Python 穆哈夫方程的解

我目前正试图找到幂律势的穆哈诺夫方程的解。我的代码现在看起来像这样

#Code to compute the power spectrum from the solution to the Mukhanov equation. 
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
import math 
plt.rcParams["font.size"] = 6
plt.autoscale(enable=True, axis='both', tight=None)

#Define Parameters.
n = float(input('Enter the value of n: '))
k_a = np.geomspace(start = 0.005, stop = 10, num = 1000)
t = np.linspace(start = 60.0, stop = 0.0, num = 1000) #t is the efold number.
k_0 = 0.005
k_max = 1e16
#########################################
#Slow-roll parameters (using phi^2 for now and t is the e-fold number)

def ep_func(t, n):
    return (n**2)/(4*n*t + n**2)

def H_func(t, n):
    return ((2*n*t + (n**2)/2)**(n/2))/3

def eta_func(t, n):
    return 2*((2*n - n**2)/(2*n*t + ((n**2)/2)))

def a_func(t):
    return np.exp(t)
##########################################
#n_s and r

def r_func(t, n):
    return (16*(n**2))/(4*n*t + n**2)

def n_func(t, n):
    return 1 -(6*(n**2)/(4*n*t + n**2)) + (2*(n*(n-1)))/(2*n*t + (n**2)/2)

##########################################
#Create the two simplified ODE's.
#This is in terms of the e-fold number not confromal time. 
def sol(Y, t, k, ep, eta, H, a):
    return np.array([Y[1], -(3 - ep_func(t, n) + eta_func(t, n))*Y[1] -((k**2)/((a_func(t)**2)*H_func(t, n)**2))*Y[0]])
##########################################
# Find the solution using Scipy odeint function while defining the I.C's.
Yr_0 = [1/(np.sqrt(2*k_max)), 0]
Yi_0 = [0, -np.sqrt(k_max/2)]
solutions = []
# Real part of the Mukhanov equation.
for k in k_a:
    asolr = odeint(sol, Yr_0, t, args=(k, ep_func(t, n), eta_func(t, n), H_func(t, n), a_func(t)))    
    solutions.append(asolr)

# Imaginary Part of the Mukhanov equation.
for k in k_a:
    asoli = odeint(sol, Yi_0, t, args=(k, ep_func(t, n), eta_func(t, n), H_func(t, n), a_func(t)))    
solutions.append(asoli)

Yr = asolr[:,0]
Yi = asoli[:,0]
########################################## 
#Plot the solution to the mukhanov equation. 
def mukh(Yi):
    return np.absolute(Yi)
m = mukh(Yi)
plt.plot(t, m, 'r-')
plt.yscale('log') 
##########################################
#Define the the power spectrum.

#Mukhanov Power spectrum
def m_spec(k_a, Yr, Yi):
    return 2.1e-9*((k_a**3)/0.005)*(np.absolute(Yr)**2 + np.absolute(Yi)**2)

#Slow-roll power spectrum 
def s_spec(k_a, t, n):
    return 2.1e-9*(k_a/0.005)**(-(2*n**2 + n)/(2*n*t + (n**2)/2))

pm = m_spec(k_a, Yr, Yi)
ps = s_spec(k_a, t, n)

#Plot the power spectrums and parameters in subplots
fig = plt.figure()

plt.subplot(2, 2, 1)
plt.plot(k_a, pm, 'r-')
plt.title('Power spectrum P(k) for n = ' +str(n))
plt.xlabel('wavenumber (k)')
plt.ylabel('P(K)')
plt.yscale('log')
plt.xscale('log')

plt.subplot(2, 2, 2)
plt.plot(k_a, ps, 'b-')
plt.title('Slow-roll Power spectrum P(k) for n = ' +str(n))
plt.xlabel('wavenumber (k)')
plt.ylabel('P(K)')
plt.yscale('log')
plt.xscale('log')

plt.subplot(2, 2, 3)
plt.plot(t, H_func(t, n), 'y-')
plt.title('Epsilon Vs e-fold')
plt.xlabel('e-fold')
plt.ylabel('$H$(N)')
#plt.yscale('log')

plt.subplot(2, 2, 4)
plt.plot(t, n_func(t, n), 'g-')
plt.title('$n_s$ Vs e-fold')
plt.xlabel('e-fold')
plt.ylabel('$n_s$')
#plt.yscale('log')

plt.tight_layout()
plt.show()

我仔细检查了一下，微分方程和它的初始条件是正确的。我相信问题在于解决方案的假想部分。我目前得到的输出如下所示

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问题是左上方的图像。我期望一条几乎恒定的梯度约为1的直线，永远不会超过10^-9，显然不是这样。

也许你对这个问题不再感兴趣，但你能通过在LaTeX代码中添加微分方程来改进帖子吗，问题中涉及的量的定义？也许你对这个问题不再感兴趣了，但是你能通过添加微分方程和问题中涉及的量的定义来改进这篇文章吗？