使用 fsolve 和不等式方程求解方程 [英] Solve equation using fsolve with inequality equations

问题描述

我无法理解如何将不等式方程添加到 fsolve 函数中.

I am having trouble understanding how to add inequality equation to fsolve function.

例如:

这是包:

import numpy as np
from scipy.optimize import fsolve

这些是我想要使用的方程:

These is the equations I want to use:

x1 >= 0.4 and x1 <= 0.7
x2 >= 0.2 and x2 <= 0.4
5x2**2 + 2x1**3 = 2 

这是我试图创建的函数:

and this is the function I am trying to create:

myFunc(z):
    x1 = z[0]
    x2 = z[1]
    
    F = np.empty((3))
    F[0] = x1 >= 0.4 and x1 <= 0.7 # <-- This is the first equation
    F[1] = x2 >= 0.2 and x2 <= 0.4 # <-- this is the second equation
    F[2] = 5x2**2 + 2x1**3 = 2 # <-- this is the third equation
    return F

然后我们调用 fsolve:

and then we call the fsolve:

zGuess = np.array([0.3,0.3])
z = fsolve(myFunction,zGuess)
print(z)

关于如何设置不等式方程有什么想法吗?

Any ideas on how to set inequality equations?

推荐答案

fsolve 方法既不能处理不等式约束，也不能处理变量的边界.您的前两个约束是简单的框约束，即变量的边界，因此您只想解决受变量边界约束的非线性方程组 2x1**3 + 5x**2 == 2.这可以表述为约束最小化问题，类似于这个答案:

The fsolve method neither can handle inequality constraints nor bounds on the variables. Your first two constraints are simple box constraints, i.e. bounds on the variables, so you just want to solve the nonlinear equation system 2x1**3 + 5x**2 == 2 subject to variable bounds. This can be formulated as a constrained minimization problem, similar to this answer:

For F(x) = 2x1**3 + 5x**2 - 2 you want to solve

min ||F(x)|| 

s.t. 0.4 <= x1 <= 0.7, 
     0.2 <= x2 <= 0.4

使用可以轻松解决这个约束优化问题scipy.optimize.minimize 如下:

from scipy.optimize import minimize
import numpy as np

def obj(x): return 5*x[1]**2 + 2*x[0]**3 - 2

# Set the variable bounds
bounds = [(0.4, 0.7), (0.2, 0.4)]

# Set initial guess
x0 = np.array([0.3, 0.3])

# Solve the problem (res.x contains your solution)
res = minimize(lambda x: np.linalg.norm(obj(x)), x0=x0, bounds=bounds)

请注意，在您的范围内没有满足非线性方程的点.无论如何，当我们最小化残差的欧几里德范数时，我们得到了最好的点.

Note that there's no point within your bounds that satisfies the nonlinear equation. Anyway, as we minimized the euclidian norm of the residual, we obtained the best possible point.

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