我想使用fsolve找出函数z(q)的两个值,其中我知道q (q_1000)的值。
我的代码:
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import fsolve
k_b = 1.38064852e-23
n_0 = 2e+25
m = 4.8e-26
g = 9.81
T = 300
I_inf = 400
sigma = 4e-22
kappa = 0.2
H = (k_b*T)/(m*g)
val = 100
z_min = 1e+5
z_max = 4e+5
z = np.linspace(z_min, z_max, val)
# ______________________________ Functions ___________________________________________
n = lambda z : n_0*np.exp(-z/H)
I = lambda z : I_inf*(np.exp(-(sigma*n(z)*H)))
q = lambda z : kappa*sigma*n(z)*I(z)
q_max = q(z).argmax() # q_max
q_1000 = q(z)[q_max]/1000 # q_max / 1000
result1 = fsolve(q,q_1000) # start at z = q_1000
plt.xscale('log')
plt.yscale('log')
plt.title('z(q)')
plt.axis([1e-11, 1e-1, 1e+5, 4e+5])
plt.plot(q(z)[q_max], z[q_max],'x', label="z*=q_max", color="red")
plt.plot (q(z),z,'k-')
plt.plot(result1,q_1000,'ro')
plt.legend()
plt.xlabel('q')
plt.ylabel('z')
plt.show()
print(q(z)[q_max])
print(q_1000)
print(result1)这就是我的结果

这就是我想要的,有两个点(假设是Z1和Z2)

发布于 2020-01-01 09:15:47
fsolve的第一个参数应该是为其寻找根q(z) = 0的函数。因此,应该将q(z) = q_1000编写为q(z) - q_1000。fsolve的第二个参数是所需根的近似值。我不确定如何获得一个好的初始值,但在这种情况下,曲线图建议1.5e5和2.5-e5。
在代码中,包括绘图:
z1 = fsolve(lambda z: q(z) - q_1000,1.5e5)[0]
z2 = fsolve(lambda z: q(z) - q_1000,2.5e5)[0]
plt.plot((q_1000, q_1000), (z1, z2), linestyle='dashed', color='dodgerblue', label='z*=q_1000')
plt.plot(q_1000, z1,'ro')
plt.plot(q_1000, z2,'ro')

https://stackoverflow.com/questions/58656612
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