如何在python中求解3个非线性方程 [英] how to solve 3 nonlinear equations in python

问题描述

我有以下三个需要求解的非线性方程组:

I have the following system of 3 nonlinear equations that I need to solve:

-xyt + HF = 0

-xyt + HF = 0

-2xzt + 4yzt-xyt + 4z ^ 2t-M1F = 0

-2xzt + 4yzt - xyt + 4z^2t - M1F = 0

-2xt + 2yt + 4zt-1 = 0

-2xt + 2yt + 4zt - 1 = 0

其中x，HF和M1F是已知参数.因此，y，z和t是要计算的参数.

where x, HF, and M1F are known parameters. Therefore, y,z, and t are the parameters to be calculated.

尝试解决该问题:

def equations(p):
    y,z,t = p
    f1 = -x*y*t + HF
    f2 = -2*x*z*t + 4*y*z*t - x*y*t + 4*t*z**2 - M1F
    f3 = -2*x*t + 2*y*t + 4*z*t - 1
    return (f1,f2,f3)

y,z,t = fsolve(equations)

print equations((y,z,t))

但是问题是，如果我想使用scipy.optimize.fsolve，那么我应该输入一个初始猜测.就我而言，我没有任何初始条件.

But the thing is that if I want to use scipy.optimize.fsolve then I should input an initial guess. In my case, I do not have any initial conditions.

还有另一种方法可以在python中求解具有3个未知数的3个非线性方程吗?

Is there another way to solve 3 nonlinear equations with 3 unknowns in python?

原来，我有病！条件是HF> M1F，HF> 0和M1F> 0.

It turned out that I have a condition! The condition is that HF > M1F, HF > 0, and M1F > 0.

推荐答案

@Christian，与您建议的帖子不同，我认为方程组不容易线性化.

@Christian, I don't think the equation system can be linearize easily, unlike the post you suggested.

鲍威尔的混合方法(optimize.fsolve())对初始条件非常敏感，因此，如果您能提出一个很好的初始参数猜测，这将非常有用.在下面的示例中，我们首先使用Nelder-Mead方法(optimize.fmin()，对于像OP这样的小问题，这已经足够了)最小化所有三个方程的平方和.然后将所得的参数向量用作optimize.fsolve()的初始猜测，以获得最终结果.

Powell's Hybrid method (optimize.fsolve()) is quite sensitive to initial conditions, so it is very useful if you can come up with a good initial parameter guess. In the following example, we firstly minimize the sum-of-squares of all three equations using Nelder-Mead method (optimize.fmin(), for small problem like OP, this is probably already enough). The resulting parameter vector is then used as the initial guess for optimize.fsolve() to get the final result.

>>> from numpy import *
>>> from scipy import stats
>>> from scipy import optimize
>>> HF, M1F, x=1000.,900.,10.
>>> def f(p):
    return abs(sum(array(equations(p))**2)-0)
>>> optimize.fmin(f, (1.,1.,1.))
Optimization terminated successfully.
         Current function value: 0.000000
         Iterations: 131
         Function evaluations: 239
array([ -8.95023217,   9.45274653, -11.1728963 ])
>>> optimize.fsolve(equations, (-8.95023217,   9.45274653, -11.1728963))
array([ -8.95022376,   9.45273632, -11.17290503])
>>> pr=optimize.fsolve(equations, (-8.95023217,   9.45274653, -11.1728963))
>>> equations(pr)
(-7.9580786405131221e-13, -1.2732925824820995e-10, -5.6843418860808015e-14)

结果非常好.

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