简单微分方程的问题求解和绘图解

Question

I'm trying to solve the differential equation R^{2} = 1/R with initial condition that R(0) = 0 in python. I should get the solution that R'(t) = (3/2 * t)^(2/3) as I get this from mathematica. Plot of solution to R'[t]^2 = 1/R with initial condition R(0) = 0

I used the following code in python:

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

sqrt = np.sqrt

# function that returns dy/dt
def model(y,t):
    #k = 1
    dydt = sqrt(1/y)
    return dydt

# initial condition
y0 = [0.0]

# time points
t = np.linspace(0,5)

# solve ODE
y = odeint(model,y0,t)

# plot results
plt.plot(t,y)
plt.ylabel('$R/R_0$')
plt.xticks([])
plt.yticks([])
plt.show()

however I get only 0 as I'm apparently dividing by zero at some point python plot of differential equation R'[t]^2 = 1/R, which is not correct . Could someone point out what I could do to get the solution and plot I am expecting.

Thank you

Answer 1

Your model needs to be changed.I get your equation for the solution would be something like thishttps://www.wolframalpha.com/input/?i=R%27%28t%29%5E2+%3D+1%2FR

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# function that returns dy/dt
def model(y,t):
    #k = 1
    dydt = 1*y**-1/2
    return dydt
# initial condition
y0 = 0.1

# time points
t = np.linspace(0,20)

# solve ODE
y = odeint(model,y0,t)

# plot results
plt.plot(t,y)
plt.ylabel('$R$')
plt.xlabel('$t$')
plt.xticks([])
plt.yticks([])
plt.show()

Answer 2

Your starting value is 0, which results in the derivative being zero ( y * -1 , or simpler -y ), which means 0 will get added to your current y-value, thus remaining at zero for the whole integration. Your code is correct, but not your formulation.

I see a 1/R in your link, so use that, eg dydt = 1/y ; which will fail, because it results in division by zero, so don't start at zero , because your derivative is not defined there. There also appears to be a square root somewhere, that you imported, but never use.