Hello I would like to solve the following first ODE:
dt/dr = +- cos(t)^2/cos(r)^2
I know the solution is : t(r) = t(r) = arctan(tan(r)+_C1), with: pi/2 < t< pi/2 and 0< r< pi/2.
I would like to know how could I improve the code below such my solution resembles the curve that is tending to + infinity on t axis in the image:
My code is:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
"""
Equations to be solved:
boundary conditions:
-pi/2 << t << pi/2
0 <= r <= pi/2
Equation:
dt/dr = +- cos^2(t)/cos^2(r)
Solution :
t(r) = arctan(tan(r) +_C1)
"""
def dt_dr(t,r):
return (cos(t)**2)/(cos(r)**2)
rs = np.linspace(0,pi/2,1000)
t0 = 0.0 #the initial condition
ts = odeint(dt_dr,t0,rs)
ts = np.array(rs).flatten()
plt.rcParams.update({'font.size': 14})
plt.xlabel("r")
plt.ylabel("t")
plt.plot(rs,ts);
and my current graphical output is:
Not sure if I understand your problem, however it seems that the solutions in the two plots are the same but they are plotted differently. In the "desired solution for one C value"-plot, the y-axis is stretched when compared to the x-axis. In your "current_solution", they are equal.
The two graphs in the question are the same, however they both have different limits. To change the limits you need to do plt.xlim()
and plt.ylim()
. Setting them to be the same as the desired outcome will give you the same result.
There is one addition to the desired outcome in that the y axis crosses the x axis at 0, rather than the left hand edge of the axis (the default). You can change this by moving the left hand spine:
rs = np.linspace(0, pi/2, 1000)
t0 = 0.0 #the initial condition
ts = odeint(dt_dr, t0, rs)
plt.rcParams.update({'font.size': 11})
plt.xlabel("t")
plt.ylabel("r")
plt.plot(ts, rs)
# Change axis limits
plt.ylim(0,0.6)
plt.xlim(-1.5,1.5)
# Move left spine to x=0
ax = plt.gca()
ax.spines['left'].set_position('zero')
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_position('zero')
ax.spines['top'].set_color('none')
plt.show()
Which gives:
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