python - 求解有条件的非线性方程组

Question

我想求解以下非线性方程组。是否可以设置所有变量都大于或等于零并且所有参数都为正的条件？变量是 (x1,x2,x3,x4,y1,y2 )，其他只是参数。

Maple 比 sympy 更好地解决这个系统吗？

from sympy.interactive import printing
printing.init_printing(use_latex=True)
from sympy import *
import numpy as np
import matplotlib.pyplot as plt
import sympy as sp


x1, x2, x3, x4, y1, y2 = sp.symbols('x1, x2, x3, x4, y1, y2')
N, c1, c2, c3, c4 = sp.symbols('N, c1, c2, c3, c4')
r1, r2, r3, r4 = sp.symbols('r1, r2, r3, r4')
f11, f21, f31, f41 = sp.symbols('f11, f21, f31, f41')
f12, f22, f32, f42 = sp.symbols('f12, f22, f32, f42')
eta11, eta12, eta13, eta14 = sp.symbols('eta11, eta12, eta13, eta14')
eta21, eta22, eta23, eta24 = sp.symbols('eta21, eta22, eta23, eta24')
eta31, eta32, eta33, eta34 = sp.symbols('eta31, eta32, eta33, eta34')
eta41, eta42, eta43, eta44 = sp.symbols('eta41, eta42, eta43, eta44')
epsilon1, epsilon2, K11, K22 = sp.symbols('epsilon1, epsilon2, K11, K22')
omega1, omega2, gamma12, g12 = sp.symbols('omega1, omega2, gamma12, g12')
beta11, beta21, beta31, beta41 = sp.symbols('beta11, beta21, beta31, beta41')
beta12, beta22, beta32, beta42 = sp.symbols('beta12, beta22, beta32, beta42')

F2 = x1 * (r1 * (1 - (eta11 * x1 + eta12 * x2 + eta13 * x3 + eta14 * x4) / N) - \
   f11 * y1 - f12 * y2)
F3 = x2 * (r2 * (1 - (eta21 * x1 + eta22 * x2 + eta23 * x3 + eta24 * x4) / N) - \
   f21 * y1 - f22 * y2)
F4 = x3 * (r3 * (1 - (eta31 * x1 + eta32 * x2 + eta33 * x3 + eta34 * x4) / N) - \
   f31 * y1 - f32 * y2)
F5 = x4 * (r4 * (1 - (eta41 * x1 + eta42 * x2 + eta43 * x3 + eta44 * x4) / N) - \
   f41 * y1 - f42 * y2)

F6 = y1 * (-epsilon1 * (1 + (y1 + omega2 * y2) / K22) - g12 * y2 + beta11 * f11 * x1 + \
   beta21 * f21 * x2 + beta31 * f31 * x3 + beta41 * f41 * x4)

F7 = y2 * (-epsilon2 * (1 + (omega1 * y1 + y2) / K11) +gamma12 * g12 * y1 + \
   beta12 * f12 * x1 + beta22 * f22 * x2 + beta32 * f32 * x3 + beta42 * f42 * x4)              

equ = (F2, F3, F4, F5, F6, F7)
sol = nonlinsolve(equ, x1, x2, x3, x4, y1, y2)   

print(sol)

Answer 1

这是一个多项式系统，我们可以将其转化为标准形式

In [2]: equ = [eq.as_numer_denom()[0].expand() for eq in equ]                                                                                                 

In [3]: for eq in equ: pprint(eq)                                                                                                                             
                                                2                                             
-N⋅f₁₁⋅x₁⋅y₁ - N⋅f₁₂⋅x₁⋅y₂ + N⋅r₁⋅x₁ - η₁₁⋅r₁⋅x₁  - η₁₂⋅r₁⋅x₁⋅x₂ - η₁₃⋅r₁⋅x₁⋅x₃ - η₁₄⋅r₁⋅x₁⋅x₄
                                                               2                              
-N⋅f₂₁⋅x₂⋅y₁ - N⋅f₂₂⋅x₂⋅y₂ + N⋅r₂⋅x₂ - η₂₁⋅r₂⋅x₁⋅x₂ - η₂₂⋅r₂⋅x₂  - η₂₃⋅r₂⋅x₂⋅x₃ - η₂₄⋅r₂⋅x₂⋅x₄
                                                                              2               
-N⋅f₃₁⋅x₃⋅y₁ - N⋅f₃₂⋅x₃⋅y₂ + N⋅r₃⋅x₃ - η₃₁⋅r₃⋅x₁⋅x₃ - η₃₂⋅r₃⋅x₂⋅x₃ - η₃₃⋅r₃⋅x₃  - η₃₄⋅r₃⋅x₃⋅x₄
                                                                                             2
-N⋅f₄₁⋅x₄⋅y₁ - N⋅f₄₂⋅x₄⋅y₂ + N⋅r₄⋅x₄ - η₄₁⋅r₄⋅x₁⋅x₄ - η₄₂⋅r₄⋅x₂⋅x₄ - η₄₃⋅r₄⋅x₃⋅x₄ - η₄₄⋅r₄⋅x₄ 
                                                                                                                               2
K₂₂⋅β₁₁⋅f₁₁⋅x₁⋅y₁ + K₂₂⋅β₂₁⋅f₂₁⋅x₂⋅y₁ + K₂₂⋅β₃₁⋅f₃₁⋅x₃⋅y₁ + K₂₂⋅β₄₁⋅f₄₁⋅x₄⋅y₁ - K₂₂⋅ε₁⋅y₁ - K₂₂⋅g₁₂⋅y₁⋅y₂ - ε₁⋅ω₂⋅y₁⋅y₂ - ε₁⋅y₁ 
                                                                                                                                   2
K₁₁⋅β₁₂⋅f₁₂⋅x₁⋅y₂ + K₁₁⋅β₂₂⋅f₂₂⋅x₂⋅y₂ + K₁₁⋅β₃₂⋅f₃₂⋅x₃⋅y₂ + K₁₁⋅β₄₂⋅f₄₂⋅x₄⋅y₂ - K₁₁⋅ε₂⋅y₂ - K₁₁⋅g₁₂⋅γ₁₂⋅y₁⋅y₂ - ε₂⋅ω₁⋅y₁⋅y₂ - ε₂⋅y₂

SymPy 将尝试使用 Groebner 基础来解决这个问题，但需要很长时间来计算:

In [4]: groebner(equ, [x1,x2,x3,x4,y1,y2]) # Not sure how long this takes

我预计，即使它完成了，结果也不会承认解析解，因为求解可能会导致大于 4 阶的多项式。

如果您用具体的有理数替换所有参数，则可能会找到解决方案，但否则就任意符号(r3 等)而言，我不希望有一个封闭的解决方案表单解决方案将存在 - 如果这是真的，那么无论您使用 Maple 还是 SymPy 还是其他任何东西都没有关系。

编辑:我现在知道您的系统是什么。每个方程的形式为 x1 * (a*x1 + b*x2 + ...)，因此它是一个线性方程乘以一个未知数。这意味着有两种可能性:x1 = 0 或满足线性方程。因此，一种解决方案是 x1 = x2 = ... = 0，然后还有另一种解决方案，其中都不为零。对于 6 个未知数，有 64 种可能的解决方案，但有些解决方案可能不满足非负假设。您可以通过以下方式找到它们:

from sympy.interactive import printing
printing.init_printing(use_latex=True)
from sympy import *
import numpy as np
import matplotlib.pyplot as plt
import sympy as sp


x1, x2, x3, x4, y1, y2 = sp.symbols('x1, x2, x3, x4, y1, y2', nonnegative=True)
N, c1, c2, c3, c4 = sp.symbols('N, c1, c2, c3, c4', positive=True)
r1, r2, r3, r4 = sp.symbols('r1, r2, r3, r4', positive=True)
f11, f21, f31, f41 = sp.symbols('f11, f21, f31, f41', positive=True)
f12, f22, f32, f42 = sp.symbols('f12, f22, f32, f42', positive=True)
eta11, eta12, eta13, eta14 = sp.symbols('eta11, eta12, eta13, eta14', positive=True)
eta21, eta22, eta23, eta24 = sp.symbols('eta21, eta22, eta23, eta24', positive=True)
eta31, eta32, eta33, eta34 = sp.symbols('eta31, eta32, eta33, eta34', positive=True)
eta41, eta42, eta43, eta44 = sp.symbols('eta41, eta42, eta43, eta44', positive=True)
epsilon1, epsilon2, K11, K22 = sp.symbols('epsilon1, epsilon2, K11, K22', positive=True)
omega1, omega2, gamma12, g12 = sp.symbols('omega1, omega2, gamma12, g12', positive=True)
beta11, beta21, beta31, beta41 = sp.symbols('beta11, beta21, beta31, beta41', positive=True)
beta12, beta22, beta32, beta42 = sp.symbols('beta12, beta22, beta32, beta42', positive=True)

F2 = (r1 * (1 - (eta11 * x1 + eta12 * x2 + eta13 * x3 + eta14 * x4) / N) - \
   f11 * y1 - f12 * y2)
F3 = (r2 * (1 - (eta21 * x1 + eta22 * x2 + eta23 * x3 + eta24 * x4) / N) - \
   f21 * y1 - f22 * y2)
F4 = (r3 * (1 - (eta31 * x1 + eta32 * x2 + eta33 * x3 + eta34 * x4) / N) - \
   f31 * y1 - f32 * y2)
F5 = (r4 * (1 - (eta41 * x1 + eta42 * x2 + eta43 * x3 + eta44 * x4) / N) - \
   f41 * y1 - f42 * y2)

F6 = (-epsilon1 * (1 + (y1 + omega2 * y2) / K22) - g12 * y2 + beta11 * f11 * x1 + \
   beta21 * f21 * x2 + beta31 * f31 * x3 + beta41 * f41 * x4)

F7 = (-epsilon2 * (1 + (omega1 * y1 + y2) / K11) - gamma12 * g12 * y1 + \
   beta12 * f12 * x1 + beta22 * f22 * x2 + beta32 * f32 * x3 + beta42 * f42 * x4)              


equ = ((x1, F2), (x2, F3), (x3, F4), (x4, F5), (y1, F6), (y2, F7))

from itertools import product
for eqs in product(*equ):
    sol = solve(eqs, [x1, x2, x3, x4, y1, y2])
    pprint(sol)

这给出了:

$ python t.py 
{x₁: 0, x₂: 0, x₃: 0, x₄: 0, y₁: 0, y₂: 0}
[]
[]
⎧                                      ε₂⋅(K₁₁⋅(K₂₂⋅g₁₂ + ε₁⋅ω₂) - K₂₂⋅ε₁)                ε₁⋅(-K₁₁⋅ε₂ + K₂₂⋅(K₁₁⋅g₁₂⋅γ₁₂ + ε₂⋅ω₁
⎨x₁: 0, x₂: 0, x₃: 0, x₄: 0, y₁: ───────────────────────────────────────────────, y₂: ──────────────────────────────────────────
⎩                                ε₁⋅ε₂ - (K₂₂⋅g₁₂ + ε₁⋅ω₂)⋅(K₁₁⋅g₁₂⋅γ₁₂ + ε₂⋅ω₁)      ε₁⋅ε₂ - (K₂₂⋅g₁₂ + ε₁⋅ω₂)⋅(K₁₁⋅g₁₂⋅γ₁₂ + ε

))   ⎫
─────⎬
₂⋅ω₁)⎭
⎧                          N               ⎫
⎨x₁: 0, x₂: 0, x₃: 0, x₄: ───, y₁: 0, y₂: 0⎬
⎩                         η₄₄              ⎭
⎧                            N⋅ε₂⋅(K₁₁⋅f₄₂ + r₄)                 K₁₁⋅r₄⋅(N⋅β₄₂⋅f₄₂ - ε₂⋅η₄₄)⎫
⎪x₁: 0, x₂: 0, x₃: 0, x₄: ──────────────────────────, y₁: 0, y₂: ───────────────────────────⎪
⎨                                      2                                       2            ⎬
⎪                         K₁₁⋅N⋅β₄₂⋅f₄₂  + ε₂⋅η₄₄⋅r₄              K₁₁⋅N⋅β₄₂⋅f₄₂  + ε₂⋅η₄₄⋅r₄⎪
⎩                                                                                           ⎭
⎧                            N⋅ε₁⋅(K₂₂⋅f₄₁ + r₄)          K₂₂⋅r₄⋅(N⋅β₄₁⋅f₄₁ - ε₁⋅η₄₄)       ⎫
⎪x₁: 0, x₂: 0, x₃: 0, x₄: ──────────────────────────, y₁: ───────────────────────────, y₂: 0⎪
⎨                                      2                                2                   ⎬
⎪                         K₂₂⋅N⋅β₄₁⋅f₄₁  + ε₁⋅η₄₄⋅r₄       K₂₂⋅N⋅β₄₁⋅f₄₁  + ε₁⋅η₄₄⋅r₄       ⎪
⎩                                                                                           ⎭
... (continues)

空解[]对应于已知不满足非负性要求的情况。