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I am having trouble following the derivation in this paper https://arxiv.org/abs/1810.07775 using the method of characteristics. By using the method of characteristics, they derive the following ODEs that are satisfied along the characteristic curve:

$$ \frac{dr}{dt} = 2\gamma r - K(t)\\ \frac{dK}{dt} = w^2r(t) $$ The solution they give for $K(t), r(t)$ is a bit complicated, but does not exactly satisfy the ODEs above. I get different expressions for $K(t),r(t)$, which seem to satisfy the above equations but produce different characteristic curves than the paper.

Specifically, I would like to know the following:

  1. What is the exact solution to the above equations?

  2. If your exact solution is different than the one in the paper, why?

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    $\begingroup$ please expand in a few lines to explain your problem, it is good that you gave a reference but some added verbiage would make it a better question. $\endgroup$
    – hyportnex
    Commented Mar 31 at 20:32

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Just a quick numerical test, to solve for $r(t),K(t)$ with Python (I assume $\gamma = 1/2, \omega=1$):

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

def coupled_ode(y, t):
    r, K = y
    drdt = r - K
    dKdt = r
    return [drdt, dKdt]

t = np.linspace(40, 60, 1000)  
sol = odeint(coupled_ode, [1.0, 1.0], t)

plt.plot(t, sol[:, 0], label='r(t), numerical')
plt.plot(t, sol[:, 1], label='K(t), numerical')
plt.legend()
plt.show()

Shows :

enter image description here

So I think it could be fluctuations in the form as $\propto~ e^{\lambda_1 t}- e^{\lambda_2 t}$

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    $\begingroup$ This code/graph was very helpful to check my analytical results; there are small differences between them which I assume are due to numerical error. The issue is that your numerical solution here seems different than the solution in their paper, and I am unsure why that is the case. $\endgroup$
    – Idieh
    Commented Mar 31 at 22:15
  • $\begingroup$ I'm not quite sure if it's different than the original solution. I would rather say that it should be the same. Just maybe, solution is not unique. $\endgroup$ Commented Apr 1 at 5:40

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