解决R^2中非线性方程系统的解决方案

Question

我正在尝试找到以下系统的解决方案，其中F和G为R^2 - ＆GT; R^2 functions:

f(x1,x2) = (y1,y2)
g(y1,y2) = (x1,x2)

I tried solving it using scipy.optimize.fsolve as follows:

def eqm(vars):
    x1,x2,y1,y2 = vars
    eq1 = f([x1, x2])[0] - y1
    eq2 = f([x1, x2])[1] - y2
    eq3 = g([y1, y2])[0] - x1
    eq4 = g([y1, y2])[1] - x2
    return [eq1, eq2, eq3, eq4]

fsolve(eqm, x0 = [1,0.5,1,0.5])

Although it is returning an output, it does not seem to be a correct one as it does not seem to satisfy the two条件，并且指定的X0似乎有很大差异。还要发出警告： “迭代并没有取得良好的进步，这是通过过去十次迭代的改进来衡量的。”我确实知道存在一个独特的解决方案，我已经通过代数获得了。

Not sure what is going on and if there is a simpler way of solving it, especially using just two equations instead of splitting up into 4. Something like:

def equations(vars):
    X,Y = vars
    eq1 = f(X)-Y
    eq2 = g(Y)-X
    return [eq1, eq2]

fsolve(equations, x0 =[[1,0.5],[1,0.5]])

Suggestions on other modules eg sympy are also welcome!

Answer 1

首先，我建议使用Numpy阵列，因为操纵这些阵列比列表更简单。

I've slighlty rewritten your code:

import scipy.optimize as opt

def f(x):
  return x
def g(x):
  return x

def func(vars):
  input = np.array(vars)
  eq1 = f(input[:2]) - input[2:]
  eq2 = g(input[2:]) - input[:2]
  return np.concatenate([eq1, eq2])

root = opt.fsolve(func, [1, 1, 0., 1.2])

print(root)
print(func(root)) # should be close to zeros

What you have should work correctly, so I believe there is something wrong with the equations you're using.如果您提供这些，我可以尝试查看可能出了什么问题。

Answer 2

这似乎更像是数学数学的问题，而不是Python编码。您的功能可能在解决方案周围具有“丑陋”的行为，可能是非常非线性的或包含奇异性。如果不看到功能，我们不禁进一步。 One thing you might try is to instead solve a system

g(f(x)) - x = 0

and simplify g(f(x)) as much as possible analytically.然后在求解方程后计算y = f（x）。