Question

我想使用<代码>fsolve,在数字上找到一条非线性超标的根基。

下面的法典做了这项工作。

import numpy as np
from scipy.optimize import fsolve
import matplotlib.pyplot as plt

kappa = 0.1
tau = 90

def equation(x, * parameters):
    kappa,tau = parameters
    return -x + kappa * np.sin(-tau*x)

x = np.linspace(-0.5,0.5, 35)
roots = fsolve(equation,x, (kappa,tau))

x_2 = np.linspace(-1.5,1.5,1500)
plt.plot(x_2 ,x_2 )
plt.plot(x_2 , kappa*np.sin(-x_2 *tau))
plt.scatter(x, roots)
plt.show()

我可以将以下两条图表编成图,对解决办法进行双轨制:f1(x) = x和f2(x) = k*sin(-x*tau),我也列入该守则。 <代码>fsolve向我提供了一些错误的答复,但没有提出任何错误或趋同问题。

问题在于,我要将不同条码(<>kappa>/code>和tau的程序自动化,而无需我核实哪些答案是错误的,哪些是正确的。但是,用错误的答复作为产出,我无法使用这种方法。是否有其他方法或选择,我可以使用,坐在安全方面?

Answer 1

根据scipy.fsolve , 您认为应增加初步的猜测,作为第二点,而不是像你在<代码>上所做的那样的范围。

不过,你当然可以在座右边通过,在座标上每个数值的座右铭(0.5、0.5、25)。虽然我不理解你试图通过不同的 k子和 t子实现什么,但如果我认为,对于这些参数,你有兴趣寻找根源,这里我会怎样做。

import numpy as np
from scipy.optimize import fsolve
import matplotlib.pyplot as plt

# Take it as it is
kappa   = 0.1
tau     = 90

def equation(x, parameters):
    kappa,tau = parameters
    return -x + kappa * np.sin(-tau*x)

# Initial guess of x = -0.1
SolutionStack   = []
x0              = -kappa
y               = fsolve(equation, x0, [kappa, tau])
SolutionStack.append(y[0])

y               = fsolve(equation, SolutionStack[-1], [kappa, tau])
SolutionStack.append(y[0])
deltaY = SolutionStack[-1] - SolutionStack[0]

# Define tolerance
tol = 5e-4
while ((SolutionStack[-1] <= kappa) and (deltaY <= tol)):
    y = fsolve(equation, SolutionStack[-1], [kappa, tau])
    SolutionStack.append(y[0])
    deltaY = SolutionStack[-1] - SolutionStack[-2]
    # Obviously a little guesswork is involved here, as it pertains to 0.07
    if deltaY <= tol:
        SolutionStack[-1] = SolutionStack[-1] + 0.07

# Obtaining the roots
Roots = []
Roots.append(SolutionStack[0])
for i in range(len(SolutionStack)-1):
    if (SolutionStack[i+1] - SolutionStack[i]) <= tol:
        continue
    else:
        Roots.append(SolutionStack[i+1]

也许你现在可能有一种想法。

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