (P_zero/P_eq) * N1 * (B[1] * (Y1 * Y3) + 2 * B[2] * (Y1 * Y3**2) + B[5] * ((Y2**2) * Y3)) - a_atoms[2]
    F[3] = (P_zero/P_eq) * N1 * (2 * B[6]**2) - a_atoms[3]
    F[4] = (P_zero/P_eq) * N1 * (B[8] * ((Y2**2) * Y5)) - a_atoms[4]
    F[5] = N3 * (B[7] * Y5)  - a_atoms[5]
# 
    F[6] = (P_zero/P_eq) * (B[1] * (Y1 * Y3) + B[2] * (Y1 * Y3**2) + B[3] * Y2**2 + B[4] * (Y1 * Y2**4) + B[5] * ((Y2**2) * Y3) + B[6] * Y4 + B[8] * Y5) - 1 
    F[7] = B[0] * Y1 - 1 
    F[8] = B[7] * Y6 - 1
    return F
#
zGuess = np.ones(9)
z = scipy.optimize.fsolve(GibbsEq, zGuess)
end_time = time.time()
time_solution = (end_time - start_time)
print('Solving time: {} s'.format(time_solution))
#
n1 = z[7] * B[0] * z[0]
n2 = z[6] * B[1] * z[0] * z[2]
n3 = z[6] * B[2] * z[0] * z[2]**2
n4 = z[6] * B[3] * z[1]**2
n5 = z[6] * B[4] * z[0] * z[1]**4
n6 = z[6] * B[5] * z[1]**2 * z[4]
n7 = z[6] * B[6] * z[3]**2
n8 = z[8] * B[7] * z[5]
n9 = z[6] * B[8] * z[1]**2 * z[4]
N_T = [n1, n2, n3, n4, n5, n6, n7, n8, n9]
print(z)
print(z[6],z[7],z[8])
print(N_T)
for n in N_T:
    if n < 0:
        print('Error: there is negative values for mass in the solution!')
        break

如何在 fsolve 中添加约束？
python 中是否有其他求解器具有更多约束选项以获得稳定性和初始猜测的更多独立性？

谢谢！

Answer 1

两个问题一个答案。

fsolve 不支持约束。您可以将初始估计值作为正值提供，但这并不能保证正根。但是，您可以将问题重新表述为优化问题，并使用任何优化函数（例如 scipy.optimize.minimize.

最小化施加约束的成本函数

作为一个最小的例子，如果你想找到方程 x*x -4 的正根，你可以像下面那样做。

scipy.optimize.minimize(lambda x:(x*x-4)**2,x0= [5], bounds =((0,None),))

采用(min,max)对的bounds参数可用于对根施加正约束。

输出：

 fun: array([1.66882981e-17])
 hess_inv: <1x1 LbfgsInvHessProduct with dtype=float64>
      jac: array([1.27318954e-07])
  message: b'CONVERGENCE: NORM_OF_PROJECTED_GRADIENT_<=_PGTOL'
     nfev: 20
      nit: 9
   status: 0
  success: True
        x: array([2.])

据此，您的代码可以修改如下。只需添加边界，更改函数 return 语句，并使用 bounds 将 fsolve 更改为 scipy.optimize.minimize。

import numpy as np
import scipy
from scipy.optimize import fsolve
import time
#
# "B" is the energy potentials of the species [C_gr , CO , CO2 , H2 , CH4 , H2O , N2* , SiO2* , H2S]
B = [-11.0, -309.3632404425132, -613.3667287153355, -135.61840658777166, -269.52018727412405, -434.67499662354476, -193.0773646004259, -980.0, -230.02942769438977]
# "a_atoms" is the number of atoms in the reactants [C, H, O, N*, S, SiO2*]  
# * Elements that doesn't react. '
a_atoms = [4.27311296e-02, 8.10688756e-02, 6.17738749e-02, 1.32864225e-01, 3.18931655e-05, 3.74477901e-04]
P_zero = 100.0 # Standard energy pressure
P_eq = 95.0 # Reaction pressure
# Standard temperature 298.15K, reaction temperature 940K.
#
start_time = time.time()
def GibbsEq(z):
# Lambda's exponentials:
    Y1 = z[0]
    Y2 = z[1] 
    Y3 = z[2]
    Y4 = z[3] 
    Y5 = z[4] 
    Y6 = z[5]
# Number of moles in each phase:
    N1 = z[6]
    N2 = z[7]
    N3 = z[8]

    bounds =((0,None),)*9
# Equations of energy conservation and mass conservation:
    F = np.zeros(9) 
    F[0] = (P_zero/P_eq) * N1 * ((B[1] * (Y1 * Y3) + B[2] * (Y1 * Y3**2) + B[4] * (Y1 * Y2**2)) + N2 * (B[0] * Y1)) - a_atoms[0]
    F[1] = (P_zero/P_eq) * N1 * (2 * B[3] * Y2**2 + 4 * B[4] * (Y1 * Y2**4) + 2 * B[5] * ((Y2**2) * Y3) + 2 * B[8] * ((Y2**2) * Y5)) - a_atoms[1]
    F[2] = (P_zero/P_eq) * N1 * (B[1] * (Y1 * Y3) + 2 * B[2] * (Y1 * Y3**2) + B[5] * ((Y2**2) * Y3)) - a_atoms[2]
    F[3] = (P_zero/P_eq) * N1 * (2 * B[6]**2) - a_atoms[3]
    F[4] = (P_zero/P_eq) * N1 * (B[8] * ((Y2**2) * Y5)) - a_atoms[4]
    F[5] = N3 * (B[7] * Y5)  - a_atoms[5]
# 
    F[6] = (P_zero/P_eq) * (B[1] * (Y1 * Y3) + B[2] * (Y1 * Y3**2) + B[3] * Y2**2 + B[4] * (Y1 * Y2**4) + B[5] * ((Y2**2) * Y3) + B[6] * Y4 + B[8] * Y5) - 1 
    F[7] = B[0] * Y1 - 1 
    F[8] = B[7] * Y6 - 1
    return (np.sum(F)**2)
#
zGuess = np.ones(9)
z = scipy.optimize.minimize(GibbsEq, zGuess , bounds=bounds)
end_time = time.time()
time_solution = (end_time - start_time)
print('Solving time: {} s'.format(time_solution))
#

print(z.x)

print(N_T)
for n in N_T:
    if n < 0:
        print('Error: there is negative values for mass in the solution!')
        break

输出：

Solving time: 0.012451648712158203 s
[1.47559173 2.09905553 1.71722403 1.01828262 1.17529548 1.08815712
 1.00294916 1.00104157 1.08815763]

求解非线性方程：为 Gibbs 自由能问题添加约束

Solving non linear equations: add constraints to Gibbs free energy problem

python

numpy

scipy

chemistry

nonlinear-optimization