求解具有条件的非线性方程组
Solve a system of nonlinear equations with conditions
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且所有参数均为正的条件？变量是 (x1,x2,x3,x4,y1,y2 ) 其他只是参数。
Maple 比 sympy 更适合解决这个系统吗？
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)
您可以为符号添加假设。关于哪些求解器遵守这些假设，文档让我有点困惑，但从 the docs 看来 nonlinsolve 确实遵守这些假设：
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)
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 基础解决这个问题，但计算它需要很长时间：
In [4]: groebner(equ, [x1,x2,x3,x4,y1,y2]) # Not sure how long this takes
我希望即使它确实完成了，结果也不会接受解析解，因为求解可能会导致大于 4 阶的多项式。
如果你用具体的有理数替换所有的参数，那么有可能找到一个解决方案，否则就任意符号（r3 等）我不希望一个封闭的形式解决方案将存在 - 如果这是真的，那么无论您使用 Maple 还是 SymPy 或其他任何东西都没有关系。
编辑：我现在知道你的系统是什么了。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。也就是说有两种可能：x1 = 0或者满足线性方程。因此，一种解决方案是 x1 = x2 = ... = 0，然后是 none 为零的另一种解决方案。对于 6 个未知数，有 64 种可能的解决方案，但有些可能不满足非负性假设。你可以用
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：
$ 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。
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且所有参数均为正的条件？变量是 (x1,x2,x3,x4,y1,y2 ) 其他只是参数。
Maple 比 sympy 更适合解决这个系统吗？
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)
您可以为符号添加假设。关于哪些求解器遵守这些假设，文档让我有点困惑，但从 the docs 看来 nonlinsolve 确实遵守这些假设：
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)
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 基础解决这个问题，但计算它需要很长时间：
In [4]: groebner(equ, [x1,x2,x3,x4,y1,y2]) # Not sure how long this takes
我希望即使它确实完成了，结果也不会接受解析解，因为求解可能会导致大于 4 阶的多项式。
如果你用具体的有理数替换所有的参数，那么有可能找到一个解决方案，否则就任意符号（r3 等）我不希望一个封闭的形式解决方案将存在 - 如果这是真的，那么无论您使用 Maple 还是 SymPy 或其他任何东西都没有关系。
编辑：我现在知道你的系统是什么了。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。也就是说有两种可能：x1 = 0或者满足线性方程。因此，一种解决方案是 x1 = x2 = ... = 0，然后是 none 为零的另一种解决方案。对于 6 个未知数，有 64 种可能的解决方案，但有些可能不满足非负性假设。你可以用
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：
$ 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。